The Berkeley Review MCAT Physics Part 2

S1CS

Part IISections VI-X

Section VISound and

Doppler Effect

Section VIIFluids and Solids

Section VIIIElectrostatics andElectromagnetism

Section IXElectricity and

Electric Circuits

Section XLight and Optics

BERKELEYUr«e«vm»e«ws

Specializing in MCAT Preparation

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Table of Contents

6. Sound and Doppler EffectProduction and Detection of Sound

Speed of Sound

Pitch and Intensity

Doppler Effect

Resonance in Strings and Pipes

Sound and Doppler Effect Review Questions

Detailed Answer Explanations

Sound and Doppler Effect Practice Exam


7. Fluids and Solids

Fluid Properties

Fluids in Motion

Solids

Fluids and Solids Review Questions


Fluids and Solids Practice Exam


8. Electrostatics and ElectromagnetismElectrostatics

Electric Fields

Electromagnetism

Electrostatics and Electromagnetism Review Questions


Electrostatics and Electromagnetism Practice Exam


page 6

page 7

page 9

page 12

page 16

page 23

page 28

page 33

page 44

page 55

page 66

page 78

page 83

page 89

page 93

page 104

page 115

page 120

page 131

page 141

page 147

page 151

page 170

9. Electricity and Electric CircuitsCurrents page 173

Voltage and Resistance page 176

Capacitors page 184

Electric Circuits page 188

Alternating Current page 199

Electricity and Electric Circuits Review Questions page 201

Detailed Answer Explanations page206

Electricity and Electric Circuits Practice Exam page211

Detailed Answer Explanations page 222

10. Light and OpticsElectromagnetic Radiation page 233

Reflection and Refraction page 238

Optics page 246

Interference Phenomena page 259

Light andOptics Review Questions page 267


Light and Optics Practice Exam page 277


Sound and

Doppler Effect

the

Physics Chapter 6

'source A

A)Compressed Wave: Elongated Wave:

Shorter X Longer XHigher/ Lower/

Elongated Wave://;^^--^N^ComPrtesed Wave:LongerK / / /O-—0\\\ Shorter^Lower/ / / / /0-=^\\\\ HiSher/

source

by

Berkeley Review

Sound and Doppler EffectSelected equations, facts, concepts, and shortcuts from this section

O Important Equations

6= 10 log (Jsound/j )° vwave ± vsender

/beat =l/l - /2I (Heard as a"destructive beat") Fixed Strings: Xn =2L/n .\ /n =nv/2L

Open Pipes: Xn =2L/n •'• /n =nv/2L closed P*!** ^n =4L/n •'• /n =nv/4L> where nis 1. 3» 5» etc-

/shifted = /unshiftedVwave±Vreceiver (Doppler shift)

© Important Concept

Echolocation (Using the time to detect echoing sound waves to establish position of an object)

If material is not uniform, refractionof the waves can create blind spots.

Waves normal to surface minimize §§tethe impact of refraction. Non-parallel IHSfsurfaces are more difficult to analyze :M*s£than parallel ones. FS^

Echo Detector

IPtj (time for reflection to return from 1st boundary):used to determine distance to first interface

mg^|Me|ir reflection to return from 2nd boundary):^;v:u|e3^Metermine distance to second interface

•.-. .v-;;

m

© Doppler EffectCalculation Approach

1) First determine whether the objects are moving towards or away from one another

If the objects are moving towards one another, then the frequency of the wave will increase. Thiswill require either adding to the numerator in the Doppler equation (if the receiver is in motion) orsubtractingfrom the denominator in the Dopplerequation (if the sender is in motion). If the objectsare moving away from one another, then the frequency of the wave will decrease. This willrequire either subtracting from the numerator in the Doppler equation (if the receiver is in motion)or adding to the denominator in the Doppler equation (if the sender is in motion).

2) Second determine if the scenario involves one direction transmission or an echo

If it's a one-direction wave only, then you need only plug into the Doppler equation once. If it'san echo, then you'll need to plug into the Doppler equation a second time, using the first solvedfrequency as the input frequency and reversing the roles of the sender and receiver from the firstwave. In other words, the sender of the outbound wave becomes the receiver of the inbound wave.

Physics Sound Production and Detection of Sound

SoundIn Chapter 5, we studied standing waves on a string. In this chapter, we shallexpand our coverage of waves to include sound waves, one example of alongitudinal wave.

Production and Detection of Sound

What is sound and how is it produced? Sound is simply a longitudinal wave thatcan be produced from the disturbance of a solid, a liquid, or a gas. A sound wavearises from the vibrations and collisions of molecules within a particularsubstance. Even though these molecules maintain their same average positionwithin that substance, the collision of one molecule with another molecule resultsin the transmission of energy in the form of a wave. Sound waves cannot travelthrough a vacuum.

Sinusoidal waves are the simplest type of sound waves. They have a definedwavelength, amplitude, and frequency. In order to detect a sound wave, themechanical vibrations of that wave must be analyzed in terms of intensity andfrequency.

The human ear is an excellent sound detector and can distinguish sounds thatdiffer in frequency from one another by as little as 0.3%. A simplified version ofthe human ear is shown in Figure 6-1.The pinna (outer ear) collects sound wavesand funnels them into the auditory canal. As the sound waves press on thetympanic membrane (eardrum), they cause the membrane to vibrate. Thisvibrational energy is transmitted through the ossicles, three small bones(hammer, anvil, and stirrup) connecting the tympanic membrane to the ovalwindow of the inner ear.

Pinna

(outer car)

Auditorycanal

Tytripinric

(eardrum)

Codi Itii (iiiicni led)I .

Oval ^_ Basilarwindow/ membrane

Highfrequencies

Low

frequencies

Figure 6-1

As the ossicles vibrate, they amplify the vibrational energy set in motion at thetympanic membrane. This vibrational energy is passed from the oval window toa fluid in the cochlea of the inner ear. The fluctuations in pressure in this fluidcause hair cells in the basilar membrane of the cochlear duct to move. Movementof the hair cells generates a nerve impulse that is sent to the brain and isinterpreted as sound. High frequency sounds are generated by nerves closest tothe oval window, while low frequency sounds are generated by nerves farthestaway from the oval window.

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Physics Sound Speed of Sound

Speed of SoundWhen a sound wave moves forward, it compresses the medium in front of it.This results in compressional waves moving outward in all directions from thesource of the sound. Consider the compressional wave in Figure 6-2. Note that asthe wave moves outward, there are regions of high density (or high pressure)and low density (or low pressure) within the medium. Alsonote that the volumeof space associated with the regions of high density decreases relative to thevolume of space associated with the regions of low density. The region ofreduced pressure is also called a rarefaction.

Propogation of wave

•=>•/:

high lowdensity density

Figure 6-2

Solids, Liquids, and GasesThe speed with which particles can return to their original position following adisturbance in the medium dictates the speed of sound in that medium. Thestrength of the forces between the molecules in a given medium and the densityof the particles in that medium determine how fast sound will travel in thatmedium. The forces between the molecules in a solid are greater than the forcesbetween the molecules in a liquid or a gas. The stronger the force between anytwo molecules, the greater the restoring forcebetween those molecules. The morerapidly molecules are restored to their original condition, the more rapidly theycan participate in another compression wave, and the faster the propagation ofsound in that particular medium. Asa result, we see the general trend for speedorsound as vsoun(j jn solid > vsouncj mliquid > vsoun£i jn gas.

The speed of longitudinal sound waves in various ideal gaseous media can becalculated from the formula in equation (6.1):

(6.1)

In this equation, pis the density (of agas), Pis the pressure (of a gas), yis Cp/Cv(where Cp is the molar heat capacity at constant pressure, and Cv is the molarheat capacity at constant volume for a gas), R is the ideal gas constant, T istemperature in kelvins, and Mis the molecular mass of the gas.For a monatomicgas y = 1.67, for a diatomic gas y = 1.40, and for a polyatomic gas y = 1.33. Thismeans that for a given pressure and gas density, a monatomic gas will permitsound to travel through it at a greater speed than a diatomic or polyatomic gaswill. For example, if the temperature of the air is 0 °C, the pressure is 1 arm (or1.01 x105 N/m2), and the density is1.29 kg/m3, then the speed ofsound inair is331 m/s. In this case, we are making the assumption that the air is composed ofgases in the diatomicstate,becauseair is predominantly N2 and O2.

v =(1.40)(1.01 xl05N/m2

1.29 kg/m3=331 m/s

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Physics Sound

In Table 6.1, we see the speed of longitudinal sound waves in various media.These values may vary from textbook to textbook, depending on whether theauthor assumed isothermal or adiabatic conditions. For the MCAT, knowing thatthespeed ofsoundin air at roomtemperature isabout340 m/s is goodenough.

Table 6.1

Material Temperature / Pressure Speed (ms)Air (0°C, 1 arm) 331

Air (20°Q 1 arm) 340

Air (100°C, 1 arm) 386

Hydrogen (0°C, 1 arm) 1286

Helium (0°C, 1 arm) 965

Water (lake) 1497

Water (sea) 1531

Aluminum 5100

Iron 5130

Glass 5500

Granite 6000

In Table 6-1, note that the speed of sound in metals and liquids is much greaterthan the speed of sound in air. This is due to the fact that metals and liquids arenot as compressible as gases. Therefore, the restoring forces for a metal or aliquidare much greater than they are for air.

Example 6.1aWhich of the following changes will increase the speed of sound in a fixedvolume of diatomic hydrogen gas that is enclosed in a fixed-volume container?

I. Increasethe temperature of the gas.n. Replace the hydrogen molecules with diatomic oxygen molecules, while

holding the temperature constant.III. Increase the intermolecular attraction between the molecules.

A. I and II onlyB. II and III onlyC. I and III onlyD. I, II, and HI

SolutionA general rule regarding the speed of sound is the following:

v oc vRestoring Force or Molecular Kinetic Energy

Molecular Inertia

Increasing the temperature of the gas in this enclosed volume will increase themolecular kinetic energy without changing the molecular inertia (i.e., massdensity). Thus, Statement I is true, and choice B is wrong. Replacing thehydrogen with oxygen will noticeably increase the massbut, presumably, changethe molecular interaction forces little. This would reduce the speed of sound,invalidating Statement II. We are left to conclude that choice C is the correctchoice. Statement III must be valid, if the above equation is to hold true.

The best answer is choice C.

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Speed of Sound

Exclusive MCAT Preparation

PhysiCS Sound Speed of Sound

Example 6.1bAn acoustic scientist measures the properties of three unknown molecules,labeled A, B, and C. He records the following relationships for the molecularweights and cohesive forces of the molecules:

MWa > MWb = MWc and Fa < Fr < Fc

These relationships hold true, regardless of the state of the material. Noting theseresults, which of the following relationships could he predict regarding the speedof sound in these materials? (Assume all solids have the same lattice structureand all gases have the same molarity.)

I. Speed of sound is greater in Gas A than Gas B, when both are at the sametemperature and pressure.

II. Speed of sound is greater in Solid C than Solid B, when both are at thesame temperature.

III. Speedofsound in Liquid C is greater than that in LiquidA,when Liquid Ais hotter than Liquid C.

A. II onlyB. I and III onlyC II and III onlyD. I, II, and III

Solution

The speed ofsound within a medium depends on the ability ofthe particles thatmake up the medium to quickly return to their original position following adisturbance. The restoring speed of the particles depends on the force holdingparticles together (cohesive force) and the mass of the particles. Lighter particlesmove faster than heavier particles, so speed of sound is faster in mediums thatare made of lighter particles, assuming all other factors are equal. The questionalso mentions forces. The stronger the forces between particles (strongerintermolecular forces), the faster the particles can return to their original positionfollowing a disturbance, which makes the speed of sound increase. BecauseParticle Cis the lightest and has the strongest intermolecular forces, the speed ofsound will be fastest in Medium C when the temperature is uniform. This makesStatement II a true statement, which eliminates choice B.

Statement III is invalid, because without knowing the temperature (or knowingthatifs equal for all mediums), we donothave enough information todeterminethe relative speeds of sound. This eliminates choices Cand D, leaving only choiceA. Statement I is false, because Particle A has the weakest intermolecular forcesand is the heaviest particle, resulting inMedium A having the smallest speed ofsound. Thisfurthersupports theelimination of choices Band D.

This makes choice A the best answer.

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Physics Sound

Pitch and IntensityThere are many subjective terms used to describe sound, such as pitch. Thisword is used to describe an individual's impression concerning the frequency ofa particular sound. For example, if a musical instrument has a high pitch, thenthe sound waves coming from that instrument vibrate at a high frequency.Conversely, if a musical instrument has a low pitch, then the sound wavescoming from that instrument vibrate at a low frequency. Human ears processsound waves with a frequency between 20 Hz and 20,000 Hz.

Sounds can be distinguished not only by their pitch, but also by their loudness.The subjective loudness of a sound is usually expressed objectively in terms of itsintensity (I). Figure 6-3 shows a reference wave and four waves that representvariation in the properties of a sound.

Amplitude <* Loudness

Louder Sound

Quieter Sound

«

Reference Sound

2A

Frequency « Pitch

Higher Pitch

u

Lower Pitch

Figure 6-3

The intensity ofa sound wave is the power (P) that the wave transports per unitarea (A) of its wavefront, as shown by equation (6.2). Remember, power is justenergy per unit time. This means that the intensity of a wave will have units ofenergy (in joules) divided by the units of area (in m2) times the time (in seconds).Therefore, the unitsof intensity are J/(m2«s). However, recall that a watt is justaJ/s. This allows us to assign the units of intensity asthe W/m2.

1 =Energy _ Power

(Area)(Time) Area(6.2)

We can also compare the intensity (I) of any particular sound with the intensity(I0) of a reference sound, the minimum sound intensity an average human earcan detect. The intensity of the reference sound is chosen to be 10"12 W/m2. Thiscomparison of I to IQ is referred to as the intensity level (f3) and is given byequation (6.3). The reason that a logarithmic scale is used is because the humanear is sensitive to a broad range of intensities that spans many powers of 10. Theunit used toexpress the intensity level ofa sound is the decibel (abbreviated dB).

C=10 •log iIn

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(6.3)

Pitch and Intensity


Physics Sound Pitch and Intensity

The intensity of sound dissipates with distance by a squared factor. As such, theintensity of a sound drops off rapidly at first, it can still be heard from a longdistance away. You may have noticed this when stepping away from a loudstereo speaker. It quickly goes from painfully loud to acceptable within a fewsteps, but you can likely hear that speaker from blocks away. As a general rule,every time the distance from the source is increased by a factor of 3.2, theintensity drops by a factor of 10, so the dB-level decreases by 10. Figure 6-4shows this relationship.

Distance (m)

Figure 6-4

Listed in Table 6.2 are theintensity and intensity levels ofsome common sounds.

Table 6.2

Sound Intensity (W7i*2) Intensity Level (dB)

Jet taking off 103 150

Threshold ofpain 1 120

Busy street traffic 10"5 70

Normal conversation 10"6 60

Quiet room 10"7 50

A whisper 10-1020

Threshold ofhearing 10-12 0

The data in the table confirms that for every drop in intensity by a factor of 10,the decibel level decreases by 10. For instance, if the intensity of a sound'increases by 1000 (three factors of 10), then the decibel level of that sound willincrease by 30 (3 x10).

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Physics sound

Example 6.2aAn annoyingneighbor who owns a stereo system has just heard his favorite songbegin to play. He turns up the stereo so that the intensity is now 16 times louderthan it was before the song started. By how much does the intensity levelincrease?

A. The intensity level increases by 1.2 dB.B. The intensity level increases by 12 dB.C. The intensity level increases by 16 dB.D. The intensity level increases by 160 dB.

SolutionFirst, know that intensity and intensity level have different units. This rules outchoice C, since increasing the intensity of a sound by a factor of 16 would notlead to a 16-dB increase. You can use equation (6.3), if you are good at calculatingbase-ten logs. If not,here is an easyway to thinkabout these intensity scales:

If the intensity increases by a factor of 10, then the intensity levelincreases by 10dB. If the intensity increases by a factor of 100, then theintensity level increases by 20 dB.If the intensity decreases by a factor of 10, then the intensity leveldecreases by 10 dB. If the intensity decreases by a factor of100, then theintensity level decreases by 20 dB.

Here, the intensity increases by a factor of 16. The corresponding intensity levelincrease must be somewhere between +10 dB and +20 dB, but closer to 10 dBthan 20dB. Onlychoice Bis possible. To solve this moreprecisely,

AdB =10 log 16 =10 log 24 =10 (4 xlog 2)10(4xlog 2)= 10(4 x0.3) = 10(1.2) = 12dB

The best answer is choice B.

Example 6.2bWhile you are discussing the virtues of antisound in a medical library, thelibrarian asks you to be quiet. In going from a talk to a whisper, the intensitylevel ofyour voice decreases from60dB to 20 dB. By what factor doesyour voicecorrespondingly decrease in intensity?

A. The intensity decreases by a factor of 30.B. The intensity decreases by a factor of 40.C. The intensity decreases by a factor of 100.D. The intensity decreases by a factor of 10,000.

SolutionThe sound drops by 40 dB, so we need to use the decibel relationship todetermine the magnitude of the drop. Because AdB = 10logIf/I0/ we can solvefor the magnitude by dividingby 10and then taking an analog.

AdB =10 log If/iQ.-- 40 =101ogIf/Ioso4 =logIf/Io.-. lO4^/^

This means that the intensity is greater by a factor of 10,000.

The best answer is choice D.

Pitch and Intensity

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Physics Sound Doppler Effect

Doppler EffectIf you have ever been near a fast-moving car as it went whizzing past you, thenyou have probably noticed that as the car approaches you, you hear a high-pitched whining sound from the engine—but as soon as the car passes you, thepitch quickly becomes much lower and fades out. There is a change in theintensity of the sound with distance, as well as a perceived change in thefrequency of the sound. The change in pitch of a sound from a high frequency asit approaches an observer to a low frequency as it moves away from the observeris called the Doppler effect.

Let's consider what happens when the source of the sound is stationary and alistener is moving towards that sound. Suppose that you are walking towardsan idling car. You move at a speed vl ("L" refers to the listener) towards thestationary car. The engine of the car is emitting sound waves with frequency /s(where the subscript "s" refers to the source). This scenario is depicted in Figure6-5.

Note that the sound waves coming from the engine located near the back of thecar (Figure 6-5) have wave fronts all separated from one another by an equaldistance X.

wave front

/rTm////^//////////W)%////////////////////7/////^7/////////X

Figure 6-5

As you approach the car, the wave fronts appear to move towards you with aspeed of v + vl- When the wave fronts arrive at your position, you encountereach subsequent wave front at a point that is slightly closer to the source thanwhere you encountered the previous wave front. Although the wavelength of thewaves emitted from the source are constant, you perceive the wave fronts to becloser together than they really are. This leads to a perceived shorter wavelength.It also leads to a perceived increase in frequency that varies with your speed.Equation (6.4) represents the relationship between the observed frequency (/jjand the actual sender frequency (fs) for a scenario where the listener is movingtowards a stationary source.

fL =v + vL

fs (6.4)

Equation (6.4) is telling us that ifyou move towards the source of the sound (i.e.,vl > 0), then you hear a higher pitch (corresponding to a greater frequency). Ifyou move away from the source ofthe sound (i.e., vl < 0), then you hear a lowerpitch (corresponding to a smaller frequency). This should match nicely withwhat you've experienced inreal life interms of the Doppler shift.


Physics Sound

A good analogy to help comprehend the Doppler effect is to consider whatwould happen if you were to toss jellybeans to a friend at a frequency of onejellybean per second. If you were both stationary, then your friend would receiveone jellybean per second. Now if you were to walk towards them while stillthrowing one jellybean per second at the same speed, your friend would receivethe jellybeans at a frequency greater than one per second, because each time youreleased one, you would be a little closer to your friend than when you threw thelast one. While your throwing would have a frequency of one per second, yourfriend's perceived reception frequency would be higher and would depend onthe speed with whichyou walked towards them.

Let's now consider what happens when the source of the sound is moving andthe listener is stationary. The source of the sound is the car's engine, and the carmoves with a velocity vs, as shown in Figure 6-6. Even though the car is nowmoving, the speed of the sound waves does not change. The speed of thosesound waves is still v.

wave front

vL = 0

11////////////////////////////////J////////////

Figure 6-6

As the car approaches you, each subsequent wave front initiates from a pointthat is closer to you at the time it is released. When the wave fronts arrive at yourposition, you encounter each subsequent wave front at a point that is slightlycloser to the source than where you encountered the previous wave front.Although the wavelength of the waves emitted from thesource areconstant, youperceive the wave fronts to be closer together than they really are. This leads to aperceived shorter wavelength. It also leads to a perceived increase in frequencythat varies with the speed of the car. Equation (6.5) represents the relationshipbetween the observed frequency (/l) and the actual sender frequency (/s) for ascenario where the listener is moving towards a stationary source.

*-(r?d'« (6.5)

Equation (6.5) is telling us that the source of the sound moves towards you (i.e.,vs > 0), then you hear a higher pitch (corresponding to a greater frequency). Ifthe sourceof the sound moves away from you (i.e., vs < 0), then you hear a lowerpitch (corresponding to a smaller frequency). Note that the relative displacementof the source and the sound wave determines the wavelength between successivewave fronts. If we combineequations (6.4) and (6.5), we get equation (6.6). This isthe general equation for the Doppler effect. It shows how the frequency of thesound waves encountered by the observer (f\J is related to the frequency ofsound waves emitted by the source (/s).

(6.6)

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Doppler Effect


Physics Sound Doppler Effect

Since this is an important equation, let's define its variables once again:

Definition of the variables in the Doppler equation

/L Frequency of sound waves encountered by the listener

/s Frequency of sound waves emitted by the source

v Speedof the sound wave (in this casethrough air)vl Speed of the listener relative to still air

vl is positive(+), if the listener is moving towards the source.vl is negative (-), if the listeneris moving away fromthe source.

vs Speed of the source relative to still airvs is negative (-), if the sourceis moving towards the listener.vs is positive (+), if the sourceis moving away from the listener.

If the listener and the source ofa sound are moving toward one another, then fLwill always be greater than fs. However, if the listener and the source of a soundare movingawayfrom oneanother, then ft willalways be less than fs.

Example 63aThe spectra of starlight tell astronomers about the chemical makeup of stellarbodies. However, the visible spectral lines may be shifted away from their usualfrequencies. A shift towards to lower frequencies (the red end of the visiblespectrum) isknown asa red shift. Ashift towards tohigher frequencies (the blueend of the visible spectrum) is known as a blue shift. If an astronomer noticesthat the spectrum ofa particular star isblue-shifted, then the:

I. star could be moving towards the Earth.II. star could be movingaway from the Earth.III. Earthcouldbe moving towards the star.IV. Earth could be orbiting the star inperfectly circular path.

A. I onlyB. II onlyC. I and in onlyD. II and IV only

Solution

To identify a Doppler effect problem, note the following:

Any type of wave can undergo a Doppler shift-sound, light, water,earthquake, anything whatever.

Afrequency shift inany wave isprobably dueto the Doppler effect.The Doppler effect requires that the wavesource and wave detector beapproaching each otheror receding fromeachother, in relative terms.When the source and detector approach each other, the detectedfrequency is higher than the emitted one.

When the source and detector recede from each other, the detectedfrequency is lower than the emitted one.

The spectrum is blue-shifted, so the detected frequency must be higher than theemitted one. The starand the Earth mustbe getting closer to one another.


1.

2.

3.

5.

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Physics Sound

Example 6.3bA bat emits an ultrasonic squeak in the direction of an insect and notices that thereflectedsound is of a lower frequency than the emitted pulse. What could BESTexplainthis frequency shift?

A. The bat is moving towards a stationary insect.B. The insect is moving away from a stationary bat.C. Either the bat or insect is moving toward the other.D. Either the bat or insect is moving away from the other.

SolutionThe reflected ultrasonic wave has a lower frequency than the emitted ultrasonicwave, so the pulse must have undergone a Doppler shift of some kind. A lowerfrequency for the echo indicates that the bat and the insect must be gettingfarther away from oneanotheras the ultrasonic wave propagates. Both choices Band D describe a scenario where the distance between the bat and the insect isincreasing. Under such a situation, it comes down to test-taking logic. You haveto askyourself what the test writer wantsyou to pick. Choice D is more generalthan choice B and accounts for more possibilities than choice B. In other words,choice Bmight explain it while choice likely will explain it. The betterchoice hereis the answer that explains it under more scenarios.


Example 6.3b introduces the concept of a Doppler shift applied to an echo. Insuch a case, the wave is actually doubly shifted rather than singly shifted. Onquestions involving the Doppler effect, it is a good idea for you to askyourselftwo questions:

1) Is the wave being analyzed an echo or a single directionwave?2) Are thesenderand receiver moving towards or awayfrom oneanother?

Consider the bat and insectfrom Example 6.3b. Theanswer to the two questionsabove is that ifs an echothat is being detected and that the two are moving awayfrom one another. Let's say for sake of argument that the bat is moving awayfrom a stationary insect at a speed of17m/s. To determine the frequency of theecho, we will do a two-step analysis determining first the frequency detected bythe insect, and then determinethe frequency of the echo detected by the bat. Weshall use f0 to represent the emitted frequency by the bat, f\ to represent thefrequency detected by the stationary insect, and /2 to represent the frequency ofthe echo detected by the moving bat.

Outbound pulse: f\ =(—^ )/o Echo pulse: f2 =(340 -vbat^v \340 + Vbat/ \ 340 /

Overall: /2 =(^M/l =^^)^^V 340 / V 340 / \340+ vbat/ \340+ vbat/

.(l-0.1)/o = 0.9/o

V 340 / V 340 / \340+ vbaf-_(340-17W -(323W -(357-34W =/1._34.Wr/2"l5io7r7)/o wr 1357 r° r 357P

You should note that the sign for the bafs speed changes from the outboundequation to the echo equation, because the direction of the wave changed fromopposing the bafs motion to complementing the bafs motion. It ends up that foran echo, wecansimply plug into theDoppler equation twice, which makes thesequestions easier to solve than we may have initially imagined. Just be sure toplug the numbers in such a way as to make the shifted frequency either higher orlower, according to their relative motion.

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Doppler Effect


Physics Sound Resonance in Strings and Pipes

Resonance in Strings and PipesThere are many sources of sound in our environment. No matter what thesources of the sound, most of them produce vibrational energy at frequencieswell within the range of human hearing (20 Hz to 20,000 Hz). Two possiblesources of the kinds ofsounds we will be considering in thissection are stringedmusical instruments and wind musical instruments. These instruments canproduce sounds with numerous natural frequencies (also called resonantfrequencies) of vibration.

Resonance in StringsConsider a guitar string of length L, which is fixed at two ends and is under acertain tension. Byfixed, we mean that it is held in place so it cannot move. If wewere to pluck this guitar string, a wave pulse would be sentdown the length ofthe string and reflect off of one ofthe fixed ends. The guitar string would beginto vibrate. If the string vibrated at a particular natural frequency, a standingwave would be produced (Figure 6-7). A standing wave is a wave that does notappear tomove ineither direction along the length ofthe string. Standing wavesare produced from waves traveling in opposite directions along a string thatcombine toproduce a wave disturbance, which appears to oscillate up and downwithout propagating.

Figure 6-7

A point along the string thatis never displaced is called a node. In other words,displacement of a string ata node is always zero. A point along the string thatreaches maximum displacement is called an antinode. The two ends of the stringin Figure 6-7 are nodes. Adjacent nodes along the string are always one-halfwavelength apartfrom each other. Ata node, thepulses ofa wave interfere witheach other in a destructive fashion. If a node on a guitar string were suddenlyfixed in place (by pressing the string against a fret with a finger), the standingwaves that had already been produced along the string would notbe affected.

There are also regions along the string where the wave pulses add in aconstructive fashion. Those regions are called antinodes, and they are locatedhalfway between two nodes. Another way to look at this is to say that anantinode and a node are one-quarter of a wavelength apart. Antinodes areregions of maximum displacement of the vibrating string. In a standing waveboth the nodes and the antinodes are atfixed points along the string.In Figure 6-8 are the first four natural frequencies for a string fixed atboth ends.In equations (5-17) and (5-18) of Section V, we derived the expression for boththe frequency and the wavelength of fixed-end strings. We can use theseequations to find the natural resonant frequencies of a fixed string and theirassociated wavelengths. The n value in each equation describes the harmonicnumber ofthe associated wave, and it must be a whole number. Every harmonicfrequency isalways a whole number multiple of the fundamental frequency, /n= n/i.

Harmonics: Frequencies that are integer multiples ofthe fundamental frequency(i.e., /i) are called harmonics. In musical terms, frequencies greater than f\ arecalled overtones. Forexample, the frequency /2 is the secondharmonic, but it isthe first overtone.

Copyright©by TheBerkeley Review 16 The Berkeley Review


Standing Wave

1st Harmonic

(Fundamental)

2nd Harmonic

(1st Overtone) *»

3rd Harmonic

(2nd Overtone)

4th Harmonic

(3rd Overtone)

Figure 6-8

Frequency Wavelength

Xi=2L

>N Mi) *-f

NN

Hi) *-f

Example 6.4aWhat is the beat frequency created upon interference ofthe third harmonic withthe fifth harmonic of waves on a fixed string?

A. The beatfrequency would equal the frequency ofthe second harmonic.B. The beat frequency would equal the frequency ofthe eighth harmonic.C. The beat frequency would equal the frequency ofthe fourth harmonic.D. The beatfrequency wouldnot correlate toanyoftheharmonics.

SolutionA beat frequency equals the absolute value of the difference between the twointerfering frequencies. Because all of the harmonics are multiples of thefundamental frequency (first harmonic), the beat frequency will also bea wholenumber multiple of the fundamental frequency. This eliminates choice D. Thebeat frequency is found as /beat = ' /5 - ft' = 15/i - 3ft I =2/i = ft. The beatfrequency is the second harmonic, making choice Acorrect.

The best answer is choice A.

Example 6.4bWhat is the fundamental frequency for a string fixed at both ends that producesconsecutive harmonics of 114 Hz, 171 Hz, and 228 Hz?

A. 114 HzB. 228 Hz

C. 57 Hz

D. 19 Hz

SolutionConsecutive harmonics differ by the fundamental frequency, ft = /n - /(n-1)-This means that we simply need to subtract171 Hz from 228 Hz, or subtract114Hz from 171 Hz. The difference is 57 Hz, so the fundamental frequency is 57 rlz.


Copyright©by The BerkeleyReview 17 Exclusive MCAT Preparation

PhyS1CS

(a)

(b)

Sound Resonance in Strings and Pipes

Resonance in PipesThe formation ofstanding waves in pipes is similar to the formation of standingwaves along a string. Instead of plucking the string as we did in the case of theguitar, we blow air across an open end in the pipe. By blowing across this openend, we produce a pressure pulse of air that travels down the pipe. Thereare twotypes of pipes to consider: an open pipe, in which both ends are open; and aclosed pipe, in which one end isclosed and the otherend open..

Open Pipe: Suppose we have a pipe that is open at both ends (like a flute,trumpet, or an oboe). If we blow across one of the open ends of the pipe shownin Figure 6-9a, a pressure pulse ofair will be sent down the length of the pipe.When this pressure pulse reaches the right end of the open pipe, part of it willleave the pipe and partofit will be reflected back into the pipe (Figure 6-9b). Thepressure pulse thatis reflected intothe pipeis a pulse ofdecreased pressure, andit travels back towards itsorigin. When it reaches theopening at the left end, partof thepulse will leave the pipe and partof the pulse will again be reflected downthe pipe (Figure 6-9c). The reflected pressure pulse again changes direction andbecomes an increased pressure pulse traveling down the pipe to the right (Figure6-9d). Upon reaching the right side, the pulse reflects again (Figure 6-9e); a lowpressure pulse now travels back to the left (Figure 6-9f). Upon reaching the leftside, the pulseagain reflects and repeats the cycle (Figure 6-9a).

Initiation of Pressure Pulse (High Pressure)

Mopen

endopen

end

Reflection of Pressure Pulse (Low Pressure)

&1reflects- •'- continues

(d) HReflection of Pressure Pulse (Low Pressure)

(e)

(c)

Reflection of Pressure Pulse (High Pressure)leaves •

Ireflects

(f) '

11<=• !•=> $

Figure 6-9

If a new pressure pulse of air were to be sent down the pipe, it could interferewith the initial pressure pulse in either a constructive or a destructive manner. Ifthe new pressure pulse interferes constructively, the pulse's amplitude increases.If the new pressure pulse interferes destructively, the amplitude decreases.

Pressure pulses can be represented as waves. The waves for the first two normalmode frequencies for an open pipe are shown in Figure 6-10. At both openings ofan open pipe there will be an antinode. All of the harmonics frequencies mustalso meet the restriction of having antinodes at each end ofthe pipe. The distancebetween two adjacent antinodes isalways one-half of a wavelength.

open

end

L =2X

First normal mode: /j = •rs- 2vSecond normal mode: ft =4f- 2ft

Figure 6-10

Copyright © by The Berkeley Review 18 The Berkeley Review

1 iiySlCS Sound Resonance in Strings and Pipes

Thus, the length of an open pipe is one-half of the fundamental wavelength andcan be given by equation (6.7). Solving equation (6.7) for the wavelength andthen substituting into the basic equation for a wave (v = Xf) gives equation (6.8).This is the lowest frequency mode (i.e., fundamental frequency mode) for a pipethat is open at both ends.

L = h. (6.7)2

/ = JL (6.8)2L

Openpipescan vibrateat all multiples of the fundamental frequency. Therefore,an open pipe can vibrate at the standing wave frequencies shown in equation(6.9). In this equation, n = 1,2,3,...

Open Pipe ^-"(i) (6.9)

Example 6.5aWhich of the following changes will NOT increase the frequency of a soundwave produced by a pipe that is open at both ends?

A. Increasing the temperature of the air in the pipeB. Using pipe material with rhombic crystallization rather than monoclinicC. Lengthening the pipeD. Shortening the pipe

SolutionAccording to equation (6.8), the frequency of a wave in a pipe can be increasedbyeither decreasing thepipe's length or by increasing the wave's speed. ChoiceD increases the frequency, so it is eliminated. Conversely, lengthening a pipe willdecrease thefrequency of the waves it produces. Lengthening the pipe does notincrease the frequency of the wave, so choice C results in a decreased frequency.

For choice A, the speed of a sound wave in a pipe increases as pipe temperatureincreases. Choice B requires test-taking logic. There is no doubt that the crystalstructure of a material impacts the wave a pipe will propagate, but there's noway 99.99% ofpeople taking the MCAT would know thedetails. It comes downto knowing that it couldbe right and not gettinghung up on not knowing a fact.


Example 6.5bWhat is the wavelength of the third harmonic produced on a 2 m open pipe?

A. 6 m

B. 1.33 m

C. lm

D. 0.67 m

SolutionThe wavelength of each harmonic is equal to 2L/n, where L is the length of thestring and n is the harmonic number. In this case, the harmonic is found as2(2)/3 = 4/3 m. The best answer is 1.33 m. You could also have reached thisconclusion lookingat Figure 6-10 and drawing the next wave in sequence.


Copyright ©byThe Berkeley Review 19 Exclusive MCAT Preparation


Closed Pipe: An example of a closed pipe is a clarinet. If we blow across theopen end of the pipe shown in Figure 6-lla, a pressure pulse of air will be sentdown the length of the pipe as it was in the open pipe. The difference with aclosed pipe compared to an open pipe is that the wave pulse will reflectoffof theclosed end. The pulse hits the closed end of the pipe (Figure 6-llb) and isreflected toward the openend with no changein phase.

As the pulse reaches the open end of the pipe, it is reflected toward the closedend again (Figure 6-llc). At this point, a changeof phase has occurred; there hasbeen a decrease in pulse pressure (Figure 6-lld). The decreased pulse returns tothe closed end of the pipe and is reflected again (Figure 6-lle). Once the pulsereaches the open end ofthe pipe, there isagain a change in the phase ofthe pulse(Figure 6-llf); the pressure increases, and we are backat Figure 6-lla. Therefore,thephase ofthepulse is inverted once witheach round trip.

Initiation of Pressure Pulse (High Pressure)

HReflection of Pressure Pulse (Low Pressure)

(a) openend

(b)

(0

<=" J

closedend

(d)

hi

(e) B(f)

Figure 6-11

The first two normal mode frequencies for a closed pipe are shown inFigure 6-12.At the opening of the pipe there will be an antinode, while at the closed endof the pipe there will be a node. The antinodes are where the air pressurefluctuates maximally; the nodes are where the air pressure does not vary at allfrom atmospheric. The distance between a consecutive antinode and a node isalways one-quarter ofa wavelength.

closedend

openend

First normal mode: /j = •— Second normal mode: /3=-ff-= 3f1

Figure 6-12

Therefore, the length of the closed pipe is one-quarter of the fundamentalwavelength and can be given by equation (6.10). A harmonic frequency can berelated to the corresponding harmonic wavelength by using the basic waveequation, v = Xn-/n. Solving equation (6.10) for the wavelength and thensubstituting into /n = v /Xn gives equation (6.12). As we see, the fundamentalfrequency for a closed pipe turns outtobe one-half ofthe fundamental frequencyfor a pipe ofequal length that is open at both ends.

Copyright©by The BerkeleyReview 20 The Berkeley Review


L = ^4

/ = JL4L

(6.10)

(6.11)

A closed pipe cannot vibrate at twice the fundamental frequency. Why? If wewere at twice the fundamental frequency, then a pressure pulse would havetraveled to the closed end of the pipe and back to the open end before anotherpressure pulse couldbe initiated. Recall that the pressurepulse was inverted as itcame off of the closed end of the pipe. If this inverted pressure pulse encountersa new pulse, there will be destructive interference. Even though a closed pipecannot vibrate at even multiples of the fundamental frequency, it can vibrate atodd multiples of the fundamental frequency. Therefore, a closed pipe can vibrateat the standing wave frequencies shown in equation (6.12). In this equation, n =1,3,5,....

Closed Pipe /n=n(i) (6.12)

Example 6.6aIn a study of organ pipes, an engineer charts the frequencies of two equally longpipeswhen they are made to resonate in various harmonic modes.

n /PipeA(Hz) /PipeB(Hz)

1

2

3

411

822

1233

396

792

1188

The MOST likelyexplanation for the data is that:

A. the air was warmer when Pipe A was tested.B. the air was warmer when Pipe B was tested.C. PipeA was tested in a lower gravitational field.D. Pipe Bwas tested in a lower gravitational field.

SolutionThe frequency of a sound wave in both an open pipe and a closed pipe dependson the speed of the wave and the length of the pipe.Thequestion states that thepipesare equally long, so the observed differences in frequencies must be due toa difference in the speed of the waves. A variation in the gravitational fieldwould not impact the wave speed (or the wave frequency), eliminating choices Cand D. This means that the difference in wave speed stems from a difference intemperature. Gases travelfaster in warmer air than cooler air,so the speed on thesound wave through the pipe will be faster in the warmer pipe. Warmer airproducesa higher frequency sound wave,so Pipe A must have the warmer air.


For many of the questions dealing with harmonic frequencies, you will notice apattern in the numbers. Look for differences between numbers or commondenominators when dealing with harmonic frequencies of standing waves. If youcan find a highest common denominator, then you have likely found thefundamental frequency.


1 JiySlCS Sound Resonance in Strings and Pipes

Example 6.6bIn studying the relationship between thevariable lengths of twoopenresonatingpipes and the number ofharmonics eachpipe can produceat different lengths, amusicologist notices the following trends in her data:

4i

3n

2

11

• Pipe A

a PipeB

L(m)

Pipes A and Bare both telescoping pipes. What could explain the difference inslopefor the lines representing eachpipe?

A. The speed ofsound isfaster for Pipe A thanfor Pipe B.B. The speed ofsound isslower for Pipe A than for Pipe B.C. Pipe A is wider than Pipe B.D. Pipe A is narrower than Pipe B.

Solution

Using the equation for the resonant frequencies of an open pipe (/n = nv/2L),you can relate n and Lasn=/n(2L/v). You'll note thatthe width ofthe pipe hasno impact on the frequency of the resonance waves, so choice C and D areeliminated. The pipewidth does notaffect frequencies and resonances other thanto decrease the overall intensity of the resulting sounds. If you didn't rememberthe precise equation, atleast know the basic variables used init (e.g., realize thatwidth is not an important factor). This way, you've increased your chances ofguessing, should you need to. The equation shows that n and L are linearlyrelated, and more importantly thata bigger sound speed v results in a smallerslope in the n-L line. This makes choice B correct and choice A incorrect.



25 SfjtindMetDoppler E$|e<b§ Reviiew Questions

I. Violin Strings

II. Train Whistle

III. Submarine Sonar

Questions Not Based on a Descriptive Passage

(1-7)

(8-14)

(15 - 21)

(22 - 25)

The main purpose of this 25-question set is to serve as a review of the materialpresented in the chapter. Do not worry about the timing for these questions.Focus on learning. Once you complete these questions, grade them using theanswer key. For any question you missed, repeat it and write down your thoughtprocess. Then grade the questions you repeated and thoroughly read the answerexplanation. Compare your thought process to the answer explanation and assesswhether you missed the question because of a careless error (such as misreadingthe question), because of an error in reasoning, or because you were missinginformation. Your goal is to fill in any informational gaps and solidify yourreasoning before you begin your practice exam for this section. Preparing for theMCAT is best done in stages. This first stage is meant to help you evaluate howwell you know this subject matter.

Passage I (Questions 1-7)

A violin produces sound because of the transversevibrations of its strings, which cause the air around them tovibrate. These longitudinal waves generate the sound that isassociated with a resonating cord. The speed of the waves ona violin string is given by the formula:

Equation 1V M

where T is the tension in the string, L is the length of thestring, and M is the mass of the string. This means that themass of the string actually affects the speed of the waves,which is unusual in that mass usually has no effect on speed.

Each string of the violin produces standing waves with manyfrequencies, given by:

Equation 2 f = ny_

2L

where n = 1, 2, 3,.... By adjusting the strings, one can alterand control the sound generated by any stringed instrument,such as the violin.

1.

2.

3.

If the length of a typical violin string is 30 cm, what arethe wavelengths of the first three harmonics?

A. 30 cm, 60 cm, 90 cmB. 30 cm, 15 cm, 10 cmC. 60 cm, 30 cm, 20 cmD. 60 cm, 40 cm, 20 cm

What is the fundamental pitch of a string that has alength of 30 cm, a mass of 30 g, and a tension of 9000N?

A. 300 Hz

B. 400 Hz

C. 500 Hz

D. 600 Hz

The pitch of a standing wave on a violin string:

A. increases as the tension is increased.

B. decreases as the tension is increased.C. increases as the tension is decreased.

D. is unaffected by the tension in the string.

Copyright ©by TheBerkeley Review® 24

4. Violin players can press down on the violin stringseffectively to shorten the length of a particular string.The shorter string will produce standing waves whichare:

A. higher in frequency and shorter in wavelength.B. higher in frequency and longer in wavelength.C. lower in frequency and shorter in wavelength.

D. lower in frequency and longer in wavelength.

5. Which of the following changes would double thefundamental frequency produced by a violin string?

A. Doubling the tension on the string.B. Quadrupling the tension on the string.C. Halving the tension on the string.D. None of the above changes would double the

fundamental frequency.

6. Two strings have the same length and fundamentalwavelength, but different fundamental frequencies.What could account for the difference in thefundamental frequency of the twostrings?

I. The two strings could have different tensions.

II. The two strings could have different masses.

III. The two strings could have different mountingangles.

A. I onlyB. II onlyC. I and II onlyD. I and III only

7. When comparing the standing waves produced on aviolin string with the sound waves produced by thatstring, we find that:

A. the waves on the string and the sound waves aretransverse.

B. the waves on the string and the sound waves arelongitudinal.

C. the waves on the string are longitudinal, while thesound waves are transverse.

D. the waves on the string are transverse, while thesound waves are longitudinal.

GO ON TO THE NEXT PAGE

Passage II (Questions 8-14)

A train that is approaching a station has two whistles.The first whistle has frequency f\, while the second whistlehas frequency ft The whistles emit sound waves, which are

longitudinal waves. Sound waves, like any harmonic waves,may interfere with each other, if they arrive at the same placeat the same time. The intensity level of a sound wave is givenby:

Equation 1 6= 10-log10|—

where I = intensity, ID = reference intensity, and 6 is the

intensity level in dB. Assume that the speed of sound is 330m/s in all questions and that f\<ft

8. If the train is in the station and blows Whistle 1 with

frequency f\ = 440 Hz, what is the wavelength of the

emitted sound?

A. 0.75 m

B. 1.00 m

C. 1.33 m

D. 0.67 m

9. Suppose /i = 440 Hz and /2 = 450 Hz. If both whistlesare blown at the same time while the train is in the

station, what is the beat frequency?

A. 445 Hz

B. 890 Hz

C. -10 Hz

D. 10 Hz

10. Sound wave A has an amplitude of 1 mm and Soundwave B has an amplitude of 3 mm. What are themaximum and minimum possible amplitudes, if A andB interfere?

A. 3 mm and 1 mm

B. 2 mm and -2 mm

C. 3 mm and -1 mm

D. 4 mm and 2 mm

11. Sound wave A has an amplitude of 1 mm and SoundWhistle 1 has an intensity twice that of Whistle 2.Whistle 1 has an intensity level that is how many dBgreater than the intensity level of Whistle 2?

A. 3dB

B. 10 dB

C. 20 dB

D. 30 dB

Copyright ©byThe Berkeley Review® 25

12. As the train approaches a station, it blows both whistles.A person standing on the platform hears both whistleswith apparent frequencies f\% and ft. Which of thefollowing would be TRUE?

A. fl<f2<fl'<ft

B. fl<f2>fl'>ft

C /i</i\/2</2'

D. f\<f\\fi>ft

13.

14.

If the train's whistles are treated as pipes that are openat one end and closed at the other, which of thefollowing plots could represent the air column pressurein the pipe as it oscillates in its fundamental mode?

A.

B.

D.

The train approaches a tunnel at 20 m/s and blowsWhistle 1 when it is 660 m away. How much timeelapses between the moment when the whistle is blownand the moment when its echo is heard by the engineer?

A. 2.0 s

B. 2.6 s

C. 3.8 s

D. 4.0 s


Passage III (Questions 15-21)

Since radar is ineffective underwater, navies around theworld use sonar (SOund NAvigating and Ranging) forunderwater detection. There are two types: passive andactive. Passive sonar listens for underwater sounds. Activesonar emits a sound signal and listens for the echo from thebottom of the ocean or another vessel. If your vessel or theenemy vessel is moving, the frequency of the echo will beDoppler-shifted. The Doppler shift equation for frequencywhen both source and observer are moving is:

\v ± vs/

where v = speed of sound, v0 = speed of observer, vs =speed of source, / = original frequency, and /' = observedfrequency. For ease in calculations and measurementsunderwater, the speed of sound is taken to be 1500 m/s.

Active

Sonar

Passive

Sonar

Figure 1

In Figure 1, assume Submarine USA moves with speedVa and has an active sonar that emits signals with frequency500 Hz. Assume Submarine CIS moves with speed Vr and

has a passive sonar.

15. Submarine USA uses active sonar to measure its heighth above the ocean floor. What is the value of h, if attime t - 0 a sonar pulse is emitted, and four secondslater the echo is heard?

A. 1500 m

B. 3000 m

C. 4500 m

D. 6000 m

16. Bats use active sonar to detect small insects. How does

the bat's signal compare to the submarine's sonar?

A. Bat sonar must have a longer wavelength.B. Bat sonar must have a shorter wavelength.C. Bat sonar must have a lower frequency.D. Bat sonar"musthave a higher frequency.

Copyright ©by The Berkeley Review® 26

17. What is true of the sound?

I. The speed of sound is faster in water than in air.

II. When sound bounces off of an object that ismoving away from the detector, the sound willreflect with an increased frequency.

III. For a fixed frequency, increasing wavelengthrequires increasing the wave speed.

A. I onlyB. I and II onlyC. I and III onlyD. I, II, and III

18.

19.

20.

If Submarine CIS is moving at 20 m/s towardsstationary Submarine USA, and the passive sonaroperator in Submarine CIS is listening for SubmarineUSA's active sonar pulse, what is the frequency f\ of

the active sonar pulse, as measured by passive sonar onSubmarine CIS?

A. 491Hz

B. 496 Hz

C. 501Hz

D. 507 Hz

Submarine USA is moving at 10 m/s toward a secondactive sonar source (SASS). Both Submarine USA andSASS emit the same standard source frequency. IfSubmarine USA sends out a signal to SASS and thenlistens to the echo, how would the echo frequencydetected by Submarine USA compare to the sourcefrequency?

A. /echo > /source

"• /echo < /source

C. /echo = /source

D. It will vary with water temperature.

Starting with Submarine USA on the left andSubmarine CIS on the right, which condition will leadto the LOWEST observed frequency for SubmarineCIS?

A. Submarine USA moving to the right at 10 m/s,with Submarine CIS moving to the left at 20 m/s.

B. Submarine USA moving to the right at 10 m/s,with Submarine CIS moving to the left at 10 m/s.

C. Submarine USA moving to the left at 10 m/s, withSubmarine CIS moving to the right at 10 m/s.

D. Submarine USA is stationary, with Submarine CISmoving to the left at 10 m/s.


21. In Figure 1, Submarine CIS makes a sudden coursechange so that now both vessels are moving in the samedirection with the same speed va = vjj = 10 m/s.Submarine USA emits a sonar signal with frequency /,and that signal is detected by Submarine CIS withfrequency f\. How does f\ compare to /?

A. />/i

B. /!>/

C / = /i

D. / and f\ are unrelated.

Copyright ©byTheBerkeley Review® 27

Questions 22 through 25 are NOT based on a descriptivepassage.

22. The fundamental frequency of a pipe closed at one endis equal to the third harmonic of a pipe open at bothends. How are the lengthsof the two pipes related?

A.

B. L

^open = ""-closed

= ±L.open closed

*-"• ^open ~^closedD. '"open ~ "'"closed

23. The waves used in sonification can best be describedas:

A. traveling transverse waves.

B. standing traveling waves.

C. traveling longitudinal waves.

D. standing longitudinal waves.

24. The observed speed of sound is the SLOWEST in:

A. a vacuum.

B. air.

C. water.

D. salt water.

25. A surgical use for sound waves is to destroy gallstoneswith the vibrations. Two ways to measure the loudness

of a sound wave are intensity, I in W/m2, and intensitylevel, 6 in dB. The loudness in decibels is given by:

6=KMog10 [-where I0 is a reference intensity. Ultrasonic surgical

tools operate at 170 decibels, whereas imagers operateat 150 decibels. How much more or less intense are the

surgical devices than the imaging devices?

A. About 1000 times more intense.

B. About 100 times more intense.

C. About 10 times more intense.

D. About Vioas intense.

1. C 2. C 3. A 4. A 5. B

6. C 7. D 8 A 9. D 10. D

11. A 12. C 13. B 14. C 15. B

16. B 17. C 18. D 19. A 20. C

21. C 22. D 23. C 24. B 25. B

YOU ARE DONE.

Answers to 25-Question Sound and Doppler Effect Review

Passage I (Questions 1 - 7) Violin Strings

1. Choice C is the best answer. If the length of a typical violin string is 30 cm, what are the wavelengths of the first three

harmonics? In the passage, we were given the formula for the frequencies, which is: / = nv/2L- Usin8 tne 8olden ru,e for

waves, v=fX, we can solve for the wavelengths: X=VU. We find that: vlf =2L/n, so the wavelengths are given by: X=2L/n. Plugging in values for Land ngives:

Xi=2(30cm)=60cm


x 2(30cmi=30cm2

X3=2(30cm)=20cm3

Choice C is the best answer. What is the fundamental pitch of a string that has a length of 30 cm, a mass of 30 g, and atension of 9000 N? To start, pitch is just the perceptual correlate of what a physicist calls "frequency." Using the frequency

formula in the passage, / = nv/2L» we can plug in the fundamental pitch value n = 1 and the length value of 0.3 m after wecalculate the speed v. To calculate that, we have to use the other formula given in the passage:

_ JTL _


(9000)(.30) .-. v = V90,000 = 300 m/M V .03

Now that we know the speed, we can calculate the fundamental pitch, where n is equal to one:

/=nv =300m/{5- = 500 Hz

2L 2(0.3) m

Choice A is the best answer. What happens to the pitch (frequency) of a standing wave on a violin string when the tensionon that string is changed? To answer that question, we must combine the two formulas given in the passage:

/ = nv and v -2L

T-L

M••• /= •£,

2L

I^= nM 2

T

M-L.-. /« Vf

We can see from the formula that as the tension increases, the frequency must also increase. Notice that choices B and C givethe same relationship between/and T, invalidating both of them. The best answer is choice A.

Choice A is the best answer. Violin players can press down on the violin strings effectively to shorten the length of aparticular string. What happens to the frequency and wavelength of the standing waves when this occurs? According to the

rule for waves, X= vle. As/increases, Xmust decrease and vice versa. Since frequency is proportional to V^, decreasinglength means increasing frequency. Since the frequency is increasing, the wavelength must be decreasing. The best answeris choice A.

Choice B is the best answer. What change would double the fundamental frequency produced by a violin string? All of ouranswer choices involve changing the string tension, which means we need to know how frequency depends on tension. Toget this relation, we must combine the two formulas given in the passage. The result is:

2L

T-L

M/« Vf"

In order to double the frequency, we must quadruple the tension. The best answer is choice B.

Choice C is the best answer. Two strings have the same length and fundamental wavelength, but different fundamentalfrequencies. How is this possible? The wavelengths are given by X = 2L/n. The wavelengths are affected only by the length

VT'L . The factors that influence the frequency are T, L, and M. Since LM

is the same for both strings, a different T or a different M must be creating different frequencies, which is choice C. Themounting angle of the string will have no bearing on the frequency, according to the equation. The mounting angle wouldsimply affect the ability of the string to be displaced. The best answer is choice C.

Copyright ©byThe Berkeley Review® 28 MINI-TEST EXPLANATIONS

7. Choice D is the best answer. Whether it's standing waves produced on a violin string or sound waves produced by thatstring, sound waves are longitudinal. As part of your general physics knowledge, you should know that sound waves arelongitudinal. This was alluded to in the first paragraph of the passage. Based on this, we can eliminate choices A and C. Youshould also know that standing waves on strings are transverse. The best answer is choice D.

Passage II (Questions 8 -14) Train Whistle

8. Choice A is the best answer. If the train is in the station and blows Whistle 1 with frequency f\ = 440 Hz, what is thewavelength of the emitted sound?This is a straightforward application of v = fX, where v is 330 m/s, and / is 440 Hz. Thismeans that: >. = (330 m/s)/(440 Hz) = 0.75 m. If you remembered the proper equation, v = fX, which you should havememorized, you should have noticed that the wavelength had to be less than 1 m. The only two choices less than 1 m werechoices A and D. The best answer is choice A.

9. Choice D is the best answer. Suppose f\ = 440 Hz and fi = 450 Hz. If both whistles are blown at the same time while therain is in the station, what is the beat frequency? The beat frequency is given by /b = I/] - ft\. This means that/b = I440 Hz- 450 Hz I, which is 10 Hz. The equation for the beat frequency is another equation that should be tucked away in yourmemory. Choice A, 445 Hz, is the average frequency. Choice B, 890 Hz, is the sum of the two frequencies. Choice C, - 10Hz, is not possible, because frequency is by definition a positive quantity. The best answer is choice D.

10. Choice D is the best answer. Sound wave A has an amplitude of 1 mm, and Sound wave B has an amplitude of 3 mm. Whatare the maximum and minimum possible amplitudes, if A and B interfere? If A and B are in phase when they interfere, theiramplitudes will sum to give the maximum amplitude. If A and B are 180°out of phase when they interfere, their amplitudeswill cancel each other out.

11.

•=> A + B

W4 mm

B=10-log10

B-A

2 mm

A and B interfering in-phase A and B interfering out-of-phase

When we subtract the amplitudes, the result must be positive, because amplitude is by definition a positive quantity. Basedon that, choices B and C are eliminated. The math gives: Max = 3 mm + 1 mm = 4 mm, and Min = 3 mm - 1 mm = 2 mm.The best answer is choice D.

Choice A is the best answer. Whistle 1 has an intensity twice that of Whistle 2. Whistle 1 has an intensity level that is howmany dB greater than the intensity level of Whistle 2? The intensity level is given by:

(l\

Let's say the intensity of Whistle 2 is I, which means the intensity of Whistle 1 is 21.The intensity level of Whistle 1 is:

=> B=10-logI0W+ 10-logI0(2)*o/ v*o/

The first term is the intensity of Whistle 2. So Whistle 1 is 10 log 2 decibels greater than Whistle 2. Log 2 is 0.3, so Whistle 1is about 3 decibels louder than Whistle 2. To increase the intensity level by +10dB, the intensity must be increased by afactor of 10! The best answer is choice A.

6=10-log10 pi-

12. Choice C is the best answer. As the train approaches a station, it blows both whistles. A person standing on the platformhears both whistles with apparent frequencies f\' and ft. How are the frequencies related? When the train is approaching thestation, the perceived frequencies heard by the listener are greater than the actual frequencies at the sound source because ofa Doppler shift. We can definitely conclude that f\ < f\' and f2 < ft, which is choice C. Choice A is not the best answer,

because we cannot say for sure that J2 < /l'- Choice B is not the best answer, because we can't say for sure that /2 > /]'.

Choice D is incorrect, because /2 < ft, not J2 > ft- The best answer is choice C.

13. Choice B is the best answer. If the train's whistles are treated as pipes that are open at one end and closed at the other, whichcould be a picture of the air column in one pipe, if it is oscillating in its fundamental mode? When a pipe is open at one end,that means the particle displacement must be at a maximum at that end. When a pipe is closed at one end, that means theparticle displacement must be at a minimum at that end. The only picture among the answer choices that has a maximum atthe open end and a minimum at the closed end is choice B. The best answer is choice B.


14. Choice C is the best answer. The engineer blows Whistle 1 when the train is 660 m away from the tunnel (see diagrambelow). How much time elapses between the moment when the whistle is blown and the moment when its echo is heard bythe engineer? If the train were stationary, the sound of the whistle would take 2 seconds to reach the tunnel (660 maway),and the echo would take another 2 seconds to return to the engineer (assuming that the speed of sound in air is 330 m/s). Thetotal time for the initial sound wave of the echo to reach the ear of the engineer sitting in a stationary train would be 4seconds. But the train is not stationary. It is moving towards the tunnel at 20 m/s. Therefore, we can quickly figure out thatthe echo will require fewer than 4 seconds but more than 2 seconds to reach the engineer. This allows us to eliminate choicesA and D.

Where will the train be when the sound of the whistle reaches the tunnel? Remember that the sound waves takes 2 seconds to

reach the tunnel, during which time the train travels (20 m/s x 2 s) or 40 m, placing it at a position 620 m from the opening ofthe tunnel. The echo 1 second later will be 330 m away from the tunnel and approaching the oncoming train. The train willbe 20 m closer to the tunnel (i.e., 600 m away from the tunnel). The echo and the engineer are still 270 m apart from oneanother-and exactly 3 seconds have elapsed. The echo time must be greater than 3 s, which allows us to eliminate choice Band select choice C as the best answer.

The problem is also solved using algebra. When the echo and the engineer meet: (dwave - 660 m)+ dlrain = 660 m, and:dwave = vwave 'l > ^train = vtrain **• If vve let x eclua' tne l'me elaPsed untiI tne whistle is heard by the approachingengineer, then: (330 m/s • x) - 660 m + (20 m/s • x) = 600 m. Solving for x, vve get:

x = '320 m = 377 s350 m/s

While the algebraic answer may offer some reassurance that our intuition was correct, there is just not enough time to do thismath during the MCAT. As you prepare for your MCAT, be sure to emphasize visualization and intuition when determiningthe best answers to your questions. The best answer is choice C.

Passage HI (Questions 15 - 21) Submarine Sonar

15. Choice B is the best answer. Four seconds is the total time it takes for the signal to leave the submarine, hit the ocean floor,and travel back to the submarine.

d= vt = (1500 m/ )(4 S) =6000 m .-. h=i =600Q m = 3000 m2 2

Be wary of questions where the time for a round-trip is given. A standard trick question involvesasking for one-waydistanceafter giving you the round-trip time. The best answer is choice B.

16. Choice B is the best answer. Because the target for the bat is considerably smaller than for a submarine, the wavelengthmust be shorter for the bat's radar signal than the submarine's sonar signal. The wavelength of the signal must be roughlyequal in size to or shorter than the object. Regarding the higher or lower frequency, remember that the frequency will alsodepend upon the wave speed, which is considerably different for the sub's waterborne sonar and the bat's airborne sonar. Thebest answer is choice B.

17. Choice C is the best answer. The speed of sound is greater in water (the denser medium) than in air. 1500 m/s is roughly3000 miles per hour. Mach I (the speed of sound in air) is just above 700 miles per hour. This makes statement I true.According to the equation, when the object moves away from the observer, the frequency decreases, so statement II is false.The speed of any wave depends on its wavelength and frequency. Specifically, v = fX requires that wave speed increase, ifthe wavelength is to increase and the frequency is to remain fixed. This makes statement III valid. Choice C is the bestanswer. Notice that statement I is part of every choice, so you do not need to spend time considering that statement. The bestanswer is choice C.

Copyright © by The Berkeley Review® 30 MINI-TEST EXPLANATIONS

18. Choice Dis the best answer. There are two methods ofsolving this problem. Method #1: Plug the numbers into the Dopplerformula, noting that v0 = vb = +20 m/s (+because the observer moves towards the source) and v$ = v^ =0 m/s.

f, =f(v+Ift.] =500 Hz(1500 m/s +20 m/s|m 507 Hzv v / I 1500 m/s I

Method #2: Understand the physics and eliminate the impossible answers. The physics: When an observer approaches thesource, theobserver mustmeasure a frequency f\ greater than theoriginal frequency /. The impossible answers: 491 Hz and496 Hz. The only answer significantly greater than 500 Hz is choice D, 507 Hz. The best answer is choice D.

19. Choice A is the best answer. Submarine USA emits the signal with frequency / = 500 Hz. The active sonar source measuresthe frequency to be:

f, =f(l±lo) =500 Hz (1500 m/s +10 m/s) a 503 Hzv v / \ 1500 m/s /1500 m/s

The active sonar source is now the source of the echo sent back to Submarine USA. The echo will undergoa second Dopplershift as follows:

h =f1(—*—) =503 HzV - Vc

1500 m/s

1500 m/s- 10 m/s507 Hz

The standard emitted pulse is shifted only once: /echo > /source- The most difficult part of this problem is realizing that wemust apply the Doppler shift twice, with Submarine USA and the stationary source exchanging roles as source and observerafter the first Doppler shift. The best answer is choice A.

20. Choice C is the best answer. The lowest frequency, according to the equation, is observed when the source and the observerare moving away from each other at the greatest relative speed. In choices A and B, the two submarines are moving towardseach other, so in each case the Doppler shift will result in frequencies greater than 500 Hz. In choice D, the distance betweenthe two submarines is increasing (reducing the frequency), but the distance between the submarines does not increase at thesame rate as it does in choice C. Choice C is best.

500 Hz f1500 m/s + 10 m/s\ >^qq j_jz [1500 m/s + 10 m/s\ >*jqq ^z / 1500 m/s \ >^qq j_jz /1500 m/s - 10 m/s \\ 1500 m/s - 20 m/s/ \ 1500 m/s - 10 m/s I M500 m/s + 10 m/s/ M500 m/s + 10 m/s/


21. Choice C is the best answer. The Doppler shift depends on whether there is relative motion between the source and theobserver. Since both submarines are moving in the same direction with the same speed, there is no relative motion betweenthem. Thus, there is no Doppler shift; f\=f. Using the Doppler formula:

fl=f_ffv-vo\_f/i500 m/s- lOm/s _= f = fv - Vc 1500 m/s - 10 m/s

Since the observer is trying to move away from the source, you must choose the minus sign for v0. Since the source is trying

to move towards the observer, you must choose the minus sign for vs. The best answer is choice C.

Questions 22 - 25 Not Based on a Descriptive Passage

22. Choice D is the best answer. For a pipe that is open at one end and closed at the other end, the possible frequencies are:

/ = JL¥- (pipe closed at one end)4L

For a pipe that is open at both ends, the possible frequencies are:

/ = 1LY_ (pipeopen at both ends)2L

In the case where the pipe is closed at one end, n can only be an odd integer. The fundamental frequency occurs when n = 1.For a pipe open at both ends, n can be any integer. The third harmonic occurs when n = 3. Setting these two equations equalto each other, we solve for the relationship between the lengths of the pipes:

v _ 3v

4 Lclosed 2 L0pen•'• Lopen —6 Lciosecj

Note that the wave speed is the same for both pipes, since the waves produced are sound waves in air. The best answer ischoice D.


23. Choice Cis the best answer. If you are aware that sonification uses sound waves, then you can easily answer this knowingthat sound is alongitudinal wave that travels. Otherwise, we need to analyze each choice. Transverse waves cannot travel insimple fluids, so choice Ais incorrect and thereby eliminated. Awave is either traveling or standing, but not both. Thisallows us to eliminate choice Bimmediately. Traveling waves are those waves where the wave pattern moves through themedium. This is what happens to a sound wave sent out by a sonification source. Standing waves are formed by theinterference oftwo traveling waves, traveling in opposite directions, usually because of boundary conditions. Sonificationemploys sound waves traveling through mediums, so choice Cis the best answer and choice Dis eliminated. The bestanswer is choice C.

24. Choice B isthebest answer. Just like sound intensity, the speed ofsound also increases asa medium becomes denser. If weare looking for the medium in which the speed of sound is the slowest, we would be looking for the medium with the leastdensity. The medium ofleast density might at first appear to be avacuum (zero density); but avacuum is the absence of anymedium. Sound waves need a medium in order to propagate, because sound is transferred via molecular vibrations andcollisions. After eliminating choice A(avacuum), we should then notice that air is the least dense ofthe remaining materials.The best answer is choice B.

25. Choice B is correct. This question tests your skills with logarithms more than your skill with conceptual physics. Theformula you are to decipher is:

B=KMog10(£We are told the intensities in decibels (dB) and are asked to find the intensities in W/m2.1 is in W/m2, and Bis in dB. We arebasically being asked tosolve for I.Recall the rules for logs: x= log ymeans y = 10*.

i=i0-io(b/io)

Now, if we want to find a ratio of intensities in W/m2, we just plug the given values into the equation. For the imager, wehave:

lj= Io1015 and for surgery: ls=Io1017

I I 1017The ratio ofIs to lj isjust: ^. = o " _ 100ili Io1015

Notice that IQ dropped out of the equation. Try to avoid plugging in values for variables in a problem. The variable may justdrop out, savingyou time and helping you avoid an error.The best answer is choice B.


52-Question Sound and Doppler Effect Practice Exam

I. Sound Intensity

II. Propagation of Sound

III. Bat Echolocation


IV. Sound Experiment

V. Jet Radar


VI. Organ Pipes

VII. Sound Resonance


Sound and Doppler Effect Exam Scoring Scale

Raw Score MCAT Score

42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3

(1-6)

(7 -11)

(12 -17)

(18 - 21)

(22 - 27)

(28 - 32)

(33 - 36)

(37 - 42)

(43 - 48)

(49 - 52)

Passage I (Questions 1 - 6)

A study was done in order test thechange in intensity ofsound waves in different media. A source capable of emittingfrequencies of 200, 1000, and 5000 Hz was placed in avacuum, air, water, and liquid mercury. At a fixed distancefrom the source, a sound detector was also placed into themedium. The detector was moved to different points forvarious readings. Already knowing that the intensity of soundwaves in a liquid is higher than in a gas, the experimenterswere trying to confirm that Equation 1 accurately describesthe intensity of sound waves in a liquid.

Equation 1 I=i/pBo)2A22

In addition to testing the intensity of the sound waves atdifferent points, the study was also meant to measure thepenetration depth of different frequencies. A commonlyaccepted theory is that although sound is observed as a wavephenomenon, it also has particle properties. Drawing aparallel to light theory, vve know that light can be brokendown into corpuscles called photons. Similarly, it is believedthat sound can be broken down into small units called

phonons. According to this theory, a single phonon has anenergy given by Equation 2, where h is Planck's constant

(6.63 x 10"34 J-s), and / isthe frequency of the sound wave.

Equation 2 Ephonon - "f

A phonon is the smallest amount of energy that can becarried by a sound of a particular frequency. The reason wedo not notice this quantization in life is because the amountof energy of one phonon is minuscule, and it would be quitedifficult to isolate its energy.

1.

2.

The intensity of sound at a particular distance from itssource will be GREATEST in:

A. a vacuum.

B. air.

C. water.

D. mercury.

In a simple liquid thicker than air, a sound wave of afixed frequency is:

A. longitudinal with ^-Medium > Âir-

B. longitudinal with ^.Medium < Âir-

C. transverse with A-Medium > Âir-

D. transverse with ^.Medium < Âir-


3. Sound waves having which of the following frequenciesmoved FASTEST through water?

A. 200 Hz

B. 1000 Hz

C. 5000 Hz

D. Sound waves of frequencies 200 Hz, 1000 Hz, and5000 Hz move at the same speed through water.

4.

5.

6.

As we move away from the source of a sound wave,from a distance r to a distance 3r, the intensity of thatwave:

A. decreases by a factor of 3.

B. decreases by a factor of 9.

C. increases by a factor of 3.

D. increases by a factor of 9.

If all of the frequencies in this experiment were emittedat the same time into air along with a frequency of 1010Hz, which of the following frequencies would also bedetected?

A. 10 Hz

B. 810 Hz

C. 3990 Hz

D. All of the above.

If the known energy of a phonon with frequency 1500Hz is about 1.0 x 10"30 J, which of the followingamounts of energy carried by a 1500-Hz sound wavecould NOT be observed?

A. 1.0xl0-30J

B. 5.0xl0-30JC. 7.5xlO"30J

D. 1.0xlO-20J



Sound travels in the form of a wave, and is thereforegoverned by the laws of wave motion. One of these laws isthat the speed of a wave, v, is related to the sound wave'sfrequency and wavelength by:

v = Xf

Another law of wave motion deals with how sound

waves reflect when they hit an interface between twodifferent media. The ratio of the reflected wave's intensity tothe incident wave's intensity is given by:

^reflected _

incident

Plvl-P2v2\2Plvl + P2v2/

where pi and p2 are the respective densities of the incident

and reflecting media, and vj and V2 are the respective speeds

of the sound waves in the two media. Typically, waves willreflect more if pi < p2 or if P2 < pi, than if pi is

approximately equal to p2-

Sound waves come in a variety of frequencies, eachfrequency range having a different degree of importance inour lives. For instance, the frequency range betweenapproximately 20 Hz and 20,000 Hz is where human beingsdetect sounds.

Above 20,000 Hz, our ears fail us. However, this doesnot mean that these sounds are useless. People use theseultrasonic sounds in the medical field daily. They allowdoctors to "see" into the bodies of patients, where X-rayswould be useless and/or dangerous. The theory behindultrasound imaging is this: Send a sound pulse into the bodyand measure how long it takes for a reflected pulse to return.Knowing the sound wave's speed, a computer can turn theseechoes into pictures.

Imaging depends upon a sound wave's speed, so it helpsto know how fast sound travels in a given medium. Sometypical numbersfor speed in various media are the following:

7.

Medium p (kg/m3) v (m/s)

Air 1.05 344

Water 1000 1480

Flesh 1047 1570

Table 1

If, when using an ultrasound imager on flesh, the

reflected pulse is detected 10"^ sec after the emittedpulse is sent, the distance between the detector and thereflecting medium is:

A. .01570 m

B. .03140 m

C. .007850 m

D. .001570 m

Copyright ©byThe Berkeley Review®

8.

9.

10.

11.

35

Which of the following images would be MOSTdifficult for an ultrasound device to "see"?

A. Bone in flesh

B. Gas-filled stomach and stomach wall

C. Bladder wall and urine in bladder

D. Gallstones in the gallbladder

Which list of numbers represents the respective times ittakes for a sound wave to travel through 5 cm of air,water, and flesh?

A. 1.5 x 10-4 sec,3.4 x 10'5 sec, 3.2x 10"5 sec.

B. 3.2x 10"5 sec,3.4 x 10"5 sec, 1.5 x 10"4 sec.

C. 1.5 x 10"1 sec,3.4x 10"2 sec, 3.2x 10"2 sec.

D. 3.4x 10"5 sec,3.4 x 10"5 sec, 3.4x 10-5 sec.

When applied to the side of the head, sound wavesbounce off of the sides of the skull, and off of the midline of the brain. Two oscilloscope traces show timeversus sound intensity in the detector, when the detectoris placed on the left side of a patient's head and on theright side of a patient's head.

detector on left side detector on right side

k ilt t

From these plots, a specialist was able to conclude:

A. that the left side of the brain was larger than theright side of the brain.

B. that the right side of the brain was larger than theleft side of the brain.

C. that both sides of the brain were the same size.

D. nothing about the relative sizes of the brain'shemispheres.

If the ultrasonic imager operates at a frequency of157,000 Hz, what is the size of the SMALLEST objectthe imager could detect?

A. About 1 mm

B. About 1 cm

C. About 1 m

D. There is no minimum size.



Batsand dolphinsare examples of animals that use sonarto locate food in their environments. Sounds that a bat emits,for example, reflect off nearby insects and return to theanimal. The elapsed time of these echoes allows the bat todetermine the distance to its food.

It is believed that bats know not only the distance to agiven insect, but also the relative speed at which it moves.Such information comes from a Doppler shift in thefrequency of the emitted sound. The relationship betweenthe reflected frequency fr and the emitted frequency fe isshown in Equation 1.

_(v +vt) f

Equation 1

when the target is moving at a speed vt relative to the bat.Here, v, > 0 means that the bat and target are approaching

each other; v is the speed of sound. (Unless otherwise stated,the speed of sound is 340 m/s.)

12. A bug and a bat are initially 17 m apart, when the batemits a sonic chirp. The creatures approach each otherat a relative speed of 0.5 m/s. How long after emittingthe chirp does the bat hear its reflection?

A. 0.0498 second

B. 0.0502 second

C. 0.0999 second

D. 0.1001 second

13. Assuming that the bat's ultrasonic chirp is emitteduniformly in all directions, and that the sound intensitylm from the bat is 25 dB, what is the approximatesound intensity 2m from the bat?

A. 6dB

B. 12 dB

C. 19 dB

D. 25 dB

14. A bug approaches a stationary bat at 0.1% of the speedof sound. If the bat emits a chirp at some ultrasonicfrequency, the reflected frequency will be:

A. 122.2% of the emitted frequency.

B. 100.2% of the emitted frequency.

C. 99.8% of the emitted frequency.D. 81.8% of the emitted frequency.


15.

16.

When a wave travels from air into water, its speedincreases by a factor of four. Compared to the wave inair, the transmitted wave has:

A. the same frequency, but a longer wavelength.

B. the same frequency, but a shorter wavelength.

C. the same wavelength, but a higher frequency.

D. the same wavelength, but a lower frequency.

All of the following play a role in echolocation whendetermining position and speed of an object EXCEPT:

A. a difference in frequency between the emitted andreflected waves.

B. a difference in wave speed between the emitted andreflected waves.

C. the speed of the object

D. a difference in wavelength between the emitted andreflected waves.

17. The HIGHEST reflected frequency is observed with a:

A. stationary bat and a bug flying towards itB. stationary bat and a bug flying away from it.

C. bat flying towards a bug that is flying away fromthe bat

D. bat flying towards a bug that is flying towards thebat


Questions 18 through 21 are NOT basedon a descriptivepassage.

18.

19.

20.

For a sound wave traveling through air at roomtemperature and pressure, which plot BEST representsthe relation between the wave's frequency and itsspeed?

A. B.

8.

020 20,000

Frequency (Hz)

C.

8.

020 20,000

Frequency (Hz)

D.

20 20,000Frequency (Hz)

K

020 20,000

Frequency (Hz)

Which relationship BEST relates the pitch of the sirenfor an ambulance relative to a stationary observer?

A. /approaching > /stationary > /leavingB. /leaving >* /stationary > /approachingC /stationary > /approaching > /leavingD. /stationary > /leaving > /approaching

Two cars have identical horns with a frequency of 400Hz. Which situation will NOT generate a beatfrequency when both cars simultaneously honk theirhorns?

A. Car I is stationary; car II is moving left at 20 m/s.

B. Car I moving right at 20 m/s; car II is stationary.

C. Car I moving left at 20 m/s; car II is moving rightat 20 m/s.

D. Car I moving right at 20 m/s; car II is moving rightat 20 m/s.


21. Which of the following graphs BEST represents thefrequency of a sound wave emitted from a closed pipeas a function of the length of the pipe?

A. B.

C.

oc

3

i"

Length (L) Length (L)

Length (L) Length (L)


Passage IV (Questions 22 - 27)

In an experiment to demonstrate how resonances work,physics students gather five tuning forks of frequencies 1020Hz, 850 Hz, 680 Hz, 510 Hz, and 340 Hz. The students alsohave a glass tube of length 1 m and diameter 8 cm. The glasstube is placed on a table and completely filled with water.Starting with the tuning fork with the highest frequency, eachtuning fork is struck and placed over the open end of thewater-filled tube, as in Figure 1. While each tuning fork isheld over the tube, the water level is slowly lowered until aloud resonance is heard. Once the resonance is heard, the

water height is recorded.

Tuningfork

Tihei

1

beghtm

1Waterheight

h

>

1 I

Figure 1. Experimental Apparatus

Theoretically, resonance should occur whenever thefollowing condition is met:

f _ nv

4L

where n = 1,3,5,..., f is the frequency of the tuning fork, v isthe speed of sound in air, and L is the length of the aircolumn. The students also measure the fundamental mode, n

= 1. The experimental data are summarized in Table 1.

Trial number Frequency (Hz) Water height (cm)

1 1020 91.7

2 850 90.0

3 680 87.5

4 510 83.3

5 340 75.0

Table 1

The experiment was carried out with a constant roomtemperature of 20 °C.

22. What is the wavelength of sound produced by thetuning fork in Trial 3?

A. 87.5 cm

B. 50.0 cm

C. 25.0 cm

D. 12.5 cm


23. How would the experiment need to be changed, if onedesires to produce the same frequencies as thosemeasured in the experiment, but when the roomtemperature is 25 °C? (Ignore thermal expansion of thepipe.)

A. The frequencies of all of the tuning forks wouldneed to be increased.

B. The frequencies of all of the tuning forks wouldneed to be decreased.

C. The water height would need to be lowered.

D. The water height would need to be raised.

24. How does the speed of sound in Trial 5, V5, compare

with the speed of sound in Trial 3, V3?

A. V5 = 2V3

B. v5 = V2v3C. v5 = v3

D. V5 = ^v3

25. The experiment will NOT work, if the tube space abovethe water is filled with:

A. 0°Cair.

B. 0°C ice.

C. 20°C water.

D. 100°C steam.

26. In Trial 5, the height of the water was 75.0 cm. Thetuning fork from which other trial could be used toproduce a resonance for a water height of 75.0 cm?

A. Trial 1

B. Trial 2

C. Trial 3

D. Trial 4

27. Suppose the tuning fork in Trial 4 were measured tohave an intensity level of 50 dB. If a second, identicaltuning fork were struck and brought near the first fork,together they would have an intensity level of:

A. 50 dB.

B. 53 dB.

C. 60 dB.

D. 100 dB.


Passage V (Questions 28 - 32)

In the landing of jets, radar systems are employed totrack and analyze the jet's pathway. In earlier years, this wasaccomplished when the jet emitted a radio signal of the samefrequency as a stationary beacon positioned at some distancefrom the runway. The motion of the jet affects the frequencyof the emitted signal, according to the Doppler equation:

Equation 1v± vs

In Equation 1, ft is the frequency received by thelistener, /s is the frequency emitted by the sender, v is thevelocity of the transmitted wave, vl is the velocity of thelistener, and vs is the velocity of the source. The ± varies

with the direction of the motion of the source and listener.

Because the signal from the beacon is not shifted, thedifference in frequency between the jet's emitted signal andthe beacon's emitted signal can be used to determine thevelocity of the jet. The difference in frequency between thetwo signals is known as the beatfrequency. This is given inEquation 2:

Equation 2 /beat ='/1-/2'

In Equation 2, /beat is me Deat frequency, f\ is the

frequency of Wave 1, and /2 is the frequency of Wave 2.

Combining the two equations for the airport scenario, with amoving emitter and a stationary emitter, the beat frequencyfor the jet signal and beacon signal is given in Equation 3:

Equation 3 /beat /l

where f\ is the frequency of the beacon's and jet's emittedwave, without accounting for Doppler-shifting.

Modern systems employ radar guns, where a signal issent out and the reflected signals are analyzed. The beatfrequency is obtained when one wave reflects off of astationary object, while the other wave reflects off of amoving object.

28. When using active radar, where radio waves are emittedfrom the same point at which the reflected waves aredetected, all of the following should be consideredEXCEPT:

A. the reflected wave is double Doppler-shifted by themoving object.

B. a stationary reference beam is needed.

C. the gun operates when stationary and in motion.

D. the best electromagnetic radiation is in theultraviolet range.


29. Will radar function in a vacuum?

A. Yes, because longitudinal waves do not need amedium.

B. No, because longitudinal waves need a medium.

C. Yes, because electromagnetic waves do not need amedium.

D. No, because electromagnetic waves need amedium.

30. Radar guns emit radio waves. The radar gun measuresthe beat frequency between the returning waves. Howdoes the radar gun determine velocity?

A. By measuring the beat frequency between twoconsecutive reflected waves.

B. By measuring the beat frequency between a wavereflected from a stationary object and one reflectedfrom a moving object.

C. By measuring the beat frequency between a wavereflected from two separate objects moving at thesame speed.

D. By measuring the beat frequency between a wavereflected from two separate objects moving atdifferent speeds.

31. Using radar, for which object would it be hardest toevaluate its velocity?

A. A moving train

B. A moving car

C. A flying bird

D. A flying insect

32. How can the frequency shift of a radio wave reflectedoff of a moving object be determined?

A. A/ = /0(v)

B. A/ = /0(c)

C. A/ = /0(v/c)

D. A/ = /0(2v/c)


Questions33 through 36 are NOT based on a descriptivepassage.

33. For the following harmonic frequencies, what is true?

53 Hz, 159 Hz, 265 Hz, 371 Hz, etc...

A. They are from a closed pipe with a fundamentalfrequency of 53 Hz.

B. They are from an open pipe with a fundamentalfrequency of 53 Hz.

C. They are from a closed pipe with a fundamentalfrequency of 106 Hz.

D. They are from an open pipe with a fundamentalfrequency of 106 Hz.

34. The depth finder on a boat sends a pulse of soundtoward the bottom and measures the time for the echo

to return. Could such a depth finder distinguish betweenthe following three things?

I. A rocky sea floor

II. A rocky sea floor, covered with a watery silt

III. A school of fish, swimming near the bottom

A. I and II can be distinguished from III, but not fromeach other.

B. I, II, and III can all be distinguished from eachother.

C. I, II, and III cannot be distinguished from eachother.

D. II and III cannot be distinguished from each other.

35. Red-shifting occurs when astronomers observe planetsmoving away from their telescope. It is attributed to theDoppler effect What causes a blue shift?

A. A planet moving towards the observer, where thelight is shifted to a lower frequency.

B. A planet moving away from the observer, wherethe light is shifted to a lower frequency.

C. A planet moving towards the observer, where thelight is shifted to a higher frequency.

D. A planet moving away from the observer, wherethe light is shifted to a higher frequency.


36. Sound waves are:

A. transverse.

B. longitudinal.

C. either transverse or longitudinal, depending on theair temperature.

D. either transverse or longitudinal, depending on thealtitude.


Passage VI (Questions 37 - 42)

Some organ pipes are open at both ends, while others areopen at one end and closed at the other. They amplifyselected frequencies (harmonics) based on the dimensions ofthe pipe and speed of the wave. An open pipe is illustrated inFigure 1 below.

air in- =r^<— lip

Figure 1

They emit sound when air is blown around a lip near oneend. This air turbulently enters one end of the pipe, and cancause air in the pipe to resonate at various frequencies.These frequenciesare given by Equation 1.

f _ nv

Equation 1

where n = 1,2,3... and v is the speed of sound in the pipe.

Pipes typically resonate at their fundamental frequencyof n = 1, but can also resonate in higher modes dependingupon the turbulence of the incoming air. As a general rule,the lowestresonantfrequency is of the highest intensity.

37. All of the following are true of harmonics in an openpipe EXCEPT:

A. the beat frequency of any two harmonics is also aharmonic.

B. that /n = n(/0), when n is the harmonic numberand/0 is the fundamental frequency of the pipe.

C. that X.n = n(XQ), when n is the harmonic number andXq is the fundamental wavelength of the pipe.

D. the length of a pipe impacts the harmonics morethan the diameter of the pipe.

38. Two pipes with identical dimensions differ only at theirends: pipe #1 is open on both ends and pipe #2 is openonly on one end. What is the ratio of the fundamentalfrequency of pipe#1 to that of pipe #2?

A. 2:1

B. 3:2

C. 2:3

D. 1:2


39. To most effectively increase the frequency in the pipe,one should replace the N2 in the air with:

A. He, and lengthen the pipe.

B. H2, and lengthen the pipe.

C. He, and shorten the pipe.D. H2, and shorten the pipe.

40. If you increase the temperature of the air by 44%, thespeed of the sound wave increases by:

A. 6.6%

B. 20%

C. 44%

D. 107%

41. For an open pipe that is 15 cm long, what is thefrequency associated with the third harmonic, assumingthe speed of sound in air is 340 m/s?

A. 3.4 Hz

B. 34 Hz

C. 340 Hz

D. 3400 Hz

42. The frequency spectrum of one organ pipe is measuredand plotted below. (The dominant frequencies and zeroare enumerated.) What is the pipe's length?

dB

A. 0.3 meter

B. 0.5 meter

C. 1.0 meter

D. 2.0 meter

lllll IT

85 170

.1.255 Hz


Passage VII (Questions 43 - 48)

In order to better understand sound waves, a group ofstudents performed a set of experiments using strings andpipes of different lengths. All of the strings were tied downat both ends. Some of the pipes were open at both ends,while some of the pipes were closed at one end and open atthe other end. To generate waves on the strings, the stringswere attached to a mechanical wave generator. Waves in thepipes were generated by means of different tuning forks. Theforks and pipes were arranged so that they were parallel toone another.

For strings that are tied down at both ends and for pipesthat are open at both ends, the possible frequencies forvibration are:

/= nv;n = l, 2,3,...2L

For pipes that are closed at one end, the possiblefrequencies for vibration are:

/=nv;n = 1,3,5,...4L

In all cases, v is used to represent wave speed and L isused to represent the pipe length.

43. Consider a pipe open at both ends and a string tied atboth ends. If the length of each is 1 meter, which of thefollowing statements is true about the waves generatedin each?

A. The waves will have the same wavelength butdifferent frequencies.

B. The waves will have the same wavelengths and thesame frequencies.

C. The waves will have the same frequencies butdifferent wavelengths.

D. The waves will always have different wavelengthsand different frequencies.

44. If the temperature of the room changes from 30°C to33°C, which of the following statements is true?(Ignore any thermal expansion of the pipe and string.)

A. The pitch of the wave produced by the stringchanges; the pitch of the wave produced by thepipe does not.

B. The pitch of the wave produced in the pipechanges; the pitch of the wave produced by thestring does not.

C. The pitchof both waves changes.D. The pitch of neither wave changes.

Copyright©by TheBerkeley Review® 42

45. A pipe open at both ends produces a frequency of 440Hz. A string tied a both ends vibrates with a frequencyof 430 Hz. Which of the following statements is true?

A. The two waves will never produce a beatfrequency, since the waves come from differentsources.

B. The waves will produce a beat frequency undermost circumstances.

C. A beat frequency of 10 Hz will be produced.

D. The two waves will exhibit almost completelydestructive interference.

46. Which will produce the higher frequency, a string tieddown at both ends or a pipe closed at one end?

A. The string.

B. The pipe.

C. They produce waves with theequal frequencies.D. Notenough information given.

47. How could thestudents generate a wave ona string thathad the same shape as the wave generated in a pipeclosed at one end?

A. It is not possible to generate an identical wavebecause the nodes and antinodes of the two waveswould be reversed.

B. The waves could be identical when the string istied at one end, and untied at the other end.

C. The waves could be identical when the string istied at both ends, and the string is the same lengthas the pipe.

D. The waves could be identical when the string istied at both ends, and the string is twice the lengthof the pipe.

48. If the students were able to perform their experiments ina vacuum chamber:

A. wavescould be produced in the pipes.B. wavescould be produced on the strings.C. waves could be produced in the pipes and on the

strings.

D. waves couldnot be produced in eitherthestrings orthe pipes.


49.

50.

A 50-dB noise is how much farther away from itssound source than a 100-dB noise?

A. 40 times farther away.

B. 160 times farther away.

C. 316 times farther away.

D. 10,000 times farther away.

A bat uses a sonic chirp to echolocate nearby smallinsects in the hopes of feeding upon them. What wouldbe the effect of putting a piece of glass between a batand its prey? The bat would:

A. not perceive a difference in the prey's location.B. perceive the prey as farther away.

C. perceive the prey as closer.D. detect a reflection of itself.

51. When sound initially traveling from a source on the leftstrikes the following barriers and echoes back, what isobserved for the various echoes?

h—a—*+Air

AAAA/1Echo I

Ro( Water

mmmmEf-;\

vAAA/fclKEcho II

A. techo II = 2(techo i); echoes I and II have the sameintensity.

B. techo II = 2(techo l)5 echo I has a greater intensitythan echo II.

C. teCho II > 2(techo i); echo I has a greater intensitythan echo II.

D. techo II < 2(techo l)J echo I has a greater intensitythan echo II.


52. A 450 Hz tuning fork shows a beat frequency of 12 Hzwith another tuning fork. The other tuning fork has:

A. no definite frequency.

B. a frequency of 438 Hz and 462 Hz.

C. a frequency of 444 Hz or 456 Hz.

D. a frequency of 438 Hz or 462 Hz.

1. D 2. A 3. D 4. B 5. D 6. C

7. C 8. C 9. A 10. B 11. B 12. C

13. C 14. B 15. A 16. B 17. D 18. D

19. A 20. D 21. C 22. B 23. C 24. C

25. C 26. A 27. B 28. D 29. C 30. B

31. D 32. D 33. A 34. A 35. C 36. B

37. C 38. A 39. D 40. B 41. D 42. D

43. A 44. B 45. C 46. D 47. B 48. B

49. C 50. C 51. D 52. D

YOU ARE DONE.

Answers to 52-Question Sound and Doppler Effect Exam

Passage I (Questions 1 - 6) Sound Intensity

Choice D is the best answer. It was stated in the passage that the intensity of sound in a liquid is higher than the intensity ofthe same sound in a gas. Since sound does not exist in a vacuum, there is no sound intensity in a vacuum, either. Taking bothstatements into account, it is apparent that either choice C or choice D is the best answer, because they both involve liquids.Looking at the intensity equation given in the passage, we should notice that sound intensity is proportional to the square rootof the density of the medium through which it travels. A medium with higher density transmits sound with a higher soundintensity as well. Since mercury is denser than water, the intensity is greater in mercury than in water. The best answer ischoice D.

Choice A is the best answer. Sound waves are always longitudinal waves, so choices C and D can be eliminatedimmediately. When sound (longitudinal) waves enter a medium that has a larger bulk modulus, the velocity of these wavesincreases. If you recall the wave equation v = Xf, then you know that as the velocity of a wave increases, so does itswavelength, assuming that frequency is constant. Both the speed and the wavelength of sound waves must be greater in theliquid than in air. The best answer is choice A.

Choice D is the best answer. Although one frequency may penetrate farther into a medium than another, sound waves of allfrequencies move at the same speed in a given medium. For this reason, sound waves of all of the frequencies mentioned inthe question move at the same speed in water. The best answer is choice D.

Choice B is the best answer. The intensity of a sound wave decreases as the inverse squareof its distance from the source.Looking at the equation I = P/A = P/4jrr2, you should be able to deduce that as the radius increases by a factor of 3, theintensity decreases by a factor of 9. The best answer is choice B.

Choice D is the best answer. If a frequency of 1010 Hz were also emitted, all of the frequencies mentioned would be beatfrequencies: /Beat = l/l - ft- Typically, beat frequency problems discuss only waves of nearby frequency, which may haveled you to pick choice A. A beat is just an interference effect between two waves, regardless of their frequency. The bestanswer is choice D.

Choice Cis the best answer. 7.5 x 10"30 Jof sound energy could not be measured in asound wave of 1500 Hz if the energyof one phonon were 1.0 x lO'30 J, because 7.5 x 10"30 J is not a whole number multiple of 1.0 x 10"30 J. All of the otheramounts of energy are whole number multiples of the energy of one phonon, 1.0 x 10"30 J and thus could be measured in asound wave of 1500 Hz if the appropriate number of phonons were present. For instance, choice B is possible, because five1500 Hz phonons collectively could sum to 5.0 x 10"30 J, making choice Ba viable amount of total energy. The bestanswer is choice C.

Passage II (Questions 7-11) Propagation of Sound

7. Choice C is the best answer. Let's consider what is happening when a sound wave, incident in some Medium 1, reflects offof some Medium 2:

Medium 1

'incident ) J

'refelcted

-* ci

Medium 2

((

We hear the echo after the reflected wave returns to the incident wave's starting point. By the time t has elapsed, the waveshave traveled a distance 2d at some speed v. We are told v= 1570 m/s (for flesh), and t = 10"5 s. We are solving for d in anequation that reads: "total distance = speed x time." This is 2d = vt. Solving for d yields:

^3 "»/sXld_(l.57x ii;-' m if) 5s) _(.0157 m)2 2

The answer has to be less than .0157 m (choice A), so rule it out. The only possible remaining choices are C and D, butchoice D is .00157 (a factor of 10 smaller the .0157). Thus, choice D is also eliminated. The best answer is choice C.

Copyright ©by The Berkeley Review® 44 REVIEW EXAM EXPLANATIONS

8. Choice C is the best answer. In the passage, we are told that sound (incident in a Medium 1) will not reflect much fromMedium 2, if the densities of the two media are close in value to each other:

Reflected _[Plvl -P2V2Wident Wl+P2V2

But we can answer this problem without resorting to the equation. The conceptual information in the passage is sufficient.Why is choice C the best answer? Urine's density is close to that of water. The bladder's density is close to that of flesh.From the chart in the passage, water and flesh also have densities that are close to each other. This should make sense, sincehuman bodies are just "bags of mostly water." Thus, the reflections between the urine and bladder wall are tiny (even thoughthey are detectable). All of the other choices have two media of very different densities: bone and flesh, gas and flesh, stoneand flesh. These media all give good reflections, so none of them is the best answer. The best answer is choice C.

9. Choice A is the best answer. Sound waves travel faster through flesh than they do through water, and faster through waterthan they do through air. Thus, the numbers we want are in decreasing order (for air, water, and flesh), eliminating choice D.Choice B is incorrect, because the numbers are arranged in an increasing order. Choice C has the correct order, but themagnitudesof the numbers are wrong. For example, it should take very little time for sound to travel 5 cm in air. This choicesays that it takes 1.5 x 10"1 sec. If that were true, it would take around 1 second for sound to travel 1 m. That violatesphysical intuition. Thus, rule out choice C. We could have multiplied out all of the numbers, but it takes a longer time thanusing physical intuition. The best answer is choice A.

10. Choice B is the best answer. The oscilloscope traces look like this:

I detector on left side I detector on right side

11.

h aThe detector on the left side of the head: We see the midline echoafter a short time. If the midline echo were in the middle ofthe plot (i.e., halfof the time to hear the"right side of the skull echo"), then the midline would be in the middle of the head.But since we see the echo sooner than that, the midline must be closer to the left side of the skull than to the right side.

The detector on the right side of the head: This plot confirms our suspicions about the first plot. Since vve see the midlineecho at a fairly late time in this plot, it must mean than the midline is farther from the right side than it is from the left side.The conclusion is that the patient has brain hemispheres of unequal size, the right hemisphere being larger than the lefthemisphere. The best answer is choice B.

Choice B is correct. To understand this answer, let's see how a wave behaves when it hits some object. Here are two cases:a boat being hit by waves, where the wavelength of the waves is smaller than the size of the boat; and a buoy being hit by thesame waves, where the buoy size is smaller than the wavelength of the waves.

boat'sshadow

d>X

waves

buoyO-

6<X

X

buoy'sshadow

Notice how you can see a "shadow" of the boat in the water, but you cannot see much of a "shadow" for the buoy. Thisbehavior applies to all waves: water, sound, or light. Ultrasonic waves do not allow you to see objects that are smaller thanthe wavelength of the sound, which is found using the equation v = fX.Thus, X= \/f. Bysubstitution, we get:

1570 m/s _ 1.57 x 103 m/sX =

157,000Hz L57x 105 1= 0.01 m= 1 cm

Using the math works fine here. However, we could have eliminated some of the answers without math. Choice Dis invalid.It states that there is no limit to what the ultrasound at 157,000 Hz can "see." If that were true, you could see individual

atoms with your eyes. Your eyes see light with wavelengths of around 10"7 m, and atoms are about 10"10 min size.Choice C is incorrect. It says that this typical ultrasound device cannot resolve anything with a diameter less than 1 m. If thiswere true, then ultrasound devices at this frequency would be useless. The best answer is choice B.


Passage HI (Questions 12 -17) Bat Echolocation

12. Choice C is the best answer. Let's find the approximate time by assuming the bat and bug are not moving. We ignore theirmotion, because incorporating it into a calculation is hard, and because their slow motion (relative to the sound speed) won'taffectour answer much. Using Decho travels = Vsoundtecho travels- and noting that Dech0 travels = 2(17 m) = 34 m gives an echotime of 0.1 sec. The actual time should be slightly shorter because the two creatures are approaching one another. This ischoice C. If you forgot that the echoing sound must travel to the bug and back, you probably picked choice A. The bestanswer is choice C.

13. Choice C is the best answer. When doubling the distance r from an isotropic sound source, the intensity / becomes 74 as

strong because / « 7r2. However, the intensity level, fi, does not drop by this amount. Because intensity should drop, choice

D is incorrect. To figure out the amount by which fi drops, recall that dividing the intensity by 10 is equivalent to subtracting

10 from the intensity level. If / became /io,h as its original value, fi would be 25dB - lOdB = 15dB. Since / doesn't dropquite that much,fi should be larger than 15dB. This points to choice C. If you want to be a little more quantitative, we knowthe intensity drops by a factor of four. Decibels are measured on a log scale, and log 4 is 0.6, which means 10 (log 4) = 6. Theintensity should drop by 6dB, resulting in an intensity of 19dB. The best answer is choice C.

14. Choice B is the best answer. Since the bug and bat approach each other, the reflected frequency should exceed the emittedfrequency. This rules out choices C and D. Since the bug speed is very small compared to the speed of sound, the increase inthe reflected frequency should be very small. This points to choice B. The math is as follows:

(v-vt)


ft - (v +v0 fc ... L_ = v+vt = v+0.001 v = 1.001 v/e v - v, v - 0.001 v 0.999 v

1.002, a 0.2% increase

15. Choice A is the best answer. When arw. wave travels from one medium into another, its frequency remains the same. Theonly casewhere thefrequency of the wavechanges is when the sender or receiver are moving relative to one another, which isnot the case here. This rules outchoices C and D. Since v =/A, an increased speed v means an increased wavelength A(when/is fixed). The best answer is choice A.

16. Choice B is the best answer. Echolocation, as described in the passage for a bat, involves analyzing the reflected wave forfrequency shifts and echo time. Because the wave reflects off a moving target, it experiences a Doppler shift. This means thatthe frequency and wavelength are varied. This eliminates choices A and D. The target must be in motion relative to theanalyzer, so the speed of the object definitely plays a role. This eliminates choice C. The only thing that does not occur is ashift in the wave speed. That is observed when the medium changes, which is not the case in echolocation for a bat (whichremains in the air during the process). In reality, because temperature fluctuates in fluid mediums, wave speeds do change asthey pass through thermal pockets within a medium. However, that is not described in the echolocation procedure used by abat. The best answer is choice B.

17. Choice D is the best answer. The highest reflected frequency results from the largest Doppler shift. The Doppler shift isgreatest when the objects are moving towards oneanother at the highest possible speed. This eliminates choices A and B. Inchoice D, the two creatures are flying towards one another, so their separation distance is decreasing faster than in choice C,where they are flying in the same direction. The best answer is choice D.


18. Choice D is the best answer. Using a bitof common sense and common experience will lead you directly to choice D. Allof the frequencies on the plots are within our hearing range, but they are different in describing how sounds of differentfrequencies are propagated. Choice C is incorrect: It says that at some frequency ranges, v = 0 (i.e., sound waves do notevenpropagate). If that were the case, then as someone sang a series of notes ranging from a low pitch toa high pitch, there wouldbe a range of sounds not heard. This does not happen. Choice B is incorrect: It says that sounds of low pitch travel muchfaster than sounds ofhigh pitch. If this were true, then you would hear sounds having a higher frequency first when listeningto a distant sound. Since you can hear and usually recognize someone speaking from a distance, and since voices combineseveral frequencies, this variation in wave speed with changing frequency must not happen. Choice A is also incorrect: Itsays that the speed varies with frequency in some periodic fashion. This would garble talk and other sounds, and it istherefore not likely to be a genuine physical phenomenon. The best answer is choice D.

Copyright©by The Berkeley Review® 46 REVIEW EXAM EXPLANATIONS

19. Choice A is the best answer. The pitch of a sound you hear is the perceptual correlate of the frequency of that sound wave.The siren at rest has a fixed frequency for the sound it emits. When the source is approaching, the waves get compressed, andthus its perceived frequency increases. This means that an approaching siren has the greatest frequency (highest pitch) of thedifferent frequencies in the question. Only choice A lists the approaching frequency as the highest. The best answer ischoice A.

20. Choice D is the best answer. When the cars are moving in the same direction with the same speed, their relative speed iszero. Therefore, each car hears 400 Hz coming from its own horn and also from the other car's horn. Because there is nodifference between the frequencies, there will be no beats. In choices A, B, and C, the cars have different speeds, so theirhorns will emit waves that are shifted differently, and thus there would be beats and thus a beat frequency. The best answeris choice D.

21. Choice C is the best answer. The question asks for the graph that BEST represents the frequency of a sound wave emittedfrom a closed pipe as a function of the length of the pipe? Let's consider the answer choices:

A. L B. C. I D.

>>

au3~

-J

X

Length (L) Length (L) Length (L) Length (L)

Frequency,/, and pipe length, L, are related through / = nv/4L- This means that/is proportional to 7^, so as one increases,the other value must decrease, eliminating choice B. The relationship is an inverse one, which is represented by anasymptotic curve,choice C. The best answer is choice C.

Passage IV (Questions 22 - 27) Sound Experiment

22. Choice B is the best answer. What is the wavelength of the sound produced by the tuning fork in Trial 3? To get the

wavelength of this sound, we could resort to the golden rule for waves, v= fX. Solving for wavelength gives X= vlj. Theonly problem is that vve do not know the sound speed, v. We could combine ?*. = v/fwith / = nwl^.Taking n = 1and solvingfor the wavelength gives X = 4L. L is the length of the air column, which is the height of the tube minus the height of thewater. For Trial 3, L = 12.5 cm. Therefore, the wavelength is X = 4L = 50 cm. The best answer is B.

23. Choice C is the best answer. How would the experiment have to be changed, if the same results were desired in a roomheated to a temperature of 25 °C? Since the same measured frequencies are desired, the same tuning forks should be used(heating the forks by 5 °C would not noticeably change their frequencies). This rules out choices A and B. Heating the room

increases the speed of sound in air. Using the equation given in the passage (/ = nv/4L, with n set equal to 1) shows that theair column length L must increase by the same percentage as the sound speed, if the same frequencies are to resonate. Thismeans that the water height must be lowered. The best answer is choice C.

24. Choice C is the best answer. How does the speed of sound in Trial 5, V5, compare with the speed of sound in Trial 3, V3?The speed of sound is governed by the medium, which in this case is air. Since the air temperature is a constant 20°Cthroughout the experiment, we conclude that the speed of sound is also constant for the experiment. This means V5 = V3. Theonly way to change the speed of sound in an ideal gas is to change its temperature or pressure. We are assuming air behavesas an ideal gas at room temperature. The best answer is choice C.

25. Choice C is the best answer. The important feature that allows the experiment to work is that there is a phase difference thatcreates an interface at which the sound wave reflects. Choices A and D are both gases. The original experiment is carried outwith a gas (20 °C air) above the water and should work whether the gas is 0 °C or 100 °C, invalidating both choices. If thespace above the water is filled with either a solid or liquid, sound can still propagate through both media. However, if themedium is 20 °C water, there will be no interface, and thus there will be no wave reflection at the water level measured in theexperiment. Wave reflection requires an interface between two media that propagate sound at different speeds. If thatdifference is not present, there is no reflection, which precludes any resonance other than that involving the entire tubelength. The best answer is choice C.

Copyright ©byThe Berkeley Review® 47 REVIEW EXAM EXPLANATIONS

26. Choice A is the best answer. In Trial 5, the height of the water was 75.0 cm. The tuning fork from which other trial could beused to produce a resonance for a water heightof 75.0 cm? If the water height is 75.0 cm, the heightof the air column must be25.0 cm. Which other tuning fork could have a resonance with an air column of 25.0 cm? The relation that must be satisfied is:

/ = nv/4j. The tuning fork for Trial 1 had an aircolumn of L= 8.3 cm. If this length were multiplied by 3, we would get 25cm. This means the tuning fork for Trial 1 can resonate with a water height of 75.0 cm. A picture might clarify the situation:

air column vibratingin mode n = 1

air column vibratingin mode n = 3

The frequency in Trial 1 happens to be 3 times the frequency in Trial 5. The tuning fork in Trial 1 fits three-quarters of awavelength into the air column, while the tuning fork for Trial 5 fits one-quarter of a wavelength into the air column. The bestanswer is choice A.

27. Choice B is the best answer. Suppose the tuning fork in Trial 4 were measured to have a sound intensity level of 50 dB. If asecond, identical tuning fork were struck and brought near the first fork, together they would have an intensity that is doubledthat of a single tuning fork. The question is how this will affect the intensity level in terms of decibels. Because their combinedintensity I is twice that of a single tuning fork, we can use a value twice as largein the intensity level calculation.

Intensity level = (10 dB) log f-L

Now we double the intensity and see how it changes the intensity level:

Intensity level =(10 dB) logftL] =(lOdB) (log-L+log 2)Vol 'o

Intensity level = (10 dB) log-L + (10 dB) log 2

The intensity level for two tuning forks is the intensity level for one tuning fork plus (10 dB)(log 2), which is about 3 dB. Sothe intensity level is 50 dB + 3 dB = 53 dB.

Such math can be avoided, ifyou remember the following trick regarding decibel levels: If the intensity ofa sound increases bya factor of 10, the decibel level increases by +10 dB. Since the intensity increases by a factor ofonly 2, the intensity level mustincrease by less than +10 dB. This eliminates choices C and D. Choice A is out, because we know the intensity level had toincrease to some degree. The best answer is choice B.

Passage V (Questions 28 - 32) Jet Radar

28. Choice D is the best answer. When using active radar (such as a radar gun), the wave reflected from a moving object is infact doubly compressed, which leads to a double Doppler shift.This makes choice A valid, so eliminate it. A reference beamis needed to compare how much the wave has shifted. This makes choice B valid, so eliminate it. The gun operates whilestationary or in motion. But when in motion, the speed of the emitter must be factored in. If the speed is analyzed bycomparing the wave reflected from a moving object with the wave reflected from a stationary object, then the motion of thegun subtracts out. This makeschoice C valid, so eliminate it. The bestelectromagnetic radiation is one that can travel far andnot be much affected by absorbance, refraction, or dispersion. The longer the EM wavelength, the better it is for use in radar.This means that UV, with a short wavelength, would not work well. In addition, it would be expensive and dangerous to sendout a signal of such high energy. For practical purposes, communication bands and radar are found at the low-energy/long-wavelength end of the electromagnetic spectrum. The best answer is D.

29. Choice C is the best answer. As stated in the passage, radar uses radio waves, so radar waves they are electromagneticwaves (which are transverse waves, not longitudinal waves). This eliminates choices A and B. Electromagnetic waves cantravel through a vacuum, so radarcan function ina vacuum. In fact, it will operate very well in a vacuum, because there is norefraction, absorption, or reflection to affect the signal. The best answer is choice C.

Copyright© by The Berkeley Review® 48 REVIEW EXAM EXPLANATIONS

30. Choice B is the best answer. The velocity of an object reflecting the wave emitted by a radar gun can be determined bycomparing the frequency of the emitted wave with the frequency of the reflected wave. Note that the returning wave isdoubly compressed by the moving object when it is reflected. This means that the velocity can be found by knowing theincident frequency and the reflected frequency. None of the choices say this outright, as they all refer to beat frequency. Twoconsecutive reflected waves have the same frequency, so there is no beat frequency. This eliminates choice A. The wavereflected from a stationary object will not be Doppler-shifted, so it will have the same frequency as the emitted signal. Thissignal can be compared to the compressed reflected wave to obtain a beat frequency, and thus the velocity of the object.Choice B is the best answer. Choice C is eliminated, because the beat frequency would again be zero. Choice D iseliminated, because the velocity determined from the comparison of the waves reflected from two separate, moving sourcesis the relative speed. This can also be applied to eliminate choice C. The best answer is choice B.

31. Choice D is the best answer. The smallest object would be hardest to observe in motion, because the wavelength may be toolong to be reflected by the object. The smallest of the objects is the flying insect, so its velocity is the hardest to measure. Asa note, bats use a radar homing system: the squeak emitted by a bat has a frequency that is high enough to detect flyinginsects. The best answer is choice D.

32. Choice D is the best answer. This question requires memory, derivation, or intuition. The shift in frequency is going to beless than the frequency itself in most cases, so choices A and B seem unreasonable. Choice D is better than choice C, giventhat the wave is doubly compressed (double Doppler-shifted) when it is reflected from a moving object. If you do not buythis approach, then a derivation must be used.

For the first shift, when the wave strikes the movingobject, theobject will encounter a wave that has been shiftedas follows:

_. ,.., c vo + vobject rFirst shift: /object= /incident

vo

For the second shift, when the wave leaves the moving object, the object reflects a compressed wave that has been Doppler-shifted as follows:

, ,-r r vn , vn /v0 + Vobject\t _/vo + v0bject\ ,Second shift: /detected = J object = " J incident - (- — J incident

vo " vobject vo " vobject V v0 / \ v0 - v0bject/

The shift in frequency is the difference between the doubly shifted frequency and /incident- where v0 isc, the /incident is /o.and v is the speed of the object.

A/ =(^)/o-/o=/o(^-l)=/c(^-|tf)=/c(^)-/o(2f)The best answer is choice D.


33. Choice A is the best answer. This isa question of pattern recognition more than anything else. Let's start by considering thedifferences between the frequencies rather than the frequencies themselves. 159 differs from 53 by 106 Hz, 265 differs from159 by 106 Hz, and 371 differs from 265 by 106 Hz, so we see a repeated difference of 106 Hz. From this point, let's lookfor a common denominator for the values as well as 106. All of the numbers are divisible by 53 and not 106, so 53 Hz mustbe the fundamental frequency. This eliminates choice Cand D. If we divide 53 Hz into the four values listed, vve get 1, 3, 5,and 7. Theseare all odd numbers, so we mustbe dealing with a closed pipe rather than an open pipe. Choice B is eliminated.The best answer is choice A.

34. Choice A is the best answer. The depth finder can detect differences in the objects, if their echoes are different in transittime and duration. Bouncing off a rocky bottom, the signal will return to the boat with approximately the same duration asthe emitted sound. Rocky bottoms are hard, and they reflect sound waves well. The addition of a "watery silt" does notchange this reflection much. Remember that above the silt is water. Since sound reflections occur when there are abruptchanges in the sound speed between two adjacent media, there is little reflection at the water-silt interface. Therefore, ObjectsI and II cannot be easily distinguished, ruling out choice B. As for the school of fish, sound would reflect from all of them.The echo, therefore, will bea diffuse reflection of the emitted pulse-now longer in duration anddifferent in timbre. It will bea noticeably different echo than that arriving from the rocky bottom. The bestanswer is choice A.

35. Choice C is the best answer. Red shift implies that the visible light is shifted toward the red end of the visible spectrum,radiation ofa lower-energy frequency, and therefore a longer wavelength, than the light. As stated in the question, red shift isattributed to a planet moving away from the observer. Blue shift implies that the light is shifted to a more energeticfrequency, which implies that the wavelength has been compressed. This is because the planet (object) is moving towards theobserver. The best answer is choice C.


36. Choice B is the best answer. A longitudinal wave is one in which the vibrating motion of the particles in a medium isparallel to the motion of the wave itself. For a sound wave in air, the air molecules vibrate back and forth in the samedirection as the direction in which the wave is traveling. The best answer is choice B.

Passage VI (Questions 37 • 42) Organ Pipes

37.

38.

39.

40.

Choice C is the best answer. The harmonics of a pipe are found as /n = n(/j), so harmonics are even multiples of the firstharmonic (fundamental frequency). A beat frequency is the difference in frequency between two waves, so if we subtractmultiples from one another, the resulting number is also a multiple. It comes down to this: /beat = '/a • /b' = na(/l) - nb(/l)

= (na - nb)(/i). This means that the beat frequency is a whole number multiple of the fundamental frequency, so it too is a

harmonic. Choice A is a valid statement and thereby eliminated. Choice B is a valid statement, because it defines theharmonics based on their fundamental frequency. Choice C is invalid, because as the harmonic number increases, thewavelength gets shorter, not longer. The equation is incorrect for wavelength, so choice C is invalid and the best answer. Asseen in Equation 1, the length of the pipe definitely impacts the harmonics. The diameter may or may not according to thelack of information in the passage, so we can only conclude that length means more than width. Choice D is a validstatement, and thereby eliminated. The best answer is choice C.

Choice A is the best answer. The fundamental mode of a pipe open on both ends has a wavelength that is shorter than thatin a pipe of equal length open only at one end. Since v = /A, the doubly open-ended pipe (pipe //l) should have a higherfrequency than pipe 1/2. This rules out choices C and D. The wavelengths of the fundamental modes of the two pipes differbya factorof 2, so the ratio should be 2:1. Choice A is the bestanswer. You could also compare the formulas for closed and

open pipes and determine that /open : /closed = V/2L : V/4L = V2 : V4 =2 : 1, choice A. Athird method for solving thisquestion involves drawing the pictures. Fora pipe open at both ends (like Pipe #1), a halfof a wave fits in the pipe. Forapipe open at oneend and closed at the other end (like Pipe #2), a quarter of a wave fits in the pipe. Thus, the first harmonichas a wavelength of 2L in Pipe #1 and a wavelength of 4L in Pipe #2. Having twice the wavelength results in half thefrequency, supporting choice A as the best choice of the four. The best answer is choice A.

Choice D is the bestanswer. Changing the type of air could change the speed of sound. Lighter molecules (oratoms) leadto a higher speed of sound, assuming all other factors remain unchanged. Being that H2 (2 amu) is lighter than He (4amu),sound travels faster in H2 than it travels in He. Since a higher speed of sound means a higher frequency, choices A and Careincorrect. Based on v= /A, decreasing Awould tend to increase /. Since, shorter pipes result in a shorter sound wavelength,choice D is the best answer. From the answer choices, you should conclude that using a less massive gas results in a fasterwave speed, because all of the choices involved a gas less massive than N2 and all of the choices dealt with a faster wavespeed (correlated with a high frequency). Perhaps you have experienced this when you breathed in helium and then talked asyou normally would talk. Your voice is of a higher pitch while it is mostly helium exiting your lung. The best answer ischoice D.

Choice B is the best answer. For a gas, you might recall that v « VTemperature . This comes from setting 3/2RT equal tol/2mv2 and solving for the relationship between temperature and average particle speed. Given that the speed of sounddepends directly on the speed of the gas molecules, this means that the Tj:T2 ratio is equal to the vp:v22 ratio. Increasingthe temperature by 44% should lead to a new speed that is Vl + 0.44 (= 1.20) times the size of the original speed. Thismeans that the sound speed increases by .2 or 20%.

J_2 _ 1.44 _ v2Tl 1.00


fX. =Vl.44 = 1.20 ^ = 1.20, so v2 = 1.20(vi)vl

41. Choice D is the best answer. Meters!!! The length must be calculated using meters, not centimeters. That is the mostcommon mistake made on this problem. This question requires that you plug into Equation 1. The value are n = 3, v = 340m/s, and L = 0.15 m. The math is solved as follows:

,-!>f m3^340m/s) m3-(340s-i) = 340^1 . 3400s-, = 3400 Hz2(0.15 m) 0.30 0.10

The best answer is choice D, 3400 Hz. The number seems viable, because 15 cm is pretty short, and from our experience,shorter pipes generate sounds of a higher pitch. Sound ranges from about 20 Hz to 20,000 Hz, so a frequency of 3400 Hzseemsreasonable for somethingon the higherend of the musical rangeof pitches. The best answer is choice D.

®Copyright ©by The Berkeley Review 50 REVIEW EXAM EXPLANATIONS

42. Choice D is the best answer. According to the data in the graph, the first three resonant frequencies are 85, 170, and 255 Hz.

To make the arithmetic as easy as possible, we use the equation /„ =nv/2h which can be manipulated to give: L=nv/2f =

Vhf r When the fundamental frequency // is85 Hz, v = 340 m/s, and n= 1, we get L=2 m. The best answer ischoice D.

Passage VII (Questions 43 - 48) Sound Resonance

43. Choice A is the best answer. If a string is held rigid at both ends, only standing waves with nodes at each end can begenerated on that string. The distance between consecutive nodes on a wave is half the wavelength, so the possiblewavelengths that can be generated on a string of length L can be written as:

n

where n is an integer. In a pipe open at both ends, the boundary condition is an antinode at each end of the pipe. Thedistance between consecutive antinodes is half a wavelength, so the possible wavelengths for a pipe open at both ends arealso:

X = 2Ln

Thus, for a string tied at both ends and a pipe open at both ends with the same length, waves with the same wavelength willbe produced. Wavelength and frequency are related by wave speed, v = Xf. The wavespeed for waves in pipes is the speedof sound in air, -340 m/s. The wave speed of waves on a string depends on the tension in the string and the density of thestring. These two wave speeds are not necessarily equal. Therefore, the frequencies of the waves will likely be different.The best answer is choice A.

44. Choice B is the best answer. If the air temperature increases, the density of the air decreases (hot air is less dense than coldair). In general, wave speed, v, is related to density, p by:

46.

47.

U -

The speed of sound therefore increases if the air temperature increases. However, the wave speed of a wave on a string is notaffected by temperature (thermal expansion of the string is being neglected). Wavelength, frequency, and wave speed arerelated by v= Xf. If we ignore any thermal expansion of the pipe and the string, then the wavelength of the waves producedin the string and the pipe do not change. For a string, the changing temperature does not affect the wave speed. Thewavelength remains the same, so the frequency does not change. For the pipe, the wave speed increases with increasingtemperature. The wavelength is unaffected, so the frequency must also increase. Note that you do not need to know theexact relationship between wave speed and density to answer this question, but you do need to know that wave speed is insome way inversely related to density of the medium. The best answer is choice B.

45. Choice C is the best answer. Although the wave on the string was originally produced on the string, the reason such a wavecan be heard at all is because this wave causes the air to vibrate at the same frequency as the string wave. It is this wave thatreaches the ear. Thus, the wave on the string and the wave produced in the pipe will produce a beat frequency. The bestanswer is choice C.

Choice D is the best answer. For strings tied at both ends, and a pipe closed at one end, the respective possible frequenciesare:

/ = Jl-V- (string tied at both ends)2L

/ = -B-Y- (pipe closed at one end)4L

Because of the4 in the denominator for the pipeand the2 in the denominator for the string,one might be tempted to say thatthe pipe will always produce a higher frequency, but that is true if the wave speeds are equal, or if the wave speed on thestring is less than the speed of sound in air. We need to know something about the two wave speeds before we can answerthis question. Without knowing that information, it cannot besolved definitively. The best answer is choice D.

Choice B is the best answer. Pipes closed at one end and open at the other must have a node at the closed end and anantinode at the open end. Theses are the boundary conditions for the waves produced in pipes closed at one end. If a waycould be found to duplicate these same boundary conditions on a string, then the waves in the pipe and the waves on thestring would have the same shape. How could the boundary conditions be duplicated? Tie the string down at one end andleave the other end free to move. The tied end has to be a node and the free end has to be an antinode. The best answer ischoice B.

Copyright ©by The Berkeley Review0 51 REVIEW EXAM EXPLANATIONS

48. Choice B is the best answer. A sound wave is a propagation of energy through a medium. The waves produced in pipestravel through air. However, there is no air in a vacuum, so no wave can be produced in the pipe. A wave can be producedon the string, however, because the string is the medium. The string can be plucked and set into vibrational motion. Couldyou hear this wave? The best answer is choice B.


49.

50.

51.

52.

Choice C is the best answer. The intensity of a sound is related to the distance r of one perceiving it from the sound sourceby the equation I « 1/r2. We know from the equation in the passage that the intensity of a 100-dB noise is 100,000 timesgreater than the intensity of a 50-dB noise. V100,000 is 316.2, so choice C is the best answer. A good approximation tofollow is that the decibel level of a sound drops by 10 dB every time the distance from the source triples. A drop of 50 dBresults from anincrease indistance ofapproximately 35. The best answer ischoice C.

Choice C is the best answer. The bat detects its possible preyand the location of that prey by bouncing sound waves off it.Since sound can bounce off glass, the bat could detect the glass and believe that the prey is actually at the location of theglass. (Note that this assumes the obviously very hungry bat interprets the glass as prey.) Since the glass is closer than theactual prey, the bat believes the prey is closer than it actually is. The best answer is choice C.

ChoiceD is the best answer. This is a two-in-one question, which is quite common on the MCAT. First, let's consider theintensity of the echo. Let's start by saying that the initial sound wave has 100% intensity. At the air-rock interface, someofthat wave reflects back (generating the first echo) and some refracts into the rock. The sound that enters the rock willcontinue on andstrikethe rock-water interface, where it will partially reflectand partlyrefract. The reflected wave will travelback tothe rock-air interface, where again some of the wave will reflect and some will refract The portion that refracts goeson tobeEcho II. Because Echo I results from reflection offa surface with of greater density, thereflected wave will bemoreintense than the refracted wave. As such, Echo I must be more intense than Echo II. This eliminates choice A.

Now letsconsider the travel timefor eachecho. The total distance traveled by Echo I is 2d while the total distance traveledby Echo II is 4d. If both echoes travel at the same speed, then the time for Echo II should be exactly twice that of Echo I.However, the speed of sound isfaster ina solid than it is in air, soEcho II will have a greater average speed than Echo I. Sodespite traveling twice the distance, because Echo II travels at a greater average speed, it will take less time than double thatof Echo I. The best answer is choice D.

Choice D is the best answer. A beat frequency results from the interference of two waves with different frequencies.Fortunately, the math that goes with a beat frequency issimple. The beat frequency is simply the difference between the twofrequencies. Ifone tuning fork has afrequency of450 Hz and the beat frequency is 12 Hz, then the other tuning fork must beeither 12 Hzgreater OR12Hzless. This would lead topotential frequencies of 438 Hzor 462Hz, not 438 Hzand 462 Hz.Ifyou didn't read the answer choices carefully, which could happen when you're in a hurry oron your last question and readyfor it to be over, thenyou mighthavemistakenly chosen choice B. The best answer is choiceD.

Copyright ©by TheBerkeley Review® 52 REVIEW EXAM EXPLANATIONS

the

Fluids and

Solids

•

Physics Chapter 7

• fc I Ruid flows fastest throughthe middle of a pipe

fY"*Vw "\ With same AP between ends of1) _»> J pipe, there is less molecule-to-vf"*^ _/ molecule pushing, so Q is reduced.

• ^ \ With same AP between ends of pipe,and same L, there is a bigger wall offluidmoving, so Q is increased.

-•

£

J

by

Berkeley Review

Fluids and SolidsSelected equations, facts, concepts, and shortcuts from this section


P2 = Pi + pgAh (for a standing fluid) Ffiuoyant = Pmedium*Vmedium displaced by the objecfg

W _ PobjectV

Q =

displaced __ Pobjectv^~7 " Pmedium (for afloating object)

APttr4

8t|LA1V1 = A2v2

B Pmedium (for a sunken object)

Pi + V2PV12 + pghi = P2 + !/2pv22 + pgh2


Poiseuille's law describes the flow of an ideal fluid through a pipe(many questions involve blood flow)

Point 1 Point 2 Point 3 A1v1=A2v2 = A3v3.". v2 is fastest and vj = v3

k=Pagainst walls + /2Pv2 +Pgn.•. P2 is lowest and Pj = P3

IfAPf,thenQf and .*. v tAS V1, "agajugt wajjs j

^ ^against walls i too HlUCh, thenvessel can potentially collapse

© Buoyancy Questions ApproachFirst determine whether the object floats or sinks in the medium

Foran objectthat floats: For an objectthat sinks:Relative densities =%Submerged Relative densities =W-to-FBuoyant ratio

^displaced _ Pobject 1 W _ Pobject'Vpbject'g _ PobjectB Pmedium'Vobject'g Pmedium

W > B; Pobject > Pmedium

object Pmedium< 1

*object > vdisplaced' Pobject < Pmedium

r,_^ WJpfecfe

Physics Fluids and Solids

Fluids and SolidsIn chemistry we consider the physical state of matter when analyzing itsbehavior. Rather than considering the phase of matter as we do in chemistry, inphysics we shall consider simply whether the material can flow or not. This iswhat distinguishes a fluid from a solid.

Fluid PropertiesBy definition, a fluid is any matter that flows. This means that liquids (such aswater) and gases (such as air) are both considered to be fluids. On a macroscopicscale, these two fluids are involved in the majority of the weather patterns in ourplanet's atmosphere. On a microscopic scale, they are also involved in themaintenance of living organisms. In this section, we will look at both fluid statics(the physics of stationary fluids) and fluid dynamics (the physics of fluids inmotion). Newton's laws will help us obtain a firm understanding of the topicsthat we will discuss.

Density and Specific GravityThe density p (p is the Greek letter rho) of a material is defined as the mass m of amaterial per unit volume V. This is shown in equation (7.1). The SI units fordensity are kg/m3. When all other factors are held constant, the density ofa fluidwill remain the same as long as its volume does not change, given that the massof a material is constant unless some matter is gained or lost.

p _ mass _ m_volume v (7.1)

Table 7-1 lists the densities of various substances (both solid and liquid) inkg/m3. To express these values ing/cm3, divide by 103.

Solids Liquids Gases

Density Density Density

Material (kg/m3) Material (kg/m3) Material (kg/m3)

Aluminum 2.70 x 103 Ethyl alcohol 0.79 x 103 Hydrogen (0°C) 0.09

Iron 7.88 x 103 Gasoline 0.82 x 103 Helium (0°C) 0.18

Copper 8.96 x 103 Benzene 0.88 x 103 Nitrogen (0°C) 1.25

Silver 10.5 xlO3 Water 1.00 xlO3 Air(30°C) 1.16

Gold 19.3 x 103 Sea water 1.03 x 103 Air(20°C) 1.21

Platinum 21.4 xlO3 Whole blood 1.06 x 103 Air (0°C) 1.29

Osmium 22.6 x 103 Mercury 13.6 x 103 Oxygen 1.43

Moon 3.35 x 103 Sun 1400

Earth 5.52 x 103

Neutron Star 1.00 xlO17

Table 7-1


Fluid Properties


PhysiCS Fluids and Solids Fluid Properties

Example 7.1aTwo equal masses of liquid and solid water have volumes of 5.00 mL and 5.45mL, respectively.The ratio of the density of the:

A. liquid water to that of the solid water is 0.083 to 1.B. solid water to that of the liquid water is 0.083 to 1.C. liquid water to that of the solid water is 0.917 to 1D. solid water to that of the liquid water is 0.917 to 1.

Solution

Liquid water is denser than solid water, because it occupies less volume for anequal mass as the solid. So the bigger number should be associated with liquidwater in the ratio. That rules out choices A and C. Next, is ice that much lessdense than liquid water? Not really. If you look into a cup of iced water, the icecubes do float (indicating they are less dense than the liquid), but they floatwitha very small amount of ice above the liquid's surface. Therefore, liquid watercannot be ten times denser then ice, as choice B indicates. Their densities must beclose to 1 to 1.This points to choice D as the best choice.

A more formal way to do any ratio problem is to use the ratio technique. Since wewant a ratio ofdensities, we identify the appropriate density formula and write itonce for each situation of interest. Here, the relevant formula is equation (7.1),since we are told about mass and volume. Next, find the ratio by dividing oneequation into the other. Our division looks like this:

PflzPO) - ml™\ _ Vs _ 5.45PH20(s) ms/Vs Vj 5.00

The masses cancel each other out, because they are thesame. At this point, lookto see which answer is the closest to the ratio here. 5.45 is bigger than 5.00,indicating that the bigger number should be on the water side of the answer.This, again, rules out choices A and C. Next, the two numbers are not all thatdifferent, making choice D the more attractive of the remaining choices. Alwaysminimize the amount of arithmetic you do for a problem by conceptuallyapproximating the answer.


Example 7.1bTen mL each ofpure water and saltwater have densities of 1.00 g/mL and 1.03g/mL, respectively. The ratio ofthemass ofpure water to thatofsaltwater is:

A. 1 to 1.03.

B. 1.03 to 1.

C. 0.03 to 1.D. 1 to 0.03.

Solution

Saltwater is denser than pure water, so an equal volume of saltwater is moremassive (weighs more) than an equal volume of pure water. So the biggernumber should be on the right side of the ratio. That rules out choices B and D.Saltwateris just slightly denser than pure water, so the ratio of the masses shouldbe relatively close to1:1. The ratio of the masses could notbe as large as choiceC indicates, which points to choiceA as the best answer.




The specific gravity of a material is defined as the ratio of its density to thedensity of water at 4 °C. Specific gravity is also called the relative density(probably due to the fact that "specific gravity" really has nothing to do withgravity). If a given material has a density p, then its relative density prelative 0-e.,specificgravity) can be given by equation (7.2):

Specific gravity Prelative =-&&**&PH2Q at 4°C

(7.2)

Pressure and Pascal's PrinciplePressure is force exerted per unit area. More precisely, as shown in equation(7.3), it is the componentof a force F that actsperpendicularly to a surface of areaA, divided by that surface area. This is because the parallel component of F doesnot press into the surface, and therefore it applies no pressure.

P=^ (7.3)

For a given force, a smaller contact area produces a greater pressure. This is whya sharp knife, with its smaller contact area, cuts (i.e., exceeds a critical pressure)more easily than a dull knife. Needles are designed with this principle in mind.The SI unit of pressure is the pascal (Pa), which is equivalent to N/m2. Thepascal is an extremely small unit, so it is impractical for describing atmosphericconditions. We generally use the units of atmosphere and mmHg (also known asa torr) to measure everyday pressures.

The pressure of a fluid can vary with elevation. For example, if you hike to thetop of a high mountain, you should notice a decrease in the air pressure.Similarly, if you have ever been diving under water, you have surely felt anincrease in the water pressure on your body (especially in your ears and sinuses).

An analogy can be made between pressure variation with depth and the stackingof pennies. Suppose that each penny represents a thin layer of air. If we place asingle penny on a table, then the only pressure on that penny comes from the airabove it. If we place a second penny on top of the first penny, then the firstpenny is supporting not only the penny above it, but also the air above thatsecond penny. The more pennies we stack on top of the first penny, the greaterthe pressure that first penny experiences (Figure 7-1). This simply tells us thatpressure is additive.

Column

of air

First penny

Figure 7-1


Fluid Properties


Physics Fluids and Solids Fluid Properties

Consider the shaded area of fluid shown in Figure 7-2. Suppose the bottomsurface of this area is at a heighty\t while the top surface is at a height y2- Wecanalso define the positive y-axis to be pointed upward.

y2-yr

Y2

LyiIW

Figure 7-2

The fluid below the surface at yi pushes up on the shaded fluid with a force Fi,as given by equation (7.4). The fluid above the surface at y2 pushes down on theshaded fluid with a force F2, givenby equation (7.5).

Fj = PiA

F2 = P2A

Fj - F2 - w = 0

PjA - P2A - w = 0

(7.4)

(7.5)

(7.6)

(7.7)

There is also a third force acting on the shaded fluid. That third force is theweight w ofthe shaded fluid itself. Ifourshaded area offluid isnotmoving, thenit must be in a state of equilibrium. This means that Fi, F2, and w must sum tozero, as shown in equation (7.6). Substitution of equations (7.4) and (7.5) intoequation (7.6) gives equation (7.7).

The volume V of the shaded area of fluid can be given by equation (7.8). Theweight ofanobject is given by equation (7.9). We can solve for the weight of theshaded area of fluid by substituting equation (7.1) and (7.8) into equation (7.9).This gives equation (7.10). Inarriving at this equation, we areassuming that thedensity of the fluidremains constantas elevationchanges.

V = area x height = A (y2 - yi)

w = mg

w = mg = pVg = p [A (y2- y,)] g

(7.8)

(7.9)

(7.10)

We can now substitute equation (7.10) into equation (7.7), to produce equation(7.11). Dividing by the area givesequation (7.12).

PlA-P2A-p(A(y2-y1))g = 0

Pascal's principle Pi-Po =

(7.11)

(7.12)

Copyright©by The BerkeleyReview 58 The Berkeley Review


Equation (7.12) tells us that if we increase the pressure on the upper surface ofthe shaded area in Figure 7-3, then the pressure on the lower surface willincrease by exactly the same amount. This is called Pascal's principle. It statesthata pressure appliedto an enclosed fluid is transmitted equally throughoutthefluid and to the walls of the fluid's container.

Example 7.2aAt a specific depth in a swimming pool, a barometer measures the total pressureto be twice that of atmosphericpressure. If the barometeris now submerged to adepth that is twice its initial depth, by how much does the total pressureincrease?

A. The pressure increases by 50%.B. The pressure increases by 100%.C. The pressure increases by 200%.D. The pressure increases by 300%.

Solution

We can use the ratio technique to handle this percentage change type of problem.The relevant situation here is the relationship of pressure to depth:

Ptotal= Patm + Pgh

Applying this to the first depth implies that pghinitial=Patnv since me totalpressure there is 2Patm. If the second depth is twice that of the first, thenPghfinal =2Patm- This means that the total pressure at the second depth is SPatm-Taking the ratio of the final to the initial situation gives:

Ptotal final _ 3Patm

*total initial 2-P.atm

= 1.5

To complete this problem, just subtract 1 from the ratio to get the percentagechange of +0.5 (i.e., 50%.) Be careful of the wording in percentage questions.Here, we want a "percentage change." If the question had asked instead, "Whatpercentage of the initial total pressure is the final total pressure?", the answerwould have been: "The final total pressure is 150% of the initial total pressure."


Example 7.2bA swimmer is at the bottom of a 5-meter pool, where the gauge pressure (pgh) isapproximately 0.5Patm. If the swimmer rises to the surface, by how much doesthe total pressure change?

A. The total pressure increases by 33%.B. The total pressure increases by 66%.C. The total pressure decreases by 33%.D. The total pressure decreases by 66%.

Solution

As with 7.2a, the relevant situation here is the relationship of pressure to depth:

Ptotal = Patm + Pgh

Pinitial = 1*0arxn + 0.5 atm = 1.5 atm; Pfinal = 1.0 atm; AP = 0.5 atm

The total pressure is reduced, so choices A and B are eliminated.The ratio of APto Pinitial is 1:3, so the pressure drops by 33%.



Fluid Properties



Buoyancy and Archimedes1 PrincipleThe inventions and testaments of the Greek mathematician Archimedes markhim as a man of genius. Archimedes is the man who ran naked through thestreets of Syracuse shouting "Eureka, eureka!" (Greek for "I have found (it)!"),after he discovered the principle of hydrostatics (buoyancy). King Hiero II hadasked Archimedes to determine whether a crown was pure gold or laced withsilver. While relaxing in the public baths, Archimedes realized that he coulddetermine the volume of the crown by placing it in a container of water and thenmeasuringthe amountofwater the crown displaced.Thisvolumewas comparedto an equal volume of pure gold. After a short calculation involving density,Archimedes reported to Hiero that the crown contained some silver. The makerof the crown was executed, and Archimedes made his place in history.

So, what was Archimedes' great insight about objects in fluids? He realized thatthe water (or any fluid) below a floating object supports that object againstgravity. To get an idea of just how big this upward buoyant force B is, askyourself what the water beneath a battleship would be supporting, if the shipwere not there. The water would be supporting a column of water above it.Figure 7-3ashows a filled tank, with an object resting on the bottom of the tank.The objectappears to weigh less underwater than in air. Why?Because the watersurrounding it is trying to push the objectup and out of the way with some forceB. Since the object is stationary, the buoyant force and normal force combinedmust be equal in magnitude to the weight of the object.

(a)

Pobject> Pwater

B

''M't

irmjt

(b)

Br-. y

iii

Pot

1

ject== Pw

*

ater

(c)

tB*- - *-

imgPobject "^ Pwater

Figure 7-3

Whathappenswhentheobject has the samedensity as the surrounding water, asin Figure 7-3b? If the object has the same volume and mass as the displacedregion of water, the object should hover in the water. After all, both the objectand the displaced water have the same weightand experience the same buoyantforce. With this in mind, we can state what Archimedes discovered in the tub:The buoyant force acting upon an object is equal to the weight of the fluid itdisplaces. Mathematically, thebuoyantforce Bis found usingequation (7.13).

Archimedes' principle B= pfluid •Vfluid displaced •g (7.13)

where the massof the displaced fluid is in terms of its density and volume.

How does the density of the object in Figure 7-3c compare with the water? If weput an object of equal size, but of lesser density, into the water, as shown inFigure 7-3c, it should float above the water line, because the buoyant force mustbalance its weight. Since it has less weight than the object in Figure 7-3b, thebuoyant force necessary to make this object float will be smaller (i.e., it willdisplace less water).



What if an object's density exceeds that of the water? Its weight will exceed thebuoyant force, and it will sink (as shown in Figure 73a). As this object falls,however, it will accelerate with an acceleration smaller than g (and appear toweigh less), because the water still exerts an upward buoyant force on it. Thisbuoyant force is a property of any fluid (liquid or gas) in a gravitational field. Ifsthe reason helium balloons float in the air, ships sail on thesurface ofa body ofwater, and submarines can ascend and descend in surrounding water by takingon or bilging out water.

Example 7.3aWhich of the following statements are FALSE regarding weather balloons?

I. Balloons risewhen their total density is less than that of the air surroundingthem.

II. Balloons fall when the buoyant force of the air surrounding them exceedstheir weight.

III. Balloons stop falling when the air surrounding them is evacuated.

A. I and II onlyB. I and III onlyC. II and III onlyD. I, II, and III

Solution

To handle these multiple-concepts type of problems, go through the listeliminating or accepting one item at a time. Then cross out the incorrect choices.Let's say that you immediately decide Statement III is false. After all, balloonsfloatbecause the surrounding air buoys them up. No air equates to no buoyancy.This means that Statement III should be in the answer and that choice A isincorrect.

Let'ssay further that you know Statement I is true and therefore should not be inthe answer. After all, helium is lighter than air, and helium balloons float. Thiswould rule out any choices that contained Statement I and leaves only choice Cas the correct choice. You are now done with this question, so you can move on.Do only enough work on a problem to convince yourself you have the bestanswer and then pick the choice and move on. You could also have consideredStatement n, knowing that an object will accelerate down when the downwardforce exceeds the upward force. A balloon will not fall when the buoyant force ofthe air surrounding them exceeds its weight.


Example 7.3bWhichof the following are TRUEregarding three balls of equal size that are fullysubmerged in water, if Ball A sinks, Ball B is stationary, and Ball C rises?

I. The density of BallC is less than that of BallB.II. The buoyant force on Ball C is greater than that on

Ball A.

III. The density of Ball A is greater than that of water.

A. I and II onlyB. I and III onlyC. II and in onlyD. I, II, and III


Fluid Properties



SolutionStatement I: Ball C rises when submerged in the water, which tells us that thebuoyant force exceeds itsweight. This is possible if Ball C is less dense than thesurrounding water. Ball B floats when submerged in the water, which tells usthat the buoyantforce equals its weight. Thisis possible ifBall Bis equally denseas the surrounding water. It is true that Ball Cis lessdensethan Ball B.

Statement II: The buoyant force a fluid exerts on an object depends on thedensity of the surrounding medium, the volume of the object, and thegravitational force constant (B = pfluid ' ^displaced ' g)- Because Ball A has thesame volume as Ball C, they experience the same buoyant force. The reason BallA sinks and Ball C floats has to do with their differences in weight, despite thefact that they experience the samebuoyant force. It is false that the buoyant forceon BallC is greater than that on BallA.

Statement III: BallA sinks when submerged in the water, which tells us that itsweight exceeds the buoyant force. This is possible if Ball A is denser than thesurrounding water. It is true that the density of Ball A is greater than that ofwater.


Figure 7-4shows a summary of the mathematics associated with a floating objectand a sunken object in a tank of fluid. The conclusions can be used to quicklyanswer multiple-choice questions.

W = B

Pobject* *object*g = Pmedium"^displaced'gB=PmediuirTVdisplaced"g Pobject"Vobject = Pmedium"Vdisplaced

^displaced _ Pobject

ôbject PmediumW=mg= PobjecfVobjecfg

ôbject > Vdisplaced' Pobject< Pmediumv

Conclusion for an object that floats: Relative densities = %Submerged

Q

" —Pmedium"ôbject'g

N = Apparent Weight

Notes: B = Apparent weight loss

Specific Gravity = ——^—Pwater

W Pobject* *object'g _ Pobject

B Pmedium'Vobject*g Pmedium

W=mg =Pobject*Vobjecfg W>B; pobject > pmedium

Conclusion for anobject that sinks: Relative densities = W-to-FBUOyant ratio

Figure 7-4



Example 7.4aWhat is the density of an object that is 85% submerged when floating inpetroleum ether (p = 0.90g/mL)

A.

B.

C.

D.

0.64 g/mL0.77 g/mL0.90 g/mL1.06 g/mL

Solution

Because the object floats in the petroleum ether, it must be less dense than thesurrounding fluid. The density is less than 0.90 g/mL, so choices C and D areeliminated.

From here it is best to use the shortcut presented in Figure 7-4. The percent of theobject that is submerged is equal to the relative densities of the object andmedium. This means mat the density of the object is the product of the percentsubmerged and the density of the petroleum ether.

Pobject%submerged =Pmedium


Pobject

° 0.900.85(0.90) = pobject = 0.765 g/mL.

Example 7.4bWhat is the specific gravity of an object that weighs 36 newtons in air but only 9newtons in water at 4°C?

A. 0.25

B. 1.33

C. 3.00

D. 4.00

Solution

The first thing to do is to identify that the object is a sunken object rather than afloating object. The object must be denser than the medium, so the specificgravity must be greater than 1. This eliminates choice A. For a sunken object, theratio of the weight of the object to the buoyant force is equal to the relativedensities of the object and medium.

The objectweighs 36 N in air, so its weight is 36 N. It has an apparent weight of 9N in water, so the normal force is 9 N and the buoyant force must be 27 N. Keepin mind that a sunken object rests on the bottom of the container, where becauseit experiences no acceleration, it has a net force of zero. This means that on thebottom of the tank, mg = N + B.

To get the specific gravity (the relative densities),we simply divide 36by 27:

Specific Gravity =


W

BSpecific Gravity = — = — = 1.33

v y 27 3


Fluid Properties

B must be 27 N

N = 9N

mg = 36 N



Surface TensionWe usually drink water from a cup. The water is confined by the cup, whichkeeps the water molecules from spreading apart, and you from cleaning up aspill. What, however, keeps the molecules that make up a small water dropletfrom spreading apart? The droplets tend to stay together because their moleculesare attracted to each other-this attraction is typically a result of intermolecularforces such as hydrogen bonding or van der Waals interactions. Within the bulkof the droplet, each molecule reacts to attractive forces from all its neighbors and,on average, experiences no net change. Figure 7-5 shows a molecular view nearthe surfaceof a droplet. At the surface, however, the situation is tense. A surfacemolecule is subjected to a net force pulling it towards the center of mass, sincethere are few molecules above the surface to pull it out. On average, these surfacemolecules are pulled towards the center of mass. This reduces the number ofsurface moleculesand, as a result, reduces the surface area of the droplet.

Surface area shrinksas the molecule ispulled into the bulk.

Surface area shrunk

after the molecule ispulled into the bulk.

Figure 7-5

The force associated with this tendency of the surface of a liquid to pull inwardand shrink in area is known as the surface tension. When weight is significant,droplets tend to have flattened fluid surfaces. When the gravitational force isrelatively small, or negligible, surface tension shrinks the surface area of a liquidby forming a sphere. This is why your coffee has a flat surfaceand soap bubblesare round.

Surface tension should relate to other surface phenomena, such as evaporationand condensation, because they all occur at the fluid's surface. How do theyrelate to each other? If the surface tension is relatively strong, the molecularattractive forces are relatively strong. Because of hydrogen bonding, water has agreatersurface tension thanethanol. At a givenpressure and temperature, whichevaporatesfirst? Ethanol willevaporatebefore water evaporates under the sameconditions. It has a smaller surface tension. Which is more likely to condense?Water is more likely to condense than ethanol, because it has a greater surfacetension and, consequently, stronger attractive intermolecular forces. In thecomparison of ethanol to water, the intermolecular force of significance ishydrogen bonding.

Example 7.5aIncreasing the surface tension of a cell surface will:

A. make the cell's shape more spherical.B. make the cell membrane more pliable.C. make the cell flatten.

D. cause the cell to decrease in volume.



Solution

Surface tension is a surface effect. As such, it can affect surface properties of thecell, including possible shape changes. It will not change the volume of the cell,since the amount of intercellular material is generally constant. That rules outchoice D. As for the shape of the surface membrane, increasing tension on themembrane makes it less pliable, ruling out choice B. Imagine how the pliabilityof a balloon surface decreases as you inflate it. When surface tension increases ona liquid, its surface decreases in area (although its volume does not change).

The resulting shape always tends towards spherical. You can see this with a soapbubble. You should try to make useful analogies with everyday phenomena toanswer questions about things you can't easily see.


Example 7.5bA sealed vessel contains two chambers separated by a phospholipid bilayermembrane. After completion of an experiment on the vessel, the membranedistorts, as shown below. This distortion may have occurred because thepressure on the right side of the vessel:

.

Membrane Membrane

Before experiment After experiment

A. increased, and the membrane's surface tension increased.B. increased, and the membrane's surface tension decreased.C. decreased, and the membrane's surface tension increased.D. decreased, and the membrane's surface tension decreased.

Solution

The membrane is being forced to the left, so pressure on the right side must haveincreased relative to the pressure on the left side of the membrane. Thiseliminates choices C and D. Because the molecules of membrane are beingstretched apart, they are feeling less of an attractive force towards one another.This means that the membrane's surface tension must have decreased, makingchoice B a better explanation than choice A.



Fluid Properties


Physics Fluids and Solids Fluids in Motion

Fluids in Motion

Viscosity and Poiseuille's PrincipleWe shall now consider what occurs when a fluid is in motion. We have alreadyexamined the friction between solids. There is also friction within fluids. Frictionin fluids is called viscosity and is denoted by the Greek letter eta, rj. The SI unitofviscosity isN»s/m2.

Viscous forces retard the motion of one part of a fluid relative to another part ofthe fluid. If two layers of a fluid are held together tightly, then the viscosity of thefluid is said to be large. If the viscosity of a fluid is large, then that fluid willmove slowly. This can be seen if we consider the difference between pouringhoney into a cup and water into a cup. Since the coefficient of viscosity (r)) forhoney is greater than that for water, honey will flow at a slower rate than water.

Liquids Gases

Viscosity Viscosity

Material (kg/m3) Material (kg/m3)

Water (0°C) 1.80 x 10"3 Air (0°C) 1.71 x 10"5

Water (20°C) 1.00 x 10"3 Air (20°C) 1.81 x 10"5

Water (37°C) 0.69 x 10"3 Air (37°C) 1.87 x 10"5

Water (100°C) 0.28 x 10"3 Air (100°C) 2.18 x 10"5

Blood Plasma (37DC) 1.28 x 10"3

Blood Whole (37°C) 2.08 x 10"3

Castor® oil (0°C) 5.3

Castor®oil (20°C) 0.97

Castor® oil (100'C) 1.71 x 10"2

Table 7-2

Table 7-2 shows some common fluids and their viscosities at differenttemperatures. Generally, if we were to increase the temperature of a fluid, theviscosity of a liquid would decrease, while the viscosity of a gas would increase.You may notice that Castor® oil shows a significant drop in viscosity astemperature increases. This property makes it an ideal lubricant for an engine.

When a simple fluid flows through a pipe, the speed of that fluid is greatest atthe central axis of the pipe and essentially zero along the walls of the pipe(Figure 7-6). A common reason that the fluid along the walls of a pipe does notmove is because that fluid encounters the rough and uneven molecular surface ofthe pipe itself. Molecules of fluid in contact with this surface tend to reside in itscrevices and to not move.

Fluid at the edge of the pipe flows the slowest.This flow speed is zero for most simple fluids.

• • • • • .....-••••• -. - -i

Figure 7-6

Fluid in the center of the

pipe flows the fastest.



As a fluid is flowing through the center ofa pipeoflength Lwitha certain speedv, there is a force that tends to oppose the fluid's motion. That force is called aviscous retarding force F and it is caused by the fluid's resistance to flow. Themagnitudeof the viscous retarding force is given by equation (7.14):

F = 4jtt|Lv (7.14)

For a fluid to flow through a pipe at constant speed, the viscous retarding forcemust be balancedby an applied force. Suppose the pressure of the fluid enteringthe pipe in Figure 7-6 is Pi. The force pushing this fluid into the pipe is Fi = PiA,where A is the cross-sectional area of the pipe (i.e., A = to*2). There is also apressure P2 at the end of the pipe that tends to oppose the motion of the fluid.The force associated with this pressure is F2 = P2A. The net driving force of thefluid through the pipe is shown in equation (7.15). It is simply the difference inbetween these two forces:

Fi - F2 = (Pi - P2)A (7.15)

For the fluid to flow through the center of the pipe at constant speed, the drivingforce and the retarding force must be equal (see equation (7.16)). If we substitutetit2for the cross-sectional area of the pipe and then rearrange equation (7.16), wewill get equation (7.17). This equation allows us to relate the pressures at the twoends of the pipe to the viscosity of the fluid and the physical properties of thepipe itself (i.e., length and radius). We could also solve for the velocity v of thefluid at the center of the pipe by rearranging equation (7.17) into equation (7.18):

4jtt|Lv = (Pi - P2)A

Pl-P2=(^)vv = [P!^2)r2

4r)L

(7.16)

(7.17)

(7.18)

Example 7.6aWhich of the following graphs BEST represents the viscosity of pure H20 as itstemperature increases?

2H

u

1 > 1 111 10 20 40 60

T(°C)

B.

2-

UIncm 1*->

flj

P-

I I I I I I I0 20 40 60

T(°C)

c.

2-

U

(0

p-

I I I I I I I0 20 40 60

T(°C)

D.

u

a,.

p-

I I I I I I I0 20 40 60

T(°C)

Fluids in Motion

Copyright © by The Berkeley Review 67 Exclusive MCAT Preparation


SolutionThis is a graph-identification problem. First, think about the qualitativerelationship between viscosity and temperature. Should the viscosity of waterincrease or decrease as the temperature increases? As you heat up water, themolecules should start to spread apart from one another, which means that itshould decrease. One way to remember this is to cook some canned soup. Thesoup comes out of the can as a viscous, unappetizing gelatinous lump, but itthins as you heat it, evolving into a tasty, salt-laden meal. The soup is mostlywater, suggesting that water thins as it heats up (although the overridingdetermining factor is actually the soup's ingredients). This rules out choices Cand D, which show no change in viscosity with temperature and an increase inviscosity with temperature respectively. Todecide betweenchoices Aand B, lookat the value on the y-axis when the temperature is 20°C. How do they differ?Choice A smoothly drops from about 2 to 0.5 over a 60°C range, passing through1 at 20°C. In choice B, the number abruptly drops from about 2 to nearly 0 over arange of 10°C and more importantly, does not equal 1 at 20°C. The y-axis showsthe actual viscosity of water divided by the viscosity of water at 20°C, so thevalue of that term at 20°C must be 1. Choice B does not fit that criterion, so itcannot be the best answer.


Example 7.6bIn calibrating a flow meter, an engineer measures the total pressure drop across apipe as a function of the gas flow velocity. Which of the following plots BESTdetails his results?

v(m/s) v (m/s) v (m/s)

I I I I I I0 2 4

v(m/s)

SolutionFluids flow from regions of higher pressure to regions of lower pressure, so inorder to get a fluid (such as a gas) to flow, there must be a AP across the length ofthe pipe. It should seem intuitive that a greater pressure difference would pushfluid faster, so choice D can be eliminated. Also, choice C implies that the speedcan increase despite no increase in pressure difference past a threshold, which isnot true. Choice C is also eliminated. A change in AP from a negative value to apositive value implies that the pressure gradient reversed directions. This wouldreverse the direction of the fluid flow, and that is not what choice A shows. Therelationship between velocity and pressure difference is linear and both shouldbe equal to zero at the same time. Only choice B shows this.




The flow rate Q is the volume Voffluid that passes through a pipe perunit timet. Note thatthe flow rate and flow speed ofa fluid are not the same thing. Whattype ofpressure would be needed to maintain a certain flow ratethrough a pipe?We canrelate the flow rate to the cross-sectional area ofthe pipe andthe averagespeed v of the fluid by equation (7.19). Recall that we have mentioned that thefluid in a pipe flows the fastestat the center of the pipe and does not flowat all atthe edges of the pipe. Using this assumption, we can express the average speedof the fluid as shown in equation (7.20):

Q = Y. = vAt

v" = v2

(7.19)

(7.20)

If we substitute equation (7.20) into equation (7.19) and include the cross-sectional area nr2, we will getequation (7.21):

Q - W«a (7.21)

Substitution of equation (7.18) into equation (7.21) will give equation (7.22). Thisequation is called Poiseuille's principle, after Jean Poiseuille, a French physician.

Poiseuille's principle Q=H.-(Pl-P2)8r|L (7.22)

The important parameter in equation (7.22) is the r4 term. Note that if the radiusof the pipe is doubled, the flow rate increases by a factor of 16.

A common application of Poiseuille's principle involves the flow of bloodthrough the circulatory system. Although it applies to an ideal fluid, which bloodis not, it still gives a good approximation of the flow dynamics of blood throughthe body. When the heart contracts, it generates a large Pi, which in turnestablishes a pressure difference in the circulatory system ultimately causingblood to circulate and return to the heart. Thus, when one describes a patient ashaving "high blood pressure", it means that their heart is generating a very largePi in order to produce the necessary Q to get blood to the tissue in a timelyfashion.

Although we have presented Poiseuille's principle as a mathematical expressionto this point, at the level of the MCAT it is highly conceptual. For instance, itexplains that an extremely tall person will have a greater strain on their heartthan a person of average height, because in order to generate the same volumeflow rate (Q), a tall person requires a greater AP to offset the greater L associatedwith their longer circulatory system. The equation also explains why it takeslonger to pour syrup than it takes to pour water and why a wider straw is muchmore efficient for obtaining a chilled, carbonated beverage from a glass than athinner straw. It basically explains important things.


Fluids in Motion



Example 7.7aWhich single change, where all other factors stay constant, gives the LARGESTincrease in the volumetric flow rate ofsynthetic blood plasmathrough a tube?

A. Doubling the tube radius.B. Quadrupling thepressure difference across the tube.C. Halving the plasma viscosity.D. Halving the tube length.

SolutionThese quantitative-change problems usually require you to know the basicfunctional relationship between variables used in some relevant equation. Thatis, you need to know whether the important variables are in the numerator ordenominator, and whether they are squared or cubed, or whatever. Here, thebasic equation to use is Poiseuille's principle, equation (7.22). The relevantvariables for this problem are the tube radius r, the pressure difference Pi - P2,the tube length L, and the plasma viscosity rj. The ji and 8 are not importantquantities, because in any quantitative-change problem, constants will neverenter into consideration. Now, go through each choice, and see how that willchange the volumetric flow rate. For choice A, what would doubling the radiusdo to the flow rate? It would increase the flow rate 16 times (i.e., 24). Choice Bwould quadruple the flow rate,since the pressure difference term in the equationis not squared or quadrupled, but raised to the first power. Cross out choice B,because choice A is better. Choice C would double the flow rate, since theviscosityis in the denominator and to the first power. But choiceA is still a betterchoice. Halving the tube length would have the same effectas choice C, makingchoiceA correct. This formula is a bit daunting and may be given in the passage,but it is commonenough to make its basic functional variables worth knowing.


Example 7.7bWhichsinglechangemakesit easiest(i.e., creates the smallestpressure differencebetween your mouth and the atmosphere) to drink a milkshake?

A. Increasing the straw length by 50%, but keeping the flow rate and milkshakefixed.

B. Decreasing the straw radius by 50%, but keeping the flow rate and milkshakefixed.

C. Melting the shake until its viscositydrops by 50%, but keeping the flow rateand straw fixed.

D. Drinking the shake at 25% of the usual flow rate, without changing the strawor shake.

Solution

To keep the flow rate fixed while increasing the straw's length by 50% wouldrequire 1.5 times the pressure difference across the straw, so choice A is out. Tokeep the flow rate fixed while halving the straw's radius would require sixteentimes the pressure difference across the straw, so choice B is eliminated. To keepthe flow rate fixed while halving the shake's viscosity would require half thepressure difference across the straw, so choice C is possible. To reduce the flowrate by a factor of four while keeping other factors constant would require one-fourth the pressure differenceacross the straw, so choice D is the easiest change.




Continuity EquationIn equation (7.19), we saw how to relate the flow rate, Q, of a liquid to the cross-sectional area A of a pipe and the average velocity v of the fluid flowing in thatpipe.This relationship was for a pipe or tube whose cross-sectional area does notchange. However, not all tubes retain the same cross-sectional area. Consider thehuman circulatory system. As blood leaves the left atrium of the heart it flowsinto the aorta, which has a radius of about 1.3 cm. As the blood leaves the aorta,it passes intoarteries having smaller radii. In the case of the human circulatorysystem, the cross-sectional areas of the blood vessels vary greatly.

Suppose the aorta has cross-sectionalarea Ai and that blood flows from the aortainto an artery with cross-sectional area A2 (Figure 7-7).

Larger AreaSlower Speed

vl

Smaller AreaFaster Speed

v2

Figure 7-7

Assuming the fluid is incompressible, at any given moment in time, the fluidentering the region at Ai must equal the fluid leaving the region at A2. In otherwords, the volume flow rate into the region at Aj must equal the volume flowrate out of the region at A2. The volume flow rate for the region at Ai is given byQ = Ajv}, while the volume flow rate at region A2 is given by Q = A2v^. In bothof these equations, the average speed V of the fluid refers to that fluid passing agiven cross-sectional area. Since the flow rates for these two regions must beequal, we can write equation (7.23):

Continuity equation ^^jv^^2^J (7.23)

Equation (7.23) is called the continuity equation. It will becomeimportant whenwe discuss Bernoulli's equation. Note that if the cross-sectionalarea A2 in Figure7-4 is smaller than the cross-sectional area A}, then the speed of the fluid passingthrough A2must be greater than the speed of the fluid passing through A\. Thishas to be the case in order for the volume flow rate of the fluid to be constant.

The continuity equation can also apply to branching as long as the fluid isincompressible and that the total cross-sectional of all of the branches isconsidered. For the system shown in Figure 7-8, equation (7.24) applies.

AjVx = A2V2 + A3V3 + A4V4 (7.24)

Alvl

Figure 7-8


Fluids in Motion



Example 7.8aAtherosclerosis is the thickening and hardeningof the arterialwalls. Theaorta ofa healthy adult has a radius of about 1.3 cm and a blood flow velocity of about0.5 m/s. An atherosclerosis-stricken bacon epicure is found to have an effectiveaortic radius that is 75% of that in a healthy aorta. What is the blood flow velocitythrough the narrowed section of artery? (Assume that the blood's volumetricflow rate is constant for all adults.)

A. 0.28 m/sB. 0.38 m/sC. 0.66 m/sD. 0.89 m/s

SolutionBefore invoking math, ask yourself if the blood will flow more quickly or moreslowly through the narrowed artery. It should flow faster, ruling out choices Aand B. Remember, for a fixed volumetric flow rate (which occurs in most pipeflow problems) the smaller the cross-sectional area, the greater the flow velocity.Isolating the best answer will require a bit of math. Let's use the continuityequation (7.25) to find the final flow velocity. We get:

irr?

-=fe)^Br=&r=((#h=W(05m/s)Looking at the choicesleft, choiceD looks like the likelier of the two answers. Ifyou're doubtful, realize that (16/9) is a bit smaller than 2, which makes theanswer a bit smaller than 1.If you forgot to square the radii,you probably pickedchoice C.There will often be an answerchoice on the test for eachcommon wayof misinterpreting or miscalculatinga problem. Please be careful.


Example 7.8bTheplunger ofa hypodermic syringe hasa cross-sectional areaofapproximately0.5 cm2, whereas theattaching needle hasa cross-sectional area of0.005 cm2. Ifadoctor (you in five or six years) pushes in the plunger at a speed of 0.3 cm/s,what is theaverage flow speed ofthe medicine through the needle?

A. 0.003 cm/sB. 30 cm/sC. 300 cm/sD. 3000 cm/s

Solution

This question involves the continuity equation, which relates the average flowspeed and cross-sectional area at two different points in a pipe. The relationshipis Aivi = A2v2. Manipulating the equation shows us that the as the as the cross-sectional area decreases, the average flow speed has to increase by the samefactor. In the syringe example, the cross-sectional area decreases by a factor of100, so the average flowspeed must increase by a factor of 100. This means thatthe average flow speed will increasefrom 0.3cm/s to 30 cm/s.




Bernoulli's EquationConsider a pipe filled with an incompressible, ideal fluid. As we noted earlier, afluid flows when it has unequal pressures applied to it, flowing from a region ofhigher pressure to a region of lower pressure. Let's consider the higher pressureto be Pi and the lower pressure to be P2, and that the two pressures represent thepressure at the two ends of the pipe. As the fluid moves, its kinetic andgravitational potentialenergies change.This is due to the work done on the fluidby the pressure difference pushing the fluid in one direction. These energychanges are related to one another by the work-energy relationship shown inequation (7.25). In this equation, W is the work done on the systemby pressuresPi and P2, AKE is the kinetic energy change, and APEg is the change ingravitational potential energy.

W = AKE + APEg(7.25)

We shall consider these three aspects (work done on the fluid, kinetic energy ofthe fluid, and gravitational potential energy of the fluid) separately using thepipe shown in Figure 7-9. In this pipe, there is both an elevation change and achange in radius. What we shall consider is the pressure exerted by the fluidagainst the inside walls of the pipe. Whether the fluid flows left-to-right or right-to-left, the same result will be obtained.

Region 1 Region 2 Region 3

'

r 1 i s

A/ . ...L...r... p.

/ : >i

/ / N—;\.

: 1

Pi"....

i iv2

V1

h =yi

fyyi !

'

vfluid T:1against wall * n T: Pagainst wall *

Figure 7-9

Let's consider work first. If the fluid moves a distance Axi on the left side of thevessel, then it must movea distance Ax2 on the right side of the vessel. Recall thatW=Fj.(Ax). Ifweapply thisexpression to thefluid moving in the vessel, thenweget equation (7.26). Given thatFi = PiA and F2 = P2A, making the appropriatesubstitution into equation (7.26) gives equation (7.27). The volume of fluid thatmoves from left to right in the vessel is the same. So, AV = AiAxi = A2AX2.Substituting this information into equation(7.27) gives equation (7.28):

W = FiAxi - F2AX2

W = PiAiAxi -P2A2AX2

W = (Pi -P2)AV

(7.26)

(7.27)

(7.28)

We're doing this math to work our way to Bernoulli's equation, which relatespressuresin the way the work energy theorem relates energyterms.

Copyright ©by The BerkeleyReview 73

Fluids in Motion


PhysiCS Fluids and Solids Fluids in Motion

Now lefs consider the change in kinetic energy for the flowing fluid. Bymanipulating the relationship p = m/V, we solve this expression for the mass ofthe fluid to show that m = pV. The change in kinetic energy for the mass of amoving fluid is given by equation (7.29). Substitution of pV for m into (7.29)gives equation (730):

AKE = ^mvi2 - V£mv32 (7.29)

AKE = ^(pV)Vl2 - Vi(pV)v32 (7.30)

In Figure7-9, the fluid has moved from yi to y2overall. Therefore, the change ingravitational potential energy can be given by equation (7.31). Substitution of m= pV into equation (7.31) gives equation (7.32):

APEg = mgAh = mg(y2-yi) (7.31)

APEg = (pV)g(y2-yi) (7.32)

We are now in a position to substitute equations (7.28), (7.30), and (732) intoequation (7.25). After we divide through by V, we will get equation (7.33). Thisequation can be applied to a fluid that is incompressible, is frictionless, and haslaminar flow. We can rearrange equation (7.33) to look like equation (734). Thisis the more common form ofBernoulli's equation:

Pi -P2 = ^p(v22 - vi2) +pg(y2-yi) (7.33)

Pi + Vfcpvi2 + pgyi =P2 + V£pv22 +pgy2 (7.34)

As you view Bernoulli's equation it is unlikely that you are thinking, "wow,what a great equation. I can't wait touse it." Atface value, the equation does notlook friendly, sowe shall rewrite it ina more user-friendly format. Equation (7.35)is a more conceptual rewrite thatwill allow us to compare the impact ofvariouschanges on thepressure ofa moving fluid against theinnerwalls ofa vessel.

k=Pagainst inner walls + ^pv2 + pgh (7.35)

This version of the equation can explain some of the phenomena you may haveobserved in life such as the bottom of a shower curtain attacking your ankleswhen you turn the water on or your bike getting sucked towards a passing bus.When you turn the water on, it falls from the shower head to the bottom of theshower. As it falls, it picks up a good deal ofspeed (being that it's in free fall).According to equation (735), as v goes up and h barely changes, the pressureagainst the walls (in this case the shower curtain) must drop. The water is fastestnear the bottom of the shower curtain, so it is pushed towards your ankles by thehigher air pressure outside the shower. As soon as you turn the water off, thecurtain relaxesback to a straight down orientation.

This also explains why rolling down the windows of a moving car generatesairflow. When thewindow is opened, the rapidairflow outside thecaracross theopenspace where the glass used to becreates a lower pressure outside of thecarwindow. This causes airinside the car toflow out, thereby helping tocool the carand allowingvaluablepapers to flyfrom inside the car to outside.

A common example of Bernoulli's principle applied to engineering involves airfoils such as airplane wings, where pressure differences between above andbelow the wing can cause a plane tolift off ofthe ground at high speed.



In the case of an airplane wing, as the wing moves through the air, thesurrounding air must pass either across the top of the wing or across the bottomof the wing. A wing (or air foil as it is more formally called)is designed in such away that the top and bottom are asymmetric, resulting in different airflowdynamics on the two sides of the wing. The top of the wing is curved in such away that it is longer than the bottom, so if we compare the paths an air moleculewould take during the period of time when the wing passes by, it must movefaster to pass across the top of the wing than it would passing along the bottomof the wing. According to equation (735), a higher speed results in a lowerpressure along the surface if the wing. In essence, the air molecules by movingfaster have fewer opportunities to collide with the surfaceof the wing. The resultis a net force upward, which adds to the overall lift force on the wing. The totallift force on the wing involves the angle of the wing as well as the difference inair pressure due to air movement.This can be viewed in Figure7-10.

Assuming height doesn't change: Pagainst wing +V6pairvair2 =kAir speed above the wing is faster, so pressure is lower

Airspeedbelowthe wing isslower, so pressure is higher

Conclusion: As vair | :Pagainst wing A•'• Greater vair => Lower Pagainst wing

Figure 7-10

Example 7.9aWhile you are stopped at an intersection turn lane, a large truckzoomsby, just afew feetfromyour car.As the truck passes,your car willbe:

A. pulled towards the truck, because the increased wind speed between thetruck and the car lowers the pressure there.

B. pushed away from the truck, because theincreased wind speed between thetruck and the car lowers the pressure there.

C. pulled towards the truck, because the increased wind speed between thetruck and the car increases the pressure there.

D. pushed away from the truck, because the increased wind speed between thetruck and the car increases the pressure there.

SolutionThis isa qualitative typeofquestion, and you have to consider the local pressureon each side of the car. As the truck zooms by, the air in front of the truck ispushed out oftheway, resulting in a wall ofair rushing byyourcaron thesideofthe truck. We can see from equation (735) that as the speed of the air increases,theforce it exerts against the side of the carwill be reduced. Asa result, there is ahigher pressure against the sideof the caropposite of thezooming truckthan theside of the car closestto the zooming truck. The result is that the car is pushed inthe direction of the truck (which feels like you are being pulled towards thetruck). This eliminates choicesB and D. The air speed causes a drop in the localpressure againstthe side of the car,so choice C is eliminated.



Fluids in Motion


PhysiCS Fluids and Solids Fluids in Motion

Example 7.9bWhen blood flows through an artery, it exerts pressure on the surroundingarterial wall. Compared to a section of healthy artery of equal size, a narrowedsection of diseased artery experiences a pressure on the surrounding arterial wallthat is:

A. greater,becausethe flow velocity decreases.B. greater, because the flow velocity increases.C. smaller, because the flow velocitydecreases.D. smaller,because the flow velocity increases.

Solution

This is a qualitativetype of question, and you have two concepts to consider: thelocal pressure and the blood flow velocity. Consider first the concept youunderstand the best. For example, will blood-flow speed increase or decreasewhen blood traverses a narrow region of an artery? It should flow faster (fromthe continuity equation (7.23)). This rules out choices A and C. Next, will thepressure increase or decrease when the blood velocity increases? It shoulddecrease (from Bernoulli's equation (734)). These two equations are the mostimportant flow equations for the physics you'll do, and should therefore be atleast qualitatively understood. It might even be a good idea to commit them tomemory. By the way, the correct answer is choice D.

You may have incorrectly assumed that the pressure should increase because ofthe "high blood pressure" associated with atherosclerosis. The heart of oneafflicted with atherosclerosis pumps with a greater pressure to keep the bloodvolumetric flow rate at healthy levels, increasing the total pressure throughoutthe arteries. However, the pressure is lower in an unhealthy section ofanarterythan it is in a nearby healthy section because of the increased flow speed. Thiscomparison was thethrust ofthis question. Asan aside, the lower pressure in thenarrowing artery can be low enough to do more than offset the increasedpumping pressure of the heart, making the arterial pressure lower than that in ahealthy individual's artery.


TurbulenceWhen we examined Poiseuille's principle, we saw that adjacent layers of a fluidhave the ability to slide past one another in a smooth and uniform fashion (seeFigure 7-6). This type of flow is laminar flow. It is also called streamline flow.Poiseuille's principle holds true only for laminar flow. If the flow of a fluid issufficiently high, then a chaotic and irregular pattern develops in the fluid. Thisis called turbulent flow. Viscous frictional forces increase duringturbulent flow.

How can we determine whether a flow is laminar or turbulent? A quantity notexpressed in termsofanystandardunits called the Reynolds number Nr tells usthis. The Reynolds number isdefined by equation (736), where pis the density, r\is the viscosity, and Vis theaverage velocity of the fluid. The radius of thevesselthrough which thefluidflows is givenby R.

(7.36)

In general, if Nr < 2000, then the flow is considered to be laminar. If Nr > 3000,the flow is turbulent. Between 2000and 3000, the flow is said to be unstable. Thismeans thatflow can beeither laminar or turbulent depending on thematerial.



Example 7.10aWhich of the following will decrease the chance of turbulent blood flow in avein?

A.

B.

C.

D.

Widening the vein while maintaining the same flowspeed.Thinning the blood without changing its density.Increasing the absolute pressure on each end of the vein by the sameamount.

Lowering the blood density without thinning it.

Solution

You can apply the Reynolds number equation to each choice, crossing outchoices that either increase the chance of turbulence or leave that chanceunchanged. Choice A widens the vein and increases the Reynolds number. Thiswill increase the chance of turbulent flow, making it an incorrect choice.

One way to remember this relationship is to watch smoke rising from a candle.As it leaves the candle, it flows upward in a smooth (non-turbulent) narrowband. This band widens as it rises and becomes turbulent. So, narrow paths (orobjects) are less likely to encounter turbulence. Wider paths (or objects) are morelikely to do so. Choice B lowers blood's viscosity, increasing the Reynoldsnumber. This will make turbulence more likely. Choice B is incorrect. Choice Ccreates a greaterpressure everywhere around the vein, but this changes nothing.The onset of turbulence depends upon a sufficiently large flow rate. Flow rate inturn depends upon a pressure difference. If that difference does not change, theflow rate does not change.This rules out choice C, leavingonly choice D.

If this were the MCAT, you'd now move on to the next problem. However, ifyou'd like to know whyD is correct, then notice that lowering theblood densitywould decrease the Reynolds number. This means that turbulence is less likely.

Odds are good that the passage text would contain an equation as uncommonand unattractive as the Reynolds number equation. In fact, any time you aregiven anequation like this in thepassage, useit tohelp organize yourthoughts.


Example 7.10bWhich of the following will increase the possibility of turbulent airflow aroundthe wings of an airborne plane? (Assume the basic shape of the wing does notchangeand that all dimensions are proportional.)

A. Flying at faster speeds and using larger wings.B. Flying at slower speeds and using larger wings.C. Flyingat faster speeds and using smaller wings.D. Flyingat slower speeds and using smaller wings.

SolutionYou can apply the Reynolds number equation to each choice, and determinewhich will raise the value of Nr. The value of Nr increases with fluid speed, sochoices B and D can be eliminated. As the wings get bigger, the region in whichthe air flows around the wing increases. An increase in that area means that the Rhas increased,so Nr gets larger. This eliminates choice C.



Fluids in Motion


Physics Fluids and Solids Solids

Solids

Bydefinition, a solidis a material that can hold a definiteshape.Many solidscanbe referred to as crystalline solids (i.e., substances comprising a crystal lattice).They have a regular ordering of their constituent atoms. Solids usually have awell-defined melting temperature.

DensityAs we stated in the section on fluids, the density of a material is defined as itsmass per unit volume (see equation (7-1)). Densities vary with temperature aswell as with pressure. Note from the data in Table 7-1 that the density of a solidis (generally) greater than the density of either a liquid or a gas.

Elementary Topics in Elastic PhysicsWe have seen that gases and liquids tend to take the shape of their containers.Solids, however, have a definite shape. If a force is applied to a solid, then bondsbetween the atoms in that solid can be bent, compressed, or stretched. When theforce distorting the solid is removed, the bonds usually regain their originalconfiguration. Solids that experience these types of deformations are referred toas being elastic. A deformation is caused by a stress, whereas the deformationthat results is referred to as the strain. The modulus of elasticity is a material-dependent quantity that relates the stress to the strain. Be aware that not allsolidshave theability to returnto theirnormal configuration aftera deformation.

The stressa is defined as a force acting on a solid divided by the area on whichthe force is acting. This is givenby equation (737). The SI unit of stress is theN/m2. The three common types of stress are tension stress, compression stress,and shear stress.

Stress = a = Force = £-Area A

(737)

The strain Eis the fractional change in length of a solid and is a measure of thedeformation caused by a stress. This is given by equation (738). Note that thestrain is a dimensionless number and does not depend on the original length ofthesolid, only on the fraction bywhich thelength changes

Strain = eChange in Dimensional Length _ ALOriginal Dimensional Length L

(738)

In Figure 7-11 is a plot of the strain versus the stress in a solid. Note that forsmall values of the strain, the graphis a straight line. This is telling us that thestressis linearly proportional to the strain in that regionof the graph, sometimesreferred to as the linear region. For larger values, we find that the stress is nolonger proportional to the strain. Just past the linear region we encounter theelastic limit of the solid. If the force causing the deformation is removed whenthe material has reached its elastic limit, then the material can still return to itsoriginal shape. In other words, the deformation of the material has been elastic.A material is plastically deformed once it passes its elastic limit Any pointpastthe elastic limitresults in covalent bond breakage of the solid lattice structure. Ifstrain increases, we reach the material's ultimate tension strength, at whichpoint the material is under maximum stress. If strain is increased beyond thispoint, the stress on the material becomes reduced and eventually the materialwill fracture. Once a fracture has occurred, the material has been plasticallydeformed, and it will not return to its original shape.

Copyright © by The BerkeleyReview 78 The Berkeley Review


Ultimate tension

strength

Elastic ylimitX

Fracture0

l/l /•$(A

IS />cr> /6

Strain (e)

Figure 7-11

Materialsthat have ultimate tension strengths closein size to their fracture stressare often called brittle. They have a tendency to fracture, without deformingmuch. You see this when breaking a saltine cracker or snapping a strand of dryspaghetti. Materials that have fracture strains far above their ultimate tensionstrength are said to be ductile. Beyond the ultimate tension strength, they willdeform noticeably beforefracturing. You see this when pullingon chewing gumor stretching a rubber band.

In Figure 7-11, the slope of the linear region is the stress divided by the strain.This is called the Young's modulus Yof the solid and is given by equation (739).Young's modulus is the proportionality constant that relates the stress to thestrain. It is sometimes referred to as the modulus of elasticity, E.

E= Y = Stress _ a _ F/AStrain e AL/L

(739)

Example 7.11aFora ropeexperiencing a constantapplied force, all of the following changeswilldecrease the stretching length of the rope EXCEPT:

I. usinga shorter rope made from the same material.II. using a thinner rope made from the same material.HI. using a rope of similar size and with an equal value of Young'smodulus.

A. I and II onlyB. I and III onlyC. II and in onlyD. I, H, and in

SolutionThis is a particularly difficult question not because of the concepts, but becauseof the ease with which you can make a careless error. Lefs use physical intuitionand then the relevant stress-strain equation to solve this problem. We want toknow which changeswill either increase or leave unchangedthe absolute lengththe rope stretches.

Regarding Statement I, a shorter rope has less material to stretch (deform), sounder identical forces, a shorter rope will not stretch as far as a longer rope ofequal radius and the same material. This results in a decrease in the length, soStatement I should not be in the answer. This eliminates all choices except C.


Solids



Regarding Statement II, thicker ropes are usually used for heavier loads-presumably tokeep the rope from stretching and breaking. Therefore, switchingto a thinner rope should increase the stretchinglength of the rope.This supportsStatement II and eliminates choice B.

Statement HI should leave the rope length unchanged. After all, that dependsupon the size of the rope and its elasticity (which Young's moduluscharacterizes). Since neither one changes, Statement III must be true, and choiceA must be incorrect.

To verify our conceptual choice, lefs use the stress-strain relationship describedby equation (739). Solving for the stretching distance AL:

AL = £LAY

Using a shorter rope decreases L, which will decrease AL—remember, the othervariables are held constant for Statement I. This confirms that Statement I wascorrectly ruled out. Using a thinner rope decreases A, which will increase AL;that supports StatementII.Finally, using a rope of equal size and equal elasticity(i.e., of the same Young's modulus) will not change AL; this validates StatementIII and points to choice C. There are many ways to answer a question-two waysare using physical intuition and looking for the basic relationship in equations. Inusing equations, remember that they always tell you the relevant conceptualphysics of the question.


Example 7.11bWhich of the following changes to a cable will increase the chance that the cablewill stretch to a point where it irreversibly deforms?

A. Make the cable thicker.B. Apply a greater a compression force.C. Apply a greater tensile force.D. Use a materialwith a greater elasticity coefficient (Y).

Solution

First off, considering we are concerned with stretching the cable, a compressionforce is not applicable. Choice B can be eliminated immediately. A materialstretches beyond its elastic limit when a threshold stress is exceeded, causing astrain so great that the material permanently deforms. Thisbasically means thatthe AL is too greatfor the material to relaxback to its original state once the forceis removed. A thicker cable is harder to stretch, so choice A will reduce thelikelihood that the material will deform. Choice A is eliminated. According toequation (739), as the value of Y increases, the AL for a cable under the sameapplied force will decrease. This means that the cable will be less likely toirreversibly deform. Of the choices, only a greater tensile force (stretching force)will be likely to excessively stretch the cable.This question was more of questionof whether you knew the terminologymore so than understanding the concept.


Copyright ©by The BerkeleyReview 80 The Berkeley Review


Bulk and Shear DeformationBulkstress and bulk strain are different from the compressive and tensile stressesand strain we considered with Young's Modulus, because now the force on theobject is being applied on all sides, so it deforms in all directions rather than justone. Bulk deformation results in a change in the volume of an object. It is oftenencounteredwhen an objectis submerged deep into a body of water, such as theocean. The volume strain is shown in equation (7.40), where AV is the change involume and V0 is the initial volume of the solid object

Volume strain =AV

(7.40)

This can be applied to measure the bulk modulus, B, of an object. The bulkmodulus of an object relates how much the volume of a solid object changes asthepressure surroundingit changes. It canbe calculated usingequation (7.41).

B =AP

AV/AVo (7.41)

The negative sign is presentbecause an increase in external pressure will reducethe volume of a solid. Figure7-12shows a basic view of a solid submerged in asurrounding fluid undergoing bulk deformation.

Av /Hydraulic Compression = B / y

Compression = AV

Figure 7-12

Bulk deformation involves forces that are perpendicular to the surface of anobjecfs walls. When a force acting parallel toa surface ofa solid object is strongenough, there is the chance for a third type of deformation know as shear. Thisoccurs when one layer slides across another layer ina material. Certain materials,such as graphite, are prone to shear because their crystalline structure is one oflayers. Graphite isused to reinforce many structures inone dimension, because itis lightweight and strong in the direction perpendicular to its crystal layering.But when a force isapplied parallel to the layers, it breaks rather easily, aswe seewith pencil lead. While there are equations for shear, we shall only consider theconcept Figure 7-13 shows a basic view ofa solid undergoing a lateral shearingforce.

Copyright©by The Berkeley Review 81

Solids



'l

t = 0

t = l

t = 2

Figure 7-13

Example 7.12aWhich of the following materials likely has the greatestbulk modulus?

A. Oil

B. Rubber

C. Steel

D. Water

Solution

A large bulk modulus is associated with a solid that is difficult to deform. Wecan eliminate choices A and D immediately, because they are liquids. Based onyour real life experiences you have no doubt come to learn that it is easy tocompress a rubber item, but steel is sturdy and can withstand great stresses.Steel has the greatest bulk modulus.


Example 7.12bWhen a stack of papers is placed on a sloped table, the top papers sometimesslide off of the bottom papers, which remain stationary. This is an example of:

A. bulk deformation.B. shear deformation.C. gravitational distortion.D. tensile deformation.

Solution

Because one layer is sliding across another, it is an example of shear. Bulkdeformation involves a change in the volume of an object due to a change in theexternal pressure on the object. Tensile deformation involves linear stretching ofa material caused by a tension. There is no physics concept referred to asgravitational distortion, so choice Cshould have been eliminated immediately.



25 Fluids and Solids Review Questions

I. Hydraulic Press

II. Relax and Float

III. Stress and Strain


(1-7)

(8 -14)

(15 - 21)

(22 - 25)



A hydraulic press is an ingenious device that allows thelifting of heavy objects by the application of a small force toa small piston. A basic schematic of a hydraulic press isshown below:

F2

£=—4=~<3Ah2 A^ r

Piston 1Piston 2

Figure 1. Hydraulic Press

A force Fj is applied to Piston 1.Thiscreates a pressurePj on Piston 1, which increases the pressure uniformly

throughout the much larger oil reservoir and results in a forceF2 onPiston 2.A pressure P2 is therefore applied to Piston 2.

In the following problems, assume that the radius ofPiston 2, r2, is five times thatof the radius of Piston 1, rj.

1. The force on Piston 2 is larger than that on Piston 1because:

A. F2= F1

B. Fo>F

2.

3.

1

C. P2= P!

D. P!>P2

The fact that a hydraulic press works the way it doesmeans:

A. energy is not conserved for this process.B. momentum is not conserved for this process.C. forces are conserved for this process.D. none of the above.

If Piston 1 is moved a distance xj, by how much willPiston 2 move?

A. x2= 5xj

B. x2=10xj

C. x2 =xl1 5

D. x2 =xl1 25


4. To increase F2 by four times, one must:

A. increasePj by a factor of 2.

B. increaser2 by a factor of 2.

C. increase r2 by afactor ofV2.D. increase rj by a factor of 2.

5. Which of the following changes will NOT decrease theaverage time it takes the hydraulic press to pass backthrough its equilibrium position, if Piston 1 is subjectedto a briefly applied external force that is subsequentlyreleased?

I. Decrease the oil viscosity.

II. Increase the mass of Piston 1.

III. Increase the volume of the oil reservoir.

A. I and II only

B. II only

C. II and III only

D. I and III only

6. Could this hydraulic press work in the absence ofgravity?

A. Yes, since the pressure throughout the oil wouldstill be equal.

B. No, since there would be no need to lift anythingagainst gravity.

C. Yes, since the oil pressure would vary across thereservoir, producing a greater difference betweenFj and F2.

D. No,sincethe inputforce must be gravitational.

7. Increasing the density of the hydraulic fluid would havewhat overall effect?

A. Both Pj andP2would increase.

B. Pj would increase; P2would remain thesame.

C. P2 would increase; Pj would remain thesame.

D. Pj and P2would be unaffected.


Passage II (Questions 8 -14)

Whether an object is submerged completely in a fluid orwhether it floats on the surface, it experiences a buoyantforce. Buoyant force is the force of the fluid on the object; itresults from pressure differences. The buoyant force is:

B = Pfluid'Vdisplaced fluid'g

A researcher sets up an experiment: A wooden box ofcross-sectional area A and density p floats on the surface of afluid, so that a length L of the box is submerged (Figure la).A force F is then applied to the top of the box, so that now anadditional length x of the box is submerged (Figure lb).

(a)

Area

2S

MZ

Fluid

Figure 1

The force F is removed and the box bobs as a simpleharmonic oscillator, where the acceleration a, displacementfrom equilibrium x, and oscillation frequency / obey:

a= -4ji2/2x

Neglect the effects of fluid viscosity in these questions.

8. The three tanks shown below contain objects of thesame size and density, at the same depth in each tank.Which object experiences the GREATEST pressure?

9.

©"II

oA. Object I

B. Object II

C. Object III

D. All three objects experience the same pressure.

Buoyant force is always equal to the:

A. weight of the object.

B. weight of the fluid displaced.

C. volume of the object.

D. volume of the fluid displaced.

®Copyright © by The Berkeley Review 85

10. Three objects are placed in a tank of water, as shownbelow. The objects all have the same size, but they havedifferent densities: pj < pn < prrj < pwater-

11.

Ql On OmIf all of the objects are released at the same time, whichobject reaches the surface first?

A. Object I

B. Object II

C. Object III

D. All three objects will have the same acceleration

If salt is added to water, the water density increases.The force needed to submerge a floating, completelyincompressible object will:

A. increase as the density increases and increaseslightly as the submersion depth increases.

B. increase as the density increases and decreaseslightly as the submersion depth increases.

C. decrease as the density increases and increaseslightly as the submersion depth increases.

D. decrease as the density increases and decreaseslightly as the submersion depth increases.

12. Three objects with the same density but different sizesare placed in a tank of water as shown below:

II

si QAfter the objects are released:

A. Object I has the greatest acceleration.

B. Object II has the greatest acceleration.

C. Object III has the greatest acceleration.

D. All three objects will have the same acceleration.


13. As a solid object sinks in water, the:

I. water pressure compresses the object, raising itsdensity.

II. local water density increases slightly, due to theweight of water above that location.

A. Both I and II increase the magnitude of the buoyantforce on the object

B. Both I and II decrease the magnitude of thebuoyant force on the object

C. Only Statement I increases the magnitude of thebuoyant force on the object

D. Only Statement II increases the magnitude of thebuoyant force on the object

14. In the experiment described in the passage, what is thefrequency of oscillation of the box?

A. In V

vs1 J Pbox

2n V L + x1 WPftuidg"

2ji V AL

B. -L.

C.

D.

Pfluid g

PboxL

g_X


Passage III (Questions 15 - 21)

Matter deforms when stressed. Stress is defined as:

a = F/A

where F is the force applied to an object and A is the cross-sectional area that experiences the force. Three stresses arecommonly defined: tension stress, which elongate an object;compression stress, which compresses an object; and shearstress, which is the application of scissors-like forces.

The deformations produced are called strains. Strainsare fractional changes in lengths:

L

where L is the original length of the object.

Stress and strain are linearly related (over a given rangeof stresses) by Young's modulus, a material's physicalproperty:

Y = 2E

Table 1 lists Young's moduli for various materials, as well astheir ultimate tension stresses and ultimate compressionstresses (the maximum stress that a material can tolerate):

Material Y[N/m2l Tension

Strength(N/m2)Compression

Strength (N/m2)Aluminum 7x 1010 2xl08 2xl08

Steel 2xlOn 5xl08 5 xlO8

Glass 7xl010 5xl07 not measured

Brass not measured 2.5 x 108 2.5 x 108Copper 9.7 x 1010 not measured not measured

Concrete not measured 2xl06 2xl07

Bone 1.6 xlO10 1.3 xlO8 1.7 x 108

Table 1

A group of students decide to study the relationshipbetween stressand strain. They do two experiments.

Experiment J

A 5-meter-Iong thin steel wire is hung vertically. The wirehas a cross-sectional area of7.85 x 10"7 m2. A 10-N weightis hung at the lower end. The positionof the lowerend of thewire is read on a scale. As more load is added to the wire,the students record their data (Table 2).

After

Load

Scale

ReadingAfter

Load

Scale

Reading

ON 4.000 cm 40N 4.128 cm

ION 4.032 cm 50 N 4.160 cm

20 N 4.064 cm 60 N 4.192 cm

30 N 4.096 cm 70 N 4.224 cm

Table 2


Experiment2

Students observe the mechanical properties of thin rods ofequal size, constructed from either concrete, brass, or steel.The rods either support loads at their middle sections whenoriented horizontally, or loads at their upper end whenoriented vertically.

15. Which material requires the SMALLEST applied stressto have a 5% strain?

A. Steel

B. Copper

C. Aluminum

D. Bone

16. What is the maximum strain of the wire in Experiment1?

A. 4.48 xlO"4

B. 0.064xlO"2

C. 0.032

D. 0.224

17.

18.

In Experiment 1, a steel wire was used. What is themaximum load that can be hung from the wire?

A. 35 N

B. 275 N

C. 393 N

D. 480 N

Further experimentation measured strains for muchhigher stresses than those reported in Experiment 1. Asthe data plotted below became non-linear:

Strain (e)

A. the wire began stretching.

B. the wire began to deform permanently.

C. the wire became stiffer.

D. Young's modulus for the wire increased.


19. What was the maximum stress in the wire inExperiment 1?

A. 7x 10!N/m2

B. 4xl02N/m2

C. 9 x 106 N/m2

D. 9 x 107 N/m2

20. When comparing the ultimate compression and theultimate tension strengths given in the passage:

A. the two values are always equal.

B. the ultimate compression strengthare always largerthan the ultimate tension strength.

C. the ultimate compression strength is independent ofthe ultimate tension strength.

D. the ultimate compression force and ultimatetension force are equal only if the material ishomogeneous.

21. In Experiment 2, the students found that:

A. when horizontal, concrete supported more weightthan brass or steel.

B. when horizontal, all three materials could supportthe same maximum weight.

C. when vertical, steel supported more weight thanbrass or concrete.

D. when vertical, brass and steel could support thesame maximum weight.



22. What will be observed when a more viscous liquid, ofthe same mass density, is substituted for the lessviscous liquid in the system below?

23.

Reservoir Column 1

Point A Point B

A. A greater fluid velocity at point B, but anunchanged fluid height in Column 1.

B. A greater fluid velocity at point B, and a greaterfluid height in Column 1.

C. A lower fluid velocity at point B, but an unchangedfluid height in Column 1.

D. A lower fluid velocity at point B and a greater fluidheight in Column 1.

Comparing the average blood pressure in similar veinsin the neck and calves of a standing human reveals thatthe blood pressure will:

A. be greater in the neck.

B. be greater in the calves.

C. be the same in both places.

D. depend upon the average blood pressure in thebody.

24. As blood flows steadily through a pipe of uniformcross-sectionand length L, its speed:

A. must increase to counteract losses to viscosity.B. must decrease to counteract losses to viscosity.

C. must increase, because there is a pressuredifference between the ends of the pipe.

D. remains unchanged, as long as the pressuredifference between the ends of the pipe is constant.


25. To make the pressure in Region 1 equal to the pressurein Region 3, what can be done?

Region 1 Region 2 Region 3

1 •

1 1

1 1

1 1

i k / : a :1

! i ^ ! _-- A

•

yi :1 •

y2

—y

\ y3!1 1

1 1

1 1

: v :

I. Increase yi

II. Lengthen Region 2

III. Decrease y3

A. I only

B. HI only

C. Either I or II

D. Either I or III

i. C 2. D 3. D 4. B 5. C

6. A 7. D 8 D 9. B 10. A

11. A 12. D 13. D 14. A 15. D

16. A 17. C 18. B 19. D 20. D

21. C 22. C 23. B 24. D 25. D

YOU ARE DONE.

Answers to 25-Question Fluids and Solids Review

Passage I (Questions 1-7) Hydraulic Press

1. Choice C is the best answer. Pascal's principle states that a pressure change applied to any point of a contained fluid istransmitted undiminished in all directions to every point inside that fluid. If a pressure change Pj is applied to a containedfluid by Piston 1, then this pressure change is felt uniformly everywhere inside that fluid. P2 must equal Pj, which means:

^1 = ^2.A, A2

Because the surface areas of each piston are different, while they both share the same force/area, the forces must be different.The force on the large piston surface is greater than the force on thesmallerpiston surface, meaning that F2 is in fact greaterthan Fj. However, that is merely an observation and not a reason for the difference in force magnitudes. Choice B is

incorrect, because the question asks for the cause of the force difference; choice B merely repeats the question. Choices Aand D are incorrect relationships. The best answer is choice C.

2. Choice D is the best answer. Although a hydraulic press makes it possible to produce a large force upon the application of asmall force, no physical laws are being violated. If the fluid is incompressible, then the work applied on one side is equal towork acting on the other side. Energy is conserved, so choice A is eliminated. The fluid moves when the piston is put intomotion, which in turn moves the other piston by a distance that is proportional to their relative sizes. This followsconservation of momentum, so choice B is eliminated. The force is different on each side of each piston, because theirsurface areas are different, so the force is not conserved. Choice C is eliminated. By default, the best answer is choice D.

3. Choice D is the best answer. By conservation of energy, the work done on Piston 1 must equal the work done on Piston 2.Work is the product of force and displacement, so we can say:

Fjx, =F2x2

Since r2 =5rj and since A= jtr2, A2 =25A j.That means that F2 =25Fj, so x2 must be -L.The best answer is choice D.

4. Choice B is the best answer. From the equation:

Aj A2

vve see that in order to increase F2 by a factor of 4, we must increase Fj by a factor of 4, increase A2 by a factor of 4, or

decrease Aj by a factor of4. Since A= jtr2, then in order to increase A2 by a factor of4, we must increase r2 by a factor of2. In order to decreaseA j by a factor of 4, we mustdecrease rj by a factor of 2. The best answer is choice B.

5. Choice C is the best answer. We want to know which changes will NOT decrease the time for the pistons to pass backthrough their equilibrium positions. We must find which choices either increase or do not affect this time. Decreasing oilviscosity would allow the oil to flow more freely and the pistons to move more easily, and therefore more quickly, towardsequilibrium. This invalidates Option I and eliminates choices A and D. Of the remaining choices, vve need consider onlyOption III, because Option II is in both of the remaining choices. Increasing the volume of the reservoir does not affect howfast the pistons move. From the passage, vve know that the reservoir is substantially larger than the piston columns.Therefore, the flow of oil from Column 2 to Column 1 is not affected much by an increase in reservoir oil volume. Thisvalidates Option III, making choice C the best answer. Let's consider Option II. Increasing the mass of Piston 1 increases thegravitational force which, in turn, leads to a smaller acceleration of Piston 1 back up towards equilibrium. The gravitationalforce here acts in the same direction as the force initially applied to Piston 1. The same can be said for Piston 2 moving in theopposite direction. The basic idea is that heavier pistons move more slowly than lighter pistons. This will increase the timetowards equilibrium, validating Option II. The best answer is choice C.

6. Choice A is the best answer. The hydraulic press works when one piston is pushed in towards the oil reservoir by any kindof force, which in turn presses outward against the other piston. This process requires only that the oil be there, not that anygravity be present. Because the press does not need gravity, choices B and D are incorrect. Choice C is incorrect, because thepressure must be equal throughout the reservoir. If the pressure varied, there would be a continuous flow of oil towards theside with lower pressure until the pressures were uniform. The principle behind this is often referred to as Pascal's principle.The best answer is choice A.

Copyright ©by The Berkeley Review® 89 MINI-TEST EXPLANATIONS

7. Choice D is the best answer. Pressure is defined as force/area. Changing the density of the fluid will not change the cross-sectional areas of the pistons, and it will not change the magnitude of the force applied to Piston 1. If:

Pl = p2 => El=EiA, A2

then F2 also remains unaffected. The bestanswer ischoice D.

Passage II (Questions 8 -14) Relax and Float

8. Choice D is the best answer. Although the three tanks have three different shapes, the height of the fluid in all three cases isthe same. Pressure at a depth h below the surface of the fluid is pgh. Since the objects are all at the same depth h, they allexperience the same pressure. The best answer is choice D.

9. Choice B is the best answer. This is the definition of buoyant force; but even if you did not know this, you could get to therightanswerfrom the equationfor buoyant force given in the passage:

B = Pfluid'Vdisplaced fluid'g

Density and volume are related by mass:

B=massdispiaced fluid'g. so, the buoyant force is equivalent to the weight ofthe fluid displaced. The best answer ischoice B.

10. Choice A is the best answer. Acceleration depends on the density of the object. An object with a smaller density will have agreater acceleration. If the objects are all released at the same time, then Object I will reach the surface First. The bestanswer is choice A.

11. Choice A is the best answer. If the fluid density increases, the buoyant force formula in the passage indicates that thebuoyant force will increase. This means that more force is needed to submerge the object, in order to overcome the buoyantforce. This rules out choices C and D. As the submersion depth increases, more force will be needed to get the object to thatlower depth. Why? As depth increases, so does the local density of the fluid. Fluids, even water, are slightly compressible.The increased pressure at lower depths compresses the liquid lightly. This increases the buoyant force against the object.Normally, the object would also compress somewhat, complicating matters; but this object is said to be completelyincompressible. The best answer is choice A.

12. Choice D is the best answer. Because the objects each have a density less than that of water and are completely submerged,the buoyant force acting on the objects is greater than the weights of the objects. There will be a net force upwards on eachobject, so they will rise. Consider the free body diagram for any one of the objects:

where B= pwater^objectg ana" the weight of the object can bewritten as:

mg = Pobject Vobject 8Note that because the object is completely submerged, the volume of the object and the volume of displaced fluid are equal.The difference between these two forces gives us the net force acting on the object. Solving for acceleration:

ma = B - mg or ma= (pwater'Vobjecfg) - (PobjecfVobjecfg)

V i •_ vobject g(pwate r" Pobjectj ___ "object g ^Pwater " PobjectJ _ g \Pwater" Pobject;

Pobject* object Pobjectm

The acceleration of each object is independent of its volume (size), but depends on the density of the object. If all of theobjects have the same density, they all have the same acceleration. If the objects are all released at the same time, they willall reach the surface at the same time. If this seems long-winded, then just remember that objects with the same density aresubjected to the same buoyant acceleration in a fluid. The best answer is choice D.


13. Choice D is the best answer. Both statements describe ways in which theobject's density increases as it is submerged intoatank of water. The object's mass cannot change, as it neither gains nor loses matter, so the increase in density is associatedwitha reduction in the object's volume. Decreasingthe object's volumedecreases the buoyant force on it, by decreasing thevolumeof fluid displaced by the object. This invalidates choices A and C. Having a larger water density, on the other hand,would increase the buoyant force, as seen in the buoyant force equation given in the passage. The best answer is choice D.

14. Choice A is the best answer. Solve for frequency, using the equation given in the passage:

=74A/ =

where the minus sign has been dropped, since the magnitude of frequency depends only upon the magnitude of theacceleration. Plus and minus signs generally indicate direction, not magnitude. Let's use this equation and physical intuitionto choose the best answer. The acceleration in the numerator means that more acceleration will increase the frequency.Gravity is one force accelerating the box (through mg and the buoyant force); we therefore expect that the gravitationalacceleration g should be in the answer. This rules out choice C. Furthermore, the units should be those of accelerationdivided by length, which rules out choices B and D. This leaves only choice A, which looks curiously similar to theoscillation frequency of a pendulum of length L. Notice that the oscillation frequency would increase, if the box were madeless dense than the fluid. What happens to the frequency when the density of the box exceeds that of the fluid? The box nolonger has a frequency, and it sinks. The best answer is choice A.

Passage III (Questions 15 - 21) Stress and Strain

15.

16.

Choice D is the best answer. The material requiring the least stress for a given 5% strain must be the material with thesmallest value of Young's modulus:

Y = 28

where o is the applied stress and e is the resulting strain. From the data given in the passage, bone has the smallest modulus,makingchoice D the best answer. This is why they use metal, which is stronger than bone, to hold shattered bones together,and why they do not use bone to hold together fractured metal. The best answer is choice D.

Choice A is the best answer. The maximum strain in the wire occurs with the maximum force, 70 newtons. The change inthe wire's length of the wire with the application of this force is 0.224 cm, so the strain is:

0.224 x 10-2 m 0.448 x 10"2e =

AL

5 m 10= 0.448 x 10° = 4.48 x 10"4


17. Choice C is the best answer. The wire is made of steel, so the maximum load by using the ultimate tension strength is:

18.

19.

aT= rmax Fmax =OtA =(5 x108 N/m2)(7.85 x10-7 m2) « 40 x10140x 101 isslightly overestimated, because 7.85 isless than 8.Looking at the possible choices, choice Cseems tobe the best.Doing the arithmetic completely will confirm this; but you have only 1.29 minutes to answer each question, which makesdoing complete arithmetic a complete waste of time. Always use the four answer choices to reduce your arithmetic (e.g., askyourself if 5 x 7.85 looks more like 3.5,27.5, 39.3, or 48.0). The best answer is choice C.

Choice B is the best answer. As the nonlinear region of the graph begins, it takes very little stress to strain (stretch) the wire.That is what the flattening of the curve represents. If you have ever pulled on one a wad of chewing gum as you were bitingon the other end, you know that at some value of stress, the gum begins to stretch very easily. Beyond this point, the gumwill never elastically rebound to its original shape; it has permanently deformed. This points towards choice B. Choice A isincorrect, because the wire stretches when any stress is put on it—remember, stress gives strain. Stretching alone would notexplain the bend in the plotted data. Choices C and D say essentially the same thing-stiffer materials have a bigger Young'smodulus. This invalidates both choices C and D, leaving only choice B. The best answer is choice B.

Choice D is the best answer. To find the maximum stress, we use the maximum applied force, 70 newtons:

70 newtons _ 70 x iq7 N0= F =A 7.85 x IQ"7 m2 7.85 m-*

Since 70/7.85 is between 1 and 10, the order of magnitude of the answer must be 107, making choice D optimal. Do notwaste time by doing the division. Just convince yourself that the answer is between 1 and 10. The best answer is choice D.

Copyright ©by The Berkeley Review® 91 MINI-TEST EXPLANATIONS

20. Choice D is the best answer. An applied stress is opposed by intermolecular forces within the material. If the compositionof the material is not homogeneous, these intermolecular forces are not equal for tension and compression. This points tochoice D as the best answer. Choice C is incorrect, because most of the numbers reported in the table are strikingly similar.For two independent quantities, you should generally expect unrelated numbers. Choices A and B are wrong for two reasons:There are missing data trends in the tables, which precludes verifying any general relationship; and the data for bone havedifferent values for the ultimate tension and compression strengths. When encountering an "always" or "never" choice, justlook for or think of one example that invalidates the choice-more often than not, there is at least one. The best answer ischoice D.

21. Choice C is the best answer. When oriented vertically, the compression strength will dictate how much weight a given rodcan withstand. From the table given in the passage, steel has the highest compression strength of the three materials. Thus,steel can support the most weight; choice C is correct. Choices A and B are wrong, as you know if you have ever seen abuilding under construction. For horizontal concrete beams, there are always steel rods within the concrete. These rods arethere for support, implying that steel is better than concrete in this horizontal orientation. Why not make all the horizontalbeams out of steel? Two reasons are: First, the building would become unreasonably heavy, requiring even more verticalsupport; and second, steel costs more than concrete. The best answer is choice C.


22. Choice C is the best answer. The fluid velocity will decrease if a more viscous liquid is used, assuming that all otherconditions remain the same. This can be deduced from the definition of viscosity, which is the resistance to flow. We caneliminate choices A and B. The fluid heights in the columns depends upon the pressure below each column in the drainagetube. The pressure difference across the drainage tube depends upon the pressure at the reservoir end (which is due to theweight of the fluid) and the pressure at the exit end (which is due to the atmosphere). Since neither pressure changes, theintroduction of the new fluid (whose mass density is equal to that of the previous fluid) will not affect the fluid heights in thecolumns. The best answer is choice C.

23. Choice B is the best answer. Although blood is a viscous fluid, the effects of viscosity are small on it, and we canapproximate the blood as an incompressible, nonviscous fluid. This means that Bernoulli's equation is applicable:

P + pgh + -pv = constant

If vve compare similar veins, then the flow velocity through them should be somewhat similar. Thus, any blood pressuredifference between them arises from blood pressure being measured at different elevations separated by a distance h. Sincethe calves are closer to the ground than the neck, in a standing human being the calves experience a greater average pressure.This is why an open wound is elevated to reduce bleeding, and it makes choice B the best answer. Choice D is invalid,because any average changes in the blood pressure produce similar changes in the neck and calves. The best answer ischoice B.

24. Choice D is the best answer. As blood flows steadily through a pipe of uniform cross-section and length L, its speedremains constant, as long as the pressure difference between the ends of the pipe is constant. The best way to answer this isto eliminate the three incorrect choices. The question stated that the blood flows steadily, which means Q is a constant. Butwe also know that Q = Av. The pipe has a uniform cross-section, which means A is a constant. This all implies that v, thespeed of the fluid, is also a constant. Choices A, B, and C all say that the speed changes, so they must be incorrect. The bestanswer is choice D.

25. Choice D is the best answer. Comparing pressures between two different regions requires employing Bernoulli's equation.Bernoulli's equation relating Region 1 to Region 3 says:

1

Copyright ©by The Berkeley Review®

pi +^-p(vi)2 +pgyi =p3 +- (v3f +PSV3Since Aj = A3t the continuity equation implies \\ = V3. This means that the above relationship reduces to:

Pi +Pg>'i = P3 + P8Y3

To make P\ equal to P3, yi must be equal to y3 (given that Aj = A3). For this to occur, y\ must increase, or y3 must

decrease. This makes Options I and III valid. The lengthening of Region 2 will have no effect on the pressures against thewalls in Region 1 or Region 3. The best answer is choice D.

92 MINI-TEST EXPLANATIONS

52-Question Fluids and Solids Practice Exam

I. Stresses and Strains

II. Balloons and Dirigibles

III. Cannula


IV. Blood Flow Analog

V. Toilets and Siphons


VI. Poiseuille's Law

VII. Buoyancy Balance


Fluids and Solids Exam Scoring Scale


42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3

(1-5)

(6-11)

(12 -17)

(18-21)

(22 - 27)

(28 - 33)

(34 - 37)

(38 - 42)

(43 - 48)

(49 - 52)


Whena stress (which is pressure) is applied to a solid, itwill deform slightly. The resulting deformation is called astrain. There are several ways of stressinga solid. In general,stresses and strains are related through an elastic modulus E,as long as weremain within the elastic limitof the solid. Thegeneral relationship of stress to strain is:

E =stress

strain

The stress is always an applied force per area, and thestrain is always a fractional change in some dimension, suchas length or volume.

B =AV/V

where V is the original volume of the solid, and B is aconstant called the bulkmodulus that depends on the materialmaking up the solid.

As a second example of stress and strain, consider acylindrical wire of length L and cross-sectional area A,attached to a ceiling with one end dangling. If a force F pullsthe wire downward, perpendicular to A, it will stretch thewire by an amount AL.The amount of stretch is related to theamount of applied force through:

Y =F/A

al/l'

where Y is called Young's modulus and is a constant thatdepends on material. Another important elastic modulus iscalled the shear modulus, S. We will not define S here; but ifthe solid is isotropic, Y, B, and S are related through:

S =2(1 + o)

B =3(1 - 2a)

where a is a dimensionless constant that depends on thematerial.

Useful information:

g= 10 m/s2, pwater = 1000 kg/m3, Bai= 7.5 x 1010 Pa

1. How much pressure P must be applied to an isotropicsolid of bulk modulus B to compress it to half of itsoriginal volume, assuming you remain within its elasticlimit at all times?

A. P = 2B

B. P-=B

C. P = B4

D. P = B2

Copyright ©byThe Berkeley Review®

2.

3.

4.

5.

94

A 1-m3 block of aluminum sank to the bottom of a lakeof depth 750 m. By how much did the block shrink?

A. 0.1m3

B. 0.01m3

C. 0.001m3

D. 0.0001m3

Suppose Y, B, and S are the Young's, bulk, and Shearmoduli of aluminum, respectively. Which of the

following statements istrue, ifo = V3 for aluminum?

A. Y<B<S

B. Y>B>S

C. Y = B,S<Y

D. Y = B,Y<S

How much work is done by a force F that stretches awire of length L by AL, if the wire has cross-sectionalarea A, and Young's modulus Y?

YAL

L

YAL

L(AL)2

A.

B.

C.

D.

YA

YA(AL)22L

Suppose an analogy were made between Hooke's law,given by FSpnng = -kAx, andYoung's modulus, Stress =(Y) (Strain). Which quantity in the Young's modulusequation corresponds to the spring constant k inHooke's law?

A. YAAL

B. YAL

C. YA

L

D. YA



When an object is placed into a fluid, it displaces a setamount of that fluid. The displaced fluid exerts an upwardforce on the object This force is called the buoyant forceand is equal to the weight of the fluid displaced.Qualitatively, the buoyant force is determined using Equation1.

B = PfluidVfiuidg

Equation 1

where the mass of the fluid displaced has been written interms of its density and volume. For an object that floats, thebuoyant force is equal to the weight of the object. Thebuoyant force is determined from the submerged portion ofthe object, where V in Equation 1 is the volume of the objectthat is submerged.

Applications of the buoyantforce usually involve objectsfloating in a liquid. However, air is a fluid, and objects canfloat in air as well. Buoyancy is the principle governingbothhot-air balloons and helium-filled balloons. The balloons

displace an amount of the surrounding air, and this air exertsa buoyant force on the balloon. If the buoyant force isgreater than the weight of the balloon, then the balloon willrise. The balloon continues to be accelerated upward untilthe buoyant force exactly equals the weight of the balloon.

Helium-filled dirigibles (blimps) are used for slow airtravel and can often be seen at major sporting events. Lateralmotion in a dirigible is carried out using thrusters. A hot-airballoon can move laterally from either thrust or the presenceof wind. The pilot of a hot-air balloon must depend on aircurrents for lateral displacement.

6. A helium-filled balloon rises to a height of 50 km. Atthis height, the balloon hasa volume of 5000 m3 and thetotal load is 2000 kg. What is the density of air at 50km, if the buoyant force on the balloon is 2.4 x 102Newtons?

A (2.4 x102)(9.8) kg/5000 /m

2.4 x102 kg/' (5000X9.8) /m

(2.4 x102)(2000) kg/(5000)(9.8) 'T

(2.4 x102)(5000) kg/(2000)(9.8) /m3

7.

m-

As a hot air balloon ascends, what is observed?

A. The external pressure decreases.B. The mass of the hot-air balloon increases.

C. The center of mass for the hot-air balloon systemshifts to a higher point.

D. The surface area of the balloon decreases.


8. To lift a 2000 kg load with a hot-air balloon rather thana helium-filled balloon, what changes to the balloonsystem should be made?

A. More ballast should be used in the hot-air balloon,since air is less dense than helium at the sametemperature.

B. The volume of the hot-air balloon should be bigger,since air is more dense than helium at the sametemperature.

C. A balloon with a smaller volume should be used,since air is more dense than helium at the sametemperature.

D. No changes need to be made, since equal moles ofair and helium occupy the same volume at the sametemperature and pressure.

9. As a hot-air balloon ascends upward at a constantspeed, a package is dropped out of the balloon. At theinstant the package is released, the momentum of theballoon:

A. increases.

B. decreases.

C. is unaffected because theforce does notchange.D. is unaffected because the decrease in mass is

compensated by an increase in velocity.

10. When using hot air instead of helium in a balloon to lifta heavy load, which of the following statements shouldbe true? [Note: Tjn = temperature inside the balloon;Tout = temperature outside the balloon.]

A. Tin = ToutJ me weight equals the buoyant forceB. Tjn> Tout; the buoyantforce exceedsthe weightC. Tjn < Tout; the weight exceeds the buoyant forceD. Tjn > Tout; the density of the air inside the balloon

exceeds the density of the air outside the balloon.

11. Which of the following is NOT responsible for lateralmotion in a hot-air balloon?

A. A horizontal wind.

B. The exhaust chute of the burner oriented in a lateral

position.C. Differing local air pressures.D. The exhaust chute of the burner oriented in a

vertical position.



One method for gauging blood pressure in a vein orartery is to use a manometer. Its U-shaped tube is alwaysopen toatmospheric pressure on one end and connected to asmall glass tube, the cannula, on the other. When thiscannula is inserted directly into the side of a vein, the localblood pressure competes with the atmospheric pressure topush a liquid solution, located in the tube, one direction orthe other. The solution level is located at the dashed line,when the cannula is open to atmospheric pressure. When inuse, the solution level's height H relates to the blood pressureaccording to Equation 1:

'blood —'am + PsolutionS"

Equation 1

Figure1 showsa standard cannula.

Cannula

Figure 1

Saline, used in the cannula, is chosen for two reasons: itis non-toxic and its density is small enough to measure thepressures typically found in a vein. Average gauge pressuresare about 103 Pa in a vein and 104 Pa in an artery. Forarteries, mercury (Hg) is often used in the tube along with athin buffer zone of saline between the artery and the mercury.The density of saline is 1.02 g/mL and Hg is 13.6 g/mL.

Pressure readings aid in diagnosing arteriosclerosis, acondition where plaque builds up on an artery's inner walland restricts blood flow. Such changes in the flow speedcorrespond to changes in the local blood pressure. (Unlessotherwise stated, treat blood as an ideal fluid.)

12. When the cannula is inserted into a healthy artery,which of the following saline levels could be seen in themanometer? (the arrow indicates the fluid level)

I. II. III.

A. I and II only.

B. II and III only.

C. Ill only.

D. I, II and III only.

.1

£=L>


13. Which of the following would result in a doubling ofthe height H?

A. Halving only the atmospheric pressure.

B. Doubling only the solution density.C. Doubling only the vein's blood pressure.

D. Increasingthe vein's blood pressureby psolutiongh-

14. The average volume blood flow rate through thecapillaries is:

A. larger than that in both the arteriesand veins.B. smaller than that in both the arteries and veins.

C. equal to that in both the arteries and veins.D. smaller than that in the arteries, but larger than that

in the veins.

15. Why is a buffer region of saline used for arterialmeasurements?

A. Blood and saline are of similar density, thusallowing an accurate pressure calculation.

B. To prevent blood from mixing with mercury,because the mixing blood will give erroneousmeasurements of H.

C. To prevent blood from mixing with mercury,because of mercury's toxicity.

D. To reduce the overall density of the manometerfluid, for increased precision in measuring H.

16. Compared to a healthy neighboring section of artery, anarrowed section (as a result of arteriosclerosis) yields:

A. an increased height for the manometer reading anda larger blood flow velocity through theconstriction.

B. a decreased height for the manometer reading and alarger blood flow velocity through the constriction.

C. an increased height for the manometer reading anda reduced blood flow velocity through theconstriction.

D. a decreased height for the manometer reading and areduced blood flow velocity through theconstriction.

17. A relatively high systolic blood pressure can causearteries to expand. This expansion results from therelatively large:

A. bulk modulus of the blood.

B. Young's modulus of the blood.

C. bulk modulus of the arteries.

D. Young's modulus of the arteries.



18.

19.

When a force F is applied to an copper wire of length Land cross-sectional area A, it stretches by AL. If thesame force is applied to an copper wire of length L andcross-sectional area 2A, it stretches by:

A. AL4

B. 4L2

C. 2AL

D. 4AL

In advanced arteriosclerosis, a vascular flutter can occurwhere blood flow periodically stops and starts againwithin an afflicted artery. This flutter is caused:

A. by plaque buildup completely clogging the artery.

B. by a variation in the local pressure, beingoccasionally less than or greater than the pressurein the surrounding tissue.

C. by an increase in the blood pressure arising from adecrease in the blood volume flow rate.

D. by a decrease in the blood pressure arising from ashrinking of the artery's diameter.

20. For the following open-ended U-tube, what is true ofsolution Z?

Solution Z

A. Solution Z is more dense than

relative to water is pSolution Z =

B. Solution Z is less dense than


C. Solution Z is less dense than


D. Solution Z is less dense than


HoO

water; its density

water; its density

(b)PH2°water; its density

water; its density


21. In which of the following columns will the water levelbe the lowest when the valve is open and water isflowing out from point A?

A.

B.

Valve

Column I

Column II

C. Column III

D. Column IV

I II III IV



If we treat blood as an ideal fluid, then blood flowthrough our arteries and veins is governed by two equations.The first is the continuity equation which is:

Equation 1 Av = k

where A is the cross-sectional area, v is the velocity of thefluid, and k is a constant.

The second equation is Bernoulli's equation, whichstates:

Equation 2 p + L pv2 + pgy = k

where P is the pressure, p is the fluid density, v is thevelocity of the fluid, y is the height above some referencelevel, and k is a constant

Figure 1 shows a tube designed for the study of fluiddynamics. An ideal fluid is forced through the tube from theleft to the right, by way of an applied pressure. This tube canbe treated as analogous to an artery in your chest.

Region 1.A

Region 2 Region 3

Figure 1. Tube for Studying Flow Dynamics

Fluid flows from left to right in Figure 1. Region 1 haspressure Pj, velocity vj, height y\, and area Ai; and

similarly for Region 2, P2, V2, y2, and A2; and for Region 3,P3, V3, y3, and A3. The relative areas and heights are:

Ai=A3 = 2A2 and y{ = y2 _ ya

22. How do the pressure and velocity of the fluid leavingRegion 1 compare to the pressure and velocity of thefluid entering Region 2?

A. Velocity decreases by a factor of 2; pressuredoubles.

B. Velocity doubles; pressure decreases by a factor of2.

C. Velocity doubles; pressure decreases by an amountthat cannot be exactly determined.

D. Velocity increases by a factor of V2; pressuredecreases by an amount that varies with Ay.


23. In which region is the fluid velocity the GREATEST?

A. Region 1

B. Region 2

C. Region 3

D. The fluid velocity is the same is all three regions.

24. How can the pressure exerted by the fluid BEST bedescribed, as it passes from Region 1 through the tubeand ultimately through Region 3?

A. From Region 1 to 2, pressure increases; fromRegion 2 to 3, the pressure change is uncertain. P3

>Pl.

B. From Region 1 to 2, pressure decreases; fromRegion 2 to 3, pressure increases. P3 > Pi-

C. From Region 1 to 2, pressure increases; fromRegion 2 to 3, the pressure change is uncertain. Pi

>P3.

D. From Region 1 to 2, pressure decreases; fromRegion 2 to 3, pressure increases. Pi > P3.

25. To increase the pressure in Region 2, what can bedone?

A. Increase A2

B. Decrease A2

C. Increase y2

D. Decrease yi

26. If the fluid behaves in an ideal fashion, in which regionis the pressure GREATEST?

A. Region 1

B. Region 2

C. Region 3

D. It cannot be determined without knowing the fluidvelocity.

27. If the cross-sectional area in Region 2 is reduced, then:

A. P2 increases.

B. P3 increases.

C. V2 increases.

D. V3 increases.



One thing that staff members at The Berkeley Reviewhave come to greatly appreciate is indoor plumbing. Theporcelain wunderchair operates based on the fundamentalprinciples of fluid dynamics. The toilet is a bowl to which adrainage tube, known as an Erkel tube, is connected. Waterstored in an elevated tank can flow into the bowl when the

flap is opened. The flowing water serves to both increase thehydrostatic pressure in the bowl, and reduce the air pressurein the Erkel tube via an aspirator tube, causing the Erkel tubeto act as a siphon tube. Figure 1 shows the basic schematicof a toilet.

Aspiral oiTube

Erkel tube

A siphon functions by having the two ends of a tubeat different local pressures (typically because they are atdifferent heights.) This is done so that fluid may flow fromhigher pressure to lower pressure. The flow is driven by apressure difference between the two ends of the tube.

28. What is true of a functioning siphon during operation?

I. The intake end of the siphon tube must be at alower height than the output end of the siphon tube.

II. Water flows because hydrostatic pressure is greaterat the output end of the tube than the input end.

III. The fluid in the tube must completely fill the tube,and flow must be initiated by a pressure difference.

A. II only

B. Both I and III

C. Both II and III

D. III, and III are all true

29. A toilet will function best with what type of fluid?

A. A fluid with high viscosity and low volatility.

B. A fluid with low viscosity and low volatility.C. A fluid with high viscosity and high volatility.

D. A fluid with low viscosity and high volatility.


30. For the toilet to function, all of the following must betrue EXCEPT:

A. the tank must be at a greater height than the bowl.B. the Erkel tube must have an air space.C. the water in the bowl mustbe at the same height as

the water in the Erkel tube.

D. the aspirator tube must be filled with water.

31. What is the role of the aspirator tube in the toilet?

A. To reduce the air pressure in the Erkel tube.B. To increase the air pressure in the Erkel tube.

C. To reduce the air pressure in the bowl.D. To increase the air pressure in the bowl.

32. How can the pressure at the base of the water tank bedetermined?

A. Pbase = Patm.

B. Pbase = Patm. + Pairgntank

C. Pbase = Patm. + PwaterSntank

D. Pbase = Pwatergntank

33. The energy responsible for causing water to flow outfrom the bowl of the toilet and into the Erkel tube is:

A. the potential energy of the water in the Erkel tube.

B. the potential energy of the water in the tank.

C. the mechanical energy of the pump in the Erkeltube.

D. the mechanical energy of the pump in the tank.


Questions 34 through 37 areNOT basedon a descriptivepassage.

34. Which of the following graphs BEST representsIvolume stressl versus Ivolume strainl for an isotropicsolid within its elastic limit?

A. B.

IVolume Strainl IVolume Strainl

IVolume Strainl IVolume Strainl

35. Hydrogen is no longer used in lighter than flight,because:

A. the amount of hydrogen required to lift a balloon isprohibitive.

B. hydrogen is less dense than air or helium.C. hydrogen is more dense than air or helium.D. hydrogen is highly reactive with oxygen gas.

36. What is the apparent weight of an object in water if ithas a density of 10.0 g/mL and a weight of 60 N in air?

A. 6N

B. 30 N

C. 54 N

D. 600N


37. When a ship goes from the ocean into a lake, it will:

A. float higher in the water and require a smallerbuoyant force to stay afloat

B. float lower in the water and require a largerbuoyant force to stay afloat

C. float higher in the water and require the samebuoyant force to stay afloat

D. float lower in the water and require the samebuoyant force to stay afloat.

GO ON TO THENEXT PAGE


When a viscous fluid flows steadily through a pipe ofuniform cross-section, the volume flow rate (Q) depends onthe pressure difference between the ends of the pipe (AP), theviscosity of the fluid (ti), the radius of the pipe (r), and thelength of the pipe (L). They are related by Poiseuille'sequation:

Q =APjtr4

8r)L

Raising r to the fourth power may seem unnerving atfirst glance, but the radius impacts both the cross-sectionalarea of the pipe as well as the average speed of the particlesthat make up the fluid.

Two separate experiments are carried out to study bloodflow. In both experiments, the viscosity of blood is taken to

be2x 10"3 kg/ms.

Experiment A

Blood flows through pipes that are all 1.0 m long but havevarying radii. AP is held constant as Q is being measured.The results of four trials are summarized in the followingtable:

Trial Radius (mm) AP(N/m2) Q (m3/s)

1 10.0 10 1.96 xlO"5

2 20.0 10 3.14X10"4

3 30.0 10 1.60 xlO'3

4 40.0 10 5.03 x lO'3

Experiment B

Blood flows through pipes of constant radius 10 mm, but thelengths of the pipes vary. Q is held constant as AP is beingmeasured. The results of three trials are summarized in the

following table:

Trial Length (m) Q (m3/s) AP(N/m2)

1 1.0 1.0 xlO'3 510

2 2.0 1.0 xlO"3 1020

3 3.0 1.0 xlO"3 1530

38. If an additional trial of Experiment B were conductedwith a pipe of length 2.5 m, the expected pressuredifferential would be:

A. 765 N/m2

B. 1275 N/m2

C. 1375 N/m2

D. 1785 N/m2


39. If Q is held constant and r is decreased to half of itsoriginal value, AP must increase by a factor of:

A. 2

B. 4

C. 16

D. 32

40. If an additional trial were conducted in Experiment Awith a radius of 50 mm, the expected flow rate wouldbe:

A. 1.22 x 10"2 m3/s

B. 7.93 x 10"3 m3/s

C. 2.50 x 10"3 m3/s

D. 9.16xlO"4 m3/s

41. If water (r\ = 1 x 10"3 ms) and blood were to flowthrough identical pipes with the same flow rate, whichwould require the LARGER pressure difference?

A. Blood, because it has a lower viscosity.

B. Water, because it has a lower viscosity.

C. Blood, because it has a higher viscosity.

D. Water, because it has a higher viscosity.

42. Which of the following graphs BEST represents howthe flow rate of a fluid varies as the length of the pipe ischanged, if r|, AP, and r remain constant?

A. . B.

a

22

o

E

C.

a

o

E

Length (L)

Length (L)

Length (L)

Length (L)



The buoyancy of an object is found by comparing theobject's weight to the weightof the fluid medium it displaces(defined as the buoyant force acting on the object). Thebuoyantforce is determined using Equation 1.

Equation 1 Fbuoyant = Pmedium"Vdisp!aced'g

To determine the buoyant force exerted by various fluidmediums, a group of physics students used a pan balance.On the left side, a 50.0-g mass was dangled and on the rightside a 10.0-g pan was equipped with counterweights tobalance the left side. The weight on the left side wasbalanced in such a way that it was flush with the top the fluidin a graduated cylinder. The fluid in the cylinder was variedand the mass of the counterweights and pan necessary forbalance was recorded. Figure 1 shows the apparatus.

Figure 1 Pan Balance for Measuring Buoyant Force

One-by-one, the students measured the counterweightneeded to balance the hanging mass in four different liquids.Table 1 summarizes the results for each trial.

Table 1 Counterweight required to balance 50.0-g mass ineach liquid and the density of each liquid

Fluid Medium Counterweight Fluid Density

Air 50.0 g 1.29 g/L

Water 33.1 g 1.00 g/mL

Oil 34.8 g 0.90 g/mL

Diethyl ether 37.1 g 0.78 g/mL

Carbon Tetrachloride 24.7 g 1.49 g/mL

43. As the density of the liquid used in the experiment isincreased, what is observed?

I. More mass is needed to balance out the system.

II. The object experiences a greater buoyant force.

III. The object's apparent weight is reduced.

A. II onlyB. Ill onlyC. I and II onlyD. II and III only


44. What counterweight is needed to balance the 50.0-ghanging mass in a liquid with a density of 1.20 g/mL?

A. 22.5 g

B. 25.2 g

C. 28.9 g

D. 29.6 g

45. What is the approximate density of the hanging mass?

A. 1.25 g/mL

B. 2.00 g/mL

C. 3.00 g/mL

D. 5.00 g/mL

46. What would be observed for the experiment in anenvironment with reduced gravity?

A. More mass in the pan would be required toestablish equilibrium.

B. Less mass in the pan would be required to establishequilibrium.

C. The exact same mass in the pan would be requiredto establish equilibrium.

D. The amount of mass needed in the pan would beinversely proportional to the apparent weight of thehanging mass.

47.

48.

Which graph relates the apparent weight of an object tothe density of the medium in which it is submerged?

A. B.

C _J3

Ic

1a,a.

Density (g/mL)

Density (g/mL)

Density (g/mL)

*>• -

Density (g/mL)

A cup of water with a floating ice cube is transportedfrom the earth to the moon. On the moon, the cube will:

A. float higher, because gravity is reduced.

B. float lower, because gravity is reduced.

C. remain the same, unaffected by changes in gravity.D. float higher, because gravity is increased.



49.

50.

51.

If one makes an analogy between fluid flow andelectricity, pressure difference AP corresponds to:

A. current

B. voltage.

C. resistance.

D. capacitance.

Releasing adrenaline dilates blood vessels, making iteasier for blood to flow. By about what percentagewould the diameter of a blood vessel need to increase to

raise blood flow by 20%, at constant blood pressure?

A. 0%

B. 4.7%

C. 10%

D. 20%

An object weighing 144 N is suspended by a stringwhile completely immersed in water. The object has adensity four times the density of water. What is itsapparent weight?

A. 108 N

B. 72 N

C. 36 N

D. 576N

52. A floating object is 90% submerged in water and 60%submerged in CCI4. What is the specific gravity ofCCI4and the specific gravity of the object?

A. Object = 0.60; CCI4 = 0.67

B. Object = 0.90; CCI4 = 0.67

C. Object = 0.90; CCI4 = 1.50

D. Object = 0.90; CCI4 = 3.00


1. D 2. D 3. C 4. D 5. C 6. B

7. A 8. B 9. A 10. B 11. D 12. C

13. D 14. C 15. C 16. B 17. A 18. B

19. B 20. D 21. D 22. C 23. B 24. D

25. A 26. A 27. C 28. C 29. B 30. D

31. A 32. C 33. B 34. D 35. D 36. C

37. D 38. B 39. C 40. A 41. C 42. A

43. D 44. D 45. C 46. C 47. B 48. C

49. B 50. B 51. A 52. C

YOU ARE DONE.

Answers to 52-Question Fluids and Solids Exam

Passage I (Questions 1 - 5) Stresses and Strains

1. Choice D is the best answer. How much pressure P must be applied to an isotropic solid, of bulk modulus B, to compress itto half of its original volume, assuming that the solid remains within its elastic limit at all times? If we compress it to half of

_ V _ Vits original volume, then Vf = —. This means that the change in volume AV is equal to AV = Vf - V; = —. If we plug this2 2

result into the formula for the bulk modulus, we get:

P PB = -

Rearrange terms to solve for AV:

= - — — = 2P, which reduces to: P = &.AV/V - (V/2)/V 2

In order to squash the solid to half of its original volume, we must apply a pressure equal to half of its bulk modulus. Noticethat all three elastic moduli must be expressed in units of force/area. The best answer is D.

Choice D is the best answer. A 1-m3 block of aluminum sank to the bottom of a lake of depth 750 m. Finding out howmuch the block shrank as a result of sinking requires determining the change in its volume AV. To find the change involume, we need to know the applied pressure. Below the surface of a liquid at depth h, the pressure of any object increasesdue to the weight of the fluid column above it. So the pressure applied to the block is P = (pwater)Sn- Applying the bulkmodulus formula:

B = -AV;

/V

- __PShAV

V

^6 m3<AV =- (PlSV = - (1000)(10)(750)(lm') = _7.5 x 10" mf) , AV =_, x 1Q-4m3 . . aQ001 m3B 7.5 x 1010 7.5 x 1010

The minus sign means that the change in volume is a decrease. This question is testing whether we can work with factors of10 quickly. The best answer is choice D.

Choice C is the best answer. Suppose Y, B, and S are the Young's, bulk, and shear moduli of aluminum. Which statement istrue, ifa =J- for aluminum? Using the relations between Y, B, and Sgiven in the passage:

S =2(1+a)


B =Y

3(1 -2a)

3/ 13

With a little manipulation, B = Y, eliminating answer choices A and B. S and Y are related as follows:

Y Y

Y

3 11 - —

Y

si= Y

= t^V = - Y, which tells us that S<Yi + - 2i

4. Choice D is the best answer. How much work is done by a force F that stretches a wire of length L by AL, if the wire has across-sectional areaof A, and Young's modulus Y?Since the answer choices look pretty nasty, weshould use our knowledgeof units for this question (joules) to eliminate incorrect answer choices.

The units ofYare N/m2, the units ofLand AL are m, and the units ofAare m2. Examining the units ofeach choice leads tothe best answer. The answer in choice A is (Y)(AL)/L, with units of N»m/m2»m = N/m2, which are the wrong units. Theanswer in choice B is YAL with units of (N/m2)»m = N/m, which are the wrong units. The answer in choice C isL(AL)2/(Y)(A) with units of (m«m2)/(N/m2)(m2) = m3/N, which are the wrong units. The answer in choice D is(Y)(A)(AL)2/2L, with units of(N/m2)(m4/m) = N-m =joules, which are the correct units.

You could solve this question using the relationship of W = Fyd, where W is work. The force can be found by manipulating

Young's modulus, Y = //\AL' t0 §'ve ^ = ']_• The distance that the center of mass travels during the stretch is

actually AL/2, not AL. This means that the work done is (YAAL/l)(AL/2) =YAAL /2l- The best answer is choice D.

<s>Copyright © by The Berkeley Review 104 REVIEW EXAM EXPLANATIONS

5. Choice C is the best answer. Suppose an analogy were made between Hooke's law, F = kAx, and Young's modulus,(Stress) = (Y) (Strain). Which quantity in the Young's modulus equation corresponds to the spring constant k in Hooke's law?If we can rewrite the Young's modulus equation so that it looks like F = (constant)AL, then whatever constant we getcorresponds to k.

Y = stress _ F/A => JF _ y A-L. =>strain AL/L A L

The quantity Y-A- corresponds to the spring constant k. The best answer is choice C.

8.

9.

10.

11.

•W AL

Passage II (Questions 6 - 11) Balloons and Dirigibles

Choice B is the best answer. The buoyant force is equal to the weightof the air that is displaced. It has nothing to do withthe weight (load), so thecalculation should notinclude 2000. This eliminates choices C and D. Theairdensity is solved foras follows:

B = Pairvair*g •• Pair =B _

V*ir'S (5000m3)(9.8k8/s2)= 4.9 x 10

•3 kg/m-

Note that the volume of the air displaced is equal to the volume of the balloon. The numerator should only contain thebuoyant force, so choice A is eliminated and choice B is the best answer. If they ask for an exact value, it is sometimeseasier touse the order of magnitude rather than solve foranexact number. The bestansweris choice B.

Choice A is the best answer. As the balloon rises, the external air pressure decreases. This is due to less air being present,therefore choice A is the best answer. Because the external pressure decreases, the balloon will expand, so its surface areaincreases. This eliminates choice D. The mass of the hot-air balloon will not change or shift relative to the balloon,therefore choices B and C are eliminated. The best answer is choice A.

Choice B is the best answer. The dirigible floats because helium is less dense than air. The hot-air balloon floats becausehot air is less dense than cooler surrounding air. The difference in density between helium and air is far greater than thedifference in density between hot air and cold air. This is to say that atthe same temperature, air is denser than helium. Thismeans that to achieve neutral buoyancy, a greater volume of hot air is needed than the volume of helium for the same load.This makes choice B the best answer. Choice A is eliminated, because less mass (ballast) should be used with the hot-airballoon given that it is less buoyant than a helium balloon of equal volume. Choice C is eliminated, because the hot-airballoon should have a greater volume than the helium balloon. Choice D is eliminated, because although equal moles ofgases occupy the same volume, they do not have the same masses, so the weights are different. This results in differentbuoyancy. The best answer is choice B.

Choice A is thebestanswer. Momentum isconserved in the absence ofa net external force. The balloon-package system isinitially moving vertically upward at constant speed, so this system is experiencing no net external force. Under theseconditions, the force of gravity is equal to the buoyant force, neglecting any resistance. Throwing a package out of theballoon will slightly decrease the gravitational force on the balloon, because the balloon now has less mass. Itwill thereforeaccelerate upward, increasing its speed and thus increasing its momentum too. The solution is also found by noting that thesystem is initially in equilibrium. When the package is released ithas momentum downward, so the balloon must have anincrease in momentum upward to balance out the package's downward momentum. The best answer ischoice A.

Choice B isthebest answer. For an object to float in a fluid medium, the density of the object must be less than the densityof the surrounding fluid medium. By heating the air inside the balloon, we decrease its density. The result is that the densityof the air inside the balloon is less than the density of the air outside the balloon. This means that the temperature inside theballoon must be greater than the temperature outside (which the name hot-air balloon implies). This eliminates choices Aand C. The densities are not equal, so choice Dis also eliminated. Lift results, because the buoyant force is greater than theweight of the system.The best answer is choice B.

Choice Dis correct. The passage references that lateral wind (crosswind) and thrust (due to the correct orientation of theburner chute) will cause lateral motion. This eliminates choices Aand B. A difference in local air pressures results in airflow (wind), because fluids flow from regions ofhigher pressure to regions of lower pressure. This can cause all sorts ofmotion, including lateral motion ifthe wind is a crosswind. The hot-air balloon will flow with the air currents. The hot-airballoon will not move laterally if the burner chute is positioned in a vertical manner, because the thrust force is vertical andthere is no lateral force. The best answer is choice D.


Passage HI (Questions 12 -17) Cannula

12. Choice C is the best answer. Since the average gauge pressure in the artery is greater than zero, the average pressure in theartery is larger than the atmospheric pressure. Consequently, the blood pressure should push the cannula fluid out (as inFigure III). You can think of what happens when you get cut. Blood squirts, oozes, or leaks out (you choose the more vividword) from you, so it must be at a higher pressure than the atmosphere. Choice C is best, because the height differenceresults from the pressure in the cannula being greater than the atmospheric pressure. In Figure I, the blood pressure is lessthan atmospheric pressure and in Figure II, the blood pressure is equal to atmospheric pressure. Neither of these are validscenarios, so choices A, B, and D are all invalid. The best answer is choice C.

13. Choice D is the best answer. Equation 1, given in the passage, has three terms in it: Pbbod, Patm, PsoiutiongH- Doubling Hdoubles the last term in the equation. One way to double H is to double the other two terms. Since the first three choicesrefer to doubling or halving single variables, but no choice refers to doubling both terms, vve can eliminate choices A, B, andC. If the blood pressure is increasedsuch that the difference between the blood pressureand the atmospheric pressure (Pbiood- Palm) doubles, then the height, H, woulddouble. That in essence is what choice D is saying. The best answer is choice D.

14.

15.

16.

17.

Choice C is the best answer. Because no blood is lost as it goes through the arteries, through the capillaries, and through theveins, the blood mass flow rate must be the same through each region. Since blood is a fairly incompressible fluid and thearteries and veins undergo minimal expansion, the average blood volume flow rate must also be the same through eachregion. The best answer is choice C.

Choice C is the best answer. As stated in the passage, the saline is non-toxic. Nothing is stated in the passage aboutmercury being toxic, but it can be inferred from the fact you need a non-toxic buffer between blood and mercury. Since Hgis toxic, the saline buffer keeps the Hg out of the patient's blood stream. Choice A starts off with a true statement that salineand blood have relatively similar densities, but that is not necessary to measure blood pressure. Choice A is eliminated.Choice Dis also tempting, because using a less dense solution would result in a bigger H. This would be easier to preciselymeasure, but choice C is the better answer, because if precision of that sort was necessary, then mercury would notbe used atall. The best answer is choice C.

Choice B is the best answer. Using the continuity equation, Aiv7 =A2V2, a decreased flow area in the artery leads to anincreased flow velocity ofthe blood. This rules out choices Cand D. Using Bernoulli's equation, an increased flow velocityleads to adecreased local pressure against the walls in the narrowed region. Using Equation 1from the passage, adecreasedblood pressure leads to a decreased height for the manometer reading. The best answer is choice B.

Choice A is the best answer. If the artery expands, either the blood is relatively incompressible or the artery is relativelypliable. Since blood is a fluid, it has no Young's modulus by definition (Young's modulus refers to a solid). This rules outchoice B. Choices C and D would correspond to stiffer arteries, which would be less likely to stretch, so they are botheliminated. If the bulk modulus ofblood is large, then it is relatively incompressible, meaning that the high systolic pressurecauses thearteries to expand rather than the blood to compress. The best answer is choice A.

Questions 18 - 21 Not Based on a Descriptive Passage18. Choice B is the best answer. When a force F is applied to a copper wire of length L and cross-sectional area A, itstretches

by AL. Ifthe same force is applied to a copper wire oflength Land cross-sectional area 2A, it stretches by how much? First,let's use a little common sense to eliminate incorrect answer choices. The second copper wire is thicker than the originalcopper wire. This tells us that for the same force, the second wire should notstretch as much as thefirst wire. Soouranswerhas to be less than AL. This eliminates choices Cand D. The best way to choose between the remaining choices Aand Bis toset up a ratio, using the Young's modulus equation:

FLAL

new

AL =

YAnew

FL

YA

ALnew

= AL (-AJ = AL(_A_) = AL\Ancw) \2Al 2

The best answer is choice B. Try to answer the same question, this time assuming that the diameter of the second wire istwice that of the first wire. That is a common twist on this kind of problem. The best answeris choice B.


19. Choice B is the best answer. The phrase "periodically starts and stops", within the question, suggests that the cause of theflutter must be a periodic or varying phenomenon. Only choice B refers to a scenario where fluctuation occurs. Plaque buildup would permanently clog it up, so it would not cause a fluctuation. This eliminates choice A. Choices C and D arc truestatements that are irrelevant to the question. Neither account for a stopping and starting phenomenon. Choice B describes ascenario where the pressure fluctuates between greater than and then less than, which would correlate to the blood flowingand then not flowing. Choice B is the best answer. The best answer is choice B.

20. Choice D is the best answer. Because Solution Z is floating on water, it must be less dense than water. This eliminateschoice A. According to Archimedes Principle, the mass of water displaced is equal to the mass of the Solution Z. The

volume of water displaced is 7tr2b while the volume of Solution Z is 7rr2(a + b). This means that a column of water withheight b has the same mass as a column of Solution Z with height a + b. The value of a is less than b (as the picture shows),so the density of Solution Z is slightly less than the density of water. We know from the picture that the density of SolutionZ is more than half of the density of water, because b is greater than a. This means that the density of Solution Z is found bymultiplying the density of water by a number greater than 0.50, but less than 1.00. This is true only in choice D. Choices Band C are the result of multiplying the density of water by a number less than 0.50. The mathematics of equating thepressures is shown below:

Psolution Zghsolution z = PH20onH20PsolutionZg(a + b) = pH2Ogb

Psolution Z(a+ b) = PH20bb

Psolution Z = PH20(a + b)


21. Choice D is the best answer. This question combines thecontinuity equation with Bernoulli's principle. First off, when thevalve is opened, water will flow from the reservoir, through the open valve, and then through the outflow pipe and outthrough point A. As the water flows through the side pipe, some water will climb the various columns, according to thepressure it exerts against the walls at the point where it passes the column. The greater the pressure exerted by the movingwater against the walls, the higher that pressure will push water up thecolumn.

Because the outflow pipe is narrowing from the point where it intersects Column I to the point where it intersects ColumnIV, the average speed of the water is increasing as it flows to the right. According to Bernoulli's principle, as the flow speedof a fluid increases, the pressure it exerts against the walls decreases. This means that the moving water will exert the highestpressure against the walls as it passes the base of Column I and the least pressure against the walls as it passes the base ofColumn IV. A lower pressure manifests itself as a smaller height for the water in Column IV. Even if you were unsure of thephysics involved with this question, hopefully you saw it as an extreme question and you eliminated choices B and C,columns in the middle. The best answer is choice D.

Passage IV (Questions 22 - 27) Blood Flow Analog

22. Choice C is the best answer. Because the area is being cut in half as the fluid flows from Region 1 into Region 2, thevelocity of the fluid must be increasing, ultimately by a factor of 2 (the relative sizes of the two regions). This eliminateschoices A and D. When the velocity increases, the 1/2 pv2 term in Equation 2 must increase: so, for the overall equation tostay constant, the pressure against the walls of the pipe must be reduced. The exact amount by which the pressure isdecreased is uncertain, because the values of P and vj are not given. The best answer is choice C.

23. Choice B is the best answer. Using the continuity equation, we find that:

Aivi = A2V2 = A3V3

Because A\ = 2A2 (as given in the passage), V2 is double v\. Because A1 = A3 (as given in the passage), v\ must be equalto V3. This means that V2 is the greatest. The best answer is choice B.

24. Choice D is the best answer. As the fluid flows from Region 1 to Region 2, the cross-sectional area decreases, so thevelocity must increase, according to the continuity equation (Equation 1). Because the V2 pv2 term in Bernoulli's equation(Equation 2) increases while the pgy term is constant, the pressure must decrease. This eliminates choices A and C. ChoicesB and D differ in their comparison only of the pressures in Regions 1and 3. Because their cross-sectional areas are thesame,both regions have the same flow velocity. Because Region 3 is higher than Region 1, Region 3 must have a lower pressurethan Region 1 if Bernoulli's equation is to remain balanced. The best answer is choiceD.


25. Choice A is the best answer. Using the continuity equation, if A2 increases, then V2 must decrease. If V2 decreases while y2remains unchanged, Bernoulli's equation states that the pressure must increase. Increasing y2 (the height of Region 2) willreduce the pressure in that region, so choice C should be eliminated. Changing yi should have no direct effect on thepressure in Region 2. The best answer is choice A.

26. Choice A is the best answer. We can see from Equation 2 that the two factors affecting the pressure that vary between theregions are the height (y) and the velocity (v). The following relationshipholds true:

Pj +-p(v1)2 +pgy1 =p2 + -p(v2)2 +pgy2 =p3+^p(v3)2 +pgy3Z. £t £t

The continuity equation relates the cross-sectional area of the pipe to the average particle speed, where Av = k as long as thevolume flow rate remains constant. Because A1 = 2A2 = A3, V2 = 2v\ - 2V3. This means that V2 is the greatest velocity. As

given in the passage, y3 > yi = y2- In order for the equality to hold true, P\ must be greater than both P2 and P3. Thegreatest pressure is found in Region 1, where the area is greatest and the tube is lowest. The best answer is choice A.

27. Choice C is the best answer. According to the continuity equation: A2V2 = k. Therefore, A2jV2i = A2f\'2f- If A2 decreases,V2 must increase to satisfy the continuity equation. This makes choice C the best answer. According to Bernoulli's equation,

Pagainst walls + '/£pv2 + pgh = k, P2 should decrease if V2 increases. This rules out choice A. Velocity and pressure inRegions 3 will have no bearing on the pressure or velocity in Region 2. This makes choices B and D invalid. The bestanswer is choice C.

Passage V (Questions 28 - 33) Toilets and Siphons

28. Choice C is the best answer. A siphon works because the two ends of the siphon tube are at different heights, and thusdifferent potential energies. In order to have water flow from the intake end to the output end, there must be a net pressuredifference, where the gauge pressure is greater at theoutput end. This occurs when the outputend is at a lower height, thus itis at a point of greaterhydrostatic pressure. This makes Statement I invalid. This same reasoning makes Statement II valid.In order to get siphoning started, there must be an initial difference in pressures that is great enough to get the water to flow.The tube must also be full, otherwise the air pocket cannot be displaced, and the fluid in the tube cannot flow. This makesStatement III valid, and thus makes choice C the best answer. The best answer is choice C.

29. Choice B is the best answer. Fora toilet to function, the liquid (fluid) must beable to flow. If the viscosity is high, then theability to flow is reduced. The greater the viscosity, the greater the pressure difference that is required to get the fluid toflow. Ideally, the fluid should have low viscosity. This eliminates choices Aand C. Low volatility is ideal sothat the liquidin the tank and bowl does not evaporate away. If the liquid evaporates away from the bowl, then the fluid height in the Erkeltube will drop, and the siphon cannot be established. Ideally, the volatility of the fluid should be low. The best answer ischoice B.

30. Choice D is the best answer. Because water must flow from the tank into the bowl, the height of the water tank must begreater than the height of the bowl. Choice A is a valid requirement. The Erkel tube must have anairspace at the top of thetube so that air pressure can be reduced, and thus a pressure difference can be established between the two ends of the Erkeltube. Choice B is a valid requirement. A fundamental idea in physics is that a liquid will seek the lowest point. Becausewater in the bowl is free to fill the base of the Erkel tube, the height of water in the bowl at rest is equal to the height of thewater in the Erkel tube at rest. Choice C is a valid requirement. In order for the aspirator to reduce the air pressure in theErkel tube, it cannot haveany waterpresent. As water flows from the tank into the bowl, air pressure in the tank is reduced,so air pressure in the aspirator tube is also reduced. This generates the reduced pressure that helps water (and various solutesand possible insoluble solids-nice way to put it) to flow through the Erkel tube (siphon). Choice D is NOT a validrequirement,and picking it might just make you score like the mightyToledo Mudhens. The best answer is choice D.

31. Choice A is the best answer. The role of aspiration in general is to take advantage of fluid flow to reduce the pressure in acolumn. As a fluid flows across the top of an open-end column, the pressure inside of the column will be reduced. In thecase of a toilet, if the pressure in the aspirator tube is reduced, then it will lower the gas pressure of the air behind the Erkeltube. This makes choice A the best answer. Choices C and D can be eliminated, because the change in air pressure in thebowl is negligible. It can also be thought of in terms of PV = nRT. As water leaves the tank, the volume of the air spaceabove thewater increases, resulting in a drop in pressure. The aspirator tube is open to the air space, so it experiences a dropin pressure as well. The air space in the Erkel tube has a drop in pressure while the bowl has a gain in pressure due to theadditional water. This causes the pressure difference between the two ends of the Erkel tube. The best answer is choice A.

Copyright© byThe Berkeley Review® 108 REVIEW EXAM EXPLANATIONS

32. Choice C is the best answer. At the base of the water tank, there is a column of water and the atmospheric air above. Thismeans the pressure at the base of the water tank is due to the pressure from the column of water (pgh) and the atmosphericpressure. The best answer is choice C.

33. Choice B is the best answer. Choices C and D can be eliminated immediately, because there is no pump associated with thetoilet shown in Figure 1. This means that there is no mechanical energy associated with a toilet. Because the water tank it atthe highest point, the water in the tank has the greatest potential energy of anything in the toilet. This potential energy isused to force the water to flow through the toilet system. The best answer is choice B.


34.

35.

36.

37.

Choice D is the best answer. We are looking for the graph that best represents Ivolume stressi versus Ivolume strainl for anisotropic solid within its elastic limit. Stress and strain are directly proportional to each other. For a solid, stress = E (strain),where E represents Young's modulus. As one value increases, so should other value. This eliminates choices A and C. Theelastic modulus that relates volume stress and volume strain is B. The general relation is Ivolume stressi = B Ivolume strainl,which confirms that they are directly proportional to one another. So the graph should be a straight line with a positive slope.The best answer is choice D.

Choice D is the best answer. Hydrogen was actually used as the gas to lift balloons well into the early twentieth century.However, hydrogen is so combustible that it is dangerous to use (as most of us have learned from the Hindenberg disaster.)The reason hydrogen was used in the first place is because at 2 grams per mole (the molecular mass of H2 gas), hydrogen ishalf as dense as helium which has an atomic mass of 4 grams per mole. This makes a hydrogen-filled balloon system lessmassive than a balloon filled with helium, and therefore easier to lift off from the ground. The best answer is choice D.

Choice C is the best answer. The object is denser than water, so it will sink when added to water. Because of the buoyantforce, its apparent weight will be less than its actual weight. As such, the apparent weight is less than60 N. This eliminateschoice D. For an object that sinks in a fluid medium, the ratio of its density to the density of the fluid medium(Pobjcct/pmedium) is equal to the ratio of the weight to the buoyant force (Fweight/Fbuoyant). In this case, the ratio of thedensities is 10 g/mL to 1 g/mL, which equals 10. This means that the buoyant force of the object is 6 N (60N/10). Giventhat the magnitude of the weight is equal to the sum of the magnitude of the apparent weight and the magnitude of thebuoyant force (Fweight = Fbuoyant + ^apparent weight), the apparent weight is the difference between the actual weight (60 N)and the buoyant force (6 N). The apparent weight is 54 N.

Pobject _ ^Wright . 10 g/mL _ 60 NPwater Fbuoyant 1 g/mL ^buoyant

Fbuoyant = 60Ji = 6N10

Fweight - Fbuoyant + Fapparent weight •'• Fapparcnt weight - Fweight" Fbuoyant -60N-6N-54N


Choice D is the best answer. The density of the ocean (saltwater) is greater than the density of a lake (fresh water). Becausethe ship goes from a denser medium to a less dense medium, it will experience an increase in the submerged volume, whichoccurs to keep the buoyant force constant. You should recall that when the shipfloats, the buoyant force equals the weight ofthe ship, and that the buoyant force is found using: Fbuoyant = Pmedium'Vdisplaced'g- Fbuoyant and g remain constant, so asPmedium decreases, Vdispiaced must increase. The ship floats lower in a lake than the ocean, so choices A and C areeliminated. Next, we must consider that the weight of the ship didn't change when the medium changed, so the buoyantforceneeded to offset the weight of the boat when it floats should also remain the same. Choice B is eliminated. The best answeris choice D.

Passage VI (Questions 38 - 42) Poiseuille's Law

38. Choice B is the best answer. To approximate how an additional trial of Experiment B with a pipe of length 2.5 m wouldcome out requires interpolation. One method to solve this problem would be to find a ratio between the new trial and oneofthe trials in the table. In this case, you can almost get theanswer directly from the table. Since L for the new trial hasa valuebetween L for Trial 2 and Trial 3, AP for the new trial should also be between AP for Trial 2 and Trial 3. This means thecorrect answer must be choice Borchoice C. For each meter ofpipe, the pressure difference increases by 510 N/m2. As wego from L= 2.0 mto L= 2.5 m, the pressure should increase by 255 N/m2. So AP for the new trial should be 1020 +255 =1275 N/m2. The best answer is choice B.

®Copyright © by The Berkeley Review 109 REVIEW EXAM EXPLANATIONS

39. Choice C is the best answer. If Q is held constant and r is decreased to half of its original value, AP must increase by whatfactor?

40.

41.

42.

Q =APjtr4

8r|L

APjr(r/2) =\ 16 / =(l\APjrr48tiL 8rjL "" U6/ 8tiL

To keep Q constant, we must increase the pressure difference by a factor of 16! The best answer is choice C.

Substitute V2 for r: Q =

Choice A is the best answer. If an additional trial were conducted in Experiment A with a pipe of radius of 50 mm, whatwould the expectedflow rate be? To do a problem like this, the best methodis to find the ratio between the value of Q in thenew trial and the value of Q in one of the trials in the table. Let's find the ratio between the new trial and Trial 1:

'new

APjt(r, r and Ql=^r1rj8r|L 8r|L

so*rnew)

(ri)4

= Qi

Qi

feewL =(L96x iQ-5) (SOjnm.)4 =(1 %xiq-5) (5/\ (r,)4 / 10 mm

Qnew «(2 x10"5)(625) =1250 x10"5 =1.25 x10"2 m3/sThis is very close to choice A. If we had justlooked at the table, Qfor r =50mm had tobe larger than anything in the table.The only choices that were larger than those in the table were choices A and B. Since Q « r4, you could have reasoned thatanswer choice B is too small. The best answer is choice A.

Choice C is the best answer. If water and blood were to flow through identical pipes with the same flow rate, which fluidwould require the larger pressure difference? The viscosity of blood is larger than the viscosity of water, so vve can eliminatechoices Aand D. We can manipulate Poiseuille's equation to find out how pressure difference depends on viscosity and flowrate:

AP oc Qr|

Since both fluids have the same flow rate, the one with the greater viscosity requires the greater pressure difference. Bloodhas greater viscosity. The best answer is choice C.

Choice A is the best answer. The graph that best represents how the flow rate ofa fluid varies as the length of the pipe ischanged (if rj, P, and r remain constant) requires manipulating Poiseuille's equation to show that flow rate is inverselyproportional to the pipe length (Q « 1/L). Choice Ccan be eliminated, because itstates that the flow rate does not depend onpipe length. Choice Dcan be eliminated, because itstates that the flow rate is directly proportional to pipe length. The graphin choiceA is the only graph that plots Q versusL in an inverse manner. The best answer is choice A.

Passage VII (Questions 43 - 48) Buoyancy Balance

43.

44.

Choice D is the best answer. As the density of the fluid medium increases, the buoyant force is increased. This makesStatement II a valid statement, and eliminates choice B. Because the magnitude of the weight is equal to the sum of themagnitude of the apparent weight and the magnitude of the buoyant force (Fweight =Fbuoyant +Fapparent weight), the apparentweight is the difference between the actual weight and the buoyant force. As such, an increase in the buoyant force results inthe apparent weight being reduced. This makes Statement III avalid statement, which eliminates choices Aand C. The onlychoice remaining ischoice D, so it must bethe best answer. This implies that Statement I is invalid, which is true, because ifthe apparent weight is reduced, then less counterweight is necessary. The best answer is choice D.

Choice Disthe best answer. This question can be solved by comparing the information in the question to the data in Table1. A 33.1 gcounterweight is necessary in a fluid with a density of 1.00 g/mL and a 24.7 g counterweight isnecessary in afluid with a density of 1.49 g/mL. With a fluid ofdensity 1.20 g/mL, the counterweight should be between 24.7 grams and33.1 grams, being a bit closer to 33.1 grams than 24.7 grams. This eliminates choice A. The average of 24.7 and 33.1 is28.9, so the counterweight should be a little heavier than 28.9 grams. This eliminates both choices B and C and leaveschoice Das the best answer. This question tested your ability to work with numbers in a table, which is something that iscommon on the MCAT. The best answer is choice D.


45.

46.

47.

48.

Choice C is the best answer. The hanging mass sinks in all of the fluid mediums, so it must be more dense than each of thefluids. The densest liquid is carbon tetrachloride, with a density of 1.49 g/mL, so the hanging mass must have a densitygreater than 1.49g/mL. This eliminates choiceA. To solve for the exact density, we can compare the weight to the buoyantforce. For an object that sinks in a fluid medium, the ratio of the density of the object to the density of the fluid medium(Pobject/pmedium) is equal to the ratio of the weight to the buoyant force (Fwejght/Fbuoyant)- Because the object is at rest, themagnitudesof the forces up must equal the magnitude of the force down. In air, a counterweight of 50.0 grams is necessary,so the mass of the hanging object is 50.0 grams. Its weight is therefore 0.05g N. The counterweight needed in carbontetrachloride is 24.7 grams, about half of its weight. Because the weight is balanced out by a buoyant force and tension inthe string holding the mass (which results from the counterweight), we can set the forces equal. This means that Fweight =Fbuoyant + Fusion- The tension is 0.0247g and the weight is 0.05g, so the buoyant force is 0.0253g. The relative densitiescan be set equal to the weight-to-buoyant force ratio, which is 0.05g/0.0253g. That ratio is about 2 : 1, so the density of theobject must be about double that of carbon tetrachloride. Carbon tetrachloride has a density of 1.49 g/mL, so the hangingmass must have a density near 3.00 g/mL.

PHanging mass _ ^weight _

PCarbon tetrachloride Fbuoyant

PHanging mass2.0 =


'weight 0.05g N _ 0.05g

Fweight - Fapparent weight 0.05g N - .0247g N 0.0253g_5_

2.5

PHanging mass

PCarbon tetrachloride 1.49 g/mL••• PHanging mass = 2x 1.49 g/mL a 3 g/mL

= 2

Choice C is the best answer. In reduced gravity, the weight of the object and the counterweights would be reduced as wouldthe buoyant force. All of the forces depend directly on g, so a reduction in gravity would impact all of the forces by an equalpercentage. This means that all of the masses could remain the same, making choice C the best answer. Choice D can beeliminated, because the apparent weight is the equal to mass of thecounterweight, so they are not inversely related. The bestanswer is choice C.

Choice B is the best answer. If an object has an apparentweight in a fluid medium, it must be more dense than the medium.In such a case, the magnitude of the buoyant force and the magnitude of the apparent weight add up to equal the magnitudeof the object's weight. This means that as the magnitude of the buoyant force increases, the magnitude of the apparentweight decreases. The buoyant force is directly proportional to the density of the fluid medium (Fbuoyant =Pmedium'vdisplaced*g)» so the apparent weight must decrease as the density of the fluid medium increases. This eliminateschoices A and D. The relationship is not inversely proportional, so the graph is not asymptotic. This eliminates choice B.The relationship is linear, because they add to a sum. As one increases, the other decreases by the same amount. The bestanswer is choice B.

Choice C is the best answer. An object floating in water, such as an ice cube, does so because its weight (mg) equals thebuoyant force (Fbuoyant = Pmedium-Vdispiaced'g)- When going to an environment with a change in the gravitational forceconstant (g) such as the moon, the weight at buoyant force are impacted equally. If the object is floating in the first place, itwill continue to float in the exact same fashion when the gravitational force changes. Choice C is the only choice that fits thisinformation. Choice D should be eliminated immediately, because gravity is reduced on the moon. Moon problems are quitecommon in physics, so it's not a bad idea to contemplate what happens to various forces and systems when the value ofgchanges. The best answer is choice C.


49. Choice B is the best answer. If one makes an analogy between fluid flow and electricity, pressure difference, AP, willcorrespond to the electrical driving force. The comparison we should make involves Ohm's law:

V=IR where: R = p ^ .'. V = (I)p -^A A

Rearranging Poiseuille's equation gives:

AP _ Q8ilL _ (Q) 8r|Ljit4 Jtr4

In the analogy with electricity, AP corresponds to V, Q corresponds to I, and 8r)L/jrr4 corresponds to p(L/A). Thiscorresponds to choice B. In a more conceptual perspective, the pressure can drop as the blood flows, just as voltage may dropas current flows. Equally, opposing pressures result in no blood flow, just as opposing voltages result in no current. The bestanswer is choice B.


50. Choice B is the best answer. The volume flow rate of an ideal fluid depends on four factors: the pipe's length, the pipe'sradius, the pressure difference between the two endsof the pipe, and the viscosity of the fluid. In this question, the length,fluid viscosity, and pressure difference remain constant, so we need only consider the change in the radius of the vessel. Therelationship isQ a r4, where Q represents the volume flow rate ofblood and r represents the radius of the blood vessel. Youmay have noticed that they mentioned diameter in thequestion to throw us off a bit, but any percent-change in the diameterhasan identical percent-change in radius. If a bloodvessel dilates, then there is a greatercross-sectional area and the volumeflow rate will increase. This eliminates choice A. Because r is raised to the fourth power, a 20% change in radius would leadto a greaterincrease in volume flow rate than 20%, so choiceD is eliminated. At this point, we are faced with some dauntingmath. Rather than taking the fourth root of 1.20 (the ratio of Qfinal to Qinitial). we are better suited to pick one of the answerchoices and raise it to the fourth power to see if we get 1.20. Let's try 10%, because it looks like the math will be easier than

4.6%. A 10% increase would results inan rf,nai to rjn}tial ratio of 1.10.1.104 isgreater than 1.20 (given that 1.102 = 1.21), sochoice C is not correct. Only choice B remains, so it must be correct The best answer is choice B.

51. Choice A is the best answer. This question is best solved by first recognizing that the object is a sinker. Because it's asinker, we can start by noting that the ratio of the densities of the object and medium is equal to the weight-to-buoyant forceratio. The question tells us that the object's density is four times the density of the surrounding water, so the weight of theobject must be four times the buoyant force. One-fourth of the weight is equal to 36 N, so we know that the object has aweight of 144 N and experiences a buoyant force of 36 N. The buoyantforce is the apparent weight loss, so to determine thebuoyant force, we need only subtract the apparent weight loss from the weight This would be 144 - 36 = 108 N, choice A.Be careful on questions like this not to accidentally choose the buoyant force as the final answer, because it's highly likelythat test writerswill leave that as a potential choice. Just to be complete, let's draw a pictureof the system.

Notes:

B = Apparent weight loss and T = apparent weight

For an object that sinks:

Relative densities = W-to-FBUOyantrau0

Pobject \yPwater "* ~" B

.\4B =WCZ> B=jW

B=jxl44N =36N

T + B = W /. T = W-B = 144N-36N = 108N


52. Choice C is the best answer. First off, we can apply aconvenient short cut for an object that floats. When an object floats,the ratio ofthe densities of the object and medium isequal to the percent of the object that issubmerged. In this case, wherethe medium is water, the ratio ofthe densities is also known as the specific gravity. Because the object is 90%-submerged inwater, it has a specific gravity of 0.90. This eliminates choice A. Because the object floats higher in CCI4 than it does inwater, we can conclude that CCI4 must be denser than water, which means the specific gravity of CCI4 must be greater than1. This eliminates choice B. To decide between the remaining two choices, we'll need to compare the percent submerged inwater to the percent submerged inCCI4 inorder to determine the density ofCCI4 relative to the density ofwater. The ratio ofthe submerged percentages is90:60 =3:2 = 1.5. This means that the specific gravity ofCCI4 is 1.5, eliminating choice Dandmaking choice C the best answer. A more detailed solution is shown below.

Because the weight of theobject is equal to itsbuoyant force inwater and its buoyant force in CCI4:

PCCl4VSubmergedg = PH20'vsubmergedg

PCCl4(0.60)Vobjecfg = PH2O(0.90)Vobjecfg

PCCl4-(0.60) = PH2o(0.90)

PCCl4/PH2O =a9%.60 =9/6 =3/2=1.5The best answer is choice C.

Copyright©by TheBerkeley Review® 112 REVIEW EXAM EXPLANATIONS

B=jxl44N =36N

T = 144-36N=108N

Q

W=144N

Electrostatics and

ElectromagnetismPhysics Chapter 8

by

the Berkeley Review

Electrostatics and

ElectromagnetismSelected equations, facts, concepts, and shortcuts from this section


F =k^2; where k=9.0 * 109 N- m2/C2 = -i-r2 4jre0

Vf = "y m (for a particle in a linear Efield)

Felectric = qE W = qEAd = qAV

F = qv x B = qvB sin8 F = 1/x B

© Important ConceptCharged particles behave differently in an electric field than a magnetic field

F = qE

' y ' y * y i y * yy ! ' ' ' y

Moving charged particles are deflectedby perpendicular electric fields.

Electric Fields

Can accelerate a charged particlefrom restDo work on a charged particle

F = qvB sin9

r_ mv

Moving charged particles traverse a circlethrough perpendicular magnetic fields.

Magnetic Fields

Cannot accelerate a charged particle from restDo no workon a charged particle

© Lenz's Law SimplifiedLike Le Chatelier's principle in chemistry, Lenz's law predicts that the system will react to undo achange

"X"

xxxx xxxx

Loop enters 1B field oriented into page Loop exits 1B field oriented into pageChange: Gains Xs inside ofloop Change: Loses Xs inside ofloop

Reaction: Induce a counterclockwise current to create 0s. Reaction: Induce a clockwise current tocreate Xs.

Physics Electrostatics and Electromagnetism Electrostatics

Electrostatics and ElectromagnetismCharged particles can interact with other charged particles in two ways: (a)through direct interaction of their charges and (b) through magnetic fieldscreated when the charges are in motion. Coulombic forces (electrostatics) alwaysexist between charged particles. Opposite charges attract one another and likecharges repel one another. However, in order to exhibit magnetism, the chargedparticle must be in motion. If their magnetic fields are aligned correctly, thenthey can also interact through magnetic interactions. We shall address bothelectrostatics and electromagnetism in this chapter.

Electrostatics

Electric Charge and Charge ConservationIt was discovered during ancient Greek times that amber, when rubbed withwool, would attract other objects. We now know that in order for the amber toattract another object, it must have acquired an electric charge. What causes acharge? That is a difficult question to answer. We cannot see an electric charge,much lesssay what an electric charge is. We can only try to quantify an electriccharge by describing its behavioral characteristics and properties. Charges arisewhen electrons are either gained or lost from a neutral species.

Atoms are the building blocks of matter. They consist of electrons, protons, andneutrons. Electrons carry a negative charge, protons carry a positive charge, andneutrons carry no charge at all; they are neutral. The masses of these threeparticles are shown in Table 8-1. Note that the mass of a proton is about 2000times greater thanthe mass ofan electron. Also note that the masses ofa protonand a neutron are approximately equal to one another.

Particle Mass (kg) Charge (C)

Electron

Proton

Neutron

9.11 x 10"31

1.673 x lO"27

1.674 x lO"27

-1.6 x 10"19

+1.6 x 10"190

Table 8-1

Protons and neutronscomprise the nucleusofan atom, whereas electrons stay inorbits about the nucleus. An atom in its simplestform has an overall net chargeof zero, because the number of protons equals the number of electrons. Atomsdiffer from each other by the numbers of electrons, protons, and neutrons theycontain.

If an atom were to lose an electron (most often it loses the electron from itsvalence shell-outermost orbital), then that atom would have an overall netcharge of+1 and is said to be a positively charged ion. Similarly, if an atom weretogain an electron in its valence shell, then thatatom would have an overall netcharge of -1. This atom is said to be a negatively charged ion. When atoms gainor lose electrons, they undergo a process called ionization. You should note thatin chemistry the ionization energy is the energy associated with losing anelectron. Ifa negatively charged atomis close to a positively charged atom, therewill be an attraction between those two opposite charges that strengthens as thedistance between them decreases. However, there will be repulsion between twopositively charged atoms or two negatively charged atoms. The strength of therepulsive force also increases in magnitudeas the charged species draw closer toone another.


FhySlCS Electrostatics and Electromagnetism Electrostatics

For example, when we rub a plasticrod with fur, the rod acquiresa net negativecharge while the fur acquires a net positive charge. In the process of rubbing therod with the fur, electrons were transferred from the fur to the rod. The netcharges on these two objects are opposite in sign and alwaysequal in magnitude.Charge has been conserved. In charge conservation, the total electric charge onthe plastic rod and the fur (or any other two objects) does not change. Eventhough chargecanbe transferredfrom one object to the next, it cannot be createdor destroyed.

Conductors and InsulatorsA conductor is a material thatallows for the movement of charge. An insulatorisa material that impedes the movement of charge. Many of the metals that weknow of (e.g., copper or silver) are good conductors. These metals allow for theflow of electrons. Nonmetals (e.g., glass or plastic) are typically good insulators.Electrons generally do not flow in these materials.

Coulomb's Law and Electric ForceIn the mid-1780s, Charles Coulomb discovered that the electrical force exerted byonecharged object onanother charged object dependson the distance r betweenthose two objects, the sign of the charges on each object, and the amount ofcharge q on each object. The magnitude of the electrical force F that each of thetwo charges exerts onthe other can be expressed in terms ofequation (8.1). Thisequation is called Coulomb's law and as written applies only to charges in avacuum. The proportionality constant k depends on thesystem ofunits used inthe calculation.

F- iJqiq2lr2

(8.1)

The SI unit of charge is called the coulomb (C). The magnitude of the charge ofanelectron ora proton (denoted by e) is 1.60 x 10"19 C. One coulomb equals thecharge carried by an aggregate of 6.3 x1018 protons (or the negative of 6.3 x1018electrons, based ontheamount ofcharge that flows in 1 second when thecurrentis 1 amp). In the SI system of units we also find that the units of the constant kare newton-meter2/coulomb2. The value of k isabout 8.99 x 109 N•m2/C2.

In equation (8.1), the electrical force F is the magnitude of a vector quantity. Itwill always be a positive value. However, the product of the charges qi and q2can be either positive (if the charges have the same sign) or negative (if thecharges have opposite signs). The absolute value bars in equation (8.1) ensurethat Fwill always bepositive.

As we have mentioned, iftwo objects have charges with the same sign, they willrepel one another (Figure 8-la). If two objects have charges with opposite signs,they will attract one another (Figure 8-lb). Figure 8-1 shows thatthe direction ofthe force that each charge is exerting on the other is always along astraight linejoining the two charges. Even though the forces are opposite in direction, theyare equal in magnitude (recallNewton's Third Law of motion).

Figure 8-1


Physics Electrostatics and Electromagnetism

The magnitude of this force, when plotted as a function of the charge separation,is often shown as the upper graph in Figure 8-2. If we consider the signassociated with force, then the graph could be either the upper graph or thelower graph in Figure8-2. For the sake of this graph, we shall consider repulsionto generate a positive force (when the charges have the same sign) and attractionto generate a negative force (when the charges have opposite signs). This "1/r2dependence" of the force is similar to that for the gravitational attraction of twomasses; it is a mathematical relationship that reappears in the discussion of manyphysical phenomena. Because of its ubiquity in physics, knowing how to plot theinverse-square law and understanding it conceptually is a good idea.

o>u

clO-

Repulsive

Attractive

Both charges are positive (+)or

Both charges are negative (-)

One charge is positive (+)andthe other charge is negative (-)

Figure 8-2

Separation r

When considering electrical charges, you shouldhave a feeling for the sizeof thecharge itself. A charge equal to onecoulomb isa huge value. Using equation (8.1)to calculate the forcebetween two 1-Ccharges exactly1 meter apart, would yielda magnitude of the force is about 109 newtons. We can put this into a morefamiliar context: knowing that 1 N = 0.225 lbs and that 2000 lbs = 1 ton, we cancalculate that the magnitude of the force between these twocharges would havea gravitational force equivalent to roughly a million tons. In realistic calculations,charges typically have values that fall within the range of 10"9 to 10"6 C.

Example 8.1aBefore a voltage is applied across a wire, the wire has no net charge. Afterapplying a voltage, the net charge on the wireis:

A. negative, if the current flows fromhigher to lowervoltage.B. positive, if thecurrentflows from higherto lower voltage.C. zero, regardless of the current direction.D. a non-zerovalue that depends upon the current direction.

SolutionWhen a voltage difference is applied across a wire, a currentwill flow. By "flow,"we mean the net movement of electrons from lower to higher voltage. Current,traditionally defined as flowing positive charges, moves from higher to lowervoltage. However, for the purpose of answering the question, we need to knowthat current makes electrons flow into the wire and out of the wire. On average,the number of electrons leaving the wire equals the number entering. Thus, thenet charge on a wire will not change-it will be neutral, with or without thecurrent. But with a current, it will have an induced dipole.



Electrostatics


Physics Electrostatics and Electromagnetism Electrostatics

Example 8.1bWhen NaCl dissociatesin pure H2O, the net charge of the electrolyticsolution:

A. becomes negative.B. becomes positive.C. remains neutral.

D. becomes positive or negative, depending upon the relative Na+ and Cl"concentrations.

Solution

A crystalline salt comprised of +l-cations and -1-anions, such as NaCl,has equalnumber of (+)-cations as (-)-anions. This means that when added to neutralwater, the overallsolution charge remains neutral.


Example 8.2aWhat is the ratio of theelectrostatic repulsive force to the gravitational attractiveforce between two electrons that are 1 nm apart? (Note: The coulombic constantk=9x109 N-m2/C2 and gravitational constant G~7 xlO-11 N-m2/kg2.)

A. 1042tolB. 1025tolC. 102tolD. lO"2 to 1

Solution

Let's approach it using the ratio method. The tworelevant equations are:

F •..-, = Gmlm2 and F . - ^l^lrgravitational - ana Electrostatic ~

Because both equations share the same denominator, we can disregard distance.Taking the ratio of the electrostatic force to that ofthe gravitational force gives:

Felectrostatical _ klqUEl _ (9 *109)(l.6 x1Q-I9)(l.6 xIP"19)Fgravitational Gmim2 (7 x10"n)(9.1 x10"31)(9.1 x10'31)

Before embarking on an arithmetic adventure, notice that the choicesoffered asanswers differ by several orders of magnitude from each other. Therefore, wecould avoid worrying about how the prefactors, such as 9 and 1.6, affect theanswer. Just add and subtract the powers of 10. Doing so gives a ratio of about1044 to 1, which is closest to choice A. (The actual answer is 3.97 x1042 to 1.)

Aside from a chance topractice arithmetic approximations, this problem revealsa few facts about electrostatic and gravitational forces. First, the two forceequations look similar-just an exchange of mass for charge and an alteredconstant (being aware that gravitational forces are always attractive, whereaselectrostatic forces can alsobe repulsive). Second, their similar 1/r2 terms cancelout, meaning that this force ratio is the same whether the charges are1 nm or 1km apart. Finally, notice that the electrostatic force is much stronger than thegravitational one. This explains why electrostatic force problems almost nevermention gravitational forces when asking abouttheacceleration ofcharges.




Example 8.2bA potassium ion and a chlorine ion are separated by a membrane, yet they exertan attractive force F on each other. When the membrane is three times thicker,the attractive force is F'. What is the ratio of F to F? (Note: Neglect forceinteractions with the membrane and surrounding solution.)

A. 9:1

B. 3:1

C. 1:3

D. 1:9

SolutionAscharges growfarther apart, the force theyexerton oneanother decreases, so Fis greater than F'. The ratio of F to F' is greater than 1.This eliminates choices Cand D. Because the force depends on the distance squared, tripling the distancewill affect reduce the force by a factor of nine. The ratio of F to F is 9 :1.


Example 83aWhat is observed over time after a +1 charge and -1 charge initially 2.0 cm apartare released from rest?

A. They move towardsoneanotherat constant v along a straight line.B. They move towardsoneanotherwith an increasing v along a straightline.C. They move towardsoneanotherwith a decreasing v along a straightline.D. They both movein the same linear directionwith the samespeed.

Solution

Because theyare oppositely charged, they willaccelerate towardsoneanother,sochoice D is eliminated. As they draw closer to one another, the force increases, sothe move towards one another with increasing acceleration and increasing speed.


Example 8.3bWhichchargesexperience the greatest magnitude ofattractive force?

A. A +1 charge and -1 charge separated by 2 cm.B. A +2 charge and -1 charge separated by 4 cm.C. A +2 charge and -2 charge separated by 6 cm.D. A +3 charge and -3 charge separated by 8 cm.

SolutionTheforce depends on the magnitude of eachchargeand the distance between theparticles squared. Lefs compare each choice one-by-one to a designated choice,starting with choice B to choice A. Choice B has one charge that is doubled thatof choice A, but the distance is also doubled. Doubled distance reduces the forceby a factor of four, which outweighs the greater charge. Choice A feels a strongerforce than choice B, so choice B is eliminated. Now lefs compare choice C tochoice A. Choice C has both charges double that of choice A, but the distance istripled. This would reduce the force by a factor of nine, which outweighs theimpact of greater charges, so choice A feelsa stronger force than choiceC. ChoiceC is eliminated. Last, lefs compare choice D to choice A. Choice D has bothcharges triple that of choiceA, but the distance is quadrupled. This would reducethe force by a factor of sixteen, which outweighs the impactof greater charges, sochoice A feelsa stronger force than choiceD. ChoiceD is eliminated.



Electrostatics


Physics Electrostatics and Electromagnetism Electric Fields

Electric Fields

Electric FieldsConsider that a single charge by itself feels no electrostatic interactions until asecond charge is near. By itself, there is nothing more than a potential to feel aforce if a second charge were moved into its vicinity. An electric field is a regionof space where a charged particle would feel an electrostatic force, if it weremoved there. This imaginary charge is referred to as a test charge (denoted asq'). Electric fields can be made of a single charge or several charges. It is simplythe region of space where a test charge would feel an electrostatic force if added.

Suppose we do not know how unknown charges are distributed in space. Oneway to probe for a charge distribution is to use a test charge. Let's assume thatthe unknown charges are arranged under a box, as shown in Figure 8-3. We thentake a test charge, which is positive (a negative test charge could work as well,but by convention test charges are considered to be positive), and place it in theregion of space near the box. We have presented three arbitrary positions inFigure 8-3. If the test charge is attracted to some region of the box, then it meansthat there is a negative charge in that vicinity. Similarly, if the test charge isrepelled from a region of the box, then it means that there is a positive charge inthat vicinity. If we apply this type of analysis to every point outside the box, wecan determine the nature and location of the charges under the box. These forcesconstitute an electric field E. The unit of an electric field is the newton/coulomb.

q' (at arbitrary position 1) q« (at arbitrary position 2)net//

q' (at arbitrary position 3)

Figure 8-3

If we were to sum up all of the force vectors associated with Figure 8-3, then wewould get a picture as shown in Figure 8-4. The lines are referred to as field lines(although you'll note that most of them bend). These lines represent theorientation and strength of the electric field.

Figure 8-4

An electric field at a given point in space is defined by equation (8.2).The electricfield E is the ratio of the net electrical force F acting on a test charge in aparticular region of space, divided by the value associated with the test charge q'.Since the force is a vector quantity, so is the electric field. Because physicists havechosen test charges to be positive (+) by convention, electric field lines alwayspoint in the direction that the net force would act on a positive (+) test charge.

E = *q'

(8.2)



Field LinesElectric fields are often represented by lines of force or field lines. The field linesshow the direction of the electric field at every point in space, and they neverintersect. This can be seen with the field lines in Figure 8-4. Field lines alwaysoriginate on a positive charge and end on a negative charge. They represent thewaya (+) charge would migrate. If the field lines are close together, it meansthatthe magnitude of the electric field is strong. Similarly, if the field lines are farapart, themagnitude ofthe electric field is weak. Two representative examples ofelectric field lines associated with point charges are shown in Figure 8-5.

Figure 8-5

Electric Fields Created By Charge DistributionMany electrical systems have numerous charged particles,each of which exerts aforce on all of the other charged particles in the system. Fortunately, chargesnaturally distribute themselveson a conducting material in a way that minimizestheir repulsion, resulting in the formation of a uniform electric field.To calculatethe net force acting on just one particular charged particle in a group of them, wemust use the process of vector addition to include the effect of each electricalforce that is acting on our selected particle.

Lefs consider a simple case involving six plate charges spread between twoparallel plates (threeper plate) and a single test charge (Figure 8-6). Suppose thatsix of those charged particles comprise the positive and negative electrodes of acharged capacitor, and that the charged particle is an ion with a positive charge.Fornow, the only aspect of this situation that we are interested in is the directionin whichthe positively charged ion will movewhen influenced by the electrodes.

net

Figure 8-6

What are the forces acting on the positively charged ion? First, there is anattractive force (F.) generated by the charge of the negative electrode. Second,there is a repulsive force (F+) generated by the charge of the positive electrode. Inorder to find the net electrical force on the positively charged ion, we sum theforce vectors. Since only the net electrical force ultimately determines the motionof the test charge,we find that it moves toward the negatively charged electrode.Equation(8.2) applies here, although it might be easier to see it as Fe= qE.


Electric Fields



Example 8.4aWhich of the following electric field configurations describes the lines of force inan electric field between two parallel plates of opposite charge?

A. + + + + + + + + + B. + + + + + + + + +

Ji M

1 u y v y v ]|

+ + + + + + + + + + + + + + + + + +

Solution

An important point to remember about electric fields is that they emanate frompositive charges and point towards negative charges (they point in the way apositive charge would migrate). This means that choices A and C are incorrect.Next, we must judge the straightness or curvature of the linesbetween the plates.

Imagine a point placed directly between two parallel plates. It will have a netelectric field determined by all the individual electric fields generated by everycharge on the two plates. If these field lines are added vectorially, then all thehorizontal components will cancel each other out; a net field oriented verticallydownward will remain. The cancellation is due to the symmetry of the point inspace with respect to the plate charges. However, this horizontal componentcancellation continues only until one approaches the left and right edges of theplates. Here, the field lines bow out, like those around a dipole (the parallelplates are like several dipoles positioned next to each other). Between parallelplates of opposite charge, the field lines are straight and constant in strengththroughout the interplate region. Thus, a parallel plate arrangement is a gooddesign for capacitors,electrophoresisequipment, and charge accelerators.


Example 8.4bWhich orientation of a uniform electric field can explain the pictured motion ofan electron initially moving in a straight line from left to right?

0

A. B. C. D.

Solution

The negatively charged particle is deflected down, so it must be feeling a force inthe downward direction. But this question has a subtlety you must not forget,and that is the fact that an electric field line is based on how a positive chargewould be accelerated. To deflect a negative particle down would require a fieldthat would deflect a positive particle up. This describes choice C.



Physics Electrostatics andElectromagnetism

Equipotential LinesThe electric potential at any point in an electric field can be described byequipotential lines. The grey circles in Figure 8-5 represent equipotential lines,where a test charge at any pointon that circle would experience the same forceand have the same potential to move. An equipotential line (orsurface) connectsall points within anelectric field sharing the same potential. Ifwewere toplace atest charge at any point along an equipotential line, we would find that theelectric field does no work on that testcharge when it moved toanother pointonthe same equipotential line. Work is done on the test charge only if the electricfield has a component tangent to the equipotential surface. This tells us thatequipotential lines and electric field lines are perpendicular to one another(Figure 8-7). When testcharges move in thedirection ofa field line, theyare saidto move from higher potential to lower potential.

+ + •*- + + + -i• H- + + + + + +

1 1f I{X\ i1 1 i I I y< y ' y f i i i i } ' y ' y f y ' /r^^V.f I-*—*•

Figure 8-7

Electrical Potential EnergyWhat happens when charged particles move closer together or farther apart?Such displacements change the electrical potential energy of the system. Theelectrical potential energy of particles with like charges increases when thecharges are brought towards each other. Conversely, the electrical potentialenergy decreases when opposite charges are brought towards each other.Equation (8.3) gives the change inelectrical potential energy APEq oftwo chargesin a vacuum, as they are moved from some initial separation r{ to some finalseparation rf. This equation supports the general rule regarding the potentialenergy of a system. It always decreases as the system approaches someequilibrium state. Opposite charges naturally attract; as they approach oneanother, their potential energy decreases. The reverse applies when two likechargesare brought towards each other.

APEq =kq,*(£-JL) (83)

Generally speaking, if you must do work to move the charged particles, thenthey've been moved to a state of higher electrical potential energy. If the chargedparticles move naturally, then they have moved to a state of lower electricalpotential energy.


Electric Fields



Electrical Potential DifferenceAs charged particles move through space, they encounter other chargedparticles. Each of these charged particles creates a little electrical environmentthat influences charges in itsvicinity. A chargedparticletends to move towardsaregion of space where its electrical potential energy decreases (much like waterflowing downhill to a region of lower gravitational potential energy). Theelectrical potential difference a moving charged particle is subjected to as itmoves from some region A to some region B(denoted as Vab) is the differencein electrical potential energy APEq at those two regions, divided by the valueassociated with the charged particle q. Thisis shown in equation(8.4).

v APEAB_PEqB-PEqA_APEqABq q q

The unit of charge is the coulomb C and the unit of electrical potential energy isthe joule J, so the final result of equation (8.4) gives us units of joules divided bycoulombs (or J/C). A joule divided by a coulomb is defined as a volt V, the unitof electrical potential difference. A more common version of equation (8.4) isequation (8.5). It calculates the change in electrical potential energy as a chargedparticle moves through a voltage difference:

APEq = qV (8.5)

Absolute PotentialWhen doing electrical calculations, it is often useful to have a point of reference.That point of reference in electrical circuits is referred to as the ground point. Bydefinition,the voltageat the ground point is zero. Measuring the voltageat otherpoints in space relative to the ground point describes an absolute potential.

Example 8.5aAn electron gun usually consists of two electrode plates, one more negativelycharged than the other. If an electron of mass m is allowed to accelerate betweenthe two electrodes separatedby a voltage V, which equation BEST represents thefinal speed of the initially stationary electron upon striking the more positiveelectrode?

A. v=*pELV:

C.

2eV

B. v=\l*&-m

-./mV

2ev=V

-vD. v = 2e

mV

SolutionLet's answer this question in two ways: using physical intuition and energyconservation. First, physical intuition. How will m, e, and V affect the electron'sfinal velocity? The potential difference supplies the energy to accelerate theelectron. If mere is a bigger voltage, the electron should get more energy and,hence, more speed. Because speed depends on the voltage, V should be in thenumerator, eliminating choices A and D. The final speed should be greater forlighter particles (since it has less mass to accelerate).Therefore, the mass shouldbe in the denominator—cross out choices A (again) and C. Only choice B has V inthe numerator and m in the denominator, so it is the best answer. Notice that we



don't even need to think about where the e should go. When approachingquestions in this way, always start by thinking about the physical variable thatmakes the mostsense to you.Theparticularstartingpointis up toyou.

Let's use energy conservation to solve this problem. Identify the initial and finalenergies. Hasany external work is done on the electron? Finally, plug into:

^initial ofobject + "0n object = Efinal of object

Initially, the electron has no energy, because it is at rest. We are neglectinggravitationalenergy, because it is relatively small (seeExample 8.2a) and becausethere are no g's in the answer choices. When the electron hits the other electrode,its energy is all kinetic. It moves because moving across the potential differencedoes work on the electron. This work and the kinetic energy fit into the energyconservation equation as follows:

0 + eV = V£mv2 .-. v2 = 2eV/m

Solving for the speed v gives choice B. Using energy conservation is a powerfultechnique. Energy conservation does not require us to know about the nature ofthe interactions (be they mechanical, gravitational, thermal, or electrostatic). It isconcerned onlywith the energies. Whenyou are in doubt abouthow to approacha problem analytically (i.e., using equations), try energy conservation first. Thisparticular problem type is a popular one. Many scientific machines and naturaleventsinvolve chargesaccelerating through a potential difference.


Example 8.5bAn ionic surfactant molecule is a long, carbon-based polymer with a chargedend. If the uncharged end is attached to a plate and the molecule is immersed inan electric field, what is the maximal amount by which the molecule wouldstretch? (Assume that interactions with the plate and surrounding solution arenegligible and that the molecule obeys Hooke's law with its spring constant k.)

A.

B.

V 2qV

x-/m:

C. x _ /kV"V^q"

D'X=V2q_kV

Solution

Intuition tells us that as the charge q increases, the force acting on the polymerwill be stronger and the polymer should stretch more. This means that xincreases as q increases,so q should be in the numerator. This eliminates choicesA and C. As the voltage, V, increases, the force on the polymer also increases, soV should also be in the numerator as well. This eliminates choice D, leaving onlychoice B. The spring constant, k, should be in the denominator, because ithinders stretching. If you prefer to use conservation of energy, the equation is:qV =Vfckx2. Solving forx will confirm thatchoice Bis the best answer.



Electric Fields



Dielectric ConstantSo far we have been considering charges in a vacuum, but they are also found inmedia. The effect that a medium can have on charges is taken into account by aparameter called the dielectric constant K, which is the ratio of the electricalforce between two charges when they are in a vacuum compared to when theyare in a medium. This is shown in equation (8.6). Dielectricbehavior results froma medium's polar nature, including its ability to assume an induced dipole.

F _j. _ r vacuum'•medium ~ (8.6)

A strongly polar medium responds to an external electric field by reorienting itspolarmolecules, so that thenet electric field within the medium becomes weakerthan the external field. This occurs when the 6+ charges of the medium alignnear the negative plate and the 5- charges align near the positive pate. Such afield strength reduction diminishes the interactive forces between all chargeswithin the medium. Figure 8-8 shows the electric field associated with a parallelplate capacitor before and after a polar medium has been added. The polarmedium aligns in such a way as to offset, and in doing so, diminish themagnitude of chargeon eachplate.

+10 -10

+10

+76'

+8ish

6--6+6--6+6— 6+6--5+

6— 6+6-"6+6— 6+5— 6+

6—6*6— 6+6— 6+6— 6+

6— 6+6— 6+6— 6+6—6+

6—6+6—6+5—6+6—6+

6— 6+6— 6+6— 6+6— 6+

6— 6+6— 6+6— 6+6— 6+

-10

+76*

-8ish

With a vacuum: +10 and -10With a polar medium: platecharges are partially cancelled

Figure 8-8

Coulomb's law applies to charges in a vacuum. We can modify that law bysubstituting equation (8.6) into equation (8.1) to give equation (8.7):

p _ Fyacuum _ fclqiq2lK Kr

(8.7)

Equation (8.7) applies to charges separated by a distance r in a medium withdielectric constant K. Note that the smaller the value of K, the larger the forcebetween charges.Table 8-2 provides a list of some common dielectricconstants.

^•^^&|erial::.::,::; Dielectric constant, K

Vacuum 1.0000

Dry Air 1.0006

Paper 3.5

Benzene 4.0

Glass 5-10

Axon membrane8(unmyelinated)

Water 80

Table 8-2



Example 8.6aThe dielectric constant K of an unknown material is measured as a function oftemperature. What can explain thesharprise and subsequent slow decay ofKasthe temperature increases?

A. The solid material melts at Ti, increasing its polarizability. Thermal motionsubsequently increases its polarizability.

B. The solid material melts at Ti, increasing its polarizability. Thermal motionsubsequently reduces its polarizability.

C. The gaseous element ionizes at Ti, increasing its polarizability. Thermalmotion subsequently reduces its polarizability.

D. The gaseous element ionizes at Ti, increasing its polarizability. Thermalmotion subsequently increases its polarizability.

Solution

Materials that are polar, or that are able to polarize easily (i.e., have an highpolarizability), typically have stronger dielectric properties than nonpolarmaterials. The basic relationship between polarizability and the dielectricconstantis the following: If the polarizability increases, then the dielectric constantincreases. With this in mind, we need to consider what impact a change intemperature would have on the polarizability of a material.

The first thing to consider is the drastic increase in polarizability at Ti- Choices Cand D state that the gaseous element ionizes at temperature T\. Ionization woulddecrease the polar nature of the gas, as the gas would become canonic and thuslack a negative pole. This eliminates choices C and D. It must be due to melting.As a material melts, its polarizability increases, because the molecules can moreeasilyorient themselves in the liquid state than they can in the solid state (wherethey exist in a crystal lattice). For example, H2O is a relativelypolar molecule. Inthe solid state, water (ice) molecules point in many directions. Thus, an externalelectric field will not be reduced much within the ice, because the net electricfield of the molecules is zero. As liquid water, the molecules can more easilyorient themselves in much the same direction when an external electric field ispresent; the net field of the molecules in no longer zero, and the total electricfield within the water is reduced. This makes the induced polar properties ofliquid water greater than those of ice (i.e.,K increases when ice melts).

From this point, ifs a matter of observing the trend more so than evaluating thereasons for the trend. As the temperature continues to increase, the thermalmotion of the molecules increases. According to the graph, as the temperatureincreases from Ti, the polarizability decreases. This means that as thetemperature increases, the polarizability of the material decreases, making choiceB the best answer. Continuing with the water example, when water heats up,thermal motion disrupts this polar alignment, and K decreases.



Electric Fields



Example 8.6bTwo experiments measure the electrostatic force between two cationic particlesin different media, A and B. What is true of the two media?

A. Medium A has more positivecharges than Medium B.B. Medium Bhas more positivecharges than Medium A.C. Medium A has a largerdielectric constant than Medium B.D. MediumBhas a larger dielectric constant than Medium A.

SolutionThe electrostatic force between the two cationic particles is reduced in medium Brelative to medium A.Based on the possibleanswer choices, it is either caused bya differencein the number of positive charges in each medium or a difference inthe dielectric constants of the two media. The presence of positive charges in themedium would not impact the force between two particles, although it wouldimpact the net force. This is because the force vectors in a charged environmentcan all be treatedas independentof one another.This eliminates choices A and B.The difference in the forces must be due to the dielectric nature of the two media.According toequation (8.6), as the dielectric constantincreases, the force betweentwo charged particles decreases. The graph shows that the force is less inmedium B than in medium A, so medium B must have a greater dielectricconstant than medium A. Physically, this means that medium B is more polarthan medium A and that it can reorient itself to reduce the strength of the electricfield between the two charged particles.


Electric DipoleAn electric dipole is established when two charges of opposite signs areseparated from one another by some distance L (Figure 8-9).

Figure 8-9

An electric dipole moment can then be definedby equation (8.8), where p is thedipole moment (a vector quantity), q is the product of the charge, and L is thedistance separating the charges.

P = qL (8.8)

What happens when we place this electric dipole in a uniform electric field, asshown in Figure 8-10?


Physics Electrostaticsand Electromagnetism

F. = -qE -q LcosO

Figure 8-10

From equation (8.2), we know that the force F+ has a magnitude +qE, while theforce F- has a magnitude -qE. Adding these forces together, we find that the netforce on an electric dipole in a uniform external electric field is zero.

However, note that the forces have different lines of action. This means that thenet torque on an electric dipole is not zero. In Figure 8-10, the torque is directedinto the page. Thedipole will therefore rotateclockwiseuntil it linesup with thefield. If we let the angle between the dipole axis and the electric field be 0, thenthe torque created by the dipole couple can be quantifiedby equation (8.9).

x = (qE)(Lsin 0) = (pE)(sin 6) (8.9)

The torque is at its minimum when p and E are parallel (or antiparallel) to oneanother (Figure 8-lla). The torque is at its maximum when p and E areperpendicular to one another (Figure 8-llb).

(a)

L

•

e A— ^ *

+q -q

•

(b)

-q

l<

^—-Figure 8-11

If the dipole were to change directions in an electric field, not only would workbe performed, but there would also be a change in potential energy (PE). If 0 =90° in Figure8-10, the electric dipole would be perpendicular to the electric field(see Figure 8-llb). In this case, the potential energy of the system is zero. If 0 =0°, the electric dipole would be parallel to the electric field (see Figure 8-lla).The potential energy of this system is at its most negative value (i.e., the stablestvalue). If 0 = 180°, the electric dipole would be antiparallel to the electric fieldand the potential energy would be at its most positive value (i.e., the mostunstable value). The potential energy of the system can be given by equation(8.10).

PE = - (pE cos 0) (8.10)


Electric Fields



Example 8.7aWhen placed in theelectric field gradientshownbelow, the dipoleis subjected:

A. to a clockwise torque and a net linear acceleration.B. to a counterclockwise torque and a net linear acceleration.C. to a clockwise torque and no net linear acceleration.D. to a counterclockwise torque and no net linear acceleration.

Solution

Use the multiple-concepts approach. If a dipole is placed in an external electricfield, it can possibly experience a torque. The positive charge in the dipole istugged to the left, and the negative charge is tugged to the right; the dipolerotates. If forces cause an object to rotate, there must be a torque on it; here, thetorque is clockwise. Thus,choices Band D are wrong.

Regarding a net linear acceleration, ask yourself whether there would be a netforce on the dipole.The positivecharge experiences a force F = qE to the left. Thenegative charge experiences a force to the right. However, the force on thenegative charge is bigger. Why? The charges have the same magnitude, but theelectric field is indicated as being stronger to the right in the picture (remember,closer lines mean a stronger field). As the dipole rotates, the negative charge willbe in a stronger electric field. The dipole will experience a net force to the right,and consequently a net accelerationto the right.


Example 8.7bWhich of the following configurations of dipoles would produce the GREATESTelectrostatic potential energy?

A. B.©—® ®—<&

C. D.

Solution

By symmetry, choices A and B are equal in magnitude, so both systems shouldshare the same electrostatic potential energy. This eliminates choices A and B.The best way to solve the question from this point is to consider how stable thetwo remaining choices appear. Choice D has like charges apart and oppositecharges close, while choice C has like charges close and opposite charges apart.Choice D is more stable than choice C, which means it has less electrostaticpotential energy (more stable implies less potential energy). A greater amount ofelectrostatic potential energy is found in choice C than choiceD.




ElectromagnetismIn our study of electrostatics, we found that electric charge can create an electricfield in the space surrounding the charge. We also saw that this electric fieldexerts a force on any charge present within it. A charge moving through spacealso creates a magnetic field capable of exerting a force on any other chargemoving in that field. Magnetism arises from charges in linear or radial motion.

Magnetic FieldsA magnetic field has a magnitude and a direction. Magnetic fields are denotedby the symbol B, and they are vector quantities. The SI unit for a magnetic field isthe N's/C'm. When we study currents, we will see that an ampere A is acoulomb per second. Substituting the SI nomenclature, we have the unit N/A«m,a unit called the tesla (denoted as T). The CGS unit of the magnetic field is thegauss (denoted as G). The relationship between a gauss and a tesla is that 1 G =10"4 T. Magnetic field strengths vary greatly for different natural phenomena:The Earth's magnetic field has a strength of about 0.5 G at the Earth's surface; aneveryday refrigerator magnet has a strength ofabout 10"2 T; andan MRI field hasa strength of a few teslas.

Magnets have both a north pole and a south pole. The north pole of one magnetattracts the south pole of another magnet, and vice versa. Poles of likeorientation, however, repel each other when placed close together. Magnetsgenerate magnetic fields, which can be represented on paper in much the sameway that we represented electric fields. The vector-like lines of the magnetic fieldshow the direction of that field at a given point in space; tighter line densitiesindicate stronger magnetic field regions. Magnetic field lines always form closedloops. Because of this natural property, bar magnets are always dipolar (i.e.,there is always a north and south pole). This is unlike electric field lines, whichcan emanate from a single charge (i.e., a monopole). However, by convention wesay that magnetic field lines emanate from the north pole and re-enter throughthe south pole, as shown in Figure 8-12.

Figure 8-12

Magnetic fields can exert a magnetic force Fmagnetic on a charge moving in thatfield. If a charge q is moving with a velocity v through a perpendicular magneticfield B, then the magnitude of the magnetic force is given by equation (8.11). Themoving charge experiences a force that is perpendicular to both v and B. It is thecomponent Bj^ of the magnetic field perpendicular to the velocity v that dictatesthe size of this force. Therefore, equation (8.11) can be rewritten as equation(8.12). The angle 0 in equation (8.12) is the angle between v and B.

Fmagnetic = qvB± (8.11)

Fmagnetic = qvBsin 0 (8.12)


Electromagnetism


Physics

Test TipRight-hand Rule

Electrostatics and Electromagnetism Electromagnetism

As shown in Figure 8-13(a), the force is greatestwhen the velocity of the chargedparticle is at rightangles to the magnetic field (i.e., when sin 8 = 1).As showninFigure 8-13(b), the force is smallest when the velocity of the charged particle isparallel orantiparallel to themagnetic field (i.e., whensin 0= 0).

(a) (b)

AFmagnetic =qvB(sin 90°) Fmagnetic =qvB(sin 180°) =0

Fmagnetic =qvB(sin 0°) =0

Figure 8-13

While equations (8.11) and (8.12) can tell us about the magnitude of the magneticforceacting on a movingcharged particle,we need to employ the right-hand ruleto determine the direction of the magnetic force. As far as getting prepared forthe MCATgoes, this is the most important concept to master in this chapter. Theright-hand rule assigns three different parts of our right hand to represent eachof the three mutually perpendicular vectors associated with a charged particlemoving through a perpendicular magnetic field (as shown in Figure 8-13(a)).There are several versions of the right-hand rule taught to physics studentsaround the world, so while we will choiceone to use repeatedly in this section, ifyou have learned and mastereda different version, please use the one that worksbest for you. Keep in mind that all the right-hand rule is doing is helping us tovisualize three mutually perpendicular vectors.

In our first application of the right-hand rule, we shall consider the chargedparticle in Figure 8-13(a). We will use our thumb to represent the velocity of a(+)-charged ion in linear motion, our fingers to represent the direction of themagnetic field vectors, and our palm to represent the direction of the resultantmagnetic force. It is important to note that the right-hand rule is based on themotion of a positively charged particle, be it a single cationic charge movingthrough space or a current of positive charges flowing through a wire. There is amagnetic force associated with a negatively charged particle moving through aperpendicular magnetic field, it just points in the opposite direction of what theright-hand rule shows.

In Figure 8-13(a), a cationic particle is moving in the plane of the magnetic fieldfrom the bottom to the top, so we point or right thumb to the top. Next, with aflat, open hand and your thumb perpendicular to your index finger, point yourremaining four fingers in the direction of the B field, which is from right to left inthis case. With your thumb pointing to the top and your fingers pointing to theleft, your open hand is "palm up", which means that the magnetic field ispointing up (the direction you'd push the charge with your palm. It is helpful tokeep in mind that your right hand is simply serving as a model for the threeperpendicular axes. In this case, the axes are bottom-to-top (velocity), right-to-left (magnetic field, and below-to-above (magnetic force). We shall do a fewexamples, because the right-hand rule is mastered only through practice. Weshall create a systematic checklist of the three axes in selected examples.



Example 8.8aIf a cation enters into magnetic field from the top (as pictured below), in whichdirectionwill the cationbe deflected while traveling through the magnetic field?

3) TO© © ©t© © ©

©©©©©©

©©©©©©

©©©©©©

A. To the leftB. To the rightC. Out of the pageD. Into the page

SolutionTo determine the direction of the magnetic force deflecting the cation, and thusthe direction in which the cation will be deflected once in the field, we need touse the right-hand rule. When using the right-hand rule, keep in mind that it isused to determine the direction of the magnetic force on a positively chargedparticle in motion, which is the case here. The velocity is from the top to thebottom, sopoint your thumb downward (thumb =down, not up; |down| up). Themagnetic field is orientedout of the page, so whilekeeping your thumb pointingdown point your fingers out from the page (fingers = out, not in; ^u| m). Onceyou've done this, you'll see that your palm points to the left (palm = left, notright; Jef^ right). Because your palm points to the left, we know thatthemagneticforce on a positively charged particle points to the left. This will deflect thecationic particle to the leftonce it has entered themagnetic field.


Example 8.8bWhich combination of cation in motion and magnetic field will result in thecation being deflected out from the page?

A.

c

I Jr v I

* » » y

v * I v

xxxx

xxxx

X X

IV

X Xi!


B.

D.

® ® ® ®

-® ® ® ®

® ® ® ®

©

133

Electromagnetism


Physics Electrostatics and Electromagnetism Electromagnetism

SolutionTo determine whichparticle will get deflected out of the page, we need to applythe right-hand rule to eachof the four choices.Choice A: The velocity is from the right to the left, so point your thumb to theleft The magnetic field is pointing down, so point your fingers down. Aligned assuch, your palm points out from the page. The magnetic force deflects thecationicparticle out from the page, so choiceA is valid.Choice B: The velocity is from the left to the right, so point your thumb to theright The magnetic field is pointingout from the page, so point your fingers out.Aligned as such,your palm points down. The magnetic force deflectsthe cationicparticle down, so choice Bis invalid.Choice C: The velocity is from the bottom to the top, so point your thumb up.The magnetic field is pointing into the page, so point your fingers into the page.Aligned as such, your palm points to the left The magnetic force deflects thecationicparticle to the left,so choice C is invalid.Choice D: The velocity is from the bottom to the top, so point your thumb up.The magnetic field is pointing to the right, so point your fingers to the rightAligned as such, your palm points into the page. The magnetic force deflects thecationicparticle into the page, so choiceD is invalid.


Example 8.9aAn electron moves from left to right (on the page) through a uniform magneticfield. Which orientation of a uniform magnetic field would result in the smallestmagnitude of force acting on the electron?

A. —• B. X X C.

—• X X"^

Solution

To find the direction and magnitude of magnetic forces, use the right-hand rulewhen considering directions,and use:

F = qvBj.

when considering magnitudes. Lefs start by consideringthe terms. Eachanswerchoice involves a moving electron, so q and v are irrelevant. Only the magneticfield orientation, Bj_, varies between the answer choices. Recall that B± is thecomponentof the magnetic field perpendicular to the charge'svelocity. We wantthe choice where theBfield is leastperpendicular to the particle's velocity.

In choice A, both the velocity of the particle and the magnetic field are pointingin the same direction, so the choice with the smallest componentof B± is choiceA. Choices B and C both represent a magnetic field that is perpendicular to theparticle's velocity, so choices B and C give the same magnitude of force, eventhough the resulting force directions will differ-this assumes that both choiceshave the same magnetic field strength. Choice D shows a B field that is neitherperpendicular to nor parallel with the particle's velocity, so the force on thecharged particle will have a magnitude somewhere between zero and qvB (themaximum force possible). To determine the exact value, we would need tomultiply the magnitude of the Bfield by sin8.



PhyS1CS Electrostatics and Electromagnetism

Example 8.9bIf an electron travels to the right in the magnetic field pictured below, in whichdirectionwould the net force on the electron be pointing?

xxxxxxxxxxxx

X XX xxxxxxxxx

M

X XX xxxxxxxxx

xxxxxxxxxxxx

B.

IC.

0D.

Solution

To determine the direction of a magnetic force, we need to use the right-handrule. However, the right-hand rule is used to determine the direction of themagnetic force on a positively charged particle in motion, not a negativelycharged particle. For this question we will need to reverse the answer we getusing our right hand. The velocity is from the left to the right, so point yourthumb in to the right (thumb = right, not left). The magneticfield is oriented intothe page, so point your fingers into the page (fingers = in, not out). Once you'vedone this, you'll see that your palm points up to the ceiling or sky (palm = up,not down). Because your palm points up, we know that the magnetic force on apositively charged particle points upward. The magnetic force on an electronmust be pointingdownward. Only choice Bshows this.


Examples 8.8a through8.9b emphasize a few concepts. First, thereis the fact thatmagnetic fields are generated by charged particles in motion and that two suchmagnetic fields when aligned correctly (perpendicular to one another) cangenerate a force. Second, the direction of the magnetic force caused by theinteraction of a charged particle in motion with an external B-field can beobtained using the right-hand rule, a technique fordetermining the alignment ofthree perpendicular vectors.

Let's apply this concept to a conducting rod moving through an externalmagnetic field. Because the rod is conducting, it has loose charges that canmigrate through the rod if a force is applied. The impact on each individualcharge will be the same as we observed in Example 8.8b, so they will all migrate.However, because the charges are bound to the conducting rod, they can onlymigrate along the rod. Figure 8-14 showsthisbehavior.

x 5+ x x

x v x

x

X X

x 0- X X

X

X

X

—•X X

Figure 8-14


te

(+) charges feel amagnetic force up

(-) charges feel amagnetic force down

135

Electromagnetism


Physics

t = 0

No current

Electrostatics and Electromagnetism Electromagnetism

Lenz's LawLenz's law takes the example shown in Figure 8-14 and applies it to a conductingloop instead of a conducting rod. As a consequence, the charges do not reach adead end at the ends of the rod, and instead are able to flow completely aroundthe loop. The result is that the magnetic force can induce a current in theconducting loop. What we learn is that the loop only experiences an inducedcurrent when it is entering or exiting the magnetic field. The exact wording ofLenz's law is as follows:

When a conducting loop experiences a change in the magnetic fluxwithin its boundaries, a current will be induced in the loop thatcreatesa magneticflux opposing the change in theexternal magneticflux.

As worded above, the law doesn't necessarily make sense at first reading. Heck,reading it a second or third time doesn't really help either. To make things easierto understand, lefs compare Lenz's law to Le Chdtlier's principle. Consider thesystem shown in Figure 8-15. As the loop moves into the field, it experiences again of Xs within the inside of the loop. According to Le Chdtlier's principle, if achemicalsystem at equilibriumgains reactants, then it shifts to make products; ifthe equilibrium system gains heat, then it shifts in a way to absorb neat (makecold). The system shifts in a way to reduce the stress. The same is true withLenz's law. The stress is that the loop is gaining Xs within the loop as it entersthe field shown in Figure 8-15, so it must generate the opposite of an X. Giventhat an X represents a magnetic field going into the page, the loop willexperience an induced current that creates a magnetic field coming out of thepage within the loop. According to the right-hand rule for magnetic fieldsemanatingfromcurrent carrying wires, a magnetic field pointing out of the pageoccurs with a counterclockwise current. Hence, as the loop enters the field at t =1, it experiences a counterclockwise, induced current.

y. j< x x.—x—>x x

x j< x Xs x Jk x

t = l

Counterclockwisecurrent

(Gains Xs, creates 0s)

t = 2

No currentt = 3

Clockwisecurrent

(Gains ©s, creates Xs)

t = 4

No current

Conductingloop movingleft to right through B± at constant v

Figure 8-15

There is no current induced once the loop is completely immersed in themagnetic field at t = 2, because the loop is feeling a constant flux (there's nochange in the number ofXs inside the loop). However, when the loopleaves thefield, it experiences a drop in the number of Xsinside the loop.To offset this lossin Xs, an induced current in the loop is generated in such a direction that itcreates a magnetic field going into the page. According to the right-hand rule,this occurs with a clockwise current. This is why as the loop exits the field at t =3, there is a clockwise, induced current. Hopefully taking a Le Chatelier'sperspective on Lenz's law simplifies it.



The magnitude of the current depends on the rate at which the flux changes. Thisis known as Faraday's law, equation (8.13), which states that the em/causing theinduced current is proportional to the rate of change of the flux within the loop.

Faraday's law em/=.AOB/At (8.13)

Faraday's law explains that as you increase the rate at which the loop enters a Bfield, the magnitude of the induced current increases, as shown in Figure8-16.

Slower immersion

Faster immersion

_j;I U ?™Counterclockwise current

(upon entering the Bfield)

Figure 8-16

Clockwise current

(upon exiting the B field)

t

Example 8.10aWhat is true of a conducting loop that spins within a magnetic field about an axisperpendicular to the magnetic field?

A. It experiences an induced alternating current.B. It experiences an induced direct current of constantmagnitude.C. It experiences an induced direct current of changing magnitude.D. It experiences no current.

SolutionAs the loop rotates about an axis that is perpendicular to the magnetic field, itgains Xs some of the time followed by a period of time where it loses Xs. Thismeans that the current could be either counterclockwise or clockwise at differentpoints in its revolution, so the current is alternating. To generate a direct current,the loop would need to always be entering or always be exiting the field, whichis physically impossible. Spinning loops in an external magnetic field generateinduced, alternatingcurrents.This is the principlebehind electrical generators.


Example 8.10bA metal loop movingleft to right across the following Bfield would induce a:

00000000

©0000000

00000000

A. clockwise current once it was fully immerse in the field.B. counterclockwise current once it was fully immerse in the field.C. clockwisecurrent upon entry and a counterclockwise current upon exit.D. counterclockwise current upon entry and a clockwisecurrent upon exit.

SolutionAn induced current only occurs upon entering or exiting the field, so choices Aand B are eliminated. Upon entering the field from the left, the loop wouldexperience a gain in dots, which would induce a clockwise current to generate Xs.



Electromagnetism



SolenoidA solenoid is a helical winding of a conducting wire, such as a coil of copperwire wound on a cylindrical tube. A solenoid should have a length that is at leastfour times its radius. Upon passing current through a solenoid, a magnetic fieldcan be generated that depends on the magnitude of the current and dimensionsof the solenoid. A solenoid can serve as an electromagnet when a constant directcurrent generates a steady magnetic field of fixed strength and orientation.Figure 8-17 shows a solenoid and a corresponding graph relating the strength ofthe magnetic field to the position in the solenoid. The n0 term is a constant formagnetic fields, n is the number of turns in the solenoid, and J is the current.

Figure 8-17

(Direct Current) Electric MotorConducting loops are an integral part of electric motors. The idea is that currentis passed around a conducting loop, causing a local magnetic field within theloop. If the loop is placed ina linear external magnetic field at such anangle thatit feels a torque and it is free to rotate, then the torque will cause it to rotate.Equation (8.14) describes the torque on the loop, where N is the number of turns,Iis the current, Ais the area of the loop, Bis the strength of the external magneticfield, and $ isthe angle between the external field lines and the loop's field lines.

x = NL4Bsin<j>

External B field

Figure 8-18

(8.14)

By having several loops arranged symmetrically, there will always be loopsfeeling some magnitude of torque. The result is that the symmetric collection ofloops spins. Figure 8-18 shows one loop inanexternal magnetic field.

Induced B field



Velocity SelectorAmongst the differences between an electric force and a magnetic force is thatonly the electric force can accelerate a charged particle from rest (speed up orslow down a particle for that matter) and only the magnetic force depends on thespeed of the charged particle in motion. For this reason, electric fields are foundin particle accelerators. However, the combination of an electric field andmagnetic field, when aligned correctly, can serve to segregate particles ofdifferent velocities. By arranging an electric field perpendicular to a magneticfield the right way, an electric force can be aligned in the opposite direction ofthe magnetic force. Figure 8-19, shows the basic schematic of a velocity selectorwhere the electric field points down and the magnetic field points into the page.

Deflects up if Fb> FE .'. v is too fast

v>B

Electromagnetism

E Travels straight*"V=B" ifFB =FE

Fdown = FE = qE

v<B

Deflects down if FB < FE .*. v is too slow

Figure 8-19

Mass SpectrometerA mass spectrometer takes advantage of the deflection of a moving chargedparticle in a perpendicular magnetic field. Once a charged particle enters amagnetic field, it will move in a circular path at a constant speed. The radius ofthe circular path depends on the mass of the particle, amongstother things. If allother factors are equal, then the relative masses of particles in the spectrometerwill correlate to their relative radii. Heavier particles traverse bigger circles andthus have a greater radius. The correlation between the radius and mass can befound by setting Fra(jial = mv2/r (equation (2.31)) equal to Fmagnetic = qvB(equation(8.12)) to generate equation (8.15).

mv /qB (8.15)

Figure 8-20 shows the top view of the three regions of a Bainbridge massspectrometer. The multiple semi-circle curves representdifferent cations.

X = B field into page I- E field pointing down CollisionDetector

+

+

+

X

X

X

X

X

X

X

X

X

X

*xX

>x+

+ ! x ir x i r X , r X i r xX

Region Iacceleration

Region IIVelocity selector

Figure 8-20


X

X X

Region IIIdeflection

(F = qvB sin6)



Example 8.11aIn a velocity selector, which of the following changes will lead to an increase inthe speed of the exitingparticle?

A. Decreasing the mass of the particle being ejected.B. Increasingthe mass of the particlebeing ejected.C. Decreasing the strength of the electric field.D. Increasing the strength of the electric field.

Solution

In a velocity selector, the particle that travels straight through and therebyescapesthe velocity selector, has a speed v = E/B. In order to increase the speedof the ejected ion, either the magnitude of E must increase or the magnitude of Bmust decrease. Changing the mass of the particle will change the speed of theparticle, but not the escape speed. All that will occur with a change in mass isthat it will be different cationic particles that achieve the correct exit speed.Choices A and B are eliminated.

Because the exit speed is v = E/B, a stronger E field, rather than a weaker E field,would result in a greater exit speed.


Example 8.11bIn a Bainbridge mass spectrometer, what could explain a greater radius ofcurvature in the deflectionregion for one cation versus a second cation?

A. An isotope of the element with fewer neutrons was used for the secondcation.

B. The electric field in Region I was decreased.C. The magneticfield in RegionIIIwas increased.D. A less ionized particle was used for the second cation.

Solution

According to equation (8.15), a greater radius of curvature could be the result ofa particle with a larger mass, a greater speed, a reduced charge, or the magneticfield couldhavea diminished strength.

Anisotope withfewer neutrons would resultin a lighter particle. Because radiusis directly proportional to mass (r « m), this would result in a smaller radius ofcurvature, so choice A is eliminated. If the magnitude of the electric field inRegion I weredecreased, then the particles leaving Region I would havea loweraverage speed. However, because of the velocity selector in Region II, theparticles leaving Region II wouldhave the samespeed, so the radii shouldnot beaffected byspeed. Choice Bis eliminated. If the magnitude of the magnetic fieldin Region III were increased, then the particles leaving Region II experience astronger deflecting force and be bent more. Because radius is inverselyproportional to the magnetic field strength (r « 1/B), a stronger B field wouldresult in a smaller radius. Choice C is eliminated.

A less ionized particle means that the cation has a lower magnitude of charge. Ifthe second cation has a reduced q, then it will feel less ofa force deflecting it inRegion III, and thus, it will traverse a greater circular radius. This is becauseradius is inversely proportionalto charge (r « 1/q).


Copyright ©byThe Berkeley Review 140 The Berkeley Review

25 Electrostatics and Electromagnetism Review Questions

I. Electromotive Force

II. Gelo-O Brand Electrophoresis

III. Electrostatics


(1-7)

(8 -14)

(15 - 22)

(23 - 25)



In order to produce a current in an electric circuit, apotential difference or emf (electromotive force) is required.This emf may arise from a chemical process, an externalelectric field, or magnetic induction. Dragging a conductingrod through a magnetic field induces an em/between the endsof the bar:

X

X

X

X

X

X

X

X

Figure 1

This induced emf is given as: e = BLv, where B is the

magnetic field, L is the length of the bar, and v is the speedof the bar. e is measured in terms of volts. This method of

inducing a voltage can then be applied in a circuit:

X

X

X

X

X

X

X

X

X

X

X

Figure 2

Faraday found that in general, an induced emf can beproduced by changing a quantity called magneticflux:

(j> = BiA

where B± is the strength of a magnetic field passingperpendicularly through an area A, enclosed by a wire loop.The rate at which this quantity changes:

At

determines the induced emfacross the ioop. This induced emfthen drives a current through the circuit, which produces itsown magnetic field that opposes the initial flux change.

A current moving through a magnetic field experiences aforce of magnitude:

F=ILB±

where Bi is the component of B perpendicular to the currentI running througha region (or wire) of length L.

In the following problems, let L = 1 meter, B = 5 T, v = 2m/s, and R = 4 Q. The electron's charge is given by -e.


1.

2.

3.

4.

5.

For the moving rod in Figure 1, where are the electronslikely to cluster?

A. At the top.

B. In the middle.

C. At the bottom.

D. The electrons will be spread out uniformly alongthe length of the bar.

How much work is done on an electron in moving itacross the bar in Figure 1?

A. 0

B. ^

C. evLB

evB

LI).

The likely explanation for the induced emf shown inFigure 2 is an increasing:

I. magnetic field as the rod moves to the right.

II. area as the rod moves to the right.

III. circuit resistance as the rod moves to the right.IV. power loss in the resistor as the rod moves to the

right.

A. II only

B. I and II only

C. Ill and IV only

D. I, II, III, and IV

For the circuit in Figure 2, which way does the inducedcurrent flow?

A. Counterclockwise

B. Clockwise

C. Clockwise, and then counterclockwise

D. No current will flow.

For the circuit in Figure 2, what is the magnitude of thecurrent?

A. 20 amps

B. 10 amps

C. 5 amps

D. 2.5 amps


6. If the rod in Figure 2 is released with zero velocity, andif the rails on which the rod sits are frictionless, the rodwill:

A. not move when a current is running through theresistor.

B. move to the left, to minimize the area of the circuit

loop.

C. move to the right, to maximize the magnetic fluxthrough the circuit loop.

D. remain stationary, to keep the magnetic fluxthrough the circuit loop constant.

7. Which of the following will NOT reduce the powerdrain in the resistor?

I. Using a larger resistance with the same currentII. Increasing the rod velocity.

III. Increasing the magnetic field strength.IV. Increasing the rod length.

A. I only

B. I and III only

C. II, III, and IV only




Electrophoresis involves the separation of chargedmacromolecules by the application of an electric field. Thecharged molecules in the field are subjected to anelectrostatic force given by:

F = qE

where q is the charge of the molecule, and E is the electricfield.

Gel electrophoresis utilizes a gel-like matrix to separatemacromolecules. This apparatus is an important tool in thestudy of DNA, a macromolecule that has an intrinsicallynegative charge due to the presence of phosphate groups. Notonly can gel electrophoresis be used to determine the basesequence of a DNA polymer, but it can also be used toseparate DNA polymers by size.

For many years after the introduction of gelelectrophoresis, it was nearly impossible to separate wholechromosomes. However, in the mid-1980s a new approachwas taken that allowed entire chromosomes to be separatedby a technique called pulsedfield gel electrophoresis. Insteadof applying one electric field to a gel containing DNA, twoelectric fields are set perpendicular to each other. Rather thanrunning one electric fields continuously, the two fields arepulsed. One field is turned on for a short time and then turnedoff. Next, the second field is turned on for a short time andthen turned off. If the pulsed fields have the same magnitude,same duration, and are set up as shown in Figure 1, the DNAwill run in a vertical line down the gel.

Electrodes

Gel Box

Figure 1

The idea is that the fields pull the DNA first one wayand then another way. The effect is the separation of largeDNA strands that run in a straight line.

Electrophoresis has been employed to separate the 16chromosomes of Saccharomyces cerevisiae, or baker's yeast.The sizes of some of the chromosomes are listed below:

Chromosome Number Size (kD)

I 198

III 311

VII 1285

XVI 864


In a typical experiment to separate chromosomes,the DNA is loaded into wells in an agarose gel. The gel isthen immersed in a tank that contains a buffer. The buffer isrun through a cooler to maintain a constant temperature ofabout 12°C-15°C. Electric fields form between the chargedplates of capacitors, and the plates are arranged around thetank to produce the fields shown in Figure 1. A constant-voltage power supply is used to monitor both the constantvoltage and the current through the buffer. The distance Dthat a particular chromosome will migrate through a gel isrelated to the molecular size M of that chromosome by:

D = a-b(logM)

where a and b are constants that depend on theelectrophoretic conditions.

8. Which of the following changes will affect themigration rate of a given band of DNA through a gel?

A. Increasing the plate separation while keeping thevoltage constant.

B. Decreasing the voltage across the plates whilekeeping the plate separation constant.

C. Changing the gel viscosity.D. All of the above.

9.

10.

Occasionally, the cooler breaks down and thetemperature of the buffer increases by as much as 20°C.Why does the current also increase?

A. An increase in temperature causes the resistance toincrease.

B. An increase in temperature causes the voltageacross the plates to decrease.

C. An increase in temperature releases more ions intothe buffer.

D. An increase in temperature causes the gel toharden.

Consider the electrophoresis setup in the passage thathas been modified so that Electric field A becomes half

as strong as Electric field B. An experimenter loads agel as shown below, but is unable to get a run. How canthis result be explained?

Gel Box

A. The bandsran diagonally off the gel's right side.B. The bands ran diagonally off the gel's left side.C. The bands ran in a straight line off the gel, from

top to bottom.

D. The DNA was loaded backwardsin such a way thatit could not migrate.


11.

12.

13.

14.

A plot of the migration distance D versus log M shouldlook like:

A. B.

Before the introduction of gels to hold the DNA duringthe electrophoresis process, DNA was placed in a liquidsucrose density gradient. However, this was notsuccessful in separating DNA by size. DNA moleculesof different sizes tended to run as one indistinguishableband. Why?

A. All DNA fragments have the same charge.

B. Diffusion causes DNA to clump together.C. The liquid sucrose density gradient neutralizes the

effects of the applied electric field.

D. Viscous drag forces generally increase withincreasing molecular size.

From the top to the bottom of the gel, the yeastchromosomes listed in Table 1 should be ordered inwhich of the following ways?

A. I, III, VII, XVI

B. VII, XVI, III, I

C. I, III, XVI, VII

D. Ill, I, VII, XVI

Suppose in the electrophoresis set-up in the passage,electric field A is half as strong as electric field B. Ifboth fields are turned on at the same time, which of thefollowing pictures BEST represents the sum of the twoelectric fields?

A. ^ B.



A group of students decide to study electrostatics. Theyhave the following items at their disposal: glass rods, hardrubber rods, silk, fur, metal rods with glass supports, andmetal spheres on glass supports.

Experiment J

The students attempt to demonstrate that there are twokinds of charge. First, they rub a glass rod with silk. This rodis hung from a long silk thread. A second glass rod is rubbedwith silk and held near the rubbed end of the first glass rod.The two glass rods are observed to repel each other. Next, ahard rubber rod is rubbed with fur. When the hard rubber rod

is held near the rubbed end of the glass rod hanging from thethread, the two rods are observed to attract each other.However, two hard rubber rods rubbed with fur are observed

to repel each other.

Experiment 2

Two neutral (uncharged) and identical metal spheres onglass supports are brought close to each other but are notallowed to touch. A glass rod is rubbed with silk and therubbed end is brought close to one of the spheres, Sphere I,but does not touch it (as shown below). With the glass rodstill held near Sphere I, a conducting wire connects the twospheres. Finally, the wire is removed and the glass rod istaken away.

glass rod \

glass support

Figure 1.

rubber end

glass support

Experiment 3

A glass rod is rubbed with silk and the rubbed end isheld near one end of a metal rod, resting on a glass support.The students predict what will happen to the charge in themetal rod.

15. How do the masses of the spheres in Experiment 2compare before and after the experiment?

A. The masses remain the same.

B. The masses of both spheres increase.

C. The masses of both spheres decrease.

D. One mass increases, and the other decreases.


16. In Experiment 1, suppose that the rubbed glass rodbecame positively charged. Then the silk that rubbed itwould:

A. acquire an equal amount of negative charge.

B. acquire an equal amount of positive charge.

C. acquire a positive charge, but not the same amountas the glass rod has.

D. remain neutrally charged.

17. In Experiment 2, why is it important that the glass rodnot touch Sphere I?

A. To maximize the amount of charge transferred tothe sphere

B. To maximize the rate of charge transfer to thesphere

C. To prevent induction of charge separation onto thesphere

D. To prevent charge transfer directly to the sphere

18. Upon completing Experiment 2, the students found:

A. that both spheres have acquired the same type ofcharge as that found on the rubbed glass rod.

B. that the spheres have remained neutrally charged.

C. that Sphere I has acquired the same type of chargeas that on the glass rod, and Sphere II has acquiredthe opposite type of charge.

D. that Sphere I has acquired the type of chargeopposite to that found on the glass rod, and SphereII has acquired the same type of charge.

19. In Experiment 3:

A. the electrons flow continuously in the metal rod, aslong as the glass rod is held near the metal rod.

B. the electrons eventually cease to flow in the metalrod.

C. both positive and negative charges flowcontinuously in the metal rod.

D. the positive charge flows continuously, but theelectrons remain stationary in the metal rod.

20. Suppose now the students repeat all of the experimentson a humid day. They find that:

A. the results remain unchanged.

B. Experiment 1 and Experiment 2 no longer work aswell.

C. Experiment 2 and Experiment 3 work better.

D. none of the experiments works as well.


21.

22.

In Experiment 3, with the glass rod held near the metalrod:

A. there is no net electrical force on the metal rod.

B. there is a net force on the metal rod; the metal rodis attracted to the glass rod.

C. there is a net force on the metal rod; the metal rod

is repelled by the glass rod.

D. there is a net force on the metal rod; the neutrons inthe metal are repelled by the glass rod.

Which of the following statements is NOT a validdescription of the role of corresponding material in theexperiments?

A. Fur is used to transfer electrons to the material that

rubs against it.

B. The conducting wire serves to transfer charge fromone surface to another when contact is made.

C. The spheres serve to create electrons.D. The glass rods serve to transfer electrons to the silk

cloth that is rubbed against them.



23. A charged particle is suspended between twohorizontally aligned parallel plates carrying oppositecharges of equal magnitude. If the separation betweenthe plates is doubled while the voltage is held constant,whathappensto the electricforce on a particulardrop?

A. It is doubled.

B. It is halved.

C. It is reduced to one-fourth of its original value.D. It remains the same.

24. For a positively charged particle moving left to rightthrough a linear magnetic field oriented into the page,whatpath will be observedfor the particle?

A. Path I

B. Path II

C. Path III

D. The particle will bedeflected outof the page

25. If two parallel wires cany currents in the samedirection:

A. the wires attract

B. the wires repel.

C. the wires rotate such that they becomeperpendicular.

D. the wires feel no net external force.

1. C 2. C 3. A 4. A 5. D

6. D 7. D 8 D 9. C 10. D

11. A 12. D 13. B 14. A 15. D

16. A 17. D 18. D 19. B 20. D

21. B 22. C 23. B 24. A 25. A

YOU ARE DONE.

Answers to 25-Question Electrostatics and Magnetism Review

Passage I (Questions 1 - 7) Electromotive Force

1. Choice C is the best answer. The electrons in the bar experience the magnetic force: F = qvB±. The direction of the forcecan be determined using the right-hand rule. There are many versions of this, so use whichever one is familiar to you. If youdo not know any right-hand rule, here's one: Hold your right hand flat, with your thumb perpendicular to your fingers. Yourfingers point in the direction of the magnetic field lines (remember: many lines, many fingers). Your thumb points in thedirection in which the charge moves (remember: hitchhiking charge). Your palm will, as a result, point in the direction inwhich the induced force would push a positive charge (remember: palms push positive charge). If the charge is negative, as itis in this case, just reverse the direction of the force.

Here, the electrons are moving towards the right, and the magnetic field points into the page. Using your right-hand ruleshows that the electrons experience a net force downwards and will cluster at the bottom of the rod. The best answer ischoice C.

2. Choice C is the best answer. The work required to push an electron from one end of the moving rod to the other end isequal to the potential energy change:

W = AU = qAV

From thepassage, we are told that the induced emfacross theends of the bar is given by: e = BLv.

The work required to move anelectron is given by: W= BLve. The best answeris choice C.

3. Choice A is the best answer. As the rod is pulledalong the rails, the area enclosed by the rod and the rails grows largerandlarger. The flux changes because the area changes, and the change in flux induces an emf, according to Faraday. Thisvalidates choices A, B, and D. Since the magnetic field is not changing in the setupfor Figure2, an increasing magnetic fieldis not the reason for the emf. This rules out choices B and D, leaving only choiceA. Regarding statements III and rv, theresistance of the circuitdoes not change as the rod moves, because the only source of resistance is the resistor (which isfixed). Therefore, assuming a constant emf, the overall resistance and power loss areconstant. The best answer is choice A.

4. Choice A is the best answer. Electrons will be pushed toward the bottom of the rod. Current, by definition, is the flow ofpositive charges. Since the electrons flow clockwise, the current will flow counterclockwise. Thebest answer ischoice A.

5. Choice D is the best answer. From Ohm's law, we know that e = IR. Rearrangement of terms gives:

I = iLR

Weare told in the passage that s = BLv. For thisproblem, wesubstitute as follows:

e=BLv so: I=S^- =<5 "HO mX2 m/s>/4 q =2.5 amps


Choice D is the best answer. Let's start by considering A. Using theright-hand ruleon a current running through the wire inFigure 2 shows you that the rod is pushed either to the right or to the left (depending upon the direction of the current). Thisinvalidates choice A. Magnetic flux does not tend tochange. That is why, as stated in the passage, the current generated inawire loop will orient itself sothat its magnetic field opposes the induced change in flux~a condition described by Lenz's law.This invalidates choices B and C, and favors choice D. The best answer is choice D.

Choice D is the best answer. Asking which change will not reduce thepower drain can be restated as asking which changewill either increase the power drain or leave it unchanged. Power drains from in a resistor because a current moves throughits resistive material, dissipating energy as heat. Increasing the resistance would therefore increase the power drain, makingOption I valid and choice Cinvalid. Increasing the rod velocity and magnetic field strength are both ways to increase the emfinduced across the circuit. This will increase the current and the resulting power loss, making only choice D valid. Noticethat increasing the rod length does not change the size of the rest of the circuit. The important part of the rod (as far asinduction is concerned) is still the same size. What is meant by the important partis thesegment of the rod between the twocontact points. Making the rod longer simply adds additional length out side of the enclosed region of flux. You might betempted to think that L in the relationship e = BLv has increased, but it is in fact the same. As a result, the power drain doesnot change. Taking the rod length into account was not necessary, assuming that the first three statements are interpretedcorrectly. The best answer is choice D.


Passage II (Questions 8 -14) Gel-O Brand Electrophoresis

8. Choice D is the best answer. Increasing the plate separation while keeping the voltage constant will change the magnitudeof the electric field between the plates. Recall that voltage and electric field are related by: V = Ed, where d is the distanceseparating the plates. Changing the value of d will change E, if the voltage is kept constant. Changing the voltage across theplates without changing d also changes E. The gel acts as a matrix to hold the DNA in place. Changing the gel viscosityalters the rate of DNA migration, since the DNA experiences forces from interactions with the gel, as well as the electricfield. The best answer is choice D.

9. Choice C is the best answer. This question can be answered by knowing definitions and does not require a completeunderstanding of the experiment. The current increases, which according to Ohm's law, V = IR, occurs with increasing V ordecreasing R. Although increasing temperature generally increases resistance, this should not cause a current increase. Infact, for a constant voltage, the current should decrease with increasing resistance. Therefore, choice A is incorrect. Thevoltage across the plates is held constant by the power supply and remains unaffected by the temperature of the buffer, sochoice B is incorrect. As temperature increases, we should expect the gel to begin to disintegrate or melt, not harden. If thegel did harden, this would increase resistance to ion migration, which results in a slower charge movement, and thus areduced current. For that reason, choice D is also incorrect. Choice C must be the best answer by the process of elimination.

In a buffer, a weak acid can dissociate so that ions are released into the solution:

H|A] + H20 H30+ + |A]-

As the temperature increases, the reaction must be driven to the right (dissociation reactions are generally endothermic), somore ions will be released. This will increase the current. The best answer is choice C.

10. Choice D is the best answer. Theexperimenter sees nothing on the gel, because he placed it in the tank incorrectly! DNA isnegatively charged. This means the gel should be placed in the tank so that the wells containing the DNA are near thenegative electrode. That way, when the electric fields are turned on, the DNA will run from the negative electrode to thepositive electrode. By loading the gel in the way indicated, the experimenter has inadvertently arranged for the DNA strandsto stay close to the (+) plates rather then migrate across the gel! The best answer is choice D.

11. Choice A is the best answer. Smaller strands of DNA (size or mass of the strand is indicated by M) will migrate farther(distance is indicated by D). The graphs in choices A and C both express this relationship. Considering the relationshipbetween M and D more carefully, it is best shown by choice A (noting that we are plotting logM and not M on the verticalaxis). The best answer is choice A.

12.

13.

14.

Choice D is the best answer. As mentioned in the passage, DNA has an intrinsic negative charge, due to the presence ofphosphate molecules. DNA polymers of different sizes should run asdistinguishable bands, since DNA polymers ofdifferentsizes have different charges (the larger the DNA molecule, the more phosphates it has, so the larger the charge it has).However, as the molecules increase in size, they will be more affected by viscous drag forces. The viscous drag forces tendto oppose the electrostatic forces. Consequently, DNA polymers of different sizes run as one indistinguishable band in aliquid. Choice A is incorrect, for the reasons just mentioned. Choice Bis incorrect; diffusion should spread the DNA out, notclump it together. Choice C is also incorrect; the liquid gradient is there simply to keep the DNA from diffusing. Sucrose iselectrically neutral and therefore has no effect on the electric field. The best answer is choice D.

Choice B is the best answer. From the passage, we are given the empirical relationship between migration distance D andmolecular size M: D = a - b (logM), where a and b are constants that depend on the electrophoretic conditions. As Mincreases, D decreases. Therefore, smaller molecules migrate farther. Since the DNA strands are negatively chargedmolecules, they generally migrate from the top towards the bottom of the gel (as the passage states). Thus, the longest strandsshould be closest to the top of the gel. Strand VII is the longest and therefore the first in the list. Only choice Bbegins its listwith Strand VII, making it the best answer. The best answer is choice B.

Choice A is the best answer. Electric fields are vectors. If the two fields are on at the same time, the net electric field is thesuperposition of the two individual fields. Electric field B has twice the magnitude of Electric field A, so the net field shouldpoint in the direction of Electric field B. The best answer is choice A.

Passage III (Questions 15 - 22) Electrostatics

15. Choice D is the best answer. Except in special cases, the actual charge carriers in conductors are electrons. Electrons havemass as well as charge, so the sphere that acquires the electrons will have slightly more mass, and the sphere that lost theelectrons will have slightly less mass. The best answer is choice D.

Copyright© by The Berkeley Review® 148 MINI-TEST EXPLANATIONS

16. Choice A is the best answer. Charge is a conserved quantity, which means that it can be neither created nordestroyed. Itcanonly be transferred from one object to another. The process of charging a glass rod (an insulator) is poorly understood,butit is believed thatfriction between thesilkand the rodstrips electrons from the glassand deposits them onto thesilk. Thesilk has a higher affinity for electrons than does theglass. Thisleaves a deficit of charge on the glass rod. Thus, thecharge onthe rubbed glass rod and the silk have to be equal in amount, but opposite in sign. The best answer is choice A.

17. Choice D is the best answer. When two objects carrying different charges come into contact with each other, the chargedensities will balanceout between the objects through the transferof electrons. To prevent charge transfer, the objects shouldbe separated from one another by an insulator, such as air. Preventing charged objects from touching each other preventscharge transfer, making choice D the best answer. The best answer is choice D.

18. Choice D is the best answer. The rubbed glass rod now has a net charge on it. Let the net charge of the rubbed glass rod bepositive. A metal is a conductor, which means that charge is free to flow through it. When the rubbed glass rod is broughtclose to Sphere I, negative charge is attracted to the glass rod, and positive charge is repelled. The same is true for Sphere II.Negative charge will be attracted to the rubbed glass rod, while positive charge will be repelled. If the two spheres are thenconnected by a conducting wire, charge is free to flow from one sphere to another. The positive charge on Sphere I will flowto Sphere II, trying to get away from the glass rod. Negative charge will flow from Sphere II to Sphere I, to get closer to theglass rod. When the conducting wire is removed and the glass rod taken away, the metal spheres are each left with a netcharge. Sphere I, the sphere closer to the glass rod, will have acquired a charge opposite to the charge on the glass rod.Sphere II, the sphere farther away from the glass rod, will have acquired a charge that is the same as the charge on the glassrod.This process of charging a conductor without actually touching it is called induction. The best answer is choice D.

19. Choice B is the best answer. Although there is a nearly inexhaustible supply of electrons in the metal rod, there are twocompetingeffects going on. Suppose the rubbed glass rod has a net positive charge. The electrons are attracted to the rubbedend of the glass rod, so they flow to the end of the metal rod closer to the glass rod. However, electrons are also repelled byeach other. Eventually, there are so many electrons at the end of the metal rod closer to the glass rod that the attractive forceexperienced by the positive charge is balanced by the repulsive force from other electrons. At this point, the electrons willcease to flow. Choice A is incorrect, for the reasons just mentioned. Choices C and D are incorrect: Electrons are the chargesthat move in conductors, not positive charges. The best answer is choice B.

20. Choice D is the best answer. These electrostatic experiments depend on insulators being able to acquire a net charge andhold it.The charging of an insulator occurs by friction-stripping electrons fromatoms, thereby ionizing them. A humid daymeans a greater amount of water is in the air than on an arid day. Water is a better conductor than air, so in the presence of abetter conductor, the ions are less likely to stay ionized. It is easier for them to recombine into a neutral atom, reducing theinitial charge buildup on the rods. This is why you canmost aptly shock yourself or your neighbor on a dry, windy day. Thebest answer is choice D.

21. Choice B is the best answer. Choice D can be eliminated immediately, because neutrons do not respond to an electrostaticforce. The question is reduced to determining whether there isanattractive force, a repulsive force, or noforce atall. Let therubbed glass rod have a net positive charge. Then the end of the metal rod closer to the glass rod would have a negativecharge, and the end farther away from the glass rod would have a positive charge. The negative charge in the metal rod andthe positive end of the glass rodexert an attractive force on each other. Thepositive charge in themetal rod and the positiveendof theglass rodwill exerta repulsive force on each other. Electrostatic forces are distance-dependent:

F « -Lr2

Since the negative endof the metal rod is closer to the glass rod, the attractive force on the metal rod is slightly larger thanthe repulsive force. Thus, there is a netattractive force on the metal rod. The best answer is choiceB.

22. Choice C is the best answer. Because glass and hard rubberare able to exert forces (attractive in some cases and repulsiveinothers) after being rubbed with fur, we canconclude that rubbing with fur is responsible for charging thespecies. Chargesaccrue when electrons are transferred, so it is reasonable to assume that fur is used to transfer electrons to the material thatrubs against it. Choice A is a valid statement and thus eliminated. A conducting rod by definition can transfer electrons, sochoice B is a valid statement, which eliminates it. Because the glass rods repel one another after being rubbed by silk, theymust be transferring electrons to the silk. Choice D is a valid statement and thus it's eliminated. The spheres do not createelectrons; they simply hold thecharge that isadded to them. Choice C is aninvalid statement. The bestanswerischoice C.



23. Choice B is the best answer. Conceptually, we know that if the plates are moved farther apart while the voltage is heldconstant, then the force acting on any particular drop will be reduced, because force depends on distance. This eliminateschoices A and D. If the plate separation is doubled while the voltage across the plates is kept constant, then according to therelationship AV = EL, the electric field must be reduced by half. Since the electric force is given by: F = qE, cutting thestrength of the field in half will reduce the force at any point in the field by one-half. For any particular drop at any pointbetween the two plates, the force will be reduced to half of its original value if the distance between the charged plates isdoubled. The best answer is choice B.

24. Choice A is the best answer. A charged particle moving through a magnetic field will feel a force that is both perpendicularto the orientation of the B field and perpendicular to the direction of motion. The particle enters moving from the left to theright. The B field is oriented into the page, so the direction of the force must be either upward or downward. This results ineither path I or path III being correct. Path II and deflection out of the page are not possible, so choices B and D areeliminated. According to the right hand rule, move your fingers in the direction of the motion, bend your fingers in thedirection of the field, and the direction of the force on a positive ion is in the direction of your thumb. This is upwards, sothe best trajectory is Path I. The best answer is choice A.

25. Choice A is the best answer. This is similar to the previous question, but involves the right hand rule. If you think that therewill be any motion of the wires at all, then cross out choice D. To be more specific, look at the magnetic field lines and howthey interact with the wires. The magnetic field lines due to the current in wire #1 are drawn (wire #2 also has field lines,they just aren't drawn.) The magnetic field due to wire #\ points out of the page where wire #2 is. The resulting force onwire#2 is to the right using the right-hand rule. If you repeat the process for wire #1, you will find it feels a force towardsthe left. Thus, the wires attract.

1

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Copyright © by TheBerkeley Review®

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150 MINI-TEST EXPLANATIONS

52-Question Electrostatics and Electromagnetism Practice Exam

I. Millikan Oil Drop Experiment

II. Lenz's Law

III. Cathode Ray Tube


IV. Velocity Selector

V. Faraday's Law


VI. Mass Spectroscopy

VII. Cyclotron


Electrostatics and Electromagnetism Exam Scoring Scale


42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3

(1-5)

(6 -10)

(11 -17)

(18 - 21)

(22 - 26)

(27-32)

(33 - 36)

(37 - 41)

(42 - 48)

(49 - 52)

Passage 1 (Questions 1 - 5)

In 1906, Robert Millikan devised an experiment thatallowed him to determine the charge of an electron. Aschematicof Millikan'sset-up is shown in Figure 1 below:

Oil Can

4B =\

o o

o o

!jS«

—

^^\ ^^

c ^^>

chargedoil drop

Figure 1

Two metal plates are connected by a series of batteries toform a capacitor. There is an electric field betweenthe plates.The metal plates are inside an insulated cylindrical containermade of clear glass.

Oil drops are introduced into the container through asmallhole in the top.The oil drops acquirea negative chargeas they pass through the nozzle of the oil can. Some of thedrops fall through a hole in the upper plate. By adjusting thevoltage between the plates, certain drops can be sloweddown, and in some cases suspended between them. Therelationship between the electric field between the plates andthe voltage across the plates is:

AV = EL

where E is theelectric fieldand L is the plateseparation.

Millikan chose oil because of its relatively low vaporpressureand high charge-holdingability.

1. Inorder for an oildrop of mass m, radius r, and chargedensity p, to be suspended between the plates, themagnitude and direction of the electric field must be:

a 3m8

B.

D.

4jtr3p3mg3mg

4jtr3 p

3mg

-; downward.

; downward.

; upward.

; upward.


2.

3.

4.

5.

If the voltage across the plates were reversed, whatwould happen to the oil drops between the plates?

A. They would accelerate upward.

B. They would accelerate downward.

C. They would move with a constant velocity upward.

D. They would move with a constant velocitydownward.

Suppose there are two oil droplets between the plates. Ifthe distance between the droplets is halved, whathappens to the force between the droplets?

A. It doubles.

B. It is reduced by half.

C. It is quadrupled.

D. It remains the same.

Suppose the original oil droplet were replaced with apositively charged one that had twice the charge andthreetimesthe massof the original droplet How wouldthe magnitude of the electric field have to be changedin order for the drop to remainsuspended?

A. No change is necessary.

B. It must be increased by a factor of —.J 3

C. It must be decreased by a factor of =•.3

D. It must be increased by a factor of *-.2

Millikan chose oil instead of water because:

A. oil is denser than water.

B. oil holds more charge than water.C. oil is less viscous than water.

D. oil is more compressible than water.



A metal bar with mass M, length L, uniform density p,and resistance R is placed on parallel rails that slope up witha gradually increasing tangent, as shown in Figure 1. Thebar, represented by the segment from X to Y, has a higherconductivity than the rest of the apparatus. Points a and frepresent connections from the base of the rails to a box thatcan serve as either a power supply or an ammeter. The railsfrom point e to point f and point a to point b are made of aconducting metal. The support structure from points e and bto the base and the base itself are insulators. For simplicity,assume the rails have negligible electrical resistance andfriction and that the resistance in the bar is negligible. Twoexperiments were conducted on this apparatus.

Experiment I

A uniform magnetic field of adjustable magnitude wasoriented upward, perpendicular to the base of the apparatus.The adjustable bar was then placed at the top of the rails andreleased. The induced current through the bar was measuredas it fell. The current induced in the bar was due to a

phenomenon known as Lenz's law. Whenever there is achange in magnetic flux through a loop, a current is inducedin the loop which fights this change in magnetic flux. ForExperiment I, the loop being studied is the cyclic path from f-a-»X-*Y — f.

Figure 1

ExperimentII

In the absence of any external magnetic field, a voltage wasapplied between points a and f to produce a current throughthe bar. The bar gradually climbed the rails until eventuallycoming to rest. It was assumed that the gravitational force onthe bar was equal to the magnetic force generated by thecurrent Different voltages were applied and the bar wassuspended at various points along the rails. The voltage andresting height were recorded for each trial.

In order to support the idea of Lenz's law, when theswitch between points a and f was open, the bar slid downthe rails unimpeded in Experiment I. In Experiment II, thebar did not move unless the voltage source was connected.


6.

7.

8.

9.

10.

In Experiment I, after the bar was released from the topof the rails (points e and b), what was true of the currentproduced as the bar slid down the rails?

A. No current is produced.

B. Direct current is produced that travelsfrom X to Y.C. Directcurrent is produced that travelsfrom Y to X.D. Alternating Current is produced.

A bar made of which of the following materials wouldNOT have worked in Experiment II?

A. Copper

B. Iron

C. Graphite

D. Calcium oxide

In Experiment II, when voltage is applied, the bar canbe suspended. Where does most of the current flowwhen this is observed?

A. From a-*X-+Y-*f-»a

B. From a-»b-»c--*c-*e-+f

C. From c-*X-*Y-*d-»c

D. From c-»b-*a-»f^e-*d

In Experiment I, releasing the bar from a higher pointon the rails will result in all of the following EXCEPT:

A. a greater induced current when the bar reaches thebottom of the rails.

B. a greater kinetic energy when the bar reaches thebottom of the rails.

C. an increase in the external magnetic field when thebar reaches the bottom of the rails.

D. an increase in the induced magnetic field when thebar reaches the bottom of the rails.

In Experiment II, starting when the bar is suspended,what will be observed when the switch between pointsa and f is opened?

A. The current through the bar remains the same, sothe bar remains stationary.

B. The current through the bar goes to zero, and thebar descends to the bottom of the rails.

C. The current through the bar decreases slightly andthe bar descends the rails.

D. The current through the bar increases slightly andthe bar ascends the rails.



Although solid state electronics has largely replacedvacuum tubes in the design and operation of computers andtelevisions, an important vacuum tube still very much in useis the cathode-ray tube, or CRT. For years, this was the mainkind of picture tube used for oscilloscopes and televisions.The basic design of a CRT is diagrammed inFigure 1.

Electromagnet

Anode

Cathode Electron beam

Figure 1.

Electrons are boiled off of a hot metal filament, thecathode. (To a good approximation, this thermionic emissionof electrons can be predicted, if we treat the electrons like themolecules of an ideal gas.) The ejected electrons are thenaccelerated by a high voltage applied to the anode. Theelectrons pass outof theanode through a small holeandentera region of electric and magnetic fields. The electric field isproduced by two parallel charged plates, whereas themagnetic field is produced by charge carrying coils near theplates. These twofieldsexert forces on the electron.

The electric force is given by: F= qE, where E is theelectricfield, and the magnetic force is given by: F = qvBi,where v is the velocity of the electrons after leaving theanode, and Bi is the component of the magnetic fieldperpendicular to the electron velocity.

The electric and magnetic fields are set up so that theforces on the electrons will point in opposite directions. Thehorizontal and vertical deflections of the electrons are

controlled by adjusting these two fields. (Some CRTs usetwo sets of parallel plates-one set to control horizontaldeflection and one set to control vertical deflection.)

Once the electrons leave the region of the fields, theycontinue until they strike the inside of the tube, which iscoated with a fluorescent substance. The tube glows at thepoint where it is struck by an electron, producing a tinyvisible spot Adjusting the voltage across the parallel platesor adjusting the magnetic field allows the point to be placedanywhere on the tube.

11. On occasion, the electrons are made to pass through theplates undeflected. If the electrons enter the regionbetween the plates with the same direction, but with anexcessive velocity:

A. they will be deflected upward.

B. they will be deflected downward.

C. they will remain undeflected.D. they will be turned back the way they came.


12. Suppose that protons were boiled off of the filamentinstead of electrons and that they, too, could beconsidered to obey the ideal gas law. Assume thetemperature of the cathode remains the same. Thekinetic energy of the ejected protons compared to thekinetic energyof the ejected electrons would be:

A. greater, due to the greater mass of theprotons.B. smaller, due to the greater mass of the protons.C. smaller, due to the different magnitudeof charge of

the protons.

D. the same.

13. Assume that an apparatus similar to the one shown inFigure 1 could accelerate protons rather thanelectrons,and at the same velocity. If we wanted the undeflectedprotons to follow the same path as the undeflectedelectrons in Figure 1, then:

A. the direction of the electric field would have to be

reversed.

B. the direction of the magnetic field would have tobe reversed.

C. the directions of both fields would have to be

reversed.

D. neither field would have to be changed in any way.

14. When electrons travel from the cathode to the target,passing through the electric and magnetic fields, theelectrons:

A. generate their own magnetic field, and this must betaken into account when selecting the magneticfield between the plates.

B. generate their own magnetic field, but this does nothave to be taken into account

C. do not generate their own magnetic field.

D. generate an electric field, and this must be takeninto account when choosing an electric fieldbetween the plates.

15. If an electron is observed to travel through a certainregion of space undeflected, what can be concluded?

A. There is definitely no electric field in that region,but perhaps there is a perpendicularmagneticfield.

B. There is definitely no magnetic field in that region,but perhaps there is an electric field.

C. No field of any kind is present.

D. Both an electric and magnetic field may be presentparallel to one another.


16. When the beam of electrons strikes the target they arebrought to a sudden stop, generating X-rays. Thewavelengths of these rays depend on the:

17.

I. accelerating potential of the electrons

II. target material.

III. speed of the X-rays.

A. I only

B. II only

C. I and II only

D. I and HI only

Instead of passing electrons through the plates, supposethat we now pass X-rays. In order to have the X-rayspass through the plates undeflected, we must:

A. change the electric field.

B. change the magnetic field.

C. change both fields.

D. Any changes to the fields will not affect the X-rays.



18. Assume that a cation enters an electric field oriented asshown below. Which path BEST represents the path ofthe ion?

I

E H H V if V ^f^V^II

III

A. Path I

B. Path II

C. Path III

D. Whether it is Path I, PathII, or Path III, depends onthe particle's mass and spin.

19. Which of the following will not be deflected by amagnetic field?

A. A moving electron

B. An orbiting proton

C. A fixed anion

D. An accelerating cation

20. Which of the following are needed to calculate the forcea magnetic field exerts on a moving electron?

I. The electron charge

II. The electron mass

III. The electron speed

IV. The magnetic field strength

A. I, II, and IV only

B. I, III, and IV only

C. II and III only


21. If an electron, proton, and neutron are moved throughthe same potential difference, what is true regarding themagnitudes of their potential energy changes?

A. APEeiectron > APEproton > APEneutronB. APEproton > APEeiectron > APEneutron

C. APEeiectron = APEproton > APEneutronD. APEeiectron > APEproton - APEneutron



In a velocity selector, charged particles are shotbetweentwo parallel plates. These plates, having different voltages,maintain a steady, uniform electric field between them. Thereis alsoa static magnet in the velocity selectorthat generates amagnetic field between the plates. The electric and magneticfields are aligned with opposing forces such that whencharged particles enter the velocity selector, they will bedeflected unless they havea specific velocity, v. Therefore, avelocity selector allows only moving ions with a selectvelocity to pass undeflected through the plates, regardless ofthe ion's mass (m) or charge (q). Upon adding a filter thatallows only undeflected particles to pass, the device forms abeam of ions that move at one speed.

The theory behind the velocity selector was formulatedby Lorentz, who specified the net force on a charged particlethat moves through an electric and magnetic field:

F = qE + qvBi,

Equation 1

where E is the electric field strength, Biis the magnetic fieldstrength perpendicular to the charge velocity v, and q is themagnitude of charge.

22. What is TRUE for an undeflected particle in a velocityselector?

I. The work done on the particle is non-zero.

II. The effects of electric fields and magnetic fieldscan treated as independent of one another.

HI. The particle could be neutrally charged.

A. II only

B. I and II only

C. I and III only

D. II and III only

23. A magnetic field pointing into the page (each field lineis represented by an V) will deflect upward all of thefollowing particles, EXCEPT a:

XXX XXX

B X X X XXX

XXX XXX

A. proton moving from the left to the right.

B. fluoride ion moving from the right to the left.

C. sodium ion moving from the left to the right.

D. titanium ion moving from the right to the left.


24. Which procedure would most easily send a negative ionthrough thefullyoperational velocity selector?

A. Reverse the direction of E only.

B. Reverse the direction of B only.

C. Reverse the directions of both B and E.

D. Change nothing in the system.

25. For a selected ion passing through the velocity selectorundeflected, how does its speed, v, relate to E, B, and

q?

A. v = £B

B. v = BE

C. v=£B

D. v= <§-E

26. When a charged particle is subjected to a Lorentz force,the particle's net:

A. force and acceleration depend upon its mass.

B. force depends upon its mass, but acceleration doesnot

C. acceleration depends upon its mass, but force doesnot.

D. force and acceleration are independent of mass.



In an experiment designed to study Faraday's Law, astudent observed the induced current generated uponmigrating three separate loops at uniform speed through auniform perpendicular magnetic field. The field was createdby placing a cylindrical magnet directly above a secondcylindrical magnet, such that the loop could be movedthrough the field between the poles, as shown in Figure 1.

Figure 1 Magnetic field used in the Experiment

In Experiment I, the student pushed through threeseparate copper loops of varying dimensions all in a way thatthe area within the loop was parallel with the surface of themagnets. The three loops have the following dimensions.

Loop C: rectangular with edges of6cm and 8cm. A=48 cm2Loop D: square with edges of7cm. A=49cm2Loop E: circular with a radius of4cm. A=50 cm2

The student pushed the loops through at constantvelocity, despite the increase in opposing force as the loopentered and exited the field. Each loop was attached to agalvanometer, used to measure the induced electromotiveforce, e. The emfcan be calculated using Equation 1:

e = - d(BA cos $)/dlEquation 1

where e represents the emf, B is the strength of theexternal field, A is the area in side the loop, and <j) is theangle between the field vectors and the vector perpendicularto the surface of the area within the loop.

In Experiment II, the student repeated the steps ofExperiment I, but used loops made of tin wire instead ofcopper.

27. In which experiment did the student observe theGREATEST electromotive force?

A. In Experiment I, because copper is a betterconductor than tin.

B. In Experiment II, because copper is a betterconductor than tin.

C. In Experiment I, because tin is a better conductorthan copper.

D. Both experiments produced the same emf


28. Asa conducting loop firstenters a magnetic field, it:

A. feels a force that pulls it into the field.B. experiencesa torque that rotates it by 90°.C. speeds up with constant acceleration.D. experiences an induced current.

29.

30.

31.

32.

Which graph accurately reflects the current associatedwith Loop E in the experiment?

A. < ^^I A Time (s)

g 0-fc3

u

B. <

3

u

c. <

3

u

D. <

3

u

r\ r\ Time (s)

Time (s)

Time (s)

What is true for Loop C in Experiment I?

A. It experiences the same emf whether it enters the Bfield with the 6cm side first or the 8cm side first

B. It experiences a greater emf when it enters the Bfield with the 6cm side first than the 8cm side first

C. It experiences a greater emf when it enters the Bfield with the 8cm side first than the 6cm side first

D. It only experiences an emf if it enters the B fieldwith the 6cm side first.

Which change will NOT increase the magnitude of thecurrent induced into a loop entering a B field?

A. Increasing the area inside of the loop

B. Moving the loop into the field at a faster rate

C. Moving the two magnets closer to one another

D. Using a loop made from a thinner wire

If Loop C is moved from the right side of the magneticfield in Figure 1, through the field and then on to theleft side, what type of current does it feel upon exitingthe field while moving to the left?

A. A clockwise alternating current

B. A counterclockwise alternating current

C. A clockwise direct current

D. A counterclockwise direct current


Questions 33 through 36areNOT based on a descriptivepassage.

33.

34.

35.

Which of the following arrows BEST represents thepathway of a negatively charged particle moving lefttoright through a magnetic field pointing into thepage?A. B.

C. D.

If two wires have current in the direction as shown,

what is true of the force between the wires?

•twire I wire II

A. The wires repel one another.

B. The wires attract one another.

C. The wires exert no force on one another.

D. The wires experience a force that oscillatesbetween attraction and repulsion.

When a charged particle is added to the followingelectric field, it will accelerate. Which particle willexperience the greatest average magnitude ofacceleration when added to the field at point A?

A. A negatively charged particle with low mass

B. A positively charged particle with low mass

C. A negatively charged particle with high mass

D. A positively charged particle with high mass


36. Which of the following graphs BEST represents therelationship betweenthe magnetic field strength and thearea of the coil in a system where e0 = BAoo, if themaximum voltage and coil rotation frequency are heldfixed?

A. B.

B

D.

B



Mass spectrometry works by correlating the curvature ofa charged particle moving through a perpendicular magneticfield to the mass of the particle. Figure 1 shows a basicschematic of a simple mass spectrometer. The apparatus iscomprised of two separate regions combined in sequence. Anelectric field accelerates a cationic particle in Region 1.Figure 1 shows the particle accelerating to the right. Thedouble filter ensures that the particle is going straight when itenters Region 2, where it is subjected to a perpendicularmagnetic field. This results in a radial force exerted on themoving charged particle. As such, the particle travels in asemicircle, before striking the detector.

Region 1 Region 2X X X X X X

X X X X X X

XXX

XXX

XXX XXX

magnetic field into page

double filter

Figure 1 Simple schematic of a mass spectrometer

The speed of the particle may be controlled by a velocityselector, which is comprised of a magnetic field and electricfield, both of which are perpendicular to the motion of theparticle. The two fields are aligned so that the forces opposeone another. The particle will continue in a straight path onlyif qE = qvB. This means that the particles that pass into thedeflecting region of a mass spectrometer have a velocity ofE/B. The mass spectrometer in Figure 1 does not have avelocity selector. Once in Region 2, the radius of the arc canbe used to determine the mass to charge ratio of the particle.The radiusof the arc in Region 2 equates to the mass, charge,magnetic field, and velocity according to Equation 1.

r_mv

Bq

Equation 1

37. Assuming that the particles enter Region 2 with thesame velocity, which of the following cationic particleswill have the GREATEST radius of curvature, r, inRegion 2?

40Ca2+

7Li+24Mg2+

A.

B.

C.

D. 23N£


38. All of the following changes will lessen thedisplacement between the point where particles enterRegion 2 and the point where particles collide with thedetector plate EXCEPT:

A. decreasing the accelerating voltage in Region 1.B. using a dication in lieu of a monocation.

C. using an isotope with more neutrons.D. increasing the magnitude of the magnetic field in

Region 2.

39. The double filter between Regions 1 and 2 ensures that:

A. a uniform ion beam, perpendicular to the plates ofthe filter, is formed.

B. a beam of perpendicular particles with the samevelocity is formed.

C. the beam consists of only cations.D. the beam consists of only singly ionized particles.

40.

41.

Doubling both the magnetic field in the velocityselector and in the deflecting zone of a massspectrometer would have what affect on the radius inthe deflecting region?

A. The radius would decrease by a factor of 4.B. The radius would decrease by a factor of 2.

C. The radius would decrease by a factor of \fl.D. The radius would not change.

To accelerate anion particles from left to right inRegion 1, and deflect those particles out of the page inRegion 2, how must the E and B fields be aligned inFigure 1?

A. Anode plate left, cathode plate right, B field upB. Anode plate left, cathode plate right, B field downC. Anode plate right, cathode plate left, B field upD. Anode plate right, cathode plate left, B field down



The cyclotron (Figure 1) is a device used by physicists toaccelerate ionsto high speeds and high kinetic energies. Thecyclotron consists of two semicircular regions Si and S2 ofthe same radius R with a small gap d between them. S\ andS2 are connected to an alternating voltage source ofamplitude V.

Ionized particlesenter here

Gap(d)(

Ionized particlesleave here

Figure 1

Ions are injected near the center of the cyclotron. Theyare accelerated across the gap by an electric field until theyenter the magnetic field region. Once there, they move in acircular path due to a uniform magnetic field which pointsinto the page. As the ions circulate, they are sped up by theelectric field that exists in the gap. As an ion increases itsspeed, it spirals outward. When the radius of the orbitmatches the radius of the cyclotron, the ion is ejected througha small outlet hole.

The frequency at which an ion circles is called thecyclotronfrequency and is given by:

f- qb2 31 m

where B is the magnetic field, q is the charge on ion, and m isthe ion's mass.

42. How many orbits are required for an ion to gain a totalkinetic energy of6 x 10~14 J, if the ion gains 1x 10"16J every time it crosses the gap?

A. 2(lxl0-16)(2xl0-14)B. (Ixl0-16) + [2(2xl0-14)]C. (6xl0-14)+ [2(lxl0-16)]D. 2*[(1 xl0-16)(2xl0-14)]

43. If the cyclotron frequency is 10 MHz, how long will ittake an ion to complete 200 orbits?

A. 0.2 ^s

B. 2/iS

C. 20 ps

D. 200 ps


44.

45.

46.

What is the kinetic energy of an ion of mass m andcharge q as it is ejected from the cyclotron?

qBA.

B.

C.

D.

2Rm

IqBR2

qBR

2m

cfBV2m

Which of the following statements would NOT be true,if the voltage amplitude across the gap were doubled?

A. An ion would spiral out in less time and be ejectedsooner.

B. An ion would be emitted with a greater kineticenergy.

C. The cyclotron frequency would remain unchanged.

D. The magnetic field would remain unchanged.

In comparing the work done on an ion by the electricfield and the magnetic field in a cyclotron, we find that:

A. the electric field does more work than the magneticfield.

B. the magnetic field does more work than the electricfield.

C. the electric field and the magnetic field do thesame amount of work.

D. the neither the electric field nor the magnetic fielddoes any work.

47. How do the cyclotron frequencies for singly ionizedH2 (m = 2) and singly ionized He (m = 4) compare?

48.

A. H2 2B. fH2 = fHeC. % = 2fHeD. fH2 = 4fHe

The cyclotron frequency of a neutron must be:

A. the same as the frequency of a hydrogen nucleus.

B. the same as the frequency of an ionized deuteron.

C. the same as the frequency of an ionized heliumnucleus.

D. zero.


Questions 49 through 52 areNOT based on a descriptivepassage.

49. Which of the following changes does NOT decrease thework required to move a negatively charged particleagainst an electrostatic force, from a positively chargedplate to a negatively charged one?

I. Increasing the battery voltage

II. Decreasing the particle charge

III. Increasing the particle mass

A. I and II only

B. II and III only

C. I and III only

D. I, II, and III

50.

51.

In which of the following ways does a magnetic forceNOT differ from an electric force?

A. A magnetic force varies with particle speed, whileand electric force does not.

B. A magnetic force is perpendicular to its field, whilean electric force is aligned with its field.

C. A moving particle can only be turned by amagnetic force and not by an electric force.

D. A charged particle can experience zero magneticforce in the middle of a uniform magnetic field,while a charged particle in the middle of a uniformelectric field will experience an electric force.

What is NOT true when moving a positively chargedparticle in the following electric field?

A. No work is done when moving from Point a toPoint b.

B. Point b is at a higher potential than Point c.

C. Natural flow for the particle is from Point b toPoint c.

D. Point b is at a lower potential than Point a.


52. When a velocity selector is fully operational, it allowsonly an ion with a specific velocity to pass through theplates undeflected. Which orientation of electric andmagnetic fields would allow a positive ion to passundeflected when traveling from left to right? (Note:The electric field points either up or down in thepictures. The magnetic field points either into "x" or outof "•" the page.)

A.

B.

C.

• y r . \ r • \ f . \ f. y'•

X I

X

X

IX i

X

X

X

X

X

X

X

X

IX

X

X

X

X

x y

X

X

rx y

X

X

rx y

X

X

f x y

X

X

X

X

'X

D. All of the fields will work.

1. A 2. B 3. C 4. D 5. B 6. B

7. D 8. A 9. C 10. B 11. B 12. D

13. D 14. B 15. C 16. C 17. D 18. C

19. C 20. B 21. C 22. D 23. D 24. D

25. A 26. C 27. D 28. D 29. A 30. C

31. D 32. C 33. D 34. A 35. A 36. A

37. D 38. C 39. A 40. A 41. D 42. C

43. C 44. D 45. B 46. A 47. C 48. D

49. C 50. C 51. D 52. C

YOU ARE DONE.

Answers to 52-Question Electrostatics and Magnetism Exam

Passage I (Questions 1 - 5) Millikan Oil Drop Experiment

1. Choice A is the best answer. We must consider both magnitude and direction. First, let's consider magnitude. Thinkingabout the basic relationship between the required electric field and the mass of the drop, we see that they should be directlyproportional to each other. That is, if the drop is more massive, a stronger electric field is required. This rules outchoices Band D. To decide direction, remember that gravity points downwards, which requires the electric force to point upwards.Since the charges being supported are electrons, and since electrons are always forced in the direction opposite that of theelectric field, the electric field must point down. You must keep in mind that convention for electric fields is to draw themaccording to the way a positive chargewould naturally flow. The best answer is choice A.

2. Choice B is the best answer. In order for the drops to be suspended, the electric force must point upward to oppose thegravitational force downward. By reversing the voltage, the direction of the electric field—and therefore the direction of theelectric force-would also be reversed. Now the electric force and the gravitational force act in the same direction(downward), so the droplets will be accelerated downward, and with a magnitude of acceleration greater than g. The bestanswer is choice B.

3. Choice C is the best answer. This question is asking us to consider the force between two charged particles. The coulombforce between two charged particles is:

F = k^1r2

This force is inversely proportional to r2. If the distance r separating the two charges is reduced by one-half, then the forcebetween the charges is quadrupled. The best answer is choice C.

4. Choice D is the best answer. The question asks about the magnitude of the forces, so we can ignore that the negativelycharged oil droplet has been replaced by a positively charged species. We simply need to consider the magnitudes of theelectric force and the gravitational force. We know that qE = mg, and that:

E = ^q

in order for the oil drop to remain suspended. If m increases by a factor of 3 and q increases by a factor of 2, then the newelectric field would be:

E = ^Hi = 3^2q 2

where E' is the new electric field, and E is the original electric field. At the very least on this question, you could haveeliminated choices B and C, as they are the exact same answer, just worded differently. The best answer is choice D.

5. Choice B is the best answer. Oil floats on water, telling us that oil is less dense than water. We can rule out choice A. Oilpours more slowly than water, which you know from personal experience, so oil is more viscous than water. This eliminateschoice C. In the passage, vve read that Millikan chose oil because it has a relatively low vapor pressure (i.e., it does notevaporate quickly), and because oil drops can hold charge well. Although we know nothing about choice D, because thepassage addressed oil's ability to hold charges, choice B is the most likely answer the test-writer wanted you to pick. Thebest answer is choice B.

Passage II (Questions 6 -10) Lenz's Law

6. Choice B is the best answer. In Experiment I, the external magnetic field points vertically upward, so the direction ofinduced current can be obtained using the right hand rule. Before considering that, we should consider which choices can beeliminated without knowing the direction of the current. The bar slides down the rails, so there will be a current. Thiseliminates choice A. Because the bar slides down the entire time, the current will remain in the same direction, making itdirect current. This eliminates choice D. When the bar slides down the rails, the magnetic flux through the loop created bythe rails and the bar decreases, because the area inside of the loop decreases. Lenz's law states that a current will be inducedin this loop to fight the change in flux, so current will definitely be induced. Now we need to determine the direction of thecurrent. If you point your thumb in the direction the bar is sliding (angular and down) and put your index finger in thedirection of the magnetic field (up), then your middle finger points from X to Y, so that is the direction of current. Sincechoice B is a direct current going from X to Y, it is the best answer. The best answer is choice B.


7.

10.

Choice Dis the best answer. Any material which conducts electricity would work in Experiment II. Since copper and ironare pure metals, they both conduct electricity. This eliminates choices Aand B. Neither graphite nor calcium oxide are puremetals, so we cannot automatically say that either of them conduct Although graphite is a nonmetal, it has a crystallinestructure that allows for its electrons in the Jt-level to have some freedom to move, so it actually conducts. Taking this intoconsideration, one should see that the only substance thatwould nothaveworked in Experiment II is calcium oxide, becauseit doesn'tconduct electricity. Salts, suchas oxides, do notconduct electricity unless they are dissociated in water. The bestanswer is choice D.

Choice A is the best answer. The current can only flow through a region where it conducts electricity. It is stated in thepassage that thesegment from pointe to pointb is for support and is made of an insulating material. This means thatcurrentcannot pass through points b, c, d, and e. This eliminates all of the choices except for choice A. If you spent time trying todetermine the direction (whether it was a-to-X-to-Y-to-f-to-a or a-to-f-to-Y-to-X-to-a), you wasted time. Occasionally aquestion may look complicated but it is in fact simple to find the best answer. The best answer is choice A.

Choice C is the best answer. In Experiment I, if the barwas released from a higher point on the rails, it would definitelyhave more kinetic energy upon reaching the bottom. In addition, there would be a larger induced current and thus inducedmagnetic field due to the fact that there was a greater change in flux through the loop than if the bar had been let go from alesser height. On the other hand, since there is no relationship between the external magnetic field and the height of the bar,it would be unchanged. The best answer is choice C.

Choice B is the best answer. In Experiment II, before the switch between a and f is opened, there is current flowingbetween points a and f that is in the oppositedirection of the currentbetween points X and Y. Once the switch is opened,thecurrent stops and there is no longer a force suspending the bar. Without any force holding the bar up, it will slide down therails due to gravity, eventually reaching the bottom. The best answer is choice B.

Passage III (Questions 11 - 17) Cathode Rav Tube

11. Choice B is the best answer. If the undeflected electrons have an increased magnitudeof velocity compared to what theyhad when they passed through in a straight line, then the magnetic force is greater than the electric force. The electrons willbe deflected in the direction of the magnetic force, so they will be deflected downward. Just as motion is necessary togenerate a magnetic field but not an electric field, the magnetic force is affected by velocity, while the electric force is not.The best answer is choice B.

12. Choice D is the best answer. According to the ideal gas law, the kinetic energy of particles is related to temperatureby:

KE = 2-kT2

where k is the Boltzmann's constant. This says that the kinetic energy of a particle depends only on its temperature, and noton its mass or charge. Thus, the kinetic energy of the protons is the same as the kinetic energy of the electrons. The bestanswer is choice D.

13. Choice D is the best answer. This question helps to reiterate the point that MCAT questions require you to choose the bestanswer and not necessarily the correct answer. Protons and electrons have opposite charges. This means that if electronsmove downward in the electric field, protons would move upward. If electrons move upward in the magnetic field, protonswould move downward. Thus, the field directions still oppose each another and therefore would not have to be changed. Inorder for the protons to travel through the plates undeflected, the electric and magnetic forces must still be equal inmagnitude:

qE = qvB or E = vB

This relationship does not depend on the mass or the charge of the particle involved, only on the speed of the particles.Provided we do not change the speed of the protons compared to the speed of the electrons, we do not even need to changethe magnitudes of the fields. We could reverse the direction of both fields (choice C), but it is not necessary, which makeschoice D a better answer than choice C. The best answer is choice D.

14. Choice B is the best answer. Electrons in motion definitely generate a magnetic field, so choice C is eliminatedimmediately. The electric field they generate does not affect itself, so choice D is eliminated. Although an electron is amoving charged particle (which produces a magnetic field), a charged particle never responds to its own fields, electric ormagnetic. The magnetic field it generates interacts with the external magnetic field, thereby causing a force that can beanalyzed using the right-hand rule, but that is not a response to its own magnetic field. The best answer is choice B.

Copyright ©byTheBerkeley Review® 163 REVIEW EXAM EXPLANATIONS

15. Choice C is the best answer.Theelectron is undeflected, which can beexplained by the absence of a netforce acting on theelectron. This is the case if (a) the electric and magnetic forces areof equal magnitude but opposite direction or (b) there areno electric and magnetic forces acting on the electron. Because there could be both an electric field and magnetic field, asopposed to just one or the other, choices A and B are eliminated. In order to produce a force on a moving electron, themagnetic field must be aligned perpendicular to the velocity of the particle, resulting in a force that ismutually perpendicularto both the magnetic field and the velocity. If the electron moves in the same direction as the magnetic field, it will feel nomagnetic force. This means that the force exerted by the magnetic field is perpendicular to the magnetic field itself; so if themagnetic and electric fields are parallel, then the magnetic force is perpendicular to the electric force. This would not allowthe forces to cancel, so the electron would be deflected as it traveled. Choice D is not possible. The best answer is that thereis no field of any kind present. The best answer is choiceC.

16. Choice C is the best answer. X-rays can be produced one of two ways. When the electrons are brought to a sudden stop,they lose all of their kinetic energy. Conservation of energy says this energy must go somewhere, and where it goes is intothe production of x-rays. In addition, some of the electrons that strike the target may interact with the atoms of the target, byknocking electrons out of their inner-shell orbits. That leaves an unfilled space in an inner-shell orbit, which is filled by anouter-shell electron making a transition. This process releases energy in the form of x-rays called characteristic x-rays, sincetheir wavelengths depend on the electron transitions made, and these depend on the atoms involved. This means that thewavelength (and thus energy) of the x-rays depend on the accelerating potential of the electrons (which affects the electron'sspeed) and on the target material (and their electronic energy levels). Options I and II are valid. The speed of the x-raysdepends on the medium and has nothing to do with the wavelength. Option III is invalid, confirming that choice C is the bestanswer. The best answer is choice C.

17. Choice D is the best answer. X-rays are electromagnetic radiation, and as such they have no charge. They will notexperience any force from either the electric or magnetic fields as they pass through them. Any change in the surroundingfields would have no impact on the propagation axis of an x-ray. The best answer is choice D.

Questions 18-21 Not Based on a Descriptive Passage

18. Choice C is the best answer. An electric field affects a charged particle by accelerating it in the direction of the field, if theparticle is positively charged. Remember that all definitions in physics are based on positive charge. The particle isaccelerated (deflected) by the electric field in the downward direction, with a force of F = qE. The path that shows constantvelocity in the lateral direction coupled with acceleration in the downward direction is Path III. The best answer is choice C.

19. Choice C is the best answer. Any moving charged particle will be deflected by a magnetic field as long as the motion is notdirectly in line with the field. Considering direction is not mentioned in the question, the key feature is motion. Choice C, afixed (i.e., not moving) particle will not be deflected by a magnetic field. The best answer is choice C.

20. Choice B is the best answer. For this question we need to apply Paring the List or Multiple Concepts. First, consider anitem that is easiest for you to grasp. If you know that electron charge is relevant, then Statement I is true and choice C is not;cross choice C out. Since Statement IV is in all remaining choices, there no need to consider it. Since mass usually onlyshows up in a gravitational force, Statement II must not be relevant. This rules out choices A and D, leaving only choice B.The best answer is choice B.

21. Choice C is the best answer. To solve this question, pick two particles whose energy relationship you know, and use thisinformation to rule out incorrect choices. For example, if you know neutrons experience no change in energy when movingthrough a potential difference, then choice D is incorrect. The remaining choices ask you to compare the energy change foran electron and proton. Since APE = qV and the two particles have the same magnitude of charge, they must have the samepotential energy change. The best answer is choice C.

Passage IV (Questions 22 - 26) Velocity Selector

22. Choice D is the best answer. Work is defined as force x distance, or by the equation W = Fid. A selected ion is not forced(deflected) in any direction, so there is no force along the direction in which it travels (Ft = 0). The net work done on theundeflected particle is 0. This makes statement I invalid. The whole concept behind the velocity selector is that the effects ofthe magnetic and electric fields are independent of each other (as shown in Equation 1), so the two fields can be set to oppose(and algebraically cancel) one another. Statement II is valid, as the remaining choices (A and D) state. If a neutral chargewere to enter the velocity selector, it would pass undeflected. Remember, both electric and magnetic forces require a netcharge on the forced particle. This validates Statement III. The best answer is choice D.

Copyright ©byThe Berkeley Review** 164 REVIEW EXAM EXPLANATIONS

23. Choice D is the best answer. A charged particle moving through a perpendicular magnetic field is subjected to a force thatis both perpendicular to the orientation of the Bfield and perpendicular to the direction of motion. If a positive particle entersthe magnetic field moving from the left to the right, and the Bfield is into the page, the direction of the force is upward. Youcan confirm this by using the right-hand rule. According to the right-hand rule, if you move your fingers in the direction ofthe motion and bend your fingers in the direction of the field, the direction of the force on a positive ion will be in thedirection of your thumb. Both a proton and sodium cation are positively charged, so choices A and C are valid statementsand must be eliminated. A negatively charged ion, such as fluoride, would experience a force in the opposite direction. Thatmeans that an anion must be moving right to left to experience an upward force in the field as shown. This makes choice Bvalid, so it is eliminated. It is for choice D that the force will be downward. Titanium, ending in "-ium," is a cationic speciesthat will deflect downward when traveling from right to left. The best answer is choice D.

24. Choice D is the best answer. Recall that the forces (and fields) are defined for positive charges. A negative charge will bedeflected in the direction opposite that of a positive charge by both a magnetic and electric field. For the velocity selector towork, the forces must oppose one another. The original design (for the positive ion) has the fields generating opposingforces. If the particle is negative, the directions of the forces generated by the fields are reversed, but they will still beopposed. Eliminate choices A and B. Changing only one field would create a scenario where the forces sum together ratherthan cancel each other out. The best answer is either changing both fields or changing nothing. While a sound argument canbe made for the validity of choice C, it is not the easiest route to the solution. Choice D is a better option, in that it is better todo nothing than a double-reverse. This may seem like quibbling, but the MCAT does require that you choose the bestanswer, not the one and only correct answer. There is a big difference between the best answer and an absolutely correctanswer. The best answer is choice D.

25. Choice A is the best answer. Paragraph 1 of the passage states that mass and charge are not involved in the deflection (ornet force), so it must be true that velocity will not vary with the mass or charge. Choices C and D are unlikely to be the bestanswer. The manipulation of Equation 1 to describe a situation where the electric and magnetic forces are opposed gives us:

F = qvB-qE

Therefore, if F = 0, then qvB = qE. Canceling out the q from the equation yields: vB = E, which tells us that the speed of theparticle can be found using: v = E/B. The best answer is choice A.

26. Choice C is the best answer. Equation 1 does not contain mass, so the net force does not depend on the mass. Thiseliminates choices A and B. Because force is mass times acceleration, the acceleration equals the force divided by the mass;therefore, acceleration depends on the mass. The best answer is choice C.

Passage V (Questions 27 - 32) Faraday's Law

27. Choice D is the best answer. The electromotive force results from the induction of charge movement caused by the changein the magnetic flux. This is independent of the material of which the wire loop is made. The materials have differentelectrical resistances which impacts the current, but that has no impact on the emf. The best answer is choice D.

28. Choice D is the best answer. As a conducting loop first enters a magnetic field, it experiences a change in the magnetic flux(the magnetic field within the loop). This induces a current in the loop, making choice D the best answer. By generatingcurrent, the energy of the moving loop is being converted into current, so energy is drained as it enters the field. As such, itfeels a repulsive force as it enters the field, not an attractive force pulling it into the field. Choice A is eliminated. A torquecan be felt if the loop had an existing magnetic field (which would be possible if a current were being pushed through it byan external emf source.) That is not the case here, so a torque causing it to rotate 90° is not viable. Choice B is eliminated.Because of the resistance the loop feels upon entering the field, it actually slows down as it enters the field. This eliminateschoice C. The best answer is choice D.

29. Choice A is the best answer. As in all cases, as the loop enters the field, the magnetic flux is increasing and as the loopexits the field, the flux is decreasing. This means that the current moves in opposite directions when the loop is enteringversus when the loop is exiting the magnetic field. This eliminates choices B and D. To answer the question from this point,we must determine whether the current is constant or varies as the loop enters and exits the field. In the case of Loop E, theloop is circular. Unlike with a rectangular loop, the area of the loop entering the field (and thus the magnitude of themagnetic flux) changes as the loop enters the field. As the circular loops enters the field, the magnetic flux increases at anincreasing rate until the loop is half way in, after which the magnetic flux increases at a decreasing rate. This is bestdescribed in the graph in choice A. Even if that doesn't seem completely clear, choice C cannot be correct, because themagnetic flux does not increase at a uniform rate, so current cannot flow at a constant rate during the induction periods. Thebest answer is choice A.

Copyright© by The Berkeley Review0 165 REVIEW EXAM EXPLANATIONS

30. Choice C is the best answer. Loop C is rectangular, so it has two possible orientations for entering the field such that theplane ofthe loop is perpendicular to the magnetic field. The greatest emf is felt with the greatest change in flux. When therectangular loop enters long side (8 cm side) first, it experiences a larger change in flux than when it enters short side (6 cmside) first, assuming it enters with the same velocity in each case. This is simply due to the fact it is wider. This eliminateschoicesA and B and leaves choice C as the likely best answer. It feels an emfno matter which side enters first, so choice Dis eliminated. The best answer is choice C.

31. Choice D is the best answer. The magnitude of the current induced into the loopdepends on the rate at which the magneticflux changes. The greater the rate at which the magnetic flux changes, the greater the induced current. Increasing the areainside of the loop increases the magnetic flux as it enters a magnetic field, so choice A is eliminated. Moving the loop intothe field at a faster rate will increase the rate at which the magnetic flux increases, so the current is greater and thus choice Bis eliminated. If the two magnets are moved closer to one another, the magnitude of the external B field will increase,resulting in a greater magnetic flux within the loop. This will increase the current, which eliminates choices C. Using a loopof thinner wire causes a greater resistance. Although the magnetic flux does not change, and therefore the induced emfdoesnot change, the current is reduced. The best answer is choice D.

32. Choice C is the best answer. If Loop C is exiting the field moving to the left, the magnetic flux is decreasing the entire timethe loop is leaving the field, so the current will travel in the samedirection during the entire time it is exiting the field. Thismeans the current is a direct current, which eliminates choices A and B. Those two choices should have been eliminatedbased on the idea that alternating current goes both directions, so it cannot be assigned a clockwise or clockwise orientation.To determine the direction of the current, we can address the question in one of two ways. The first way is to apply Lenz'slaw. Lenz's law states that as the magnetic flux within a loop changes, a current will be induced in the loop that opposes thechange in the magnetic flux. In this case, the external magnetic field points down, so upon exiting the region inside of theloops experiences a reduction in the magnitude of the magnetic field pointing down. As such, the current induced in the loopwill compensate for the loss by creating a magnetic field pointing down in the region within the loop. This is done bycreating a clockwise current in the loop. Choice C is the best answer. This can be verified using the right-hand rule. As theloop leaves the magnetic field, only the back edge of the loop is completely immersed in the field. Considering motion of theconducting rod to be to the left and the orientation of the magnetic field to be into the page, the force on positively chargedparticles in the back edge of the loop will be downward (according to the right hand rule). That push will cause the positivecharges to flow in a cyclic fashion, down the back edge, to the left in the lower edge, up the front edge, and to the right in thetop edge. This describes a clockwise current in the loop, so choice C is verified as the best answer. The best answer ischoice C.


33. Choice D is the best answer. The right hand rule describes the pathway of a cationic particle moving through a magneticfield. In Figure 1, the pathway of a cationic particle through a magnetic field into the page (represented by an x) is shown.An anion will show the same shape pathway as a cation, but with deflection in the opposite direction. This is best representedby choice D. Choice C is too drastic (and non-circular) to be an observed pathway. The best answer is choice D.

34. Choice A is the best answer. Wires with currents moving in the same direction attract, while wires carrying oppositecurrents repel. Because we are only considering a direct current in each wire, there will be a force that maintains the samedirection. This rules out choices C and D. This phenomenon is caused by the magnetic force on each wire due to the B fieldcaused by the current in the parallel wire. Looking at wire 1, it creates a B field into the page at the point of wire 2. Using F= U co B and the right hand rule, the force on wire 2 due to wire 1 is found to be to the right. Similarly, the force on wire 1

due to wire 2 is found to be to the left. Thus parallel wires carrying currents in opposite directions repel one another

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35. Choice A is the best answer. First off, you need to consider that the force exerted on the particle depends on the magnitudeof the charge and the strength of the field. Assuming that the magnitude of the charge is the same in all of the choices, weneed to consider the field's strength. The lines separate on the right side, so the field is stronger on the leftside than the rightside. As a particle moves left to right, the strength of the electric force is diminishing. Conversely, as a particle moves rightto left, the strength of the electric force is increasing. Field lines are drawn the way a positive charge would migrate, sopositive charges would experience a decreasing force over time. On the other hand, negative charges would feel an increasein field strength, so negative charges will feel a greater average acceleration over the migration. Choices B and D areeliminated. To decide between choices A and C, we need to consider F = ma. For a given force, as mass increases, theacceleration of the particle will decrease. To generate the greatest acceleration, we need the lightest particle. The best answeris the light, negative particle. The best answer is choice A.

36. Choice A is the best answer. This may very well be an extremely intimidating topic, in which case many of us let panicinfluence our thought process. Even though we may have little understanding of the subject matter, sometimes a question isactually asking us to demonstrate our analysis skills and not our knowledge. The relevant equation is given in the questionas:

e0= BAco

The equation states that the magnetic field, B, is inversely proportional to cross-sectional area of the loop, A. That is, as Aincreases, B must decrease, which eliminates choices C and D. This particular plot is choice A, an asymptotic relationship.Many people incorrectly think choice B is the graph of an inversely proportional relationship, but it is not. Choice B is alinear relationship, with a negative slope. The best answer is choice A.

Passage VI (Questions 37 - 41) Mass Spectrometer

37. Choice D is the best answer. The biggest arc results from the particle with greatest mass-to-charge ratio, assuming all ofthe particles have the same velocity and the magnetic field magnitude is equal in each trial. This is because as massincreases, the radius of curvature increases and as the charge decreases, the radius of curvature increases. The greatest mass-

to-charge ratio isobserved with 23Na+, The bestanswer ischoice D.

38. Choice C is the best answer. Reduced displacement at the detector plate results from a smaller radius in Region 2. Theradius will be reduced when mass decreases, velocity decreases, charge increases, or the magnetic field increases.Decreasing the accelerating voltage in Region 1 reduces the velocity, so the radius of the arc is reduced. Choice A decreasesthe displacement, so it is eliminated. Using a dication in lieu of a monocation increases the magnitude of charge, so theradius of the arc is reduced. Choice B decreases the displacement, so it is eliminated. Using an isotope with more neutronsincreases the mass, so the radius of the arc is increased. Choice C increases the displacement, so it is the best answer.Increasing the magnetic field in Region 2 increases the deflecting force, so the radius of the arc is reduced. Choice Ddecreases the displacement, so it is eliminated. The best answer is choice C.

39. Choice A is the best answer. The double filter serves to ensure that the beam is perpendicular to the surfaces of both filters.In order to pass through both filters, the particle beam must be perpendicular to both plates. The filter does not distinguishparticles by velocity, charge, or size. This eliminates choices B, C, and D. This means that the beam is perpendicular to theplates, but it says nothing about the charges or velocities of the particles. The best answer is choice A.

40. Choice A is the best answer. The magnetic field in the velocity selector is applied to produce a force that counteracts theforce caused by the electric field in the velocity selector. Because the electric field remains the same, the increased magneticfield results in a decreased velocity associated with a cation that is allowed to pass (v = E/B, so as B goes up, v must godown). If the selector's magnetic field is doubled, then selected particle velocity is cut in half. If the velocity is cut in half,then the radius of the arc in Region 2 is also cut in half. Because the B field in Region 2 is doubled, the force deflecting theparticle is doubled, so the radius of the arc is again cut in half. This means that the overall affect of doubling the twomagnetic fields is to decrease the radius of the curvature by a factor of four. The best answer is choice A.

41. Choice D is the best answer. Anionic particles migrate from the cathode to the anode in uniform electric fields, so in orderto accelerate the anionic particles from the left to the right, the anode must be on the right side and the cathode on the leftside as in Region 1. This eliminates choices A and B. According to the right hand rule, if a positively charged particle weremoving from left to right through a B field point pointing up, then the magnetic force acting on the particle would point outof the page. This means that to deflect a cationic particle out of the page, the magnetic field must point up in the plane of thepage towards the top of the page. However, the question asks for an anionic particle, therefore the B field must haveopposite orientation, thus it points downward in the page. The best answer is choice D.


Passage VII (Questions 42 - 48) Cyclotron

42. Choice C isthebest answer. How many orbits are required for an ion to gain a total kinetic energy of6 x lO"1^ J, ifitgains1x 10-1^ J every time it crosses the gap? The ion crosses the gap twice per orbit, so itgains 2 x 10_1^J with every orbit. Thenumber of orbits is given by:

6x 10'14JNumber =


Total energy

Energy gained in 1 orbitor N =

2(1 x 10"16 j)

43. Choice C is the best answer. If the cyclotron frequency is 10 MHz, how long will it take an ion to complete 200 orbits?According to the passage, the cyclotronfrequency is the frequency at which an ion circles. To calculate the time required for1 orbit (the period of orbit), we use:

T = I = 1

f 10 x 106 Hz= 1 x 10-7 s

The period is the time to completeone orbit, so we must multiply T by 200 to determine the time required to complete 200orbits. 1x lO'7 x200 =200 x 10"7 =20 x 10"6 =20/<m. The best answer ischoice C.

44. Choice D is the best answer. What is the kinetic energy of an ion of mass m and charge q that is emitted from thecyclotron? According to the passage, an ion is emitted when the radius of its orbit is equal to the radius of thecyclotron, R.The definition of kinetic energy is:

45.

46.

Newton's second law gives:

Solving for v gives:

Plugging into the formula for KE gives:


KE = Imv22

_tnrqvB =R

v=aMm

KE =2n2=cpETR

2m

Choice B is the bestanswer. Which statement would be false, if the amplitude of the voltage in the gap were doubled? Agood way toanswer this question is to eliminate the true answers, starting with theeasier ones to eliminate. Choice C is true,because the cyclotron frequency, as given in the passage, does not depend upon the gap voltage. Choice D is true, becausethe magnetic field is generated by a device other than the voltage source and is therefore independent of the gap voltage.Regarding choice A, the ion would spiral out sooner, because each time the ion crosses the gap, it gains kinetic energy. Adoubled voltage amplitude means that it would gain twice the energy per gap-crossing that it gains under normal voltageconditions. That is, the ion would reach its emission speed sooner. Choice B is false, because the kinetic energy uponemission is related to the cyclotron frequency, which does not change here. Why does this matter? Well, the frequency tellsus how many times the ion circles the cyclotron per second. This dictates its speed, and hence its kinetic energy, uponemission. Since thefrequency does notchange, neither should the kinetic energy. Another way to think about this is that theexit point is at a fixed distance from the center, so the particle must reach an exact speed to attain the exit radius, whichmeans it reaches an exact kinetic energy to attain the exit radius. The best answer is choice B.

Choice A is the bestanswer. How does the work done by the electric field and the magnetic field compare for an ion inacyclotron? Choice C says the electric field and magnetic field dothe same amount of work, and choice D says they both dono work. If choice D is correct, then choice C must also be correct, which means neither can be correct. The best answer isthat the electric field does more work. It does work when it speeds upthe ion while the ion is in the gap. Themagnetic fieldactually does no work on an ion, because the magnetic force is always perpendicular to thedirection of travel (given by theright-hand rule). When a force is perpendicular to the path, it cando no work. This is why the particle traverses a semicircleat constant speed in a perpendicular magnetic field, gaining no speed and therefore gaining no energy. The best answer ischoice A.


47. Choice C is the best answer. How do the cyclotron frequencies for singly ionized H2 and singly ionized He compare?

2 2nmH2

2jtmHe

They both have the same charge, but He has twice the mass of H2.

2m

l*H2 = 2fHc


48. Choice D is the best answer. The questions asks for the cyclotron frequency of neutron, a species with no charge. Thecyclotron frequency is given by:

f_ qB2jtm

Since the neutron is neutrally charged, it has a cyclotron frequency of 0. This means that a cyclotron cannot be used toaccelerate a neutron. The best answer is choice D.


49. Choice C is the best answer. If the battery voltage increases, the droplet must be moved through a stronger electric field.This requires more work, which supports Option I and invalidates choice B. If the droplet's charge were smaller, lesselectrostatic force would act on it, and less work would be required to move it. This invalidates Option II, as well as choicesA and D, and leaves only choice C. Regarding the mass of the droplet, the question asks about the work required tocounteract electrostatic force, which does not depend upon mass. If you reasoned that an increased droplet mass means anincreased charge on the droplet (assuming a fixed p), that would increase the work needed to move the charge against theelectrostatic force. Either way, the work will not decrease, which validates Option III. The best answer ischoice C.

50. Choice C is the best answer. An electric force exists between any two charges, independentof their orientation in space andwhether or not they are moving. A magnetic force, on the other hand, depends on the alignment of two magnets and requiresthat charges be in motion (linear or cyclic) to create the magnetism. These basic difference are behind all of the differencesin the forces. Because motion is only necessary for a magnetic force to exist, a magnetic force varies with particle speed,whilean electric force does not. Choice A is a valid statement, so it is eliminated. As you learned from doing the right handrule, the motion vector, magnetic field vector, and magnetic force vector are all mutually perpendicular. For electric forces,they are either aligned with or against the electric field lines, so they are in fact parallel with the field. Choice Bis a validstatement, so it is eliminated. A moving charged particle can experience both a magnetic force and an electric force, so it issusceptible to either force. A magnetic force always turns a moving charged particle, because the force is perpendicular tothe particle's motion. An electric force can also turn a particle if it's aligned at an angle to the velocity vector. Chargedparticles in motion will turn until they are moving with the field lines. It is NOT true that a moving particle can only beturned by a magnetic force and not by an electric force, so choice Cis an invalid statement, and is therefore the best answer.If acharged particle is stationary, then itwill experience no magnetic force, no matter where it lies within the magnetic field.On the other hand, a charged particle will always feel a force when it is immersed in an electric field. This makes choice Davalid statement, which thereby eliminates it. The best answer is choice C.

51. Choice D is the bestanswer. As weird as this electric field may appear, this question is simply testing your knowledge ofthe fundamentals of electric field conventions. Field lines are drawn according to the way in which a positively chargedparticle would accelerate ifadded to the field. As field lines get closer together, the field is said to be stronger. As field linesmove apart, the field is said to be weaker. If we follow these rules, we should be able to work through each answer choice.When the particle is moved from Point a to Point b, it is moving perpendicular to the field lines, so no work is being done. Inother words, itdoesn't change potential, so no work was done. Choice A is eliminated. A particle would naturally flow fromPoint b to Point c, and positively charged particles flow from regions of higher potential to lower potential, so choice Bis avalid statement, and thereby eliminated. Natural flow for a positive charge is according to the field lines, so choice C is avalid statement, and thereby eliminated. Point a and Point b are found along the same potential line (perpendicular to thefield line), sothey are at the same potential. Choice Disan invalid statement. The bestanswer is choice D.


52. Choice C is the best answer. In order for the particle to pass through the velocity selector unaffected, it must experience nonet force. This will occur when forces are equal in magnitude, but opposite in direction. In this case, it means that qvB mustoppose qE. The particle is assumed to be moving from left to right The field arrows for the electric field show the directionof the force experienced by a positively charged particle. For the magnetic force, the right-hand rule must be applied. Inchoice A, the magnetic fields points out of the page, so the magnetic force on a particle moving from left to right will bedownward. This couples with downward electric force, rather than canceling it out, so the particle will be deflecteddownward. Choice A, and therefore choice D, are eliminated. In choice B, the magnetic fields points into the page, so themagnetic force on a particle moving from left to right will be upward. This couples with upward electric force, rather thancanceling itout,so theparticle will be deflected upward. Choice B is eliminated. In choice C, themagnetic fields points intothe page, so the magnetic force on a particle moving from left to right will be upward. This cancels out the electric force, ifthe two are of equal magnitude (which is possible at only one velocity), so the particle will not be deflected.

When qvB exceeds qE, the particle will be deflected in the direction of the magnetic force. When qE exceeds qvB, theparticle will be deflected in the direction of the electric force. The velocity selector, therefore, works by having the twoforcesof equal magnitude oppose one another in directionand therebycancel one another. The best answer is choice C.

Copyright ©by TheBerkeley Review® 170 REVIEWEXAM EXPLANATIONS

Electricity andElectric Circuits

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Physics Electric Circuits Currents

Electricity and Electric CircuitsIn Chapter 8, we studied voltage and potential. We learned that when there wasa voltage difference between two points in space, charges could migrate. In thischapter, we shall consider mass charge migration through a conducting medium.

Currents

In electrostatics we examined electric fields and charges, without consideringmotion. We also considered conductors and insulators. Recall that a conductor isa material that allows for the movement of charge (e.g., copper or silver), whilean insulator is a material that impedes the movement of charge (e.g., glass orplastic). If we wanted a charge to flow through a conductor, we would need tomaintain a steady force on that charge. One way to do that is with an electricfield E in the conductor, related to charge q and force F by equation (9.1). [This isa rearrangementof equation (8.2) in Section VIII.]

F = qE (9.1)

Negatively charged electrons flow through a conductor, and experience randommotion at great speeds. Yet whenan electric field is applied totheconductor, theelectrons experience not only random motion, but also a small net movement(called drift) in the direction opposite of the electric field. The drift velocity vd ofelectrons in an electric field is given by equation (9.2), where L is the distance(i.e., length of the conducting material) the electrons travelper unit time t.

v., = L (9.2)

The flow of charge from one point to another point during a certain amount oftime is calledan electric current (a scalar quantity). Current is given the symbol Iand is defined by equation (9.3). The ampere (amp or A) is the unit of current.An ampere is the flow ofone coulomb ofcharge per second (i.e., 1 A = 1 C/sec).When studying currents it is often convenient to use the milliampere (mA),which is 10"3A.

At(9.3)

Current has a direction in an electric field. It flows in the same direction in whichpositive charges flow and in the direction opposite to that in which negativecharges flow (Figure 9-1).

©

•••i Current flows in

/Tyji^^^. the same directionas positive charges

Current flows in theopposite directionas negative charges

Figure 9-1

By definition, current is said to flow from regions of higher electrical potentialto regions of lower electrical potential.


0


Physics Electric Circuits Currents

The drift velocity depends in part on the affinity of charge (electrons) for theconducting material. The drift velocity can be related to the current flowingthrough a wire. If the potential difference between two points in the wireincreases, then the drift velocity, and subsequently the current, will increase. Thisis the idea behind Ohm's law,whichwe shall address later in this chapter.Whenwe think of current we consider it to flow as a stream of charge through a wire,with a slightly less resistive path along the surface of the wire than the inside ofthe wire. Consider a wire (Figure 9-2) with a cross-sectional area A.Suppose thatan electron travels a distance L in a given time t across this segment of wire.

Figure 9-2

From this point forward, we shall refer to charge flow as current and follow thestandard pnysics conventions that currentis the flow of positive charge.

Example 9.1aWhich of the following changes will increase the transit time of an electronthrough a conductingwire?

A. Decreasing the mass of an electron.B. Increasingthe resistivity of the wire.C. Decreasingthe length of the wire.D. Increasing the current in the wire.

Solution

Decreasing the mass of an electron will make the electron more mobile. If lessmass hasanyeffect, it should increase theaverage velocity of the electron-a lessmassive charge in an electric field will have a larger acceleration. This makeschoice A incorrect.

Increasing the resistivity of the wire should make it harder for the electron totraverse the wire. Although resistivity has not been covered yet, you should beable to use the name ofa property to guess the resulting physics. The "resist" inthe name impliesa hindrance of somesort, in this case a hindrance of current, ascurrent is what flows through the wire. This makes choiceB the best answer. Atthis point, you can click on choice Band move on if you so wish or if you arepressedfor time. Forthesakeofwritinga complete answer explanation, we shalladdress the remaining choices. Ifyouarecurious aboutwhy choices Cand Darewrong, then read on.

Decreasing the length of the wire will naturally decrease the amount of time ittakes an electron to traverse the wire, assuming it travels at the same rate. Thismakes choice C incorrect.

Increasing the current through the wire can mean one or both of the following:more electronstravel through a given region in a given time,or the same numberof electrons travel through a given region in less time. Neither possibility willlead to an increased transit time for an electron. This makes choice D incorrect.

Copyright ©byTheBerkeley Review 174 The BerkeleyReview

Physics Electric Circuits

To give you an idea of the drift speed of an electron, consider a household wire(like the ones used for most appliances), but assume that it carries direct current,DC. For a 10-amp current—approximately that of a 1200-watt hair dryer—theelectrons take 40 seconds to travel 1 cm along the wire.

The correct choice is B.

Example 9.1bFor a fixed potential difference across a resistive wire, the power drain throughthe resistor can be increased by increasing the:

A. neutron density in the wire.B. length of the wire.C. temperature of the wire.D. conductivityof the wire.

SolutionThepower drain through a resistor that has a fixed potential difference dependson the voltage drop and the current. The question states that the voltage is fixed,so we needonlyconsider the current through the resistor. Toincrease the powerdrain throughthe resistor, we need to increase the currentthroughthe resistor.

The neutron density has absolutely nothing to do with the current through theresistor, so choice A can be eliminated immediately. As a wire increases inlength, the charges traversing the wire will encounter more resistance. Thisreduces the magnitude of the current, which in turn reduces the power drainthrough the resistor. This eliminates choice B. Increasing the temperature of thewire will increase its resistance, because the atoms in the wire spread apart fromone another as the temperature increases. The result is that conductivitydecreases, resulting in a drop in the magnitude of current. This will reduce thepower drainthroughthe resistor, so choice C iseliminated.

Increasing the conductivity of the wire results in a greater current passingthrough the resistor, which leads to more energy being dissipated. This meansthat the power drain through the resistor will be greater when the conductivityof the wire increases.

This makes choice D the best answer.


Currents


Physics Electric Circuits Voltage and Resistance

Voltage and Resistance

Batteries and Electromotive ForceBatteries are an important source of electric power. A common type of battery isthe zinc-carbon dry cell (Figure 9-3), used in devices that drain minimal powersuch as flashlights and transistor radios. It is a simple 1.5-V galvanic cell thatconverts chemical potential energy into an electromotive force (emf) thatgenerates electrical flow. The outer coating of a battery is generally made of inertinsulating material that serves as a protective barrier against the outsideenvironment. Immediately inside this protective outer layer is a zinc shell, whichserves as the anode (source of electrons, which in physics we shall consider to bethe negatively charged pole). Inside this cup-shaped shell is a thin layer ofammonium chloride (NH4CI) paste, which surrounds another paste comprisedof manganese dioxide (MnC>2) mixed with carbon powder. The inner pasteserves as the oxidizing agent in the chemical reaction. At the core of thebattery isa graphite rod, through which electrons may freely flow. A graphite rod, whichacts as an internal conducting rod, allows for the passage of electrons to themanganese oxide cathode (endpoint of electron flow, which in physics we shallconsider to be the positively charged pole), is positioned within this paste.

Metal cap (+ terminal)

-a

*— Graphite rod

Mn02 paste

NH4CI paste

Zn(s)

Metal contact point (- terminal)

e" flow

Figure 9-3

The cathode and anode of the battery are called terminals. If we attach wires tothese terminals and then connect those wires to a light bulb, it glows and givesoff light. Power to light the bulb is supplied by the battery. Electrons leave theanode of the battery and move through the wire leading to the bulb. Afterpassing through the filament of the bulb, the electrons will complete a circuit byreturning to the battery through the wire connected to the cathode.

Recall that we have mentioned that the electrical potential difference a movingcharged particle experiences as it moves from some region A to some region B(denoted as Vab) is the difference in electrical potential energy APEq at thosetwo regions divided by the value associated with the charged particle q. This isthe same as saying that there is a voltage change (i.e., a voltage) between regionA and region B.



If there is a separation of charge between two points on a device like a battery,and there is a force moving the charge from lower potential to higher potential,then that device is referred to as a source of electromotive force (abbreviated asemfand given the symbol e). An electromotive force is not really a "force," butrather an energy-per-unit-charge quantity (much like a potential). We canconsider an emf as a voltage source derived from a chemical reaction. The emfunit is the volt (and 1V = 1J/C).

In electric circuits, a battery is represented by parallel linesof unequal length, asshown in Figure 9-4a. This notation represents a source of emf. The longer linerepresents the positive terminal (usually at a higher potential) and the shorterline represents the negative terminal (usually at a lower potential). The cathodeis assigned a (+) charge and the anode is assigned a (-) charge. Sometimes youwill see the notation used in Figure 9-4b, which implies that there are multiplecells in series making up the emfsource. This tells us that voltages adds in series.

(a) (b)

wire wire wire wire

cathode (+) (-) anode cathode (+) (-) anode

source of emf source of emf

Figure 9-4

Resistivity and ConductivityThe resistivity p (rho) ofa material is a measure ofhow difficult it is for chargesto conduct through the material; higher resistivities are associated with betterelectrically insulating materials. The resistivity relates to the current density, J,and the electric field E through equation (9.4). The units of the resistivity are(V/m)(A/m2) which is just V»m/A. However, a volt divided by an ampere(V/A) is called an ohm Q (omega). The units of resistivity can also be written asohm-meters (Q»m). Equation (9.4) is telling us that if the resistivity ofa materialis high, then a larger electric field is needed for a given current density.Conductors areconsidered to be "perfect" when they have zero resistivity, while"perfect" insulators have infinite resistivity.

(9.4) P=J a=I (9.5)

Conductivity a (sigma) is the reciprocal of the resistivity, as shown in equation(9.5). In other words, the larger the conductivity of the material, the better thatmaterial isat conducting current. Another way to put this is tosay thata materialwith low resistivity has high conductivity. The unit ofconductivity is l/(Q»m).Table 9-1 lists some resistivities for some common conductors, semiconductors,and insulators.

ResistivityMaterial (Q-m)

Conductors

Silver 1.67x 10"8

Copper 1.74 x10-8Gold 2.24 x 10"8

Aluminum 2.68 x 10"8


Semiconductors

Arsenic 3.37 x 10"7

Silicon 2.31 x 103Boron 1.50 x 104

Table 9-1


Insulators

Glass 9.0 x 1011

Wood 108-1012Rubber 1.0 x 1016

Copyright©by TheBerkeley Review 177



Physics Electric Circuits Voltage and Resistance

The electrical resistance R of a conductor is given by equation (9.6). Resistanceis different from resistivity in that it considers both the material's conductivityand the dimensions of the resistive device. We typically want the resistance of aresistor, as that influences the current flowing through a resistor. The resistanceis the ratio of the voltage to the current for a given conductor. The SI unit ofresistance is the ohm (Q). One ohmis equal to one volt per ampere (1Q = 1V/A).

_«LR = p (9.6)

Example 9.2aFor the data given, what is the ratio of the resistance of Wire 1 to the resistance ofWire 2?

Wire p (Q-m) L(cm) A (um2)

1 1.7 x 10"8 1 6

2 1.7 x lO'8 3 1

A. 18:1

B. 2:1

C. 1:2

D. 1:18

Solution

Because we are asked to find a ratio, let's use the ratio method to answer thisquestion. The relevant equation is equation (9.6). We want the ratio of Rj: R2.Applying theresistance equation toeach resistor, and dividing, gives:

R1 =piiandR2 =c^^^ = =kA2= M)=iAi A2 R2 pL2/A L2Ai (3)(6) 18 = 1:18

This is choice D. Notice that we didn't spend time writing out the units (we alsodid not waste time writing out the constant p), since all these cancel out. Whendetermining a ratio, you can often omit units and constants from the ratio.

You could have used physical intuition to narrow the choices to C or D. Wire 1 isshorter and wider than Wire 2, so Wire 1 should have a smaller resistance thanWire 2. An analogy to remember is this: Electrons can get through a widerconductor more easily than through a narrow one, just as you can getthrough acrowded hallway more easily if it is wider. The narrower the wire, the moreresistance you encounter. Because Wire 2 is much longer and thinner than Wire1,the ratioshould besignificant; thispointsto choice D.


Example 9.2bFor the data given, what is the ratio of the resistivities of resistor A to resistor B?

A. 4:1

B. 2:1

C. 1:2

D. 1:4

Resistor

A

B

R(kQ)

2

4

Radius (mm)

1

2

L(cm)

3

6



SolutionBecause we are asked to find a ratio, let's again use the ratio method to answer thisquestion. The relevant equation is equation (9.6). We want the ratio of pi : p2-Applying the resistanceequation to each resistor, and dividing, gives:

Ra = PA-^- and RB = ps^B.AA

2RA = RB 2pA-A- = PBAA

AB

Lb.AB

PA _ LBAA - L*nr£ = LBrA2 _ (6)(1)2 6.PB 2LAAB 2LAnrs2 2LArB2 2(3)(2)2 2x3x4

= _6_24

This is choice D. Notice that just as with Example 9.2a, we ignored the units,because they cancelled out. As long as the respective variables have the sameunits, then we can ignore them in ratio calculations, as they will cancel out.

Using physical intuition on this question might have proven a bit trickier thanmostquestions, because the answer choices are all relatively close to 1. Wire Bisdouble thelength and four timesthe area ofWire A,soWire Bshouldhavea halfthe resistance of Wire A if they were made of the same material. Because Wire Bhas twice the resistance of Wire A, rather than half the resistance, Wire B musthave a greater resistivity than Wire A. This makes the ratio less than 1, whicheliminates choices A and B. To go from half as muchto twice as much, somethingmust increase by a factor of 4, so the ratio must involve a 4 rather than a 2. Thiseliminates choice C and confirms choice D as the best answer. This is one of thosetime consuming questions that no matter whether you used the ratio method orphysical intuition, it will take you time and require focus.


Before we move on from resistors, lefs consider what they do. A resistor opposesthe flow of current, and thus, drains energy from an electric circuit. A resistor isany device in a circuit that hinders currentand drains power from the circuit.Resistors can be a segment of wire or an electrical appliance like a computer.

Ohm's LawIt would be useful to have an equation that relates the current, I, in a conductorto the potential difference, V, across the ends of that conductor. Consider asection of wire with length L and cross-sectional area A. Across this length ofconducting material existsa voltage V. As Figure 9-5 shows, the electrons movein response to this potential difference-but they will move erratically. As thevoltage induces the electrons to the right, they interact with protons and otherelectrons. Such interactions jar and redirect their travel, but they do not preventthe electrons from flowing as a current.

conductor of length L and cross-sectional area A

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PhysiCS Electric Circuits Voltage and Resistance

Because the electrons flow in response to an applied voltage, the voltageand theresulting current must somehow be related. As the voltage is increased, thecurrent increases, so V must be directly proportional to I. They are relatedaccording to Ohm's law, given by equation (9.7).

Ohm's Law V = IR (9.7)

In using Ohm'slaw we find that it adequately describes the relationship betweenvoltage, current, and resistance. However, be aware that Ohm's law does notapply to all materials.

Example 9.3aThe current through a wire will be the GREATEST when the two ends of the wireare connected to terminals at:

A. -500 V and+500 V.B. OV and 1000 V.C. 10,000 V and 11,000 V.D. Allconnections resultin a current of the same magnitude.

Solution

According toOhm's law, equation (9.7), a current flows in response toa potentialdifference V across some material. Thebigger the potential difference across thewire, the bigger the current will be (V = IR, so as V goes up, I goes up). Thepotential difference across the wire is 1000 V for choices A, B, and C. Therefore,all threechoices resultin a currentof thesamemagnitude.


Example 9.3bThe voltage across and the currentthrough three different wireswere measuredand reported in choices A, B, and C. For which of the three wires was theresistance the SMALLEST?

A. V = 100 volts and I = 1 mA.

B. V = 1 volts and I = 100 mA.C. V = 10 volts and I = 10 mA.D. The resistance was the same for all three measurements.

Solution

This question requires that you apply Ohm's law, V= IR. The question isaskingfor the smallest resistance, so we need to divide the voltage by the current todetermine the resistance.

R=V/l .-. Choice A:100/10-3 =105Q;Choice B: VlO"1 =101 Q; and Choice C:10 j' 10-2 =103 Q

The smallest resistance of the choices is 10 Q.




Potential and Internal ResistanceAs we have mentioned, batteries are one particular emfsource. Let's examine theterminals of the emfsource and expand on the diagram shown in Figure 9-4a.Suppose we label the positive terminal as A and the negative terminal as B, asshown in Figure 9-6. Since the positive terminal (cathode) is at a higher potentialthan the negative terminal (anode), there will be a potential difference betweenthe two terminals. Recall that electric field lines flow from a positively chargedsource to a negatively charged source. The electric field in Figure 9-6 goes fromterminal A to terminal B. The electric force that a charge between the twoterminalsexperiencesis governed by equation (9.1). The direction of that force isfrom terminal A to terminal B.

higher VE

lower V

wire A B wire

1 cathode (+) (-) anode

source ofemf J

1 '. !Figure 9-6

The potential across the terminals from point A to point B is defined as VAb, anathis is referred to as the terminal voltage (or potential). In this case, a positivecharge moves from higher potential to lower potential.Suppose we now connecta wire across the two terminals (Figure 9-6) to complete the circuit. Once the twoterminals are connected, an electric field is established in the wire. Positivecharges can get from terminal A to terminal B in one of two ways: (1) throughthefield between the platesor (2) through the wire connecting thecathode to theanode. Because the wire has less resistance than air, the preferred pathway isthrough the wire. Current begins to flow in the wire from terminal A to terminalB. The current in the circuit is given by £ = VAb = IR. This relationship is sayingthat as a charge moves through the circuit, the potential increase in the emf e isequal to thepotential drop VAb as that charge moves through theresistor.

As a charge moves through a circuit, it encounters an internal resistance r. Thus,the current moving through this internal resistance will have a terminal voltagedrop equal to Ir. This is shown in equation (9.8). The current in the externalcircuit is still givenby equation (9.7). Substitution of equation (9.7) into equation(9.8) gives equation (9.9). We conclude that the current is equal to theelectromotive force divided by the total resistance.

(9.8) VAB = £-IrR + r

(9.9)

Electrical PowerPower is the rate of conversion of electrical potentialenergyinto someother typeof energy, defined as the ratio of APEq/At. The SI unit of power is the watt,where1W= lj/sec. We need a way to relate the power to the current and to thevoltage. Recall that we can calculate the change in electrical potential energy as acharged particle moves through a voltage difference using APEq = qV.Substitution of qV into APEq/At gives equation (9.10), which upon replacingAq/At with I becomes equation (9.11):

(9.10) P =_ AqV

AtP = IV (9.11)




PhysiCS Electric Circuits Voltage and Resistance

Equation (9.11) and Ohm's law,equation (9.7), are two very important equationsto remember when considering electric currents. Combining the two equationsleads to equation (9.12) and equation (9.13).

(9.12) P=¥i P=I2R (9.13)R

This gives us three different equations to choose from when solving for thepower drain through a resistor or a circuit. We choose the best equationbased onthe variables we are given in the question.

Example 9.4aWhich formula represents the time it takes to accelerate a car of mass m from restto a speed v, using an electric motor with voltage V and drive current I?

A. t-2ELmv2

B. t = 2hLmV

C. t =2IV

D. t = 2lV2Iv

Solution

We usually approach this kind of problem by thinking about units and limitingcases. If youare not toohandy with electrical units (especially how they relate tomechanics units), then avoid checking units-you can't check what you don'tknow. Try limiting cases.

You need to think about how I, V, m, and v affect the acceleration time t. Thinkfirst about the one that is easiest for you to grasp conceptually. For example, ifthe car is to reach a higher speed v, what should happen to t? It should go up.Therefore, the answer is either B or C. Between these two choices, I and m areswitched-Vis not.Considering voltage would be a waste ofeffort-thinkaboutIor m. If the current I increases, what should happen to t? It should decrease;afterall, there is more current to make the cargo.Thisleadsto choice C.

You could have also singled out this choice if you knew, from solving suchproblems in the past, that mand v gotogether, or thatI andVgotogether.

You can also solve this problem using energy conservation. Energy questions areubiquitous in science, meaning thatyou canrelate different physical phenomena(here, kinetic energy and electrical energy) through their energies. Specifically,theelectrical energy generates thecar's kinetic energy:

^electrical= KEcar

The electrical energy comes from the electrical power supplied over some time t(this is actually the time t needed to accelerate the car to a speed v). Using theelectrical power, equation (9.11), the energies become:

v2m

Solving for t yields choice C.


IVt = %mv2

Copyright ©byTheBerkeley Review 182 The Berkeley Review


Example 9.4bWhich equation the represents rate at which a liquid of mass m, with specificheat c, is heated by a conducting coil of current I and voltage V?

A.AT

At

_ mcV

I

B.AT

At

= IYmc

C.AT

At

AT

_ V

mcl

nr

D.At IV

Solution

As with Example 9.4a, the cumbersome units make it not so easy to approach thequestion using units. So we shall try limiting cases.

You need to think about how I, V, m, and v affect the increase in temperatureover time. Think first about the one that is easiest for you to grasp conceptually.Forexample, if the materialhas a low heat capacity c, then it should reach a hightemperature and thus experience a large AT. This means that c and AT areinversely related. Therefore, the answer is either B or C. Between these twochoices, I is the only variable that is switched. Considering voltage or masswould be a waste of effort, and not help to solve the question. If the current Iincreases, what should happen to AT? It should increase; after all, there is morecurrent to heat the solution. This leads to choice B.

You could have also singled out this choice if you knew, from solving suchproblems in the past, that m and c go together, or that I and V go together.

You can also solve this problem using energy conservation. Energy questions areubiquitous in science, meaning that you can relate different physical phenomena(here, heat energy and electrical energy) through their energies. Specifically, theelectrical energy generates the heat of the system:

Eelectrical = cjsystem

The electrical energy comes from the electrical power supplied over some time t(this is actually the time t needed to accelerate the car to a speed v). Using theelectrical power,equation (9.11), the energiesbecome:

IVAt = mcAT

Solving for AT/Atyields choiceB.





Physics Electric Circuits Capacitors

Capacitors

CapacitanceA capacitor is formed from two conductors separated by an insulator. Aninsulator is a material that does not conduct charge (e.g., rubber or air). Acapacitor is represented by equal length parallel lines, as shown in Figure 9-7a.

(a)

wire

Insulator

wire

\ /Conducting material cathode (+) (-) anode

Figure 9-7

When a voltage is placed between the parallel plates of a capacitor, chargeaccumulates on those plates (Figure 9-7b). The capacitance C ofa capacitor is theamount of charge q that can be stored per volt of potential difference V acrossthe two parallel plates. This is shown by equation (9.14). The SI unit ofcapacitance is the farad (IF = 1C/V). Capacitances in electronic devices usuallyhave values between 5000 uV and 1pF (where 1 j/F = 10"6 Fand1pF = 10"12 F).

c = iLV

(9.14)

Equation (9.14) basically tells us that the amount of charge that can be stored on acapacitor depends on howhard you push the charge (the voltage) and how muchthe plates can store (the capacitance). The capacitance will depend on the area Aof the conducting plates in the capacitor, the distance d between those plates, andthe dielectric constant Kof the insulator between them. If the area of the plates islarge, then a large amount of charge can be stored. Similarly, if the plates arecloser together, then greater stability allows more charge to be stored. Thecapacitance of a parallel plate capacitor can be calculated using equation (9.15),where k is a constant(9.0 x 109 N-m2/C2) and Kis the dielectric constant.

C K •£4rck vd

(9.15)

DielectricsA dielectric is a non-conducting material between the conducting plates of acapacitor. A capacitor's capacitance will increase by a factor K when a material(e.g., Mylar, mica, Teflon, etc.) with a specific dielectric constant is placedbetween the conducting plates. As shown in equation (9.16), the charge on theconducting plates increases by the factor of the dielectric constant K.

q = CV = KCVacuumV (9.16)

Material placed between conducting plates normally does not conduct electriccharge. However, if there is a large electric field between the plates, then it couldcause the atoms in the dielectric to ionize. This results in positive charge in thedielectric being attracted to thenegatively charged plate and negative charges inthe dielectric being attracted to the positively charged plate. Conduction ofcurrent can now occur. This phenomenon is called dielectric breakdown.



Charging a CapacitorAn emf source, such as a battery, canbe used to charge a capacitor. Suppose westartwith two parallel plates that have no charge (Figure 9-8a). Letcharge movefrom one plate to the other. As one plate gains positive charge, the other platebecomes deficient in positive charge (i.e., it will be leftwitha negative charge). Itdoes not take much energy to move the first few charges across the plates.However, as more and more positive charges flow to the positively chargedplate, they become bothrepelled by the positively charged plate and attracted bythe excess of negative charges on the negatively charged plate (Figure 9-8b). Atsome point, no more positive charge can be transferred, and the two plates willbe chargedwith some maximum amount of charge q (Figure9-8c).

(a)

e LJ

KDespite emf, no charge canflow with an open circuit.

(b)

6 ^

Charge flows and builds upon plates, causing repulsion.

Figure 9-8

(c) + p+

+

+

——— I •

Once repulsion offsets voltage,no more charge is able to flow.

The electrical potential energy needed to charge the capacitor is a modifiedversion of equation (8.5). In equation (9.17) we consider the average potentialdifference (Vavg) between the plates. The average potential difference is YiV,because the initial potential (Figure 9-8a) is 0 V and the final potential (Figure 9-8c) isV. Substituting V2V for VaVg yields a more useful version ofthe equation.

APEq = qVavg =q(ViV) (9.17)

Combining this with equation (9.14), we can express the change in potentialenergy in two other forms, as shown in equations (9.18) and (9.19):

(9.18) APEa=i/£CV2 APEn =.q22C

(9.19)

How is a capacitor charged? Consider the current over time for circuitshown inFigure 9-9, with an em/source, a switch, a resistor, and a capacitor.

fast current

slow current

Switch Switch

Figure 9-9

Capacitors


Physics Electric Circuits Capacitors

Thistype ofcircuit, witha resistor and a capacitor in series, is called an RCseriescircuit Thevoltage V across theemfsource remainsconstant. When theswitch inthe circuit is closed, the voltage across the resistor (V = IR) varies with time, asdoes the voltage across thecapacitor (V= q/C). Thesum of the voltage across theresistor and the capacitor mustbe equal to the voltage across the terminals of theemf source. This is shown by equation (9.20). Solving for the current givesequation (9.21):

V=Vr+VC = IR +-S-

i=v__q_R RC

(9.20)

(9.21)

When the switch is open at t = 0 (Figure 9-9a), no current flows in the circuit.Once the switch is closed at t = 1 (Figure 9-9b), the emfsource "pushes" currentfrom the positively charged terminal,around the circuit and on to the negativelycharged terminal in a counterclockwise direction. At the instant the switch is firstclosed, q = 0 C. Plugging in 0 for q into equation (9.21) tells us that the initialcurrent through the circuit is Ii = V/R. As charge begins to accumulate on theplates of the capacitor, the current flowing back to the negatively chargedterminalof the em/source is reduced (Figure9-9c). The value of q/RCin equation(9.21) now increases, causing the current to decrease. Once the capacitor iscompletely charged at t = 3 (Figure 9-9d), current will no longer flow in thecircuit.At that point I = 0 A,and the final charge on the capacitoris qf = VC.

Ifcurrent passes througha resistor and flows to a capacitor, then both the currentthrough that capacitor, equation (9.22), and the charge on that capacitor,equation (9.23), will change exponentially with time. The quantity RCin both ofthese equations is referred to as the time constant.

(9.22) UI^/RC) q=qf(l-e-t/RC) (9.23)

Larger capacitors require a longer time to charge and discharge, as they havemore charge to store or lose. Larger resistors increase the charging anddischarging time, because they reduce the current flow that occurs duringcharging and discharging. Graphs of the capacitor discharge current and storingcharge are plotted as functions of time in Figure 9-10, along with their timeconstants. These times represent how long it takes for the discharge current andstoringchargeto change by 1/e and 1 - (1 /e), respectively.

Discharging Capacitor Charging Capacitor

qmaxd - 7)

t = RC t = RC

Figure 9-10



Example 9.5aTwo parallel plate capacitors of equal size are charged by a power supply ofvoltage Vsuppiy. What is true of these capacitors when their maximum storagecharge is plottedas a function ofsupply voltage?

max

S*\ B

A

S^

A. Capacitor B has a stronger dielectric between the plates; Capacitor A has ahigher breakdown voltage.

B. Capacitor A has a stronger dielectric between the plates; Capacitor B has ahigher breakdown voltage.

C. Capacitor B has a stronger dielectric between the plates; Capacitor A has alower breakdown voltage.

D. Capacitor A has a stronger dielectric between the plates; Capacitor B has alower breakdown voltage.

Solution

Let's use the multiple-concepts approach to figure out dielectric strength andbreakdown voltage separately. Regarding dielectric strength, a capacitor with astronger dielectric can store more charge (all else being equal)-that's partly whydielectrics are used. Because Capacitor A reaches a greater Qmax/ it must have astronger dielectric, which rules out choices A and C. Regarding the breakdownvoltage, the plot shows that Capacitor B loses its charge at a lower voltage thanCapacitor A does. Capacitor B must have the lower breakdown voltage.


Example 9.5bTwo different capacitors, each connected to an identical voltage supply but adifferent resistor, are charged. Their charge, as a function of time, is plotted tothe left. What is true of the two RC circuits?

A. A has a larger capacitor; B has a larger resistor.B. Bhas a larger capacitor; A has a larger resistor.C. A has a larger capacitor; A has a larger resistor.D. Bhas a larger capacitor; B has a larger resistor.

SolutionLet's again use the multiple-concepts approach to figure out capacitor size andresistor size separately. Regarding capacitor size, Circuit A gains a larger totalcharge given the same voltage, so Capacitor A must be larger than Capacitor B.This rules out choices B and D. Regarding the resistance, the plot shows thatCapacitor A gains its charge at a faster rate than Capacitor B does, so it mustexperience less resistance. Resistor A has a lower resistance than Resistor B.



Capacitors


Physics Electric Circuits Electric Circuits

Electric Circuits

Kirchhoffs RulesElectric circuits can become rather complicated at times. The two Kirchhoff'srules that will help us analyzecomplicated circuits are Kirchhoff's loop rule andKirchhoff's junction rule. Fortunately, the MCAT tests only RC circuits, whichfollow those basic rules. Between Ohm's law and two Kirchhoffs rules, youshould be able to navigateany RCcircuitsyou'll encounter on the MCAT.

Kirchhoffs Loop RuleThis rule states that the algebraic sum of the potential differences (voltagechanges) in any closed circuit (loop) is equal to zero. Consider the single circuitshown in Figure 9-13. Let's follow a positive charge as it leaves the positivelycharged terminal of the emfsource.Because current is the flow of positive charge,start at the cathode.As the charge flows along the wire from the cathode to pointB, there is negligble resistance. Thus, the change in voltage (potential) here iszero (Vdrop =IR =1(0) =0). As the positive charge moves from point Btopoint Cit encounters a resistance R. There is a potential decrease across the resistor (Vbc= -IR). From point C to the anode, the change in voltage (potential) is again zero.Once the charge reaches the anode, it stops moving, so its "potential" to move iszero. All of the voltage from the emf source has been used up. Thus, the initialvoltage at the cathode is equal in magnitude to the voltage dropped as currenttraverses the circuit (Vdrop = -IR)- The energy that is expended to do this is thechemical energy of the emf, or Eceii. The sum of the voltage changes around thisclosed circuitmust add up to zero. If the chargewere to move from point D backto point A, it would encounter the emf.

vdrop ="IR

wvRi

BI C Y

<

9"T3

II D (-)O

•

(+) A

I

o

II

>

Figure 9-13 Loop Rule.

4i

wRi

t A B

Rr

4h(+) (-)

Figure 9-14 Junction Rule.

Kirchhoffs Junction RuleNot all electric circuits contain a single loop. Many electric circuits containmultiple loops. There may be times when the current from a wire reaches ajunction and divides. Thetotal currentflowing through the pathwaysleaving thejunction must equal the current that entered the junction. In other words, all ofthe electrons entering the junction must be accounted for~no electrons fall off ofthe wire or jump onto the wire.

For example, consider the diagram in Figure 9-14. As the current I leaves thepositive terminal of the emf source, it encounters a junction at point A. Thecurrent I entering that junction must equal the sum of the currents Ii and I2 thatleave the junction. In other words, I = Ii + I2. This relationship allows us todetermine currents after they have been divided through parallel pathways.



Example 9.6aWhat are the values for the various unlabeled currents in the circuit below?

6.17 ArVWS HWH

2.33 A

-JWV-HfiH

5.02 A

HMH-W-

A. I3 = 3.84A;l4 = 1.15A;l6 = 6.17AB. I3 = 3.16A;l4 = 1.15A;l6 = 6.17AC. I3 = 3.84A;I4 = 1.15A;I6 = 4.33AD. I3 = 3.16A;l4 = 7.32A;I6 = 4.33A

Solution

This question is just an application of Kirchhoff's junction rule. The currententering the first junction is 6.17amps, so the current leaving the last junction,!(,,should also be 6.17 amps. This eliminates choices C and D. From here, we cannote that 2.33 + I3 = 6.17 or that 5.02 + 14 = 6.17. Solving both relationshipsconcludes that I3 = 3.84 amps and that I4 = 1.15 amps. The value of I4 doesn'thelp to solve the question, but I3helps differentiatebetween choices A and B.


Example 9.6bWhich relationshipaccurately describes the relativecurrents in the circuitbelow?

8.23 A

-Wr- ,1 -tfhHWV-i

1.89 A

2.22 A

Rd = 2.5Q

A. Ia>Ic>2.62A

B. Ib>Id>Ic>IaC. Id>Ib>Ia>IcD. Id>Ic>2.75A>Ia

Solution

Thisquestion doesn't involve as much math as it would seem at first glance. Firstand foremost, Kirchhoff's junction rule tells us lb = Id = 8.23 A. This eliminateschoices Band C. Next we need to compare Ia to Ic. The total current splits in sucha way that Itotal = 5.61 + Ia and Itotal = 1-89 + 2.22 + l0 so setting the relationshipsequal to one another we get 5.61 + Ia = 4.11 + Ic. For that to hold true, Ic > 1&which eliminates choice A. (Ia = 2.62 A, lb = 8.23 A, Ic = 4.12A, and Id = 8.23 A)



Electric Circuits



Example 9.7aGiven a 12-volt battery in the circuit below, what is the voltage across R2 and thecurrent through R3, if eachresistoris 4 Q and Ii is 2 A?

w-Ri

vwR2

R,

vwA. V2= 4 volts, and I3= 1 ampB. V2 = 4 volts, and I3 = 2 ampsC. V2= 8 volts, and I3= 2 ampsD. V2= 8 volts, and I3 = 4 amps

Solution

Let's use Kirchhoffs loop and junction rules to solve this one. Applying the looprule on the inner loop(i.e., the one that includes the battery, Ri,and R2), we get:

V = I1R1 + I2R2

As current flows around the loop from cathode to anode, the voltage drops from12 to 0, decreasing upon crossing each resistor (e.g., -I1R1 and -I2R2)- Based onthe battery orientation, the current flows counterclockwise and passes throughRl before splitting at the junction. Based on Kirchhoffs junction rule, the currententering the junction (Ii) mustequal the sum of the currentsleaving the junction(12 + l3)- Because 2 A enters the junction and R2 = R3, both I2 and I3 = 1 A. Thismakes choice Athe best answer. To be complete, lefs solve for Vdrop 2-

V2 = -I2R2 = (1 A)(4 Q) = -4 V

Vdrop 2is4volts. This supports thatchoice Ais correct.


Example 9.7bWhat is the voltage drop across R2 and the current through R3 in the circuitbelow if the current leaving the battery is 6 A and that all resistors are 3 Q?

V

A.

B.

C.

D.

rn/W MAL— Rl Ro R2

V2 = 2 volts, and I3 = 3.3 ampsV2 = 3 volts, and I3 = 2.0 ampsV2 = 6 volts, and I3 = 4.0 ampsV2 = 12 volts, and I3 = 4.0 amps

VW1



Solution

Let's use Kirchhoff's junction and loop rules tosolve this one. Applying the junctionrule, we know thata current of6 amps enters the junction and is split betweenthe two pathways. The two pathways have resistances of6 Q (Ri + R2 = 3 Q + 3Q=6 Q) and 3 Q (R3 = 3 Q) respectively. Because the upper pathway has twicethe resistance of the lower pathway, it will have half of the current. The sum ofthe upper current and lower current must equal 6 A and the ratio of the uppercurrent to the lower current must be 1 : 2. This means that the current throughthe upper pathway is 2 amps and the current through the lower pathway is 4amps. This eliminates choices A and B.

The voltage dropcurrent across resistor 2 is -I2R2- The current is2 amps and theresistance is 3 ohms,so the voltage drop must be:

V2= I2R2 = (2 amps)(3ohms) = 6 volts

V2 is 6 volts, which eliminates choice D.

The correct choice is C.

Series versus ParallelCircuits can have several circuit elements. As current from the cathode passesthrough the wiring and to the anode, it will encounter the various circuitelements within the circuit. The relative alignment of the circuit elements can bedescribed as either in series or in parallel. In series, the circuit elements areconnected in a direct path, where current passes through each circuit element. Inparallel, the circuit elements are part of different pathways, where one circuitelement can be thought of as being a part of an alternative pathway to the othercircuit element. Circuit elements in series share the same current, while circuitelements in parallel share the voltage drop.

Resistors in SeriesIt is occasionally necessary when analyzing a circuit to replace several resistorsin series (e.g., Rj, R2, R3, etc.) with a single equivalent resistor Req. In otherwords, we need to determine what size single resistor would have the sameimpact on the current as all of the resistors in series combined. For resistors inseries this is simple, because connecting multiple resistors in series is like makinga longer resistor. Resistors connected in series are additive, as in equation (9.24):

Requivalent = Rl + R2 + R3- (9.24)

When is a resistor in series with other resistors? This is the case when the samecurrent can flow through each of the resistors without branching between them,as shown by the circuit in Figure 9-15:

WW—AV—MMRi R2

Figure 9-15


R,

191

Electric Circuits



Resistors in ParallelThere will be timeswhen it willbe necessaryto mentally replaceseveralresistorsin parallel with a single equivalent resistor. Lefs consider what happens whenwe have several resistors in parallel. Ifs like stacking several resistors in a side-by-side fashion. This would in essence make a wider resistor. Given thatresistance decreases with an increasing cross-sectional area, we at firstapproximation can assume that the resistance would decrease. This can be doneby using equation (9.25).

.=J^+J,^equivalent Rl R2 ^3

(9.25)

When is a resistor in parallel with other resistors? This is the case when bothsides of each resistor are connected to a common junction, as shown by thecircuit in Figure 9-16.

Ri

M/frRi

^vw-R,

^-AMrFigure 9-16

Capacitors in ParallelNow lefs consider what happens when we have several capacitors in parallel.Ifs like stacking several capacitors in a side-by-side fashion. This would inessence make a wider capacitor. Given that capacitance increases with area of theplates, we at first approximation can assume that the capacitance shouldincrease. The equivalent capacitance when capacitors are added in parallelinvolves the addition ofeachcomponent capacitor. Equation(9.26) allowsfor thecalculation of the equivalent capacitance (Ceq) of several capacitors in parallel.[Note that the form of thisequation is different from the form of the equation forresistors in parallel.]

^equivalent = Q. + C2 + C3... (9.26)

An example of three capacitors in parallel is shown in Figure 9-17, where eachcapacitor will share the same voltage,but not the same charge build up:

-qi .iiJL-q2 _ -LLi-q3'! -n— C2 —1— C3+ + +q1 + + +CJ2 + + +q3

Figure 9-17



Capacitors in SeriesEquation (9.27) allows for the calculation of the equivalent capacitance (Ceq) ofseveral capacitors in series. [Notethat the form of this equation is different fromthe form of the equation for resistors in series.]

.=J-+J^ +.^equivalent Cl C2 C3

(9.27)

An example of capacitors in parallel is shown in Figure 9-18:

Figure 9-18

Example 9.8aWhat is TRUE of the voltage across, current through, and power loss by twounequal resistors connected in series?

A. The larger resistor has a bigger voltage drop and dissipates more power;both resistors have the same current.

B. The larger resistor has a smaller current and dissipates less power; bothresistors have the same voltage drop.

C. The larger resistor has a bigger current and smaller voltage drop; bothresistors dissipate the same power.

D. The larger resistor has a smaller current and bigger voltage drop; bothresistorsdissipate the same power.

SolutionWhen current flows through a series resistor circuit, as in Figure 9-15, currentcan flow along only one path (just as a water current would flow, if this figurewerea top view of plumbing). The same current must flow through each part ofthe circuit; each resistor must have the same current, regardless of its size.Because all of the resistors have the same current, only choice A is possible. Butfor the sake of review, lefs consider the voltage drop and power loss for eachresistor. Given that each resistor has the same current, the larger resistor willhave a larger voltage drop, since:

Vdrop = "IR

Because the resistors have the same current, we must use a version of the powerequation with I in it. Given that we are determining power for different sizedresistors, we should choose the form of the power equation that has I and R.

P = I2R

Because the current is the same in each resistor, increasing resistance willincrease the power. The larger resistor will dissipate more power, since thepower equation shows that power is directly proportional to resistance (P « R).The relationships for resistors in series—same I, bigger R has bigger V and P—hold generally for any number of resistors in series. Know these relationships.



Electric Circuits



Example 9.8bWhat is true of the voltage across, current through, and power loss by twounequal resistorsconnected in parallel?

A. The larger resistor has a bigger voltage drop and dissipates more power;both resistors have the same current.

B. The larger resistor has a smaller current and dissipates less power; bothresistors have the same voltage drop.

C. The larger resistor has a bigger current and smaller voltage drop; bothresistors dissipate the same power.

D. The larger resistor has a smaller current and bigger voltage drop; bothresistors dissipate the same power.

Solution

When current flows through a parallel resistor circuit, as in Figure 9-16, currentcan flow along different paths, suggesting that the current is not necessarily thesame along each path. However, parallel resistors will have the same voltagedrop across them, because each end makes a direct (i.e., no intervening circuitelements) connection with the same voltage source. This means that choice B isthe only possible answer. But for the sake of review, lefs consider the currentthrough and power loss for each resistor. Given that each resistor has the samevoltage drop, the largerresistormust have a smaller current, since:

I = VR

Remember the old saying, "Choose the path of least resistance"? This applies tocurrent. The above equation is how that statement is said in math. Because theresistors have the same voltage drop, we must use a version of the powerequation with V in it. Given that we are determining power for different sizedresistors, we should choose the form of the power equation that has V and R.

P = YiR

Because the voltage drop is the same in each resistor, increasing resistance willdecrease the power. The larger resistor will dissipate less power, since thepower equation shows that power is inversely proportional to resistance (P «1/R). The relationships for resistors in parallel-same V, bigger R has smaller Iand P-hold generally for any number of resistors in parallel. Know theserelationships.


Example 9.9aWhat is true of the voltage across, charge stored on, and energy stored by twounequal capacitors connected in parallel?

A. The larger capacitor has a smaller voltage gain and stores less energy; bothcapacitors store the same charge.

B. The larger capacitor stores more charge and more energy; both capacitorshave the same voltagegain.

C. The larger capacitor stores more charge and has a smaller voltagegain; bothcapacitors store the same energy.

D. The larger capacitor stores less charge and has a larger voltage gain; bothcapacitors store the same energy.



Solution

When charge stores in a parallel capacitor circuit, as in Figure 9-17, the voltagegain across each capacitor will be the same (for the same reasons resistors inparallel have the same voltage drop). Given that the voltage is the same, thelarger capacitorwill store more charge, since Q = CV.

The larger capacitor will store more energy, since the energy equation can bewritten as:

APEn =lCV2q 2

Since each capacitor has the same voltageacross it, the largercapacitor will storemore energy. The relationships for capacitors in parallel-same V, bigger C hasbigger Q and APEq-hold generally for any number of capacitors in parallel.Know these relationships.


Example 9.9bWhat is true of the voltage across, charge stored on, and energy stored by twounequal capacitors connected in series?

A. The larger capacitor has a smaller voltage gain and stores less energy; bothcapacitors store the same charge.

B. The larger capacitor stores more charge and more energy; both capacitorshave the same voltage gain.

C. The larger capacitor stores more charge and has a smaller voltage gain; bothcapacitorsstore the same energy.

D. The larger capacitor stores less charge and has a larger voltage gain; bothcapacitorsstore the same energy.

Solution

When charge stores in a series capacitor circuit, as in Figure 9-18, the chargestored on each capacitor will be tiie same. For example, notice the section ofwire between Ci and C2 in Figure 9-18. Before charging begins, there is no netcharge on the wire; after charging, there must still be zero net charge, becausecharge cannot be created, and that section of wire is isolated from any chargesources. Thus, one capacitorgets +q, while the other gets -q from that section ofwire. This same charge separation occurs in each isolated section of wire. Sincecapacitors are connected to these sections, each capacitor must have the samecharge. Given that the charge is the same, the larger capacitor will have asmaller voltage gain, since:

V =Q

The largercapacitor stores less energy, since the energy equation is written as:

APEq =l2i4 2 C

Because each capacitor has the same charge on it, the larger capacitor must storeless energy. The relationships for capacitors in series—same Q, bigger C hassmaller Vand PEq—hold generally for any number of capacitors in series. Knowthese relationships.



Electric Circuits


Physics

\4Test Tip

Calculation Shortcut

Electric Circuits Electric Circuits

Example 9.10aWhich of the following statements are true about the circuitshownbelow?

6V

#kR1 = 2Q

R2= 3Q

I3 = 0.5A

I. Req =4ohmsII. R3= 6 ohms

III. Itotal = 1-5amps

A. II onlyB. I and II onlyC. II and m onlyD. I, n, and III

Solution

As a general rule, you should attack these questionsby starting with an analysisof the feature they give you the most of. In other words, we are given two of thethree resistances, so lefs start by analyzing resistance. Unfortunately, we can't domuch tosolve for R3 orReq. We are limited in what wesolve for, so this questionforces us to work backwards from the answer choices. Because there is no "noneof the above" answer, we have to assume that at least one statement is valid. Solefs see if any of the answers compliment one another.

We can solve for Req by solving the parallel pairfirst, and then adding that toRi.A short cut that willsavea largeamountof timeis to rewriteequation (9.25). Fortwo resistors in parallel, the equivalent resistance is their product over their sum.

6Vl! = 1.5A

R1 = 2Q

-3V


RlXRnReq =1 R2 + R2

IfR3 is6Q, then Rparallel is18/9Q=2Q. This means thatR^ is2Q+2Q =4Q.Statements I and II are complementary, so now lefs determine the total current.If Req is 4 Q and the battery is 6 V circuit, then Itotal is 1.5 amps. All threestatements match. Lefs considerthe values for the lower segment of the circuit.

3V

12 = 1.0A

R2= 3Q

13= 0.5A

•vw1R3= 6C

-3VOV



Circuit Calculation ShortcutExample 9.10a introduced a shortcut for determining the equivalent resistancefor two resistors in parallel. This same shortcut also works for capacitors inseries. The idea behind theshortcutis simple: You don't have toshow yourworkon the MCAT, so you might as well manipulate your common formulas into aneasier touse form. Solefs consider the calculation oftworesistors in parallel.

1 _ 1 + 1 _ R2 + Rl^equivalent Rl R2 Rl * R2 Rl * R2

R2 + Rl _Rl+R2Rl x R2 Ri x R2 Ri x R2

1 -Rl+R2 . p . , Rl*R2r — — • • ^equivalent - — —Requivalent Rl * R2 ^ Rl +R2

Example 9.10bWhat arethe current leaving the cathode, the current through Ri, andReq for thefollowing electrical circuit?

12 V

m/W—MA-iR1 = 2Q I2= 4A

I3 = 2A

M—A. Itotal =6A,Ii = 4 A, and Requivalent = 2 QB. Itotal =6A,Ii=2A,andRequivalent = 4QC Itotal = 10A,Ii = 4 A,and Requivalent =2 QD. Itotal =10A,Ii = 6 A,and Requivalent =4fi

SolutionAt first glance, it seems like this is a difficult question. But if we attack thequestion using Kirchhoff's Junctwn rule, we should be on our way to the answer.Resistors 1 and 2 are in series, so they will share the same current. I2 is given as 4amps, so Ii must also be 4 amps. This eliminates choices B and D. The currentsplits at the junction into two pathways: the upper path and lower path. Thecurrent through the upper path is 4 amps and the current through the lower pathis 2 amps, so the current going into the junction, Itotal/ is 4 + 2 amps. BecauseItotal is 6 amps, the best answer is choice A.

The equivalent resistance for the circuit can be found using Ohm's law. For theoverall circuit, we get:

Vcircuit = Itotal Requivalent

The total voltage is 12 V and the total current is 6 amps, so the equivalentresistance for the circuit is 2 Q.



Electric Circuits



Lefs apply this to some simple circuits. Figure 9-19 shows three examples ofsimplified calculations thatyoucando in your head.

RX = 8Q

R2 = 4Q

RaxR2

Rj + R2 8 + 4 12 3

Rlx R2 9x3Rx +R2 ~ 9+3

27 9~ = 7 = 2.25Q12 4

R1= 9Q

R2= 3Q

Ci=2C

-\\C2= 6C

C!xC2

Ci +C2 =

Figure 9-19

2 + 6 8 2

A shortcoming of this technique is that it applies only if there are two resistors inparallel or two capacitors in series. When there are three or more resistors inparallel, then you need to analyze the circuit two at a time. Figure 9-20 shows anexample of the simplified calculation applied to three resistorsin parallel.

n/WHRj = 6Q

R2= 3Q

-to-R3= 8Q

f-20

16

RjxR2 6x3

Rx + R2 6 + 3

R12 x R3 2x8

2 + 8 10Rl2 + R3= 1.6Q

Figure 9-20

Lastly, before leaving circuits and circuit calculations, lefs rehash the differencesbetween parallel circuit elements and series circuit elements.

Series Circuit Elements

Same current

Different voltage drop

Increased equivalent resistance

Reduced equivalent capacitanceLarger resistors drain more power

(P =I2R(Pf as Rf)Devices are dependent on one another

Parallel Circuit Elements

Same voltage dropDifferent current

Reduced equivalent resistance

Increased equivalent capacitance

Smaller resistors drain more power

(P =V2/R(PtasR|)Devices are independent of one another



Alternating Current and VoltageSo far, we have been discussing direct current (DC) circuits, in which a constantvoltage source produces a unidirectional current flow (flowing constantly fromthe cathode to the anode). For 99% of the electricity generated for commercialuse, however, the voltage source is not constant but instead is alternating, so itresults in an alternating current (AC). A simple AC circuit is shown in Figure 9-21; the new circuit element is the alternating voltage source (the sine wave in thecircle indicates AC).

Figure 9-21

AC is used instead of DC, owing to inherent inefficiencies (e.g., heat loss) in DCcircuits that can be overcome when the voltage alternates. In addition, with ACcircuits, there are brief instances where the current and voltage are zero, meaningthat ifs a safer circuit if one were to accidentally contact it. It is easier to let go ofa circuit when it has no current running through it. Because of this alternatingvoltage, the current also alternates with time (Figure 9-22). In the United States,this oscillation frequency is typically 60 Hz.

time

time

Figure 9-22

Recall that we defined power in equation (9.11) as P = IV. The power that we areusually interested in is the average power. The relationship between averagepower and current involves the root-mean-square current Inns quantified inequation (9.28), while the relation between average power and voltage involvesthe root-mean-square voltage Vmis quantified in equation (9.29). As plotted inFigure 9-22, the root-mean-square of something is just a way to describe a nonzero average value of a quantity that can be either positive or negative. Thisresults in an average power of Pavg = Irms^rms-


Alternating Current


Physics Electric Circuits Alternating Current

Equations (9.28) and (9.29) describe the root-mean-square current and voltagerelative to the maximum current and maximum voltage respectively.

Irms =-£r (9.28)V2

Vrms =^ (9.29)V2

Example 9.11aDoubling both the root-mean square current through and root-mean squarevoltage across a resistor in an AC circuit will increase the power drain by a factorof:

A. 1.

B. V2.C. 2.

D. 4.

Solution

This question is a great example of irrelevant information being intimidating.The fact that ifs an AC circuit as opposed to a DC circuit, with which we aremore familiar with in terms of physics exposure, means that we must considerthe Irmsand Vnns instead of just I and V. The reality is that it just doesn't matter,because it will be solved in the same fashion as a DC circuit question is solved.Doubling both factors Irms and Vrms in equation (9.11) results in a four-foldincrease in the power drain, since the current and voltage are multiplied. Don'tbe scared when you see an "rms" in a problem about AC circuits; the conceptualaspects of current, voltage, and power are still the same as they are for DCcircuits.


Example 9.11bDoubling the maximum voltage in an AC circuit will increase the root-meansquare voltage by a factor of:

A. 1

B. V2C. 2

D. 4

Solution

According to equation (9.29), the root-mean square voltage is directlyproportional to the maximum voltage. This means that if one of the two were toincrease, then the other value must increase by the exact same factor. Doublingone will therefore double the other. Because Vmax is doubled, V^s should alsobe doubled. Be careful not to fall for the trick of seeing a square root in theequation and blindly assuming that the answer must also include a square rootterm.



25 Electricity and Electric Circuits Review Questions

I. Resistors and Heating

II. Series vs. Parallel Circuits

III. Circuit City


(1-7)

(8-14)

(15-21)

(22 - 25)

The main purpose of this 25-question set is to serve as a review of the materialpresented in the chapter. Do not worry about the timing for these questions.Focus on learning. Once you complete these questions, grade them using theanswer key. For any question you missed, repeat it and write down your thoughtprocess. Then grade the questions you repeated and thoroughly read the answerexplanation. Compare your thought process to the answer explanation and assesswhether you missed the question because of a careless error (such as misreadingthe question), because of an error in reasoning, or because you were missinginformation. Your goal is to fill in any informational gaps and solidify yourreasoning before you begin your practice exam for this section. Preparingfor theMCAT is best done in stages. This first stage is meant to help you evaluate howwell you know this subject matter.


Resistors are often made of carbon-based compoundsand have a cylindrical shape. Such resistors have a resistancethat depends on their dimensions, as is shown in Figure 1.

1 1

I »1 *I »1 *I %\ *

length

0.6 cm

0.9 cm

w + m

Cylinder radius (cm)

Figure 1 Resistance as a function of resistor dimensions

Resistors are often used in Joule heating experiments toraise the temperature of liquids or gases by a desired amount.The resistance to current flow through a resistor connected toa power supply leads to heat dissipation in direct proportionto the current through the circuit The resulting temperatureincrease of a fluid depends upon this resistor, as well as thefluid's specific heat and the heating time. Figure 2 showstypical data for Joule heating experiments in various fluids.Each trial lasts 10 minutes, all fluid samples are 10 gm, andeach resistor is connected to a 12-V DC voltage source.

5

DOO 25 kQ

DO

D O

DlOOkQ

D o

D O -

C (cal/gm-K)

Figure 2 Datafor a Joule heatingexperiment

If two resistors experience the same current, then the onewith greatest resistancedissipates the most heat

1. When designing a common resistor, an engineer canmaximize its resistance by:

A. minimizing boththecylinderlength and radius.B. minimizing the cylinder length while maximizing

the radius.

C. maximizing the cylinder length while minimizingthe radius.

D. maximizing both the cylinder length and radius.

2. When a known and unknown resistor are connected in aseries DC circuit, the unknown resistor heats morerapidly than the known resistor. Comparing the tworesistors, the unknown resistorwould likelyhave:

A. a largerresistance anda largervoltage drop.B. a largerresistance anda smallervoltage drop.C. a smaller resistance anda larger voltage drop.D. a smaller resistance anda smallervoltage drop.


3. In a Joule heating experiment which of the followingchanges would increase the rate at which the solutiontemperature rises?

I. Decreasing the mass of the solution.

II. Increasing the specific heat of the solution.

III. Increasing the voltage of the battery.

IV. Increasing the resistance of the resistor.

A. I and HI only

B. II and IV only

C. I, II, and III only

D. I, HI, and IV only

4.

5.

6.

7.

Where p is the resistivityof a material, A represents thearea of a cylindrical resistor, and L represents thelength, the resistance of the resistor is found to be:

A pAL.

?A/L-

PL/A-P/AL-

A.

B.

C.

D.

For two cylindrical resistors made of identical materialand with equal radii, Ri and R2, where R\ is longerthan R2, whatis truewhen the twoare in parallel?

A. Ri dissipates moreheat than R2.

B. R2dissipatesmoreheat than R\.

C. Ri experiences a greater voltage drop than R2.D. R2 experiences a greater voltage dropthan Ri.

Which of the following systems requires the greatesttime to undergoa 10°Ctemperature increase?

A. A 0.50 cmCu resistor connected to a 6V battery in200mLof a solution with C = 0.80cal/g-K

B. A 0.50 cm Cu resistor connected to a 12V batteryin200mLof a solution with C = 0.80cal/g-K

C. A 0.50 cmCu resistor connected to a 6V battery in200mLof a solution with C = 1.20cal/g-K

D. A 0.50 cm Cu resistor connected to a 12V batteryin200mLof a solution with C = 1.20cal/g-K

Two resistors of unequal resistance aligned in parallelexperienceall of the following EXCEPT:

R1R2A. an equivalent resistance of

B. an equal voltage drop.C. an equal power drain.D. an unequal current

Rl +R2



Common electric circuits consist of batteries and

resistors. Batteries convert chemical energy, throughoxidation-reduction reactions, into electrical energy (in theform of electrical flow). Resistors convert electrical flowinto heat, light or work. Examples of resistors include aheating coil in an electric stove, the filament in anincandescent bulb, or the rotating loop of an electric motor.

Electrical circuits that consist of more than one resistor

and more than one battery can be hooked up in one of twoways: in parallel or in series. Six unique circuits withresistors and batteries in both series and parallel are shown inFigure 1. The batteries have voltage V and the resistorsresistance R. Ohm's Law relates the variables of a circuit,where I is current

V = IR

Equation 1 Ohm's law

8.

Circuit I

Circuit II

Circuit in

Circuit IV

Circuit V

Circuit VI

Vi-=F

V,"^

WW"RiR2

WW-

•vWARiR2

v2R2

WW

i—WW-RiR2

i_VWv^

VlT

V, T V2

4=- v2 r,

Figure 1

TX

v2

=r^V2

Ri

•v2

R2

R2

How do the voltages of Circuits I and II compare?

A. Vcircuit I > Vcircuit II

B. Vcircuit I < Vcircuit II

C Vcircuit I = Vcircuit II if Rl = R2

D- Vcircuit I = Vcircuit II if Rl * R2


9. If Ri < R2 and Vi < V2 in Circuit II, then how wouldthe current flow?

A. Clockwise with Ii > I2

B. Clockwise with Ii = I2

C. Counterclockwise with Ii > I2

D. Counterclockwise with Ii = I2

10. How does the internal resistance of a battery, Rbattery*affect the power drain by Ri and R2?

A. Both drain less power.

B. Both drain more power.

C. The power drain does not change in either resistor.

D. It depends upon how big Rbattery is compared toRl and R2.

11. If Ri has greater resistance than R2 in Circuit V, whichresistor drains more power?

A. Ri drains more power.

B. R2 drains more power.

C. Ri and R2 both drain the same power.

D. Ri drains more power if V > (R1HR2), but R2

drains more power if V < (R1HR2).

12. How are the total voltage and equivalent resistancefound in Circuit VI?

A. VTotaI = Vi+V2;Req = Rl + R2

B. VTotal =Vi-fV2;Req =oRrR„2Rl + R2

C VTotal= Vl'V,? ;Req =Rl+R2Vi + V2

D. VTotaI =-^_;RK,= -Vl+V2 Rl+R2

13. What is true for a circuit having two resistors, Ri andR2, of unequal resistance?

I. Resistors in series experience the same current.

II. Resistors in parallel experience the same voltage.

HI. They have equal power drain if voltage is equal.

A. I and II only

B. I and III only

C. II and III only

D. I, II, and III

14. How do total currents compare in Circuits I, II, HI, andIV, given Vi = 6V, V2 = 6V, Ri = 2Q, and R2 = 4Q?

A. Ii = Im > Iiv > III

B. Iiv > III > Ii = IlII

C. Ii > Im > Iiv > III

D. iiv > in > ii > im



Batteries, resistors, and capacitors are three of manycircuit elements that may be connected in series, in parallel,or in combinations of both to achieve a desired effect Circuit

elements in series share the same current while circuitelementsin parallel share the same voltage difference. Often,the operation of a circuit may be changed by the opening orclosing of a single switch, which can change the circuitelements from being in series to being in parallel. This isobserved in the following diagram of an RC circuit, whichcontains two switches:

Ri Switch 2

R3

Switch 1R2

Figure 1

Ohm's law is given as:

AV = IR

where AV is the potential difference across the resistor, I isthe current and R is the resistance. The capacitance of acapacitor is written as:

C=^AV

where C is thecapacitance, Q is the charge on the plates, andAV is the potential difference across the plates. Theseequations let us determine voltage, current resistance, andcapacitance at any given point within the circuit.

15. When Switch 2 is closed, is the capacitor in the abovecircuit in parallel or in series with any other circuitelement?

A. No.

B. It is in parallel with R2 and in series with R3.C. It is inseries with R2 only.D. Itis inparallel with both R2 and R3.

16. Initially, the capacitor is uncharged. Switch 2 is leftopen, while Switch 1 is closed. Immediately afterSwitch 1isclosed, what is thecharge on thecapacitor?A. Ce

0

Ce

B.

C.

(Ri +R2)D. CIR2


17. Immediately after Switch 1is closed (Switch 2 isopen)and the capacitor is still negligibly charged, what is thecurrent through Rj?

A.

B.

D.

(Ri +R3)E

RI+52±R3Ro x Rj

18. Immediately after Switch 1 is closed (Switch 2 is open)and the capacitor is still negligibly charged, what is thecurrent through R3?

19.

20.

21.

A.

B.

C.

D.

0

R,

(Ri +R3)8

Ri + R2 + R3Rt X Ra

Once the capacitor is fully charged, Switch 2 is closed(Switch 1 remains closed). Closing Switch 2 will havewhat effect on the total resistance of the circuit?

A. The total resistance will increase.

B. The total resistance will decrease.

C. The total resistance will be unaffected.

D. Whether the total resistance increases, decreases,or remains unaffected will dependon the numericalvalues of the resistors.

After Switch 1 has been closed for a long time (withSwitch 2 open), the capacitor becomes fully charged.The charge on the capacitor cannow be writtenas:

A. 0

Ce

C*6*lv2- R3

B.

C.

D.

R, + R2 + R3Oe_R3

R1 + R3

If Switches 1 and 2 are both closed and if thecapacitoris allowed to charge completely, the current through R2will exceed the current through:

A. R3, only if R3 > R2.

B. R3,onlyifR2>R3.

C. C, only if R2 > Ce.

D. C, only if Ce > R2.



22. What happens over time to the capacitance of acapacitor, as the space between its plates is filled at aconstant rate with a polar solution?

A. The capacitance remains constant until the volumeis completely filled, then the capacitanceimmediately drops to the new value of K»C0.

B. The capacitance remains constant until the volumeis completely filled, then the capacitanceimmediately increases to the new value of K»C0.

C. The capacitance gradually decreases as the solutionis added until the volume is completely filled, thenthe capacitance remains at the new value of K'Q,.

D. The capacitance gradually increases as the solutionis added until the volume is completely filled, thenthe capacitance remains at the new value of K»C0.

23. When a capacitor is fully charged and before it beginsto discharge, the potential difference across thecapacitor plates is:

A. 0 volts.

B. equal to the emf'of the battery.

C. C/Qmax*

D. QmaxC

24. The maximum charge that can be stored on a capacitordepends on:

I. the capacitance of the capacitor.

II. the emfof the battery.

HI. the equivalent resistance of the circuit

IV. the power drain of each resistor in the circuit.

A. I and HI only

B. II and HI onlyC. I and II only

D. II only


25. How is the net energy transfer in a circuit bestdescribed?

A. Kinetic to Electrical Potential

B. Chemical Potential to Heat

C. Heat to Electrical Potential

D. Electrical to Kinetic

1. C 2. A 3. A 4. C 5. B

6. C 7. C 8 A 9. D 10. A

11. B 12. B 13. A 14. A 15. D

16. B 17. B 18. A 19. B 20. D

21. A 22. D 23. B 24. C 25. B

YOU ARE DONE.

Answers to 25-Question Electricity and Electric Circuits Review

Passage I (Questions 1-7) Resistors and Heating

1. Choice C is the best answer. According to Figure 1, as the cylinder radius increases, the resistance of the resistor decreases.This rules out choices B and D. Now, for a fixed cylinder radius, the longer resistor will have the larger resistance. Thismakes choice C a better answer than choice A. Keep in mind that R= pL/j\» where Ris resistance, p is resistivity, L isresistor length, and A is the cross-sectional area of a cylindrical resistor. According to the equation, as the length of acylindrical resistor increases, its resistance increases. As the width of a cylindrical resistor increases, its resistance decreases.The best answer is choice C.

2. Choice A is the best answer. When in series, all circuit elements share the same current, so two resistors in series have thesame magnitude of current. Since P=IV and I is the same for both resistors, the resistor that uses more power (i.e., heatsmore rapidly) must have the larger voltage drop. This rules out choices B and D. Since V=IR and I is the same for bothresistors, the larger voltage drop occurs across the larger resistor. This means that choice A is a better answer than choice C.For resistors in series, the current is the same through each resistor, so the relative power can be most easily ascertainedusing P=I2R. This tells us that power drain is directly proportional to the resistance. The best answer is choice A.

3. ChoiceA is the best answer. Because all of the heating experiments are done for the same 10-minute duration, the AT axisin Figure 2 directly relates to the heating rate. Toanswer this question, therefore, let's seehow mass, specific heat, voltage,and resistance affect the temperature increase. According to Figure 2, increasing the specific heat, C, decreases thetemperature increase AT. This may seem intuitive, because a greater heat capacity requires more heat so the temperaturedoes not rise as much for a given amount of heat You may also recall that q = mCAT, where a larger C would equate to asmaller AT,q and m being constant This invalidates Statement II, and eliminates choices B and C. Between choices A andD, we need only consider Statement IV. According to Figure 2, fora given specific heat, the smaller resistor has the largertemperature increase. This makes Statement IV invalid. The best answer is choice A.

4. Choice C is the best answer. You could answer this straight from your memory, if you happen to recall the equation forresistance in a cylindrical resistor (or segment of wire). If you do not recall the equation, then Figure 1 lists the pertinentdata to deduce therelationship between thekey variables. All choices have resistivity, p, in the numerator, so to determinethe correct answer, we need only consider Aand L. The longer resistor (0.9 cm versus 0.6 cm) has higher resistance atequalradius, so resistance increases with L. Thiseliminates choices B and D. As theradius increases, theresistance decreases, soresistance must beinversely proportional to the cross sectional area. This eliminates choice A and leaves only choice C.Thebest answer is choice C.

5. Choice B is thebest answer. The longer resistor of two resistors with equal cross sectional area and equal resistivity hasgreater resistance. When the two resistors are in parallel, the voltage drop across each resistor is equal but more currentflows through the smaller resistor, so the resistor with lower resistance drains more energy and therefore dissipates moreheat. The resistors experience the same voltage drop inparallel, sochoices Cand Dareeliminated. Given that R2 isshorterthan Ri, it has less resistance and therefore drains more energy and dissipates it in the form of heat. The best answer ischoice B.

6. Choice Cisthe best answer. The greatest amount oftime required to heat a solution will be found with the solution havingthe greatest heat capacity combined with the slowest rate of heating. Choices A and B can be eliminated immediately,because they have lower heat capacities than choices Cand Dwhile having all other variables equal. The resistors are equalin all of the choices, but the voltages differ. At higher voltage, according to V = IR, there is greater current through theresistor at higher voltage. With greater current the solution in choice D heats faster than the solution in choice C. Keep inmind that we are looking for the longest heating time, sowe are looking for the slowest rate of heating. The bestanswer ischoice C.

7. Choice C isthebest answer. Inparallel, the reciprocal ofthe equivalent resistance equals the sum of the reciprocals ofeachresistor in parallel, so choice A is a valid statement and thus eliminated. In parallel, the voltage drop is equal, but thecurrents are different As a consequence, choices B and D are also valid statements. This eliminates choices B and D andleaves choice C as the lone choice standing. The power drain isV2/R, which means that with different values for resistance,the powerdrain through each resistor must also be different. ChoiceC is an invalid statement so it is the best answer. As apoint of interest, the resistor with less resistance drains more energy when they are aligned inparallel. This is the opposite ofwhat is observed in series. The best answer is choice C.

Copyright ©byTheBerkeley Review® 206 MINI-TEST EXPLANATIONS

Passage II (Questions 8 -14) Series versus Parallel Circuits

8. Choice A is the best answer. In Circuit I, the two batteries are in series and aligned so that the overall voltage is the sum ofVj and V2. In Circuit II, the two batteries are in series and aligned so that the overall voltage is the difference between Vjand V2. That is to say that the two separate batteries oppose one another in Circuit II. This means that the overall voltage isgreater in Circuit I than in Circuit II, making choice A the best answer. This is true regardless of the resistors. The bestanswer is choice A.

9. Choice D is the best answer. Current flows from cathode to anode (opposite of electrical flow). V2 is greater than Vi, sothe electrical current is dictated by V2. Circuit notation dictates that the longer lineof the battery represents the cathode, socurrent will flow in a counterclockwise fashion. This eliminates choices A and B. The resistors are in series, so the currentthrough each resistor is the same. Current varies when resistors are in parallel, because current has twopathways from whichto choose. But when the resistors are in series, current has only one way to flow, thus it will have a fixed value (Itotal- Thebest answer is choice D.

10.

11.

Choice A is the best answer. We are talking about a series circuit that now includes an extra resistor, Rbattery- You canimagine that the more resistors you have in series, the harder it is for the battery to push a current. Thus, this extra resistanceresults in a slightly smaller current. The power drained by the resistors is:

P=I2R

Thus, if the current, I, decreases a bit, then so will the powerdrained. Note that R in the equation does not change, because itis a property of the resistor itself, and not dependent on the circuit in which it is placed. The best answer is choice A.

Choice B is the best answer. The formulas of interest are: V = IR and P = IV. CircuitV has the two resistors in parallel.Resistors in parallel experience the same voltage drop, but not necessarily the same current. Thus, we should rewrite thepower formula so that it contains only V and R (which is what we will use for the comparison.) Plugging Ohm's law into thepower equation gives:

2 9 9p=Y_ SOj for Rj and R2 we get: P =y_ and p2 =y_R Ri R2

If Rj > R2, then P2 > P\. So, in a parallel circuit, a smaller resistor drains more power. This is the opposite of the series

circuit case. A bigger resistor does not always drain more energy. More power gets drained when more current runsthrough a particular circuit. In the series circuit, the same current runs through both resistors. The bigger resistor drainsmore power. However, in a parallel circuit, more current runs through the path of less resistance. More power is drained bya smaller resistor. The best answer is choice B.

12. Choice B is the best answer. In Circuit VI, the two batteries are in series, so the total voltage is found by summing theindividual voltages. This eliminates choices C and D. The two resistors are in parallel, so the reciprocal of the equivalentresistance is the sum of the reciprocals of each individual resistor. This eliminates choice A, leaving only choice B as apossible answer. To double check, you may either calculate the equivalent resistance, or verify the equation by observing theunits. As shown, choice B will have units of Q for resistance, so the equation seems viable at a glance. As a point ofinterest, when there are two resistors in parallel, the equivalent resistance is their product divided by their sum. The bestanswer is choice B.

13. Choice A is the best answer. When two resistors are in series, the same current flows through both of the resistors.Electrical current is like water flow in that the volume moving is the same at any point in the path. When there is only oneflow path, the same current must flow through every part of the path. Thus, both resistors must have the same current, makingStatement I valid. When a circuit is wired in parallel, the current splits up, so current can flow more than one way. Thecurrent need not split up evenly. In fact, more current flows along the path of least resistance - like with water in a divergingriver. This results in each parallel path having the same voltage drop (-IR).

In parallel, resistors share a common voltage. This makes Statement II valid. The voltage across each resistor and thebattery is V. This is because a line can be traced from the top plate of the battery to the top of Rj without crossing any

other electrical components. This means that the top of the resistor is at the same potential as the top plate of the battery.A line can also be traced from the bottom of Ri to the bottom plate of the battery without crossing any other components, so

the bottom of R\ is at the same potential as the bottom plate of the battery. Since the voltage across the battery is V, the

voltage across both R\ and R2 must also be V. Resistors in parallel have the same potential V across them. Because Req isdifferent in the two circuits, Ieq is also different, so the power drain (V-I) must be different. This makes Statement IIIinvalid. The best answer is choice A.

Copyright©by The Berkeley Review5 207 MINI-TEST EXPLANATIONS

14. Choice A is the best answer. The total current will be greatest in the circuit with the greatest total voltage and the leastequivalent resistance. Equivalent resistance is minimized when circuits are in parallel. Circuits I and III are equal becauseboth have the two batteries in series and the two resistors in series. This eliminates choices C and D. The current in CircuitII must be zero because the batteries are of equal voltage, and they oppose one another. The total voltage is the differencebetween Vi and V2, which is zero. With zero voltage, the current must also be zero. This eliminates choice B and makeschoice A the best answer. Circuit IV has a low current because the batteries are in parallel (minimizing total voltage) and theresistors are in series (maximizing equivalent resistance). Circuit IV has less total current than Circuits I and III. The bestanswer is choice A.

Passage III (Questions 15 - 21) Circuit City

15. Choice D is the best answer. When circuit elements are in parallel, they share the same voltage difference. The three circuitelements in this question (C, R2, and R3) are all side-bys-side and they share the same voltagedrop, which means they are allin parallel. A good way to check whether circuit elements are in parallel is to see whether their ends meet at commonjunction points. If we redraw part of the circuit:

I

R3

R2

II

we see that the two ends of each circuit element touch at points I and II. Thus, all three elements are in parallel. The bestanswer is choice D.

16. Choice B is the best answer A capacitor is a device that stores charge. When Switch 1 is first closed, the charges have notyet been pushedonto the capacitor. It takes a finite time for the charges to move onto the capacitor plates. Thus, immediatelyafter Switch 1 is closed, the charge on the capacitor is zero, but it will begin to build up very soon thereafter. The bestanswer is choice B.

17. Choice B is the best answer. C and R3 are connected in parallel. Remember that initially, there is no voltage drop across thecapacitor (as there is negligiblecharge on it), which means that there is no voltage drop across R3 at the instant that Switch 1is closed. This is to say that at the instant Switch 1 is closed, all of the current goes to the capacitor (following the path ofleast resistance) rather than through R3. From Ohm's law: AV = IR. If there is no voltage difference, there can be no currentflow. When Switch 1 is closed, initially it is as though all we have is one battery and one resistor, Rj, connected in series.Again, from Ohm's law:

e = I Rj, therefore I =- E

R,

Note that R2 does not enter into consideration here. With Switch 2 open, R2 will not have any current flowing through it andwill not havea voltagedrop. With Switch 2 open, it is as though R2 werenot there. The best answer is choice B.

18. Choice A is the best answer. When thecapacitor is negligibly charged, anda current is turned on in thecircuit, almost all ofthe current will flow toward the capacitor. The resistor offers resistance to the flow of current, but an uncharged capacitordoes not. Therefore, current will flow through the capacitor section of the circuit and not through R3. The best answer ischoice A.

19. Choice B is the best answer. With just Switch 1 closed, R\ and R3 add in series. With Switch 2 closed, current can nowflow through the R2 path as well. Because R3 and R2 are connected in parallel, Req decreases for that segment of the circuit.Therefore, Req for the entire circuit decreases. Resistors connected in parallel have an equivalent resistance that is smallerthan either individual resistor; R2 and R3 connected in parallel produce an equivalent resistance that is smaller than R3 alone.When this equivalent resistance is added to Rj, we get a resistance that is smaller than when just Switch 1 is closed. The bestanswer is choice B.

Copyright©by The Berkeley Review® 208 MINI-TEST EXPLANATIONS

20.

21.

Choice Dis the best answer. Since the two cars have the same initial kinetic energy, the work required to stop both cars is:

Q = CAVC= 0-AV

Since Cand R3 are connected in parallel, the voltage difference across Cis the same as the voltage difference across R3 (AV=I3R3 =1^3). After Switch 1has been closed for a long time, the capacitor has become fully charged, and no more currentflows to the capacitor. This means that there is now a voltage drop across C (a maximum value), so there is a voltage dropacross R3. Since Switch 2 is still open, no current flows through R2. Thus, in determining the total current flowing in thecircuit, it is as though we have a battery connected to two resistors in series:

1—vWARi

v — R3.

Resistors in series add together to give an equivalent resistance of Rserjes = Ri + R3- Using Ohm's law, wecan solve for thetotal current in the circuit:

e ~ * ^series ==> * ~R1 + R3

To solvefor the voltagedrop across R3,we use Ohm's lawagain:

AV = IR3 => AV =

Finally, we use this voltage to solve for the charge on the capacitor:

Q= CER3R1+R3

£-R3R1+R3

This answer can also be approached using a bit of physical intuition. Since Switch 2 is open, R2 will not figure into any

voltage drops or charge accumulations. Therefore, choice C can be eliminated. Choice A can be eliminated, because ifcurrent flows through the capacitor section at some point in time (as it does), then charge will accumulate on the capacitor.Choice B can be eliminated, because it says that the voltage across the capacitor plates is e. This is wrong, because there is apotential drop from the battery across Rj, which will lower the actual voltage across the capacitor. The best answer ischoice D.

Choice A is the best answer. After the capacitor is fully charged, no current will flow into the section of circuit containingthe capacitor. That may make choices C and D seem correct-but those choices compare R2 to Ce, representing resistanceand charge, respectively. Comparing unlike physical quantities is nonsensical, so choices C and D are incorrect. Between thetwo resistors R2 and R3, the smaller resistor necessarily must have more current flowing through it. Notice that these tworesistors have the same voltage across them, as they are wired in parallel. So if R2 is to have more current than R3, R2 must

have the smaller resistance. The best answer is choice A.


22. Choice D is the best answer. As the spacebetween thecapacitor plates is filled with a polar substance, the capacitance willincrease. This eliminates choices A and C. Because the fluid is being added at a constant rate, the capacitance will change ata constant (gradual) rate, not abruptly. The best answer is choice D.

23. Choice B is the best answer. When the capacitor is fully charged, it will have some voltage across it, because it storescharge and it has a capacitance. This eliminates choice A. Recalling thatQ = VC, vve know thatV = Qmax'C, notC/Qmax-This eliminates choice C. Choice D has the wrong units, leaving only choice B. Regarding choice B, the capacitor shouldhave the same voltage across it as the battery, because this circuit (battery-Rc-C) has no current flowing through it. Becauseno current flows through the resistor, there must be no voltage dropacross the resistor; this leaves a voltage dropacross thecapacitor equal to thevoltage gainacross thebattery (i.e., the emfof thebattery). The best answer is choice B.


24. Choice C is the best answer. Capacitance is defined as: C = Q/V, where Q is the charge on the plates of the capacitor, and Vis the voltage difference between the plates. The maximum charge therefore depends on option I and II only. There is nodependence on resistance of any kind. The best answer is choice C.

25. Choice B is the best answer. As current is passed through a resistor, it heats up. The energy transfer must involve heat, sochoices A, C, and D can all be eliminated. Energy is being transferred from the chemical battery to the resistors, so it startswith chemical potential. It goes through electric flow and eventually becomes heat. The best answer is choice B.


l-t|^^M-i9it^Si&'' ajtid Electric Circuits Practice Examr- - »~t; ,' J•.-*'" r-s

I. Basic Circuit

II. " Dielectric Experiment?

IQ. Resistor Alignment


IV. Wheatstone Bridge

V. Pacemaker Circuit

,Questions Not Based on a Descriptive Passage

Yt. RC Circuits

VII. Resistance is Futile


Electricity and Electric Circuits Exam Scoring Scale


:. 42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3

(i-6)

(7-12)

(13 -18)

(19 - 22)

(23 - 28)

(29 - 33)

(34-37)

(38-43)

(44-48)

(49-52)

Passage I (Questions 1-6)

In a laboratory experiment, a student is given threeresistors and a battery of voltage V. The assignment is toconstruct the circuit shown in Figure 1 and to measure thecurrents at different points within the circuit The values forI, Ii, I2, and I3 are all recorded. During the course of the

experiment, a battery with a different voltage is substitutedfor the original one, and the current is recorded at the samepoints as before.The resistors are aligned in the same fashionfor all of the trials, so they have the same resistance in eachtrial, despite the change in voltage.

Figure 1

Rj = R2 = 2 R3 = 2 ohms

In all questions that follow, ignore the internal resistanceof the battery. The only resistance to consider is that of theresistors specifically shown in the circuit.

1. If V = 12volts, then which of thefollowing changes inresistorswill increase the value of Ij?

A. Increase R\B. Increase R3

C. Increase R2, while changing R3 so that Req oftheentire circuit remains constant

D. Decrease R2, while changing R3 so that Req oftheentire circuit remains constant

2. If the voltage of the battery were 6V, how could theequivalent resistance of the circuit be determined?

A. Req = Rj + R2 + R3

B. Req =R,

R, + R2

C. Req - R3 +Rj +R2

D. Ren =(RlHR2) +RR, + R2eq


3.

4.

5.

6.

If all three resistors have equal resistances, which of thefollowing statements is true?

A. R3 will dissipate the least power.

B. Ri will dissipate more power than R2.

C. R3 will have a larger current than either Ri or R2.

D. Ri and R2 will each have a larger current than R3.

What is the power dissipated in the circuit, if Req is 2ohms and I is 6 amps?

A. 9 watts

B. 18 watts

C. 36 watts

D. 72 watts

In which of the following circuits will the energydissipated in the resistors be the GREATEST?

A. Vbattery = 6 volts, Itotal =3 amps, and t = 2seconds

B. Vbattery =9 volts, Itotal = 2 amps, and t = 1secondC Vbattery = 6 volts, Itotal = 2 amps, and t =2

seconds

D» Vbattery = 12volts, Itotal = 3 amps, and t = 1.5seconds

Which graph BEST relates the current flowing througha circuit to the power dissipated in it, if theelectromotive force in the battery is 12 volts?A. B.

Current (A) Current (A)



A capacitor builds up potential energy by storingelectrical flow. It has two plates, one rich in electrons and theother poor in electrons. The capacitance of a capacitor isaffected by the medium between its two plates. Equation 1shows the capacitance of a standard capacitor that has somemedium between the plates (i.e., the space between the platesis not a vacuum):

Equation 1 C = K'C,

where C is the capacitance, K is the dielectric constant, andC0 is the capacitance in a vacuum.

It is difficult to measure capacitance in a vacuum. Tomeasure the dielectric constant of various solutions, thefollowing apparatus was designed:

Figure 1

The coil has inductance of L henries, Q has variable

capacitance, and Cn has a fixed capacitance. The circuit hasa natural resonance frequency v0 given by:

Equation 2 v„ =1

2n YLfa +Ci,)

An experiment involves dipping Cn into varioussolutions and measuring the change in capacitance, bymonitoring the frequency of the circuit. The change incapacitance is determined by the dielectric constant (K) foreach solution. This constant is always equal to or greater thanone. As a general rule, more polar compounds have greatervalues of K than nonpolar compounds.

7. Which of the following changes will decrease thenaturalfrequency of the circuit by a factor of 2?

A. Increasingthe capacitance of Q by a factor of 4.

B. Increasing the capacitance of Q by a factor of 2,

and increasing Cn by a factor of 2.

C. Increasingthe inductance of the coil by a factor ofV2.

D. Increasing the inductance of the coil by a factor of4.


8.

9.

10.

11.

12.

Which of the following solutions will likely exhibit theGREATEST value of K?

A. Benzene

B. Cyclohexane

C. Octane

D. Water

How can the total capacitance of the circuit be found?

1A.

Qotal Ci Cn

B« Qotal = — +Ci Ch

C —L- = Ci +Q]Qotal

D. Qotal = Ci + Cn

Which of the following changes will NOT affect thecapacitance?

A. Increasing the distance between the capacitorplates.

B. Filling the inter-plate region of the capacitor with apolar liquid.

C. Changing the voltage source of the circuit.

D. Changing the materials of the capacitor plates.

Why might the solution into which Crj is dipped during

the experiment be immersed in an oil bath that is fittedwith a heating coil?

A. Polarity of the solution may be fixed, but theheating coil is needed to maintain homogeneity.

B. Temperature must be held at a fixed value, becausethe dielectric constant can vary with temperature.

C. Capacitors must be heated to compensate for thedecrease in current as the capacitor charges.

D. The resistance of the circuit is minimized at highertemperatures.

How is the value of K measured during the experiment?

A. By comparing the current flowing through Q withthat flowing through Cn-

B. By comparing the current generated by Q with thatflowing through Cn-

C. By measuring the change in inductance.

D. By measuring the change in the resonancefrequency of the circuit.

GO ON TO THENEXT PAGE

Passage 111 (Questions 13-18)

In an experiment designed to study resistance, a studentdesigns three different circuit arrangements, each consistingof three resistors and a battery of voltage V. The studentmeasures the currents at the different points within each ofthe circuits. During the course of each trial, the resistors areinterchanged, and the current is recorded at the same points.

The values for total current (I) and the currents througheach resistors (Ij, I2, and I3) are all recorded and compared.

The resistors are aligned in the separate trials as shown inFigure 1. The same resistors are used in each trial, so eachindividual resistor has the same resistance in each system.

System I

=r V

System II

I l2gR2Ri^J1

System m

TV r4I2|%J|i3

Figure 1

Rl = 1 ohm, R2 = 2 ohms, R3 = 3 ohms

13. In System II, which of the following arrangements ofthe resistors will yield the GREATEST total current, I,drawn by the circuit if V = 12 volts?

A. Ri upperleft, R2 upper right,and R3 bottom

B. R2upperleft, Rj upperright, and R3 bottom

C. R2upper left, R3 upper right, and Ri bottom

D. R3 upper left, Ri upper right, and R2 bottom


14. In a circuit arrangement like System II, if V = 12 volts,Req =4Q, and Ii = 2 amps, then which of the followingstatements is true?

A. Ri must be larger than R2.

B. R3 must be 4 Q.

C. I2 must be 1 amp.

D. Ii must be greater than I2.

15. How do the values for the equivalent resistance in eachof the three systems compare?

A. Req System I> Req System II > Req System IIIB. Req System III > Req System II > Req System IC. Req System II > Req System I> Req System IIID. Req System III > Req System I> Req System II

16. In which systems will Requivalent vary with thesequential placement of the resistors?

A. Requivalent varies inSystem IIonly.B. Requivalent varies inSystem I and System IIIonly.C. Requivalent varies inSystem IIand System III only.D. Requivalent varies inSystems I, II, and III.

17. What is the overall power dissipated in a circuitconnected toa 12-voIt battery, ifReq is4 Q?

A. 18 Watts

B. 36 Watts

C. 72 Watts

D. 108 Watts

18. What is the equivalent resistance for the circuit shownas System II in Figure 1?

A. 0.56 Q

B. 3.67Q

C. 4.50 Q

D. 6.00 Q


Questions 19through 22are NOT based on adescriptivepassage.

19. How would you describe the overall energy conversionin an electric circuit?

A. From kinetic energy to potential energyB. From potential energy to kinetic energyC. From heatenergy to electrical energyD. From electrical energy to heat energy

20. The characteristic time required for a capacitor todischarge through a resistive circuit is called the timeconstant of the circuit. The time constant of a

discharging capacitoris given by T = RC, where R is thetotal resistanceof the circuit and C is the capacitanceofthe circuit. If a third resistor were added in parallel to Riand R2, what would happen to the time constant of thecircuit?

A. The circuit's time constant, T, increases, becausethe circuit's resistance increases.

B. The circuit's time constant, T, decreases, becausethe circuit's resistance increases.

C. The circuit's time constant, T, increases, becausethe circuit's resistance decreases.

D. The circuit's time constant, T, decreases, becausethe circuit's resistance decreases.

21. An ammeter measures the current that passes through aparticularregion in a circuit With respect to any circuitwhose current it measures, the ammeter should have a:

A. larger internal resistance and be connected in serieswith the circuit.

B. larger internal resistance and be connected inparallelwith the circuit

C. smaller internal resistance and be connected in

series with the circuit.

D. smaller internal resistance and be connected in

parallel with the circuit.


22. For a simple circuit made of two resistors of unequalresistance, Ri and R2,and a battery, whatis true?

I. If the two resistors are connected in parallel, thenthey experience thesame voltage drop.

If the two resistors are connected in series, thenthey have different currents flowing through them.

The two resistors will experience the same powerdrain, whether they are connected in series or inparallel.

I only

III only

I and III only

II and III only

II.

III.

A.

B.

C.

D.



A circuit known as a wheatstone bridge can be employedto measure the resistance of a resistor. It requires havingthree resistors of known resistance, one of which has variableresistance, a resistor of unknown resistance, a galvanometer,and a voltage source. Figure 1 shows a wheatstone bridge.

*&M

i K I

"kh 7

Figure 1 Wheatstone Bridge Circuit

If the voltage across the galvanometer, G, is zero, thenthe voltage drop across Ri is equal to the voltage drop acrossR2- If this is true, then the following applies:

'lRl ='2R2 and qR3 = /2R4

*'lRl _ *2R211R3

Rj_ = R2 ., R4 =(R2XR3)12R4 R3 R4 Rl

In an experiment to compare the resistivity of threedifferent metals, students used a variation of the wheatstonebridge to study three resistors of identical dimensions, buteach made from a different metal. The device used in theirexperiment is shown in Figure 2.

Figure2 Wheatstone Bridge alternative for experiment

The slide-wire between Point b and Point z can move atPoint z. The wire is moved until the voltage across thegalvanometer is zero. The distance between Points a and zand the distance between Points z and c are measured. Theresistor for Rtj is changed in each of the three trials.

23. The resistance of the unknown resistor, Rtj, isequal to:

A (4-zX4-c)

B.

Rknown

Rknown

(4a-zX4-c)

C» Rknown ——•Oz-c

D- Rknown^,4i-z


24. The wire between Points a and c is made of a uniform

materia] and has a uniform radius to ensure that the:

A. relative resistances of segments AZ and ZC onlydepend on their respective lengths.

B. relative resistances of segments AZ and ZC onlydepend on their respective cross-sectional areas.

C. relative resistances of segments AZ and ZC areequal to one another.

D. resistances of segmentsAZ and ZC are equal to theunknown resistor, Rtj.

25. For the wheatstone bridge in Figure 1 to function, withR4 being the unknown resistor, what CANNOT betrue?

A. Either Ri, R2, or R3 must be a variable resistor.

B. Ri must equal R3 for the circuit to work.

C. The galvanometer should be able to measure bothmagnitude and direction.

D. The circuit will exhibit the same results when thepoles of the battery are reversed.

26.

27.

28.

What is the purpose of the galvanometer betweenPoints b and z in the wheatstone bridge in Figure 2?

A. To supply voltage to the circuitB. To measure the voltages at Points b and zC. To measure the potential difference between Point

b and Point z.

D. To serve as a short circuit so no current flowsthrough the unknown resistor,Rtj.

When Metal I is used as Rfc and Metal II is used in Rtj,the slide-wire is closer to Point a than Point c. WhenMetal I is used as Rk and Metal III is used in Rtj, theslide-wire is closer to Point c than Point a. What are therelative resistivities of Metal I, Metal II, and Metal III?

A« PMetal I > PMetal II> PMetal III

PMetal II > PMetal I > PMetal III

PMetal III > PMetal I > PMetal II

PMetal III > PMetal II > PMetal I

B.

C.

D.

If the voltage of the battery were doubled, how wouldthe position of Point z change?

A. The distance between Point z and the midpoint ofwire AC would increase by a factor of 4.

B. The distance between Point z and the midpoint ofwire AC would increase by a factor of 2.

C. The distance between Point z and the midpoint ofwire ACwould increase bya factor of V2.

D. The distance between Point z and the midpoint ofwire AC would remain the same.



The beating of a human heart is a regulated cycle ofcardiac muscle contractions. A group of specialized cells inthe right atrium of the heart acts as a pacemaker and beginseach cardiac cycle by generating automatic and rhythmicaction potentials that spread throughout the organ. For thosepeople whose natural pacemaker fails to work properly, anartificial pacemaker may help. An artificial pacemakeris twoRC (resistor-capacitor) circuits joined together. The capacitoris repeatedly charged and discharged. The basic pacemakercircuit is diagrammed below:

a-

QQ

HVWR„

circuit

Figure 1

When the switch (which is really a transistor) is closedto the left (as shown), the capacitor is charged through Rc

(the charging resistance). The switch then changes positionand the capacitor is discharged through R<j (the discharging

resistance). When the charge on the capacitor drops to e"1 =0.37 times its maximum value, other transistors in the formof a triggering circuit deliver a short pulse of current to theheart. The time to charge the capacitor is typically muchlonger than this pulse and is approximately equal to thenatural time between heart beats. Some pacemakers operateat a fixed frequency, while others operate only if the patient'sown heart fails to do so.

An important parameter used in characterizing an RCcircuit is the time constant, T, which obeys T = RC, where Ris the resistance (in ohms) and C is the capacitance (infarads). After one time constant has elapsed as the capacitorcharges, the current in the circuit has dropped to 0.37 timesits original value (I= I0 e"1), and the charge on the capacitoris 0.63 times its maximum value(Q = Qmax (1 - e"'))- Whendischarging the capacitor, the elapse of one time constantindicates that the charge on the capacitor has fallen to 0.37times its maximum value(Q = Qmax e"1)-

29. Ignoring the discharging part of the circuit, one wouldconclude that the maximum potential difference acrossRc occurs:

A. immediately after the switch is closed.

B. after a long time has elapsed.

C. at the same time that the maximum potentialdifference occurs across the capacitor plates.

D. after the maximum potential difference across thecapacitor plates is reached.


30.

31.

If the time to charge thecapacitor is 60 times thatof thetime to discharge the capacitor, the ratio ofRc : Rd is:A. 1:60

B. 1:1

C. 60: 1

D. 23 : 1

Which of the following graphs BEST represents thecharge on the capacitor as a function of time, over onecycle?

A. B.

Time Time

Time Time

32. For discharging the capacitor, we want Rd to be:

A. smaller than Rc, to insure a faster discharge timethan charge time.

B. larger than Rc, to insure a faster discharge time

than charge time.

C. smaller than Rc, to insure a faster charge time thandischarge time.

D. larger than Rc, to insure a faster charge time than

discharge time.

33. For the above artificial pacemaker, suppose that pulsesare triggered 60 times per minute by a 5-mF capacitor.What is the time constant of the charging RC circuit?(Neglect the time needed to discharge the capacitor.)

A. 0.25 seconds

B. 0.50 seconds

C. 1 second

D. 1.25 seconds



34. How are the electrical outlets in a typical home hookedup?

A. In series

B. In parallelC. In a combination of both series and parallel

D. Neither in series nor in parallel

35. Which of the following statements are TRUE?

I. The current passing through a conducting wire inthe presence of an electric field is zero.

II. Resistors provide electrical energy to a circuit

III. The amount of current leaving a resistor is smallerthan the amount of current entering a resistor.

IV. The current leaving the cathode is equal inmagnitude to the current arriving at the anode.

A. I only

B. I and III only

C. I, II, and III

D. IV only

36. Which of the following observations are valid?

I. If resistors are aligned in parallel, rather thanseries, then the equivalent resistance will be less.

II. When two resistors are aligned in parallel, thecurrent through the resistor of higher resistance isgreater than the current through the resistor oflower resistance.

III. Only resistors that are aligned in parallel willincrease in temperature as current passes.

A. I only

B. II only

C. I and II only

D. I and III only


37. What is the voltage drop across R3?

12V

A. 2V

B. 4V

C. 8V

D. 12V

IA

2A

1A

wvw^—'R3

4A

WW—'2Q



Physics students construct an RC circuit as shown inFigure 1. The voltage on the battery is V = 10 mV, thecapacitance is C = 2 pF, and all of the resistors have aresistance of R = 5 mQ. When switch Si is closed, thecapacitor charges up. When switch Si is open and switch S2is closed, the capacitordischarges.

Switch 1 Switch 2

V — R2.

Figure 1

38. When only switch Si is closed, how much charge willbuild up on the capacitor?

A. 2 x lO'8 C

B. 2 x 10"4 C

C. 5xlO-3C

D. 2x10'^

39. If the capacitor were initially uncharged, it would betrue that when both switches are closed simultaneously,the voltage across the capacitor:

A. increases, while the current through the resistorsdecreases.

B. increases, while the current through the resistorsincreases.

C. would shoot up rapidly and then decrease, whilethe current through the resistors increases.

D. would shoot up rapidly and then decrease, whilethe current through the resistors decreases.

40. A long time after both switches are closed, the currentthrough resistor Ri is:

A. 2 mA.

B. 50 mA.

C. 2 A.

D. 50 A.


41. If both switches were closed, and a new resistor wereinserted into the circuit before switch Si, then themaximum charge that could be stored on the capacitorwould:

A. increase, because the voltage across the capacitorwould increase.

B. decrease, because the voltage across the capacitorwould increase.

C. increase, because the voltage across the capacitorwould decrease.

D. decrease, because the voltage across the capacitorwould decrease.

42. If a steady current of 1 mA passes through resistor Rifor 10seconds, how much heat is generated?

A. 5xlO"9J

B. 5xl0-8J

C. 5xl06J

D. 5xl0"5J

43. Which graph accurately depicts the charge of acapacitorover time as it is discharging voltage?A. B.

Time

C.

Time Time



When a conducting wire has a net electricfield across itthefree charges in thewireare propelled fromone endof thewire to the other; this is the source of current in a circuit Theexternal electric field is typically provided by a battery or agenerator. The charges do not move straight through thewire, butrather proceed fromone end of the wireto theotheralong jagged paths. Thecharges suffermany collisions on theway and lose energy. The resistance of a conducting wire,measured in ohms, is a measure of how difficult it is for thecharges to move through that conductor. Ohm's law relatesthe loss of potential by the current, AV (measured in volts),to the resistance, R, by the equation AV = IR, where I is thecurrent measured in amperes (or amps).

Although conducting wires have a small amount ofresistance associated with them, in electronic devices themain source of a high resistance is due to resistors. Thepurpose of a resistor is to control or modify the currentthroughthe circuit Resistors typicallyobey Ohm's law.

There may be several resistors in the same circuitResistors can add together in one of two ways: in series,where the effective resistance is given by:

Reff = Rl + R2 + ...

or in parallel, where the effective resistance is given by:

1

R eff

--L + .L+...Rl R2

Resistors in series share a common current whereasresistors in parallel share a common potential difference.

44. A voltmeter is a device that measures the voltage dropbetween two points in a circuit In comparison to thesection of circuit between these two points, thevoltmeter should have a:

A. larger internal resistance and be connected in serieswith the section.

B. larger internal resistance and be connected inparallel with the section.

C. smaller internal resistance and be connected in

series with the section.

D. smaller internal resistance and be connected in

parallel with the section.

45. Light bulbs, such as holiday lights, can be strungtogether in series or in parallel. How should these bulbsbe connected so that if one bulb burns out the otherbulbs remain unaffected?

A. In series.

B. In parallel.

C. Either in series or in parallel.

D. As a combination of series and parallel.


46. Consider the following circuit:

r^A/WSwitch

R3

R2

WWH

VVW-

Battery

When the switch is closed, the current through Ri:

A. increases,

decreases,

is unaffected.

undergoes a changes that depends on whether R5 isless than or greater than (Ri + R2).

B.

C.

D.

47. Birds can be seen perched on high-voltage power lines,yet they seem unaffected by the high voltage. This isdue to the:

A. low resistance of the birds compared to the wire,minimizing the current through the birds.

B. birds keeping only one foot on the wire, thus notmaking a complete circuit

C. low capacitance of the birds.

D. minimal potential difference across the birds.

48. Consider the following circuit

R = 0.1Q

i-wvw—

1CQ

Switch 1 Switch 2

When both switches are closed, as compared to whenjust Switch 1 is closed, the:

A. total resistance of the circuit is unaffected.

B. total resistance of the circuit is larger.

C. voltage drop across Ri is larger.

D. power drain in the circuit is larger.


Questions 49 through 52areNOT based ona descriptivepassage.

49. Which of the following is NOT an example of aresistor?

A. Thefilament of an incandescent lightbulbB. A heating coil in a hair dryerC. The anode of a Galvanic cell

D. The magneticcoil of an electric motor

50. What are the respective currents through each of thethree resistors in the circuit below?

6V

A.

B. Ii=2amps;l2 = 4amps;l3 = 2amr.

C. Ii = 1 amp; I2 = 3 amps; I3 = 1 amp

D. Ii = 2 amps; I2 = 1 amp; I3 = 2 amps

Ri=4Q

R2 = 2Q

rn/VW—I

LyvV\MII = 1.5 amps; I2 = 3 amps; I3 = 1.5 amps

;;I2 = 4 amps; I3 = 2 amps

h = 3 amDs: H = 1 amn

51. Which statement accurately describes the impact ofadding an additional circuit element to an RC circuitwith multiple circuit elements?

A. Adding a resistor in parallel will decrease the totalcurrent

B. Adding a resistor in series will reduce the totalvoltage.

C. Adding a battery in series will increase the totalcurrent

D. Adding a capacitor in parallel will decrease thetotal voltage.


52. Given that R4 is greater than R3, which ofthe followingrelationships is NOT true?

R3

Rl

R2

rWWVHr-vW-i

R4

•WW-R5

A. Ii isdefinitely greaterthan I2

B. I2 is definitely greater than I3

C. I3 is definitely greater than I4

D. I4 is definitely greater than I5

1. C 2. D 3. C 4. D 5. D 6. B

7. D 8. D 9. D 10. C 11. B 12. D

13. C 14. C 15. A 16. A 17. B 18. B

19. D 20. D 21. C 22. A 23. D 24. A

25. B 26. C 27. B 28. D 29. A 30. C

31. A 32. A 33. C 34. B 35. D 36. A

37. B 38. A 39. B 40. C 41. D 42. B

43. C 44. B 45. B 46. C 47. D 48. D

49. C 50. A 51. C 52. D

YOU ARE DONE.

Answers to 52-Question Electric Circuits Practice Exam

Passage I (Questions 1 - 6) Basic Circuit

1. Choice C is the best answer. The current flowing through any resistor depends on the total current, the voltage drop acrossthe resistor, and the resistanceof the resistor in question. The voltage does not change in the question, so the resistance of Riand the total current I must be considered. In choice A, the resistance of R\ has increased, so the current through that resistor

II must decrease. This eliminates choice A. In choice B, the equivalent resistance of the circuit has increased, because Reqmust increase as R3 does. Because V= (I)-(Req), the total current will decrease. If Itotal decreases, then all currents l\, I2, andI3 must decrease. This eliminates choice B. To determine the effect on l\ when R2 changes, given that the change in R2 iscompensated for by a changing R3 (so that Req of the entire circuit does not change), you must apply Kirchhoffs junctionrule: I = Ij + I2.

If Rj is equal to R2, II must also equal I2; if Ri is less than R2, then \\ must be greater than I2. This is to say that currentfollows the path of least resistance. Increasing R2 will decrease I2; and since Itotal >s constant (V and Req have not changed,so Itotal must he the same), the value of Ij must increase. The best answer is choice C.

Choice D is the best answer. The first step involves adding R\ and R2, which are in parallel.

1 _

" R, R-1 ^2

The second step is to add R to R3 in series:

Copyright ©by The Berkeley Review®

R- R, R, + R-i _ (Ri)(R2)Ri)(R2) (Ri)(R2) (*i)(»J

_ (Ri)(R2

R =Rj + R2

Req = R + R3 =

222

+ R,Rj + R2

The voltage information is useless without also knowing the total current in the circuit. Also, you should note that you couldhave used your shortcut of knowing that for two resistors in parallel, their equivalent resistance the the product of the tworesistors divided by the sum of the two resistors. Take shortcuts wherever possible. The best answer is choice D.

Choice C is the best answer. We are asked to consider the amount of power drain by the current flowing through eachresistor. Current will flow from the battery, split up and move through R\ and R2, recombine and move through R3, and

finally return to the battery. Therefore, R2 will have the largest current, making choice C correct. Choice B is incorrect,

because identical resistors in parallel should have the same current, and therefore the same power drain. Choice A isincorrect, because R3 has the largest current, making I^R biggest for R3. Choice D is incorrect, because it contradicts choiceC. The best answer is choice C.

Choice D is the best answer. The question gives you the total current and the equivalent resistance, so the power equation ofchoice will include I and R. Recalling that P= I2R, where Pispower, I is current, and Ris the resistance, we find:

P=I2Req =(6 amps)2(2 ohms) =36-(2) amps2-ohms =72 wattsThe best answer is choice D.

Choice D is the best answer. First, you must recall that Energy = (Power)-(time). In other words, E = Pt = V-I-t. This meansthat the greatest energy will be dissipated in the choice that yields the greatest product of voltage, current, and time. Choice Cis eliminated, because choice A has the same voltage and time, but it has a greater current. Choice B is eliminated, because ithas a smaller voltage, current, and time than choice D. Choice D yields more energy than choice A, because although theyhave the same current, choice D has a greater voltage and a longer operating time. The energy dissipated in choice D is 48joules. The best answer is choice D.

Choice B is the best answer. The power of the circuit is found using the equation P = VI. Because the voltage is heldconstant, we can conclude that P«I. The graph for this is a linear relationship with a positive slope, so choice B is the bestanswer. Choices A should be eliminated because of the increasing relationship between power and current, which doesn't

match a linearly decreasing graph. It may be tempting to consider choice C, because P = I2R; but with V held constant, Rmust decrease with an increase in I. This means that the power has term increasing by a square factor coupled with anotherterm decreasing in inversely, which once again points to a linear relationship between P and I. Choices C and D should beeliminated. The best answer is choice B.

REVIEW EXAM EXPLANATIONS

Passage II (Questions 7 -12)

7.

10.

11.

12.

Dielectric ExperimentChoice Dis the best answer. This question requires using the equation given in the passage. According to the equation thenatural resonance frequency v0 is inversely proportional to the square root of L, the inductance, and the square root of thesum of the two capacitors. Increasing the variable capacitor by afactor of 4does not increase the sum (Ci +Cn) by afactorof 4, so the natural resonance frequency is not cut in half. This eliminates choice A. Increasing both Cr and Qi by afactor of2does increase the sum (Ci +Cn) by afactor of 2, but this will decrease the natural resonance frequency by afactor of V2,not by afactor of 2. This eliminates choice B. The inductance Lof the coil must be increased by a factor of 4 to reduce thenatural resonance frequency by a factor of 2. Choice Cis thereby eliminated, leaving only choice D. The best answer ischoice D.

Choice Dis the best answer. The most polar substance will exhibit the greatest value for K, the dielectric constant. This isto say that the molecules of a polar material can align themselves within the electric field between the plates of the capacitorand in doing so reduce the effective plate charges. This reduces each plate's repulsion to current, which in turn allows morecurrent to flow, to reestablish the maximum plate charge. The more polar the material, the more the capacitance will change.This means that the best answer is the most polar substance. Benzene, cyclohexane, and octane are all nonpolar (organic)compounds, so choices A, B, and C can be eliminated. The best answer is choice D.

Choice Dis the best answer. Choices Band Cshould be eliminated immediately, because two sides of the equation haveinconsistent units (one side is in Fand the other is 1/F). This leaves choices Aand D. The two capacitors (Cj and Cn) are inparallel, so according to circuit convention, their capacitances are added together. This makes choice D the best answer.Remember: Ctotal s Q +Cn, in parallel, because it's similar to make a wider capacitor, and wider capacitors (like plates ingeneral) can store more stuff. The best answer is choice D.

Choice Cis the best answer. Increasing the distance between the charged plates affects the strength ofthe electric field, andthus, it affects the plate charges that can be developed. This means that the distance between the plates affects thecapacitance. Choice A is eliminated. The whole idea behind the passage is to study the effect of adding a solution to thespace between the two plates of acapacitor, so choice Bcan be eliminated. No matter what voltage source is chosen-battery,inductance coil, or any other incident current source--the capacitor has a maximum amount ofcharge that it can store. Thismeans that the voltage source of the circuit does not affect the capacitance, so choice C is the best answer. Changing thematerial of the plates definitely affects the amount of charge that can be stored. For instance, if an insulator is used, the°n nocharge may be stored on the capacitor. This eliminates choice D. The best answer is choice C.

Choice Bis the best answer. The oil bath and heating coil unit keeps the solution at aconstant temperature. The heating coildoes not affect the homogeneity of the solution, so choice A is eliminated. Because the density of the solution varies withtemperature, the amount of solution that fits between the plates can vary with temperature. In addition, entropy increaseswith temperature, so the molecules ofthe solution will not align as well at higher temperatures. These effects will change thedielectric constant, making choice B the best answer. Heating the circuit adds energy to the circuit, but not in the form ofcurrent. This eliminates choice C. Resistance is observed to increase at higher temperatures. Even if you did not know this,you should not assume that it would decrease at higher temperatures. Either way, choice D is not the best answer. The bestanswer is choice B.

Choice D is the best answer. The current does not flow through a capacitor, meaning choice A can be eliminated.Capacitors store energy in the form of charge buildup, which is caused by current. The currents generated by the storedelectrical energy are never observed in this experiment. Therefore, the amount of current flow the capacitors can inducewhile discharging is irrelevant. This eliminates choice B. The inductance is held constant, because the coil does not changeduring the experiment. It is not mentioned how the current from the adjacent circuit (flowing through the other coil of thetransistor) is altered, so choice Cshould be eliminated. The only measurement mentioned in the passage (that is also offeredamong the answer choices) is the natural resonance frequency of the circuit. As Cn changes, the value of the naturalresonance frequency of the circuit will change. The change in resonance frequency isobserved, recorded, and then convertedinto the change in capacitance experienced by Cn. This makes the best answer choice D. The value of Cn can be found asfollows:

C,i =231V.

L-C,

where L and C\ are constant, and v0 is measured. The best answer is choice D.

Copyright © by The Berkeley Review® 223 REVIEW EXAM EXPLANATIONS

Passage III (Questions 13 -18) Resistor Alignment

13. Choice C is the best answer. The greatest total current is found in the circuit with the least equivalent resistance.Equivalent resistance is minimized when the larger resistors (the resistors with greatest resistance) are in parallel and thesmaller resistor (the resistor with the least resistance) is in series. You can use math to verify that this is true in the case ofthree resistors with resistances of 1Q,2Q, and 3& respectively. If the 1Q resistor is in series and the 2Q and 3Q resistors are

in parallel with one another, then the equivalent resistance is 1+(2 °° V/(2 +3) = l +6/5 =2 f5- If tne 3^ resistor is inseries and the 1Q and 2Q resistors are in parallel with one another, then the equivalent resistance is 3+0 °° 2)/(j +2) =3+2/~ - 32fy This means that the arrangement with R\ in series, and R2 and R3 in parallel minimizes equivalent resistance,and thus maximizes current. For maximum current, R) must be at the bottom, which is only true in choice C. Note thatupper left and upper right resistors can be interchanged, and the equivalent resistance will not vary. This means that choicesA and B are the same answer. The best answer is choice C.

14. Choice C is the best answer. The key values they give us are the circuit's equivalent resistance and the its voltage. Fromthese values, Ohm's law tells that the current must be V/R, which is 12/4 = 3 amps in this case. From here we can note that I3is also 3 amps and that Kirchhoffs junction rule tells us that: Itotal = Ii + Pi- If *1 = 2 amps, then I2 must be 1amp. This tellsus early on that choice C is the best answer.

But let's consider the other choices to be complete. First off, we have determined that h is 1 amp and are given that \\ is 2amps, so Ii is greater than I2, which eliminates choice D. If 11 is greater than b, then R\ must less than R2 given that theyshare the same voltage drop (-IR). This eliminates choice A. The equivalent resistance is found by adding the parallelresistance to R3, so it is not possible that R3 could be 4 Q if the equivalent resistance is 4 Q. Choice B is also eliminated.The best answer is choice C.

15. Choice A is the best answer. There is a general rule that states that the equivalent resistance is minimized when theresistors are placed in parallel. This would make choice A the best answer at initial glance, because in System III, all threeresistors are in parallel. In System II, only two resistors are in parallel, and finally in System I, all of the resistors are inseries. But to solve this thoroughly, it is necessary to identify the different circuits. In System I, the resistors are in series, soReq = Ri + R2 + R3 = 6Q.

In System II, R\ and R2 are in parallel, so the are summed in the following manner.

1 _ 1 + 1_ R2 + Ri _ Ri +R2 . Ra|-(Ri)(R2)Rpar Rl R2 (Rl>(R2) (R1MR2) (R1MR2)' ^ R1+R2

To obtain Req for System II, the value of Rpar (for Rj and R2) must be added to R3, which is in series:

R3 =(R'HR2) +r3? which is 32 QRl + R2 3

In System III, R\, R2, and R3 are all in parallel, so the equivalent resistance is found as follows:

Req - Rpar +

1-1,1,1 _(R2-R3) +(RrR3) +(RrR2) .Ri R2 R3 (R1XR2XR3)

••• Req =R eq


(R1XR2XR3)(R2-R3) + (RiR3) + (RrR2) 11

= -6_Q.

16. Choice A is the best answer. When all three resistors are in series, as is the case in System I, the equivalent resistance isfound by summing the individual resistors. No matter how the three resistors are sequenced, the sum is the same. Thismeans that in System I, the equivalent resistance does not vary with sequence. When all three resistors are in parallel, as isthe case in System III, the reciprocal of the equivalent resistance is found by summing the reciprocals of the individualresistors. No matter how the three resistors are sequenced, the result is the same. This means that in System III, theequivalent resistance does not vary with sequence. This leaves only System II, which makes choice A correct by process ofelimination. Because System II has two resistors in parallel and each of those in series with a third resistor, the sequenceimpacts the equivalent resistance. The equivalent resistance for the two resistors in parallel is found by taking their productand dividing that number by their sum. This is the mathematical fallout from the relationship between the reciprocal of theequivalent resistances and the sum of the reciprocals for the two resistors. The equivalent resistance for the entire circuit isfound by summing the parallel equivalent resistance and the third resistor. All of this bombastic banter simply confirms thatfor System II, finding the equivalent resistance is a pain in the butt. . The best answer is choice A.


17.

18.

Choice Bis the best answer. There are three versions of the equation for the power drained by a resistor in a circuit: P=IV, P=I2R, and P=V2/R, where Pis power, Vis voltage, Iis current, and Ris the resistance. In this question, we are givenonly V and R, so to determine the power, vve should use P = V2/R:

P= V2/Req =°2 V)2/(4 ohms) =144U V/ohm =72/2 Watts =36 WattsThe best answer is choice B.

Choice Bis the best answer. To determine the equivalent resistance, we should start by determining the parallel equivalentresistance for R\ and R2 combined. Because there are two resistors are in parallel, it is easy to use the shortcut that theequivalent resistance (Req) equals the product of the two resistors divided by the sum of the two resistors.

1 _ R2 A Rl _ Rl +R2 . d _(Rl)(R2)1 _ 1

Rpar Rl R2 (Rl}(R2) (R1MR2) (Rl>(R2)•'• Rpar —

R 1 + R2

This leads to a parallel equivalent of (1 x2)/(l + 2) = 2/3 =0.67 Q. Now to determine the overall equivalent resistance of theentirecircuit, we need to add the parallel equivalent to R3, because they are in series.

Requivalent = Rparallel + R3 = 0-67 + 3 = 3.67QThe best answer is choice B.


19. Choice D is the best answer. When current is passed through a resistor, the resistor heats up. This means that the energytransfer of thesystem must involve heat, so choices A and B can beeliminated. Overall, energy is being transferred from thebattery to the resistors. Thus, the best answer is that their electrical energy is being converted to heat energy. The-bestanswer is choice D.

20. Choice D is the best answer. The characteristic time that it takes for a capacitor to discharge through a resistance is calledthe time constant of the circuit. The time constant of a discharging capacitor is given by T = RC, where R is the totalresistanceof the circuit and C is the capacitance of the circuit. If a third resistor were added in parallel to R\ and R2, then theresistance of the circuit would change, which would change the time constant of the circuit. Whenever we add a resistor inparallel, the total resistance of the circuit always decreases. This eliminates choices A and B. Since R decreases, the timeconstant must also decrease. What this means physically is that a capacitor discharges faster when there is less resistance inthe circuit. That should seem physically intuitive, given that less resistance would allow the charge to leave the plates fasterand thereby take less time to completely dissipate. The best answer is choice D.

21. Choice C is the best answer. To measure current flowing through a particular part of a circuit, the ammeter must beconnected in series, so that the current can flow through it. This eliminates choices B and D. In order to facilitate this,without disrupting the current flow much, the ammeter should have a small internal resistance compared to the circuit ofinterest. The best answer is choice C.

22. Choice A is the best answer. When the resistors are arranged in parallel, then there are two different pathways current maytake from the cathode to the anode. Both pathways must result in the complete drain of voltage, so each parallel pathwayexperiences a voltage drop equal to the voltage of the battery. Although a different current will flow through each of theresistors, they will experience the same voltage drop, which is equal to I x R. Statement I is valid, which eliminates choicesB and D.

When the resistors are wired in series, then the current that passes through one must also pass through the other. There isonly one possible path, so the current is the same through each resistor. This makes Statement II invalid, which doesn't reallyhelp out with our process of elimination pathway.

In parallel, resistors share a common voltage. But because they have unequal resistances, they must have unequal currents. Ifwe use the fundamental power equation, P = IV, then we see that the resistor with lower resistance, and thus higher current,will drain the more power. In parallel, smaller resistors drain more power than bigger resistors. In series, resistors have acommon current. If we use the power equation P = I2R, then we see that the resistor with higher resistance will drain morepower than the resistor with lower resistance. In series, larger resistors drain more power than smaller resistors. The powerdrain for the two resistors differs when they are wired in series versus when they are wired in parallel, which makesStatement III invalid. The best answer is choice A.


Passage IV (Questions 23 - 28) Wheatstone Bridge

23. Choice D is the best answer. The passage provides us with math to reference if we so choose. If we assume that theunknown resistor, Rtj in Figure 2, is equivalent to R4 in Figure 1, then we simply need to match the corresponding resistorsin Figure 1 to those in Figure 2. The AZ segment of wire AC equates to R\, the ZC segmentof wire AC equates to R3, andthe known resistor, Rknown. equates to R2. Given that R4 is equal to (R2)(R3)/Ri in Figure 1, vve can conclude that Rij inFigure 2 is equal to (Rknown)(dz-c)/da-z- This makes choiceD the best answer. If you prefer, then the math can be solved asfollows.

'lRa-z = Rknown and qRZ-c = '2Ru

'lRz-c _ *'2RUn(p\vireAC

'lRa-z '2Rkr n(PwireAC)

dz-c

^wireAC / _ ^RlJ

4-z I '2Rkno\vn

wrwireAC2/You could have also solved this by inspection. We know that in a wheatstone bridge, the voltage drops through the firstresistor are equal in both pathways and that the voltage drops through the second resistor are equal in both pathways. So, theratio of the first voltage drop-to-the-second voltage drop is the same in each pathway. Given that current is constant in eachpathway, we can simplify the relationship to be that the ratio of the second resistor to the first resistor is the same for eachpathway. The ratio of Rij-to-Rk is equal to the ratio of Rz_c-to-Ra-z, which reduces to dz.c-to-da.z. Multiplying each ratioby Rk results in choice D. The best answer is choice D.

dz-c

4-z

_ Rij

Rkr- - k:::ill„a|4*

24. Choice A is the best answer. The resistance of a cylindrical segment of wire depends on the resistivity of the material, p,the length of the wire, and the cross-sectional area of the wire. In Figure 2, the wheatstone bridge functions by having aslide-wire capable of moving in such a way that wire AC is broken into two segments, AZ and ZC. The ratio of theresistancesof the two segments of wire (AZ:ZC) is equal to the ratio of the two resistors (Rknown:Runkno\vn)- If the wire ACis uniform in terms of material, then the resistivities of the two wire segments are equal. If the wire AC has a uniformradius, then the cross-sectional areas of the two wire segments are equal. The differences in the resistance between the twosegments of the wire are due exclusively to the differences in their lengths. This makes choice A the best answer andeliminates choice B. The resistances of the two segments of wire do not need to equal one another, so choice C is eliminated.The resistances of the two segments of wire do not need to be equal to the resistance of the unknown resistor, especiallysince that resistance is an unknown value. Choice D is eliminated. The best answer is choice A.

25. Choice B is the bestanswer. There must be a variable resistor amongst the three permanent resistors tooffset the variabilityassociated with an unknown resistor. If R4 represents the unknown resistor, then the variable resistor can be R\, R2, or R3,although it is best in the position of R3. This eliminates choice A. R\ does not need to equal R3, because that would meanthat R2 must equal R4 for the galvanometer to read zero. This means choice B is not necessarily true of a functioningwheatstone bridge. Choice B may be the answer. The galvanometer needs to measure the magnitude of voltage differencebetween the two sides so that adjustments to the variable resistor can be made to in turn vary the currents, i\ and *'2. enoughto get the voltage drops through the first resistor of each parallel pathway to equal one another. The sign of the voltagedifference is needed to know whether to increase or decrease the resistance of the variable resistor. The galvanometer shouldbe able to measure both magnitude and direction, which makes choice C a valid statement. This eliminates choice C. Nomatter which direction the current flows, the ratio of theresistances needed to generate a wheatstone bridge circuit with equalvoltage drops in the respective regions of the parallel pathways does not depend on the magnitude of voltage. The circuitexhibits the same results when the poles of the battery are reversed, so choice D is a valid statement. This eliminates choiceD. The best answer is choice B.

26. Choice C is the best answer. A galvanometer, as the name implies, measures the voltage (galvanic emf) of a circuit. Itsupplies no voltage to the circuit, so choice A is eliminated. In the wheatstone bridge, we need to make sure that the voltagedrops across the first resistor in both of the parallel pathways are equal. If the voltage drops are equal, then the voltage atPoint b is equal to the voltage at Point z. The galvanometer connects points b and z, so it measures the potential differencebetween the two points, not the absolute potential at each point. This eliminates choice B and makes choice C the bestanswer. The galvanometer is used to ensure that the potential difference between Points b and z is zero. The galvanometerdoes not serve as a short circuit (path of zero resistance). Even if it did, such a feature is not needed in the wheatstonebridge. This eliminates choice D. The best answer is choice C.


27.

28.

Choice Bis the best answer. It is not specified in the question, but vve can assume they are asking about the wheatstonebridge in Figure 2. With Metal I constituting Rk and Metal II constituting Rij, Wire segment AZ is shorter than Wiresegment ZC. This means that the resistance of Rk must be less than the resistance of Rij. Given that the two resistors havethe same dimensions but are made of different metals, the difference in resistance must be due to a difference in resistivityfor the two metals. The metal with lower resistivity will correspond to the resistor of lower resistance. From this, we canconclude that Metal I must have a lower resistivity than Metal II. From this point wecan reason that Metal HI must have alower resistivity than Metal I, by virtue of the slide-wire now being closer to Point c than Point a (when Metal II is replacedby Metal III). This makes choice Bthe best answer. Even ifyou are uncertain ornot reasoning clearly, you should be ableto eliminate choices Aand D, because Metal I must lie in the middle, given its opposite behavior in the two trials. Honingyour process-of-elimination skills is an asset. The best answer is choice B.

Choice D is the bestanswer. If the voltage of the battery were doubled, the current would increase, which in turn wouldincrease the voltage drop across each resistor. However, the resistance of each resistor would remain unaffected, becauseresistance depends on the resistivity and dimensions of the resistor, not the voltage of the battery. This means that the ratiosof the resistances would remain unchanged, and therefore the ratio of the wire segments would not be affected. Thiseliminates choices A, B, and C. The best answer is choice D.

Passage V (Questions 29 - 33) Pacemaker Circuit

29. Choice A is the best answer. The capacitor is initially uncharged. Immediately after the switch is closed to begin thecharging process, the currentbegins to flow. Charge has not yet builtup on the capacitor, so there is a maximum amount ofcurrent in the circuit. By Ohm's law, the potential difference across the resistor is given by: AV = IR. If the current is at amaximum, then the potential difference is at a maximum. Choice B is incorrect: As time goes on, charge builds up on thecapacitor, so there is less current in the circuit. After a long time, there is no current in the circuit. Choice C is incorrect: Ifthe potential difference is maximal across thecapacitor plates, it should be minimal across the resistor, because the voltagedrop across both of them should equal the emf of the battery. Choice D is incorrect: Current flows before charge builds upcompletely on the capacitor, so the maximum potential difference across the resistor occurs before the maximum potentialdifference across the capacitor. The best answer is choice A.

30. Choice C is the best answer. The relationship between the charge time and the resistance is given in the passage as T = RC.The ratio of Rc : Rd will therefore be equivalent to Tc : Td. Using the charging and discharging times in the questionindicates that the best answer is choice C. The best answer is choice C.

31. Choice A is the best answer. During one complete cycle of the system, the capacitor must start and finish at the same point,which eliminates choice B. Choice B can also be eliminated because it is linear in nature. Charge on a capacitor builds up onthe capacitor's plates exponentially, indicated in the passage by the repeated appearance in the charge equation of e"1, theinverse of the natural log e. When a capacitor discharges, the charge drops rapidly at first and then gradually decreases untilit reaches zero. This is expressed in the first half of the graphs in choices A and D. Choice C depicts a slow discharge at firstthat increases until an abrupt zero. This eliminates choice C. The capacitor fills rapidly at first and then slows until it reachesa maximum charge. This is expressed in choice A. Choice D shows the opposite in the second half of the graph. The bestanswer is choice A.

32. Choice A is the best answer. We want the capacitor to discharge much more quickly than the time between these discharge"heart beats." This would favor choices A and B over C and D. The resistors play a role in this discharge time, as T = RC.Therefore, the resistance Rd should be relatively small compared to Rc. The best answer is choice A.

33. Choice C is the best answer. If pulses are triggered 60 times per minute, that means there is one pulse triggered everysecond. From the passage, we know that a pulse is triggered each time the charge on the capacitor drops to e"1 of itsmaximum value, which corresponds to the time constant. Therefore, the time constant for this circuit is 1 second. Thecapacitance provided in the question is extraneous information. The best answer is choice C.


34. Choice B is the best answer. The passage states that an example of a parallel circuit is the wiring in apartments. Anapartment is a home (assuming you like your roommates) so a home is "wired up" in parallel. Another way to tell that this isthe best answer is the following: All of the components of a parallel circuit have the same voltage across them. In your home,all of the outlets have the same voltage. Thus, your house's circuits must be in parallel. If outlets were in series, then everytime someone turned on an appliance, any other appliances that are already operating would experience a drop in power. Thebest answer is choice B.


35. Choice D is the best answer. One method of inducing current flow is to apply a voltage difference across a conductor,creating an electric field across it. Thus, if an electric field is present, current can flow. This makes Statement I and invalidstatement, which eliminates all of the answer choices except choice D. Sometime you get lucky like this. Resistance in theconductor will tend to slow down the charges, draining energy from the system as the current flows through them. Resistorsdo not supply energy to a circuit. In fact, resistors are responsible for taking energy out of a circuit. This makes Statement zllan invalid statement, which vve already suspected based on our elimination of choices A, B, and C. As has already beenstated, increased resistance in a conductor means more interactions between charges. Because of such interactions (i.e.,collisions), resistors convert electric potential energy to heat. The amount of current passing through a resistor is unaffectedby the resistor. Only the energy of the current is affected. Statement III is invalid, which again vve suspected. All of theelectrons that leave one terminal must arrive at the other terminal, so vve can conclude that the current leaving the cathode isequal in magnitude to the current arriving at the anode. This is supports the notion that current is the same for all circuitelements in series. Statement IV is in fact valid. The best answer is choice D.

36. Choice A is the best answer. Statement I is valid, because resistors in series sum directly, while resistors in parallel sum asreciprocals. For instance, three resistors in series have an equivalent resistance of R\ + R2 + R3, while in parallel, theequivalent resistance is less than each of the individual resistors. This makes Statement I valid. Based on the relationship, V= IR, as resistance increases, the current must decrease, when the voltage is constant. For resistors in parallel, the voltageacross each resistor is the same, so the resistor with the greater resistance has the lesser current. This makes Statement IIinvalid, which eliminates choices B and C. All resistors will heat up when current passes, making Statement III an invalidstatement. The best answer is choice A.

37. Choice B is the best answer. The voltage drop across R3 is the same voltage dropexperienced byall three parallel resistors.To solve thisquestion, vve can apply Kirchhoff's loop rule. All 12 volts mustbe lost when charges travels from the cathodeto the anode. Because 8 volts are lostacross the series resistor (4A x 2Q = 8V), the parallel resistors must each drop 4 volts.The voltage drop across R3 is 4 volts. The best answer is choice B.

Passage VI (Questions 38 - 43) RC Circuit

38. Choice A is the bestanswer. Once Switch S\ is closed, the current will flow until the capacitor iscompletely charged. Thecharge and voltage on acapacitor are related through Q= VC. Plugging in the numbers from the passage gives:

Q = VC = (2/<F)(10mV) =(2xlO-6)(10x 10"3) =20x 10"9 = 2x 10"8CThe best answer is choice A.

39.

40.

Choice B is the bestanswer. If both switches are closed simultaneously, assuming the capacitor were initially uncharged,the circuit would include a capacitor in parallel with two resistors. When the switches are closed, the initial surge of currentwill go exclusively to the capacitor (which is the path of least resistance). But over time, the capacitor will build up charge.When charge builds up on a capacitor, so does the voltage across it—remember: Q = VC. This eliminates choices C and D,which describe the current across the capacitor, not the voltage difference. When the switches are closed, the current throughthe resistors starts at 0 and builds up to some maximum value given by V = IR. The best answer is choice B.

Choice C isthe best answer. A long time after both switches are closed, the current will travel exclusively through resistorRl, because the capacitor passes no current once it is fully charged. To calculate the current through Rj, vve use Ohm's law,V = IR. Plugging in the appropriate values, vve get 10 mV = 1(5 mQ), or:

I = iOmV = 2A5 mQ


41. Choice D is the best answer. If another resistor were inserted into the circuit before switch Si, then it would be in serieswith the capacitor. The change this would produce in the maximum charge that could be stored on the capacitor depends onthe change in voltage across the capacitor. Voltage and charge for a capacitor are related through Q = VC. If the voltageincreases, the charge will increase; and if the voltage decreases, the charge will decrease. Based on that, vve could eliminatechoices BandC. What vve now have todecide is what effect the new resistor has on the voltage across thecapacitor. Withoutthe new resistor, the voltage across the capacitor is 10mV. With the new resistor in the circuit, some of the 10 mV has to goon the resistor and some on the capacitor. Since the whole 10 mV is spread across two circuit elements instead of just thecapacitor, vve can conclude the voltage must decrease across the capacitor. The best answer is choice D.


42.

43.

Choice Bis the best answer. If asteady current of 1mA passes through resistor Ri for 10 seconds, energy will be given offin the form of heat. The amount of heat is: Heat =(Power) x (time), so Heat =(I2R) x (time). Substituting values, vve getHeat =(1 mA)2 (5 mQ) (10 sec). Solve for Heat: Heat =(1 x 10'6) (5 x 10"3) (10) =5x 10"8 J. The best answer is choiceB.

Choice C is the best answer. A discharging capacitor loses current in a characteristic fashion. When a capacitor isdischarging, it is losing charge, so charge as a function of time must decrease. This eliminates answer choices A and D. Thecharge on acapacitor decreases exponentially, not linearly, so choice Bcan be eliminated. This can be explained by the factthat as the capacitor loses charge, there is less repulsion on its plates, resulting in a smaller electromotive force pushingchargeoff the plates. The only graph showing nonlinear decay is choice C. The best answer is choice C.

Passage VII (Questions 44 - 48) Resistance is Futile

44. Choice B is the bestanswer. Devices (such as resistors) connected in parallel have the same potential difference. Inorder tomeasure the potential difference between two points in a circuit, the voltmeter must be connected in parallel. (This meansthat it does not matter which path the current takes to get from one point to another. It will lose the same amount ofpotentialalong either path.) Since vve do not want the voltmeter to affect the measurement, the voltmeter should have a large internalresistance. This will allow only a minimum amount of current to flow through the voltmeter-just enough to allow it tomeasure the voltage difference. The best answer is choice B.

45. Choice B is the best answer. In a parallel arrangement, the current that flows through one light bulb is not the current thatflows through the remaining light bulbs. Remember that a parallel arrangement means that all bulbs have the same potentialdifference across them. If one bulb burns out, then current can no longer flow through that portion of the circuit; but theremaining light bulbs are unaffected, because they still have a potential difference across them. Devices wired in parallel areindependent of one another. This favors choice B. Choice A is incorrect: In a series arrangement, the current in the circuitflows through all of the light bulbs. If one of the bulbs burns out, then current can no longer flow through that bulb or anyother bulb in the circuit. Devices wired in series are dependent on one another. This eliminates choices A, C, and D. Thebest answer is choice B.

46. Choice C is the best answer. The top branch, the middle branch, and the bottom branch of this circuit are connected inparallel. This means all three branches have the same potential drop across their ends. Closing the switch and allowingcurrent to flow through R5 does not change this. Closing the switch also does not change the total resistance of the top

branch, the branch associated with all of the answer choices. If the potential drop across the top branch remains the same, andif the resistance remains the same, then by Ohm's law (AV = IR) the current cannot change. This eliminates choices A and B.However, something has to change when the switch is closed. By closing the switch, another resistor is added in parallel tothe circuit. This will reduce the overall resistance of the circuit (the effective resistance of resistors connected in parallel isalways less than the individual resistances). That, in turn, will increase the total current drawn from the battery (again, byOhm's law). Current now can flow through R5 without affecting the current flowing through any other branch. The best

answer is choice C.

47. Choice D is the best answer. Although the voltage in the wire may be very high, the voltage is measured with respect to theground. The voltage will not vary as quickly along the wire. Thus, when the bird stands on the wire with both feet, every partof the bird is at approximately the same potential. Current flows only when there is a potential difference between two points.The bird would be in trouble only if it had one foot on the wire and the other foot on a different wire (at a different potential),or if it had one foot on the wire and the other foot on the ground. The best answer is choice D.

48. Choice D is the best answer. When two or more resistors are connected in parallel, the equivalent resistance of the circuit isless than that of any one of those resistors alone. Closing Switch 2 adds another resistor in parallel. Thecircuit's Req shouldtherefore decrease. Thus, choices A and B are incorrect. Since R\ and R2 are connected in parallel, then when both switchesare closed, R\ has the same voltage drop across it as when Switch 1 alone is closed. Therefore, choice C is incorrect. Choice

D is the only remaining choice, making it the best answer by the process of elimination. However, vve also know that choiceD is also correct because connecting resistors in parallel reduces the total resistance of the circuit. A circuit with a smallerresistance has a larger current, which consequently drains more power (P = IV). The best answer is choice D.



49. Choice C is the best answer. A resistor is anything that drains energy from electrical flow and converts it to some otherform of energy, often light, heat, or work. A filament in a light bulb converts electrical flow into both heat and light, sochoice A is a resistor. The heating coil in a hair dryer converts electrical flow into heat (and occasionally light), so choice Bis a resistor. The magnetic coil of an electric motor converts electrical flow into mechanical work, so choice D is a resistor.The anode of a Galvanic cell is the starting point for electrical flow, so it does NOT serve as a resistor. The best answer ischoice C.

50. Choice A is the best answer. All three resistors are in parallel with each other, so they all experience the same voltage drop.This means that the resistor with the greater resistance will also have lower current. The ratio of the resistances is 2 : 1 : 2, sothe ratio of the currents (I]-to-l2-to-l3) must be 1 : 2 : 1. This eliminates choices C and D. To decide between the remaining

two choices, vve first need to solve for the equivalent resistance. Let's consider the resistors in duos, starting with the R1-R3

pair first. The quickest way is to take the product and divide by the sum, which in this case means (4 x 4)/(4 + 4) = 16/8 = 2.We can now take that 2 and pair it with R2 to get (2 x 2)/(2 + 2) = 4/4 = 1Q. Because the equivalent resistance is 1Q and the

circuit voltage is 6V, the total current entering the junction is 6 amps. Adding the values listed in choice B for \\, I2, and I3

equals 8 amps, which is incorrect. Choice B is eliminated. In choice A, l\ + I2 + I3 = 6 amps, which is the value vve want.


51. Choice C is the best answer. Adding a resistor in parallel will decrease the equivalent resistance for the entire circuit.Assuming the voltage remains the same, the total current will increase, not decrease. Choice A is eliminated. Adding aresistor in series will increase the equivalent resistance of the circuit, but it will have no effect whatsoever on the totalvoltage. Choice B is eliminated. Voltages add in series, so adding a battery in series will increase the total voltage of thecircuit. Given that resistance does not chance, according to V = IR, the total current must increase. Choice C is a validstatement. Adding a capacitor in parallel will increase the equivalent capacitance of the circuit, but it will have no effectwhatsoever on the total voltage. Choice D is eliminated. The best answer is choice C.

52. Choice D is the best answer. According to Kirchhoffsjunction rule, the current splits at junctions in such a way where thetotal current entering the junction must equal the total current leaving thejunction. For the circuit in this question it meansthat I] =12 + 15 and 12 = 13 + 14- BecauseIi = I2 + I5, Ij must be greaterthan I2, making choiceA a true statement. ChoiceAis thereby eliminated. Because I2 = I3 + I4,12 must be greater than I3, making choice Ba true statement. Choice B is therebyeliminated. The voltage drop across R3 equals the voltage drop across R4, so given that R4 is greater than R3,14 must be lessthan I3. This eliminates choice C and leaves only choice D standing. Without knowing the exact values of R2, R3, R4, andR5, there is no way to compare the currents through R4 and R5. For instance, if R2 is much larger than R5, then [5 will belarger than I4and I3 combined. However, if R5 is much larger than R2, then I5 will besmall and could very likely be smallerthan both I3 and I4. Because vve have no way to know for sure, vve can't say that I4 is definitely greater than I5. The bestanswer is choice D.


Light and OpticsPhysics Chapter 10

Object Lens 1Lens 2

Image 1 = Object 2

by

the Berkeley Review

Lightand OpticsSelected equations, facts, concepts, and shortcuts from this section

v = A

1 =1 + 1f o i


Ephoton =h/ Ephoton =hc/*. nisin6i =n2sin62dt =Wo Xf)/{do . f) m=himage/hobject =-ty^


With refraction, light bends towards the normal when goingfrom a faster medium to a slower medium

Medium 2slower and

shorter wavelength

Medium 1

faster andlonger wavelength

Normal J_ to interface

Medium 1

Medium 2

As n f: v | and 0 J,

Faster-to-slower

ni

n2

9A n2>n!

© Image Estimation Trick

Single diverging mirrors and lenses generate SUVs for images (smaller, upright, virtual images)

Single converging mirrors and lenses generate images that vary with the object position and focal length

Converging Mirror Summary

nr

Object Positions

II III TV V

R

ir

IR images: i = + #

VUV image:

i = -#

Image Positions

ObjectPosition

I

II

III

IV

V

ObjectDistance

dD>R

R

R>d0>f

d0 = f

f>d,

ImageFormed

Smaller IR

Same size IR

Larger IR

No image

Larger UV

ImageDistance

R>dj>f

dj = R

ds>R

dj = oo

d; = -#

Physics Light and Optics

Light and OpticsIn Chapters 5 and 6, we studied wave related phenomena. In this chapter, weshall expand our coverage of waves to include electromagnetic radiation, oneexample of a transverse wave.

Electromagnetic Radiation

Electromagnetic waves (often abbreviated as EM waves) are electrical andmagnetic disturbances that propagate through space at the speed of light. Thespeed of light c is taken to be 2.98 x 108 m/s. There are many types ofelectromagnetic waves. Gamma rays, X-rays, microwaves, and radio waves are afew examples. All of these differ from one another in their wavelength A, andtheir frequency /. We can relate the wavelength and frequency of anelectromagnetic wave to the speed of light by equation (10.1).

c = Xxf (10.1)

The magnetic and electric fields of an electromagnetic wave are bothperpendicular to one another and perpendicular to the direction of wavepropagation. In Figure 10-1 we see a diagram of an electromagnetic wavepropagating at the speed of light from left to right. Note the alternations of theelectric and magnetic fields

Vertical plane = Electric wave


Horizontal plane = Magnetic wave

Figure 10-1

We can distinguish different electromagnetic waves by their frequency orwavelength in the electromagnetic spectrum (Figure 10-2). Low-frequency EMwaves are used primarily for communication. These frequencies range fromabout 0 Hz to 109 Hz. AM radio waves are at the lower end of this range, whileFM radio waves and television signals are at the higher end. Microwaves fallbetween 109 Hzand 1011 Hz. This typeofEM radiation isused in navigation andcommunications. The infrared region of the EM spectrum falls between 1011 Hzand 1014 Hz. This type of radiation is often involved in heating, since H2Oreadily absorbs in the infrared. It is also used in remote control units.

Visible light begins at about 4 x 1014 Hz and continues to about 7 x 1014 Hz.Within this very small range of the electromagnetic spectrum are all the shadesof colors the human eye can see, from red to violet. This defines our visiblespectrum. Ultraviolet (UV) radiation is between 8x1014 Hz and 1x1017 Hz. UVradiation causes your skin to tan after long exposure to sunlight. Extendedperiods of UV radiation exposure can also cause cancer. X-rays fall within the1017 Hz to 1019 Hz region of the EM spectrum. Gamma rays have frequenciesthat exceed 1019 Hz. In this section, we want to consider primarily that portion ofthe electromagnetic spectrum visible to the human eye.


Physics

X,(m)101 10*"

AM, "FM,TV

810

/(Hz)

10"

Light and Optics

10" 10" 10" 10' 10H 10" 10•8 -9

10


10"10 10"11 10-12

k(m)

-1310

Microwave- -^-Ultraviolet*"

-<

Gamma rays-

1010

Infrared X-rays-12

10 Visible

Red Orange Yellow Green Blue Indigo Violet

750 to 630 to 590 to 560 to 490 to 440 to 420 to Nanometers630 590 560 490 440 420 390

Figure 10-2

2010

/(Hz)

Example 10.1aIf a charge oscillates in space, it emits electromagnetic radiation. If the chargeoscillates with a fixed period and the wave speed of the emitted radiationsubsequently decreases, then theelectromagnetic wave will necessarily have:

I. a lower frequency.II. a shorter wavelength.

III. a lower energy.

A. I onlyB. II onlyC. I and III onlyD. II and III only

Solution

When an electromagnetic wave travels through space or some material, its wavespeed relates to its frequency and wavelength through the equation: v = fk. Thisrule actually holds for all types ofwaves andistherefore agood equation tokeepin mind whenever the word "wave" appears in physics problems. A decrease inthe wave speed implies that either the wave frequency or the wavelength (orboth) must decrease. The frequency of an electromagnetic wave is set by thesource ofthewave. Since the oscillating charge hasa fixed period, it musthave afixed frequency. Recall that therelation between frequency / and period T is: / =1/T. We know the frequency is fixed, so Statement I is invalid. If ifs notfrequency that decreased, then it must be the wavelength that decreased.Therefore,Statement II must be valid, since v = fk. This leaves either choiceB orD as the best answer.

To make the correct choice, consider that Statement III says that slowing thewave will necessarily decrease its energy. How a wave's energy changes willdepend upon how it is slowed down. For instance, if the wave deceleratesquickly by hitting a glass surface, some of the light will be transmitted into theglass and some will reflect from its surface. Since energy is conserved, thetransmitted wave must have less energy. However, if the wave speed isdecreased slowly enough, there will be essentially no reflection; all the energywillbe transmitted. Therefore, Statement IIIis not necessarily true.This rules outchoice D. You could also have reached this conclusion by knowing that theenergy of a photon depends on its frequency according to E = h/. The energydoes not change with wavelength or speed, as the speed-to-wavelength ratio is aconstant.




Example 10.1bTo increase the oscillation frequency of an electron (isolated in a vacuum) thatoscillates because it is stimulated by an electromagnetic wave, one must:

I. decrease the wavelength of the wave.II. increase the electric field amplitude of the wave.

III. increase the speed of the wave.

A. I onlyB. II onlyC. I and III onlyD. None of the changes will increase the oscillation frequency of the electron.

Solution

The electron is being stimulated by electromagnetic radiation in a vacuum, so weshould start by noting that the speed of EM radiation in a vacuum is constant.This means that if the frequency of the radiation were to increase, the wavelengthwould have to decrease (v = fk). This makes Statement I valid, which eliminateschoices B and D. We need to only consider Statement III at this point. Because thespeed an EM wave in a vacuum is constant, Statement III can have no impact.


PolarizationFigure 10-1 diagrams a set of transverse, sinusoidal waves moving perpendicularto the direction of a propagating light wave. If we were to consider just theelectric field vectors of that radiation, we would find they radiate outward fromtheir source in all planes. The magnetic field vectors radiate in a similar manner.Light like this, random in its orientation, is said to be unpolarized. But if wewere to pass this light through a perfect polarizing filter, the vectors wouldradiate in only one plane, because the light wave hasbeen linearly polarized. Thepolarizing filter blocks all light perpendicular to the polarizing axis, but it allowsthe light aligned with the polarizing axis to pass through (Figure 10-3). Theresulting light that does pass through the filter is called plane polarized light.Polarization can result from transmission of light through a polarizing filter orby the interaction with an interface between mediums, where some of lights ofdifferent orientation exhibit different degrees of reflection and refraction.

E waves in natural light have a randomassortment of all possible orientations.

E waves leaving the polarizerhave only one orientation.

Note that the B waves are notshown here, but they do exist

Figure 10-3




FhysiCS Light and Optics Electromagnetic Radiation

Note that the electric field vector of the light that has passed through the filter isparallel to the polarizingaxis of the filter. Measuring the intensity of this light ina photocell would reveal that it has half the intensity of the unfiltered naturallight. To demonstrate why, we could break the electric field vector ofunpolarized electromagnetic radiation into its components, one parallel and oneperpendicular to the axis of polarization. On average, these two components areequal. But since the polarizing filter allows just the electric field vectors parallelto its plane to pass through the filter, only half of the unfiltered natural lightpasses through.

Example 10.2aVertically polarized light is sent through an empty sample cell and then into ahorizontally oriented polarizer. No light gets through the polarizer. Theexperiment is repeated, this time with a dilute solution of aqueous D-Glucose inthe cell, and one-quarter ofthe incident lightintensity gets throughthe polarizer.What must be TRUE of the solution and the rotation of the polarized light?(Assume that thesolution transmits all the incidentlight to the polarizer.)

A. Thesolution is optically active and rotates the polarization by 30°.B. Thesolution is optically active and rotates the polarization by 45°.C. D-Glucose is an asymmetric molecule and the solution rotates the

polarization by 60°.D. D-Glucose is an asymmetric molecule and the solution rotates the

polarization by 90°.

Solution

We need to think about the optical properties of the solution'smolecules and thepolarization rotation. First optically active materials are materials that can alterthe polarization oflight passing through them. This activity usually arises fromanasymmetry in the structure and/or orientation of the molecules that make upthe material. Here, the solution is optically active--the activity arises from thehelical, and therefore, asymmetric structure of the D-Glucose molecules. Thismightbe illuminating, but it doesn'thelp to isolate the best choice, because all ofthe choices areoptically active. We must think about thepolarization rotation.

The light that does get through the polarizer isone-quarter as intense as the lightincident upon it. To calculate the amount of linearly polarized light gettingthrougha polarizer, useequation (10.2), Malus' law:

I=I0cos2 6 (10.2)

where I is the intensity of light getting through the polarizer, when an initialintensity I0 is incident upon the polarizer. The angle 8 is the angle between thelight's polarization direction and the polarization axis ofthepolarizer. Here,

i=I=cos2 0Io 4

Cos 60° = 0.5, so cos2 60° =0.25, meaning that 6must be 60° with respect to thehorizontal polarizer. Since the polarizer's polarization axis is horizontallyoriented, the exiting light must be at an angle 30° to the vertical. The solutionmust have rotated the polarization of the initially vertical light by 30°. This is tosay that the light exitinglight has an orientation closer to vertical than horizontal,so the angle must be less than 45°.




Example 10.2bWhen unpolarized light reflects offof a specular (mirrored) surface at a glancingangle with respect to the surface, the light can partially polarize parallel to thesurface. That is, the parallel polarized light is more intense than theperpendicular polarized light. If this surface is lying on flat ground, in whatorientation should a polarizing anti-glare filter be with respect to the ground, inorder to minimize the total light intensity through the filter? The polarizer'spolarization axis should be oriented at:

A. 0° with respect to the ground.B. 45° with respect to the ground.C. 90° with respect to the ground.D. 180° with respect to the ground.

Solution

This question is a topic unto itself. When light strikes a surface, some of the lightreflects(bounces off if you will) and some refracts (passes into the new medium).When unpolarized light strikes that surface, certain orientations of the electricwave are more apt to reflect and other orientations are more apt to refract. Thequestion states that light with an E wave parallel with the surface is more apt toreflect than refract. Hence, the reflected light is semi-polarized in a horizontalfashion. The reflected light is richer in horizontally polarized light than it is invertically polarized light. To filter out the reflected light (glare), the polarizingfilter should be a vertical polarizer.

A vertical polarizer is consider to be at a 90° angle with respectto the ground, sothe best answer is choice C. As a side note of interest, driving glasses and pilotssun glasses are vertically polarized.



Copyright ©by TheBerkeley Review 237 Exclusive MCAT Preparation

Physics Light and Optics Reflection and Refraction

Reflection and Refraction

ReflectionWhen a wave strikes the boundary between two media, we find typically thatsome light is reflected, some transmitted, and some absorbed. We want toconsider for the moment the light that is reflected from the boundary.

Suppose a light beam strikes the surface of a mirror (Figure 10-4). The incidentlight ray strikes the mirror at some angle 8i with respect to a line perpendicularto the mirror. The reflected light ray leaves the surface of the mirror at an angle8r, which is equal to Of. The law of reflection says that the angle an incident lightray makes with a surface is equal to the angle that the reflected light ray makeswith the surface.

Incident light ray

Mirror

Reflectedlight ray

Figure 10-4

RefractionElectromagnetic radiation that passes through a vacuum travels at the speed oflight, c. However, in a medium other than a vacuum the velocity of the wavedepends on the medium itself. The index of refraction n is the ratio of the speedc of an electromagnetic wave in a vacuum to the speed of that wave in a givenmedium. This is given by equation (10.3). Note that the index of refraction isexpressed without units.

n _ c_ _ vvacuumvmedium

(10.3)

The velocity of an electromagnetic wave in a medium other than a vacuum cannever be greater than the speed of light. Therefore, the index of refraction willnever be less than 1. In Table 10-1, we list the indexes of refraction for someselectedmaterials(usingyellowsodium light).

Medium

Index of

Refraction Medium

Index of

Refraction

Vacuum

Air

Carbon Dioxide

Water

1.00000

1.00029

1.00045

1.333

Fiber Optic Cable

Benzene

Light Crown GlassDiamond

1.483

1.501

1.517

2.417

Table 10-1

The source of the light wave will establish the wave's frequency. The frequencyof thiswaveis not typically altered as it travels throughsome medium; thespeedand wavelength of this wave, however, are altered. From equation (10.1), weknow that v = fk. From equation (10.3), we know that v = c/n. If we set thesetwo expressions equal to one another as shown in equation (10.4), we can see thatas the index of refraction changes, so does its wavelengthof light (remember, thespeed of light c is a constant).

(10.4) **-£ "2 ki(10.5)



Suppose that a beam of light passes from a medium with an index of refractionni (where ni = c/f-ki) to a medium with index of refraction n2 (where n2 =clf'ki). The wavelength of that light changes according to equation (10.5). Notethat the wavelength of light is shorter in a medium with a greater index ofrefraction. Refraction affectsboth the speed and wavelength of light.

Example 10.3aIf an Earthbound scientist reflects a laser beam from a mirror left on the Moon byastronauts, how long in seconds will it take the beam to make the trip from theEarth to the Moon and back? Assume that the Earth is approximately 250,000milesfrom the lunar surface and that the speed of light is 186,000 miles/second.

A. t =

B. t

C. t =

D. t =

186,000

2(250,000)

250,000

2(186,000)

2(186,000)

250,000

2(250,000)

186,000

Solution

This is a formula-identification type of problem—except with the numbers arealready pluggedin. You can use units or limiting cases to narrow the choices. Youwant units of time, so the choice must be either B or D. Using limiting cases,what should happen to the travel time, if the travel distance increases? It shouldincrease, too. This method also isolates choices B or D. Which is better? The laserbeam is making a round trip, so the 2 in the formula has to multiply the distance,not the rate of speed. The 2 must be in the numerator.


Example 10.3bNeglecting absorptionand scattering,how long does it take for light to traverse 2m of H2O, which has an index of refraction equal to 1.33?

A. t =

B. t =

C.

2(1.33)

3xl08

3xl082(1.33)

t_(3xl08)(1.33)

D. t =

(3xl08)(1.33)

SolutionIf you don't recall distance = rate x time, then you can get the relationship fromthe units. The time is equal to the distance divided by the rate. The distance is 2and the rate is (3.0 x108)/1.33. Both2 and 1.33 should be in the numerator.






Snell's LawAs a light ray moves from a medium like air to a medium like water, it isrefracted. Suppose that the indexes of refraction of the two media are ni and n2,respectively. The incident light ray has an angle 81 with respect to the normal asit strikes the water and an angle 82as it passes into the water (Figure 10-5).

Incident light ray

Air (nx)

Water (n2)

5v!~A Refractedlight ray

Figure 10-5

Therelationship between theindex of refraction ofeachmediumand theangle ofincidence and the angle of refractionis governed by Snell's law, which is shownin equation (10.6).

Snell's Law nisin8i = n2sin82 (10.6)

If n2 > ni, the light ray will bend towards the normal, while if n2 < ni, the lightray will bend away from the normal.

Example 10.4aIf the index of refraction doubles for a liquid (in which a coin is submerged), theapparent depth of the coin when viewed from the air above, will:

A. increase,and it appears at a position shallower than its true depth.B. decrease,and it appears at a position shallower than its true depth.C. increase, and it appears at a position deeper than its true depth.D. decrease, and it appears at a position deeper than its true depth.

Solution

Apparent depth is the depth at which you think an object is submerged; it mayvary from the object's true depth because of refraction. For example, if you lookat a coin in a fountain, you may think the object is higher than it actually is. Thereal coin is at the bottom; you may think ifs only a few feet from the surface.

Refracted light ray (bends away from normal)

Air (nair= 1.0003)

Water (nwater = 1.3331)

• ''~v--» Perceived coin location

Incident light ray (from coin)Actual coin location



Light coming from the surface ofthe coin bends asitgoes from the water into theair (and into your eyes). Because the index of refraction of water is greater thanthat of air, we can deduce that 8air> 6water, and thus light bends away from thenormal. Your brain will think that the coin lies on a direct path from your eye tothecoin's apparent position (asshownin the diagram above). This apparentlightpath (represented by the dashed line) is, however, higher than the true lightpath; you think that the coin ishigher than it truly is. Thus, the apparent depthofa submerged object is less than its true depth. The answer mustbeeither A orB.

Now, increasing the refractive index of the liquid medium would onlyexaggerate thebending. You would then thinkthe apparentdepth had decreased(i.e., the coin would appear to be closer to the surface). That rules out choice Aand leaves us with choice B.


Example 10.4bThe material n2 has parallel sides, and is surrounded by media with ni and n3.Under which conditions will 8 exceed (J>?

A. ni > n3

B. n3 > niC. ni > n2D. n2 > n3

SolutionBecause the two interfaces are parallel to one another, the intermediate medium(with an index of refraction of n2) does not impact the path of the light ray. Theincident ray (inmedium1)and final refracted ray (inmedium 3)will be the samewhether or not the light passes through medium 2. That can be quickly provenusing Snell's law to evaluate the two refractions and then setting them equal toone another.

nisin 8i = n2sin 82 and n2sin 82 = n3sin 83 .-. nisin 81 = n3sin 83

This means that the value of n2 has nothing to do with either angle in thequestion (8 and <J>), sochoices Cand D can beeliminated. If8>4>, thenoverall thefinal refracted ray has bent towards the normal relative to the incident ray. Thelightraybends towards the normal when it slowsdown,so we canconclude thatthe speed of light is slower in medium 3 than medium1. A slowerspeed of lightmeans that the medium must have a larger indexof refraction. This means thatn3 must in fact be larger than ni.






Total Internal ReflectionA light ray bends away from the normal as it moves from a medium of highrefractive index to a medium of low refractive index. If this refracted light raywere to keep bending away from the normal as the incident angle increases, itwould eventually reach a refracted angle that is 90° away from the normal. Thisincident angle is called the criticalangle 8C.

Suppose a skindiver shined an underwater light toward the surface above, asshown in Figure 10-6a. Ifa lightray from the underwater sourcestruck the watersurface at the critical angle, it wouldbe refracted along the surface of the water.If the lightray were to strike the water surface at an angle to tiie normal that isless than 80 the refracted ray would pass into theair. But what would happen ifthe lightraystruck thesurface at an angle largerthan 8C (Figure 10-6b)? It wouldbe reflected (not refracted) off of the surface of the water. This phenomenon iscalled total internal reflection; and we know it must work this way, becauseSnell's lawtells us thatthesine ofa refracted angle cannever be greater than1.0!

Incident

light ray

Light at crtical angle

Incident

light ray

Figure 10-6

; Refractedlight ray

Total Internal Reflection

Example 10.5aAssume that a light ray is incident on an interface between two media ofdifferent optical densities and that it travels within the optically densermaterial.For which two indexes of refraction will its critical angle for total internalreflection be the SMALLEST?

A. When the indexes of refraction are 1.00 and 1.33.B. When the indexes of refraction are 1.00 and 1.01.

C. When the indexes of refraction are 1.00 and 1.52.D. When the indexes of refraction are 1.46 and 1.88.

Solution

Total internal reflection can occur only when light starts out in an opticallydenser medium (i.e., higher n) and then hits an interface with an optically lessdense medium. The critical angle comesfrom Snell's law:

("less "dense")( (sin 8ieSs "dense") = (nmore "dense")( (sin8more "dense")

To get the critical angle, set the anglein the less "dense" (i.e., optically lessdense)material to90° (since, thatis thematerial thelightwill notgetinto). This gives:

(niess "dense")( (sin90°) = (nmore "dense")( (sin Scntical)

Solving for the criticalangle yields:

nless "dense"sin 8 critical

^ore "dense"



The important factor in determining the critical angle is the ratio of the refractiveindexes. Therefore, the choice with the smallest ratio of indexes must be thecorrect choice (i.e., will have the smallest critical angle).

L00/l.52 is smaller than im/im, L00/l.33, or L42/i.88-


Example 10.5bFiber optic cables can transmit light, because light in the cable achieves totalinternal reflection as it travels. From which material could the BEST fiber opticalcable be made for use in air, if the best cable is one that would let no light raysescape into the air, regardless of the degree to which the cable may be bent?

A. An aerogel, with n = 1.05B. A polymer gel, with n = 1.35C. Glass, with n = 1.50D. A plastic, with n = 1.60

SolutionBefore attacking the question, you might have noticed that this is an extreme typeof question. The best material will be one of the extremes, either the materialwith the smallest index of refraction or the material with the largest index ofrefraction. In order to increase the number of angles of incidence that result intotal internal reflection, we want a critical angle as close to perpendicular to thesurface as possible. A perpendicular incident ray would have an angle of 0° withrespect to the normal, so we are looking for a material with the smallest criticalangle. This can be seen in the diagram below, where we consider the range ofangles that could generate total internal reflection (TIR) for both a large criticalangle and a small critical angle.

In this region,incident rayswill result in TIR

nless "dense"^

nmore "dense"

In this region,incident rayswill result in TIR


nless "dense"

Large critical angle Small critical angle

Note from the pictures that a small critical angle allows for more potentialincident angles that result in total internal reflection. We can also get this bymath if we relate the critical angle to the index of refraction for a material. InExample 10.5a we learned that:

sin Criticalnless "dense"

nmore "dense"

This confirms that we are looking for a material with a very large index ofrefraction to be in the denominator on the right side of the equation, so that thesin Scritical can be as small as possible. The best answer is the material with thegreatest index of refraction, which is a plastic withn = 1.60.




DispersionIn Figure 10-2 we saw that the visible portion of the electromagnetic spectrumvaries from about 750 nm to about 400 nm. Electromagnetic waves of differentwavelengths travel at different speeds throughmaterial substances (i.e., anythingother than a vacuum). As we saw in equation (10.5), the index of refraction for agiven material depends on the wavelength of light passing through it In otherwords, the amount that light bends at an interface between two materials varieswith the wavelength. This gives rise to an effect known as dispersion. A goodexample of a dispersive medium is a prism. If we pass white light through theprismshown in Figure 10-7, weseethat the colors of the visible spectrum refractaccording to their wavelengths. [Note that red light is deflected the leastfromitsoriginal directionof travel,whileviolet light is deflected the most]

Red(=750nm)1 Orange

YellowGreenBlue

Violet (« 400 nm)

kf

*/Shorter k correlates to greaterbending

Figure 10-7

Example 10.6aTwo initially parallel visible lightrays, A and B, are incident upon onesurface ofa prism. Which of the following is true regarding their wavelengths and wavespeeds within the material?

A. kA>kj} and va>vbB. X,a<^b and va>vbC. Xa>^b and va<vbD. kA<k& and va<vb

Solution

When two rays bend differently upon entering a new medium, their indexes ofrefraction must be different. The ray that bends more has a higher refractiveindex; Ray Ahas the higher n. Now, consider how the wavelengths and speedsof the light rays are affected within the prism. Because v = c/n, Ray A mustmove more slowly than Ray B. This rules out choicesA and B. To determine therelative wavelengths, let'suse the principal waveequation:

v = fk



In the visible spectrum, light of a higher frequency refracts more than light of alower frequency (i.e., n is higher for higher /). Given that va is less than vb, forfrequency to remain constant as a ray changes medium, Xa must be less than k&.Ray A has a smaller wavelength than Ray B.That is, if / remains constant and vdecreases, then Xmust also decrease (in order for the equation to balance).


Example 10.6bPolychromatic light refracts through a prism and strikes a screen. Which plot ofintensity versus screen position BEST represents this set-up?

A.

Intensity

C

Intensity

B.

D.

Intensity X(nm) Color

10

3

7

488

530

633

Blue

Green

Red

Intensity

Intensity

SolutionThis question tests whether or not you know that a shorter wavelength of lightwill exhibit greater bending when undergoing refraction. This means that bluelight will bend the most of the three colors used in the experiment Because bluelight has the greatest intensity (according to the data in the table), the lowestpeak on the screen should have the greatest area (be the biggest). This eliminateschoices A and B.

Green light has the lowest intensity of the three colors, so the smallest peakshould be found in the middle of the screen, where green light would strikehaving bent an amount somewhere between the extremes of red and blue. Thiseliminates choice C.





Physics Light and Optics Optics

OpticsOptics, as we shall study it, is applied geometry. When light emanates from anobject, that light can interact with such optical devices as a mirror (reflectivesurface) or a lens (refractive object through which light transmits). Because thelight can be altered by the mirror or lens, there is the chance that it will cast animage that is altered with respect to the original object. We shall study therelationshipbetweenan object, the opticaldevice,and the image it casts.

General RulesBecause we are dealing with positions of objects and images, we will establishwhat in essence is a number line where the (+)-side represents the region of spacewhere a real image can form and the (-)-side represents the region of spacewhere a virtual image can form. A real image is one that can be projected onto ascreen, while a virtual image cannot be cast onto a surface. We also have rulesthat describe the magnification of an image relative to the object These rules willdefine the size of the objectand whether the image is upright or inverted withrespect to the object The numerical rules are shown in Table 10-2.

Magnification Real/VirtualConvention Image Convention Image

M < 0 Inverted

M > 0 Upright i > 0 Real

IMI >1 Enlarged i < 0 Virtual

IMI =1 Same size

1M1 < 1 Reduced

Table 10-2

The Lens Maker's EquationWe can relate the focal length of a lens or a spherical mirror with the distanceseparating the object and the image from the lens or mirror by equation (10.7),which is often referred toas the lens maker's equation or the thin lens equation.

Thin Lens Equation 1 = 1+1f s S'

(10.7)

When youconsider this equation, it is important to keep thesigns correctThereare a few conventions to keep in mind:

1.

2.

3.

Focal Length (f): This value is positive (+) for a converging lensor mirror and negative (-) for a diverging lens or mirror.

Object Distance (s, p, or o): If the object is to the left of the lens,this value is positive (+). If the object is to the right of the lens,this value is negative (-).

Image Distance (s', q, or i): Real images will give a positive (+)value, while virtual images will give a negative (-)value.

You may have noticed that three different letters can be used to represent theimage distance as well as three different letters representing the object distance.Because all three are seen in physics, we will interchange them betweenexamples and questions without any rhyme or reason to our usage. The goal is toget you used to more than one letter representing image and object.



MirrorsA mirror is an optical device that can alter the path of light coming from anobject Mirrors focus light by reflecting it There are three general types ofmirrors. A plane mirror is flat and resembles the one on your bathroom wall. Aconcave (converging) mirror has its reflecting surface on the inside of a curve. Aconvex mirror (diverging) has its reflecting surface on the outside of a curve.

The location of an image produced by a spherical mirror can be calculated byusing equation (10.7). Both mirrors and thin lenses use the same equation. Thelinear magnification produced by a mirror can also be calculated from theobject's height and image height, and we will do in several examples. For now,however, we'll just say mat the magnificationis a ratio of the image height to theobject's height. In Figure 10-7, we see examples of images produced by a planemirror, a concave mirror, and a convex mirror.

Plane Mirror

*—r

-*+*-

ImageALWAYS

I Upright, Virtual, and Samesize (I = - O)

•I

Optics

Convex (Diverging) Mirror (f =- #) Concave (Converging) Mirror (f =+ #)

h 2 4

Image f *"* R

ALWAYS

Upright, Virtual, and- , Smaller (I<f)

Figure 10-7

Imagecan vary depending 1^onobject andfocal length. Pf

When you look at these examples, note whether the image produce is real orvirtual. For a single plane mirror and single diverging mirror, the images willalways be inverted and virtual. For a converging mirror, we will consider whattype of image would be produced, ifthe object were placed either to the left ortothe right of the focal point Regarding the focal lengthassociated with a sphericalmirror, its size relates to the radius of curvature of the mirror as equation (10.8):

f = K2

(10.8)

where R is the radius of a sphere that would be formed if the mirror were acomplete sphere.

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While the images produced by plane mirrors and diverging mirrors arepredictable, the image produced by a converging mirror takes some effort todetermine. We will look at five different scenarios, and then generate a summarychart that will be our guide when answering questions. Consider the objectat thefollowing fivepositions relative to the R and f of a converging mirror.

I. Objectbeyond the radius of curvature.II. Objectat the radius of curvature.III. Object inside the radius of curvature but beyond the focal pointIV. Objectat the focal pointV. Objectinside the focal point.

Before we start lefs considerthe thin lens equation for an objectthat is exactly 2ffrom the mirror. According to equation (10.7), the image should also form at 2f.This can be manipulated and rewritten as equation (10.9).

I = 2_ = 1_ + _1_ Given that2f= R, we get:f 2f 2f 2f 5

t = L + L (10.9)f R R

This is going to provide a useful hint when analyzing image formationassociated with converging mirrors and lenses. Let's start by considering Case I,where the object is positioned outside the radius of curvature. In order for thethin lens equation to hold true, if the object is beyond R, then image must forminside of R. This is to say that if one of the denominators on the right in theequation (10.9) isgreater than R, then the otherdenominator on the right mustbelessthan R. Consider the raydiagramfor Case I shown in Figure10-8.

Case I: Object Beyond the Rof a ConvergingMirror

1

Object R / f^-^ 1 f # > r #< r

Real, Inverted, and Smaller Image0>R (R>I>f)

Figure 10-8

This will serve well asa short cutin thefuture. For instance, if the question wereto ask about the image formed by a converging mirror with a focal length of 15cmby an object thatis 45 cmfrom the mirror, thenwe could do a quick analysisas follows:

If f = 15 cm, then R = 30 cm. The object is beyond R, so the image will formbetweenf and R, meaning it willbe an invertedand real image foundbetween 15cm and 30 cm.This may be enough information to nail down the best answer ona multiple-choice exam. We shall do this foreach of the five positions and in theend realize that these questions can besolved withoutinvoking raydiagrams.

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Consider theray diagramforCase II shownin Figure 10-9.

Case II: Object at the R of a Converging Mirror

1=J_ -If ~ R+ R

Real, Inverted, and SameSizeImage0 = R (I = R)

Figure 10-9

When the object is located at the radius of curvature for a converging sphericalmirror, the image that forms will be an inverted, real image also located at theradius of curvature. Now lefs consider the ray diagram for Case III shown inFigure 10-10.

CaseIII: Object Insideof the R, but Beyond the fofa Converging Mirror

R>0>f

1 + lf #<R #>R

Real, Inverted, and Larger Image

Figure 10-10

When the object is located inside the radius of curvature but beyond the focallength for a converging spherical mirror, the image that forms will be aninverted, real image located beyond the radius of curvature. Now lefs considerthe ray diagram for Case IV shown in Figure 10-11.

Case IV:Objectat the f of a ConvergingMirror

No Image Forms(I = oo)

Figure 10-11


Optics



When the object is at the focal length for a converging spherical mirror, nofocused image forms, although it can be said that the image forms at infinity.Now lefs consider the ray diagramfor CaseV shown in Figure 10-12.

CaseV: Object Insideof the f of a ConvergingMirror

^nVirtual, Upright,and Larger Image

Image

Figure 10-12

Now that we've considered all five cases, we can keep a summary in our mindand apply the information to answer multiple-choice questions quickly. Try tocomplete the following twoquestions in less than fifteen secondseach.

Example 10.7aWhat is true of the image formed by a converging mirror with a focal length of 25cm by an objectthat is 40cm from the mirror?

A. It is an inverted, real image that forms 66.7cm from the mirror.B. It is an inverted, real image that forms 33.3cm from the mirror.C. It is an upright, virtual image that forms 33.3cm from the mirror.D. It is an upright, virtual image that forms 66.7cm from the mirror.

Solution

The pertinent facts are (1) a converging mirror and (2) the objectis between f andR.This makes it a scenario like Case HI, where the image forms beyond the R. If f= 25 cm, then R = 50 cm, so the image must form beyond 50 cm. ChoicesB and Care eliminated. Objects beyond the focal length of a converging mirror formupright, real images, eliminating choiceD.


Example 10.7bWhat is true of the image formed by a convex mirror with a focal length of 20 cmby an object that is 30 cm from the mirror?

A. It is an inverted, real image that forms 60 cm from the mirror.B. It is an inverted, real image that forms 12 cm from the mirror.C. It is an upright, virtual image that forms 12 cm from the mirror.D. It is an upright, virtual image that forms 60 cm from the mirror.

Solution

The pertinent fact is (1) ifs a diverging mirror. For a diverging mirror, the imageis always upright and virtual, forming inside of f. If f = 20 cm, then the imagedistance must be less than 20 cm. Choices A and D are eliminated. Objectsreflectedin a diverging mirror form upright, virtual images, ousting choiceB.




LensesLight rays coming from an object can be altered if they pass through a lens. Alens is an optical device with two curved, non-parallel refracting surfaces. Thereare twotypesof lenses to consider: converging and diverging. A converging lens(convex lens) with smooth surfaces allows parallel light rays to bend toward anaxis that passes through the center of the lens and to converge at a single pointcalled the focal point f. The focal length for a converging lens isalways a positivevalue. This is shownin Figure 10-13a. Because lightcanenter the lens frombothits rightand leftsides, the lens has two focal points.

(a)

Converging Lens (f is a + #)

(Also known as a Convex Lens)

Figure 10-13

Diverging Lens (f is a - #)

(Also known as a Concave Lens)

A diverging lens (concave lens) allows parallel light rays to move apart as theypass through the lens. As shown in Figure 10-13b, the focal lengthofa diverginglens is a negative value. This is because the diverging light rays appear to leave asingle focal point from the side of the lens that light enters. Like converginglenses, diverging lenses will also have two focal points.

As light passes through a lens, either a real image or a virtual image can beformed. The rules for single lenses are the same as they are for single mirrors,except that the real and virtual sides are flipped and convex and concave arereversed. As was the case with mirrors, diverging systems generate smaller,upright, virtual images. Converging systems, be it lenses or mirrors, also sharesimilar rules. Lefs consider the formation of a real image using a converginglens. This can only occur if the object is outside of the focal length. An object isplaced a distance s to the left of the lens, as shown in Figure10-14.

Virtual sideObject

Real side

Figure 10-14

Light rays leaving this object travel in all directions, and some of them passthrough the lens. Of all the rays that pass through the lens, we will focus ourattention on just two of them. They pass through the lens to form an invertedimage of the object at distance s' to the right of the lens. Because the light rayspass through the lens and come together to form an (inverted) image of theobject, that image is called a real image. One property of real images is that onlythey can be cast upon a screen, photographic negative, or retina. Thus, lenses incamerasand eyes have focal lengths that produce real images.


Optics



Now lefs consider the formationof a virtual image using a converging lens. Thisis only possible if we place our object between the focal point and the lens, asshown in Figure 10-15. As tiie lightrayscoming from the object pass through thelens, they will diverge on the right side. If we follow those diverging rays backthrough the lens (dashed lines), we find thatthey intersect at a point Ato the leftof the object Since no light rays from the object intersect at that pointin space,the image at that point is said to be virtual.

Image

Virtual side X Real side

Figure 10-15

MagnificationSuppose we wanted to know the height of our image h' relative to the height ofthe objecth in Figure 10-14. This is called linear magnification m and is given byequation (10.10). The linear magnification value is unitless, and the conventionsare listed in Table 10-1. In Figure 10-8, the object with height h is uprightpointing above the axis. The value of h is taken to be positive. The invertedimage with height h' is below the axis. The value of h' is taken to be negative.

mh' (10.10)

Consider s, s', h, and h' in Figure 10-14. We can relate these four variables to oneanother using geometry, where the height-to-length ratio (h : s) is the same forboth the object and the image. Both ratios need to have the same sign, so anegative sign is included. This relationship allows us to use the objectand imagedistances to calculate the linear magnificationusing equation (10.11).

m = §- (10.11)

The point of preparing for the MCAT is to be able to answer questions quickly.As such, lefs develop a plan ofattack for questions dealing with either a singlelens or a singlemirror. Tnefollowing three-stepalgorithmshould prove useful.

1.

2.

3.

Identify whether the system is diverging or converging. Keep inmind that convex lenses and concave mirrors are convergingsystems, so you must do some work to determine whether the opticdevice is converging or diverging.If tiie optic device is diverging, then the image will be smaller thanthe object, upright, virtual (an SUV), and found between the focallength on the virtual side and the optic device. If the optic device isconverging, then we need to do a positional analysis to see whichcase it fits out of I through V.Sketch a diagram of the object, optical device, and image. Doublecheckusing equation (10.12) (a shortcut version of equation (10.7).

l =oxf

o-f(10.12)



Example 10.8aWhere is an image formed when an object is 20 cm to the left of a concave lensthat has a focal length of 15 cm?

A. It is an upright, virtual image that is found 60 cm to the left of the lens.B. It is an upright, virtual image that is found 8.7cm to the left of the lens.C. It is an inverted, real imagethat is found8.7 cmto the rightof the lens.D. It is an inverted, real imagethat is found 60 cmto the rightof the lens.

Solution

Lefs use the three-step method. First off, a concave lens is a diverging system.Second, because ifs a diverging system, the image will be upright, virtual,smaller than the object, and found between the focal length on the virtual sideand the optic device. The image distance must be less than 15 cm, eliminatingchoices A and D. Because ifs a lens, the objectis on the virtual side, so a virtualimage will form to the left of the lens, eliminating choiceC.

The third step is not necessary here, because only one answer choice remainsstanding.Butlefs considershow the step for a senseofcompletion.

Object20 cm


i =o x f 20 x (-15) -300

o-f 20-(-15) 35

-300 -60— = — = - 8.somethmg

Example 10.8bWhat is true of the image formed by an object 50 cm from a convex lens that hasa radius of curvature equal to 60 cm?

A. It is an inverted, real image that is found 75 cm from the lens.B. It is an inverted, real image that is found 30 cm from the lens.C. It is an upright, virtual image that is found 30 cm from the lens.D. It is an upright, virtual image that is found 75 cm from the lens.

Solution

Lefs use the three steps listed above. First off, a convex lens is a convergingsystem. Second, because ifs a converging system, we need to determine whichscenarioit fits. The objectis between R and f, so an inverted, real image will formbeyond R The image distance must be greater than 60 cm, eliminating choices Band C. Becausetiie image is real, choice D is eliminated.

i =

Object @50 cm

cmft_60

o x f 50 x 30

o-f "50-30

30 cm•

1500

20= 75



>60cm

Image

30 cm 60 cm

V

253

Optics

Test TipCalculation Shortcut

Test TipCalculation Shortcut



Example 10.9aAn object is moved from within the focal length of a converging lens to adistance beyond the focal length. During this repositioning, the image of theobject is observed to change from:

A. uprightand virtual toinverted and real.B. uprightand realto inverted and virtual.C. inverted and virtual to upright and real.D. inverted and real to upright and virtual.

SolutionLooking at howthelight bends(and theimage forms) in Figures 10-14 and 10-15,we see that the image will go from upright to inverted. The answer must beeither choice A or B. The image is virtual when the object is inside of the focallength, but realonce the object is beyond the focal length. The object starts insidethe focal lengthand is moved to outside the focal length,so the image goes frombeingupright and virtual to inverted and real. This makes choice A correct.

Upright, virtual^ Image

Object inside f

J.Object outside f

Inverted, realImage I

Two ideas for to remember for a single lens or a single mirror, are the following:Upright images are always Virtual, and Inverted images are always Real. Justremember UV-IR. This relationship works for any single lens or single mirror(whether it be convex, concave, or flat).


Example 10.9bWhen an object is moved from within tiie focal length of a diverging lens to adistance beyond the focal length, the image is observed to:

A. changefrom upright to inverted, and move away from the lens.B. changefrominverted to upright, and move closer to the lens.C. remain upright and moveaway from the lens.D. remain inverted, and move closer to the lens.

SolutionThis questionis testing you on pure memorization from this section. Because thevirtual rays for a diverging optical device start from virtual focal point, the rayswill always intercept at some point between the surface ofthe optical device andthe focal point on the virtual side.This means that for a single diverging lens or asingle diverging mirror, the image will alwaysbe an upright, virtual image thatis smaller than the objectand is found between -f and 0. Keep in mind that for adiverging lens or mirror, we use the negative focal length. This makes for a niceshort cut. The image will remain upright, so choices A, B,and D are eliminated.




Example 10.10aA diverging lens has a focal length of 10 cm. If an 8-cm tall vertical arrow isplaced 25 cm to the left ofthe lens, the resulting image will be approximately:

A. 7 cm to the left of the lens, upright, and reduced.B. 15cm to the right of the lens, inverted, and reduced.C. 35 cm to theleftof the lens, upright, and enlarged.D. 250 cm to therightof the lens, inverted, and enlarged.

Solution

Lefs solve this problem in three ways: using the shortcut first, the picturemethod second, and then equation (10.12) last. Because ifs a diverging lens, theimage will be upright and virtual, formed between 0 and -f. Because the focallength is 10cm, the image must form at some distance less than 10 cm. Onlychoice A fits this restriction, so it must be the best answer.

If you opt not to use the shortcut method, then tracing is often fast for problemsof this nature; make a quick picture before doing long math. The two rays todraw are shown below.One ray goes through the centerof the lens, as it does fora converging lens. Theother is parallel to the optical axis, diverging from the leftfocal pointas it passes throughthe lens. Since the tworeal rays (on the rightsideof the lens) do not intersect, they must be extrapolated backwards to a commonpoint-this point locates the image. The image is to the left of the lens, within the10-cmfocal length, and reduced; this makes choice A the best choice.

Object25 cm

For the algebra fans, lefs use the modified version of the thin lens equation(equation (10.12)) to solve this problem without resorting to all of that slow andpainful drawing. Using the sign conventions, the problem asks for i, given o =+25cm and f = -10cm. A quick bit of calculation gives:

i =

oxf _ 25 x(-10)o-f 25-(-10)

-250

35--= * -7«-ta,

This again supports choice A. You have three methods, and which you chooseshould be both fast and comfortable for you. Just to confirm the fact that theimage is reduced, which occurs whenever the image is closer to the opticaldevice than the object, lefs apply equation (10.11) and solve for themagnification. This gives M = +7/25. The image will be 7 cm to the left (i < 0) ofthe lens, reduced (M < 1), and upright (M > 0). This is choice A. Although themath can produce the correct answer, it requires you to know the signconventions for thin lenses (a knowledge that can easily slip away under thepressure of a timed test).



Optics



Example 10.10bA diverging lens hasa focal length of10 cm. Ifa 4-cm tall vertical arrow isplaced7 cm to the left of the lens, the resulting image will be approximately:

A. 3 cm to the right of the lens, inverted,and reduced.B. 4 cm to the left of the lens, upright and reduced.C. 17cm to the left of the lens, upright, and enlarged.D. 70cmto tiie right of the lens,inverted,and enlarged.

SolutionLefs again solve this problem in three ways: using the shortcut first, the picturemethod second, and then equation (10.12) last. Because ifs a diverging lens, theimage will be upright and virtual, formed between 0 and -f. Because the focallength is10cm, the image must form at some distance less than 10 cm and ontheleft side of the lens. Only choice B fits this restriction.

We can also use a ray trace.

For those fond of algebra, lefs use the modified version of the thin lens equation(equation (10.12)) to solve this problem. Using the sign conventions, the problemasks for i, given o = +7cm and f = -10 cm. A quick bit of calculation gives:

i =oxf _7x(-10) __70o-f 7-(-10) 17

z68 =-417

This again supports choice B. Just as we saw with Example 10.1a, you have threemethods, and which you choose should be both fast and comfortable for you.Just to confirmthe fact the image is reduced, which occurswhenever the image iscloser to the optical device than the object, lefs apply equation (10.11) and solvefor the magnification. ThisgivesM = +4/7. The image willbe 4 cm to the left (i <0) of the lens, reduced (M< 1),and upright (M > 0), which fits only choiceB.




Combinations of LensesThings get a bit more complicated when there are multiple lenses within anoptical system. The basic rule to multiple lens systems is that the image of thefirst optical device is the object of the second optical device and so on for anyadditional lenses. Most multiple-lens devices have two lenses total. Dependingon the goals, the optical system can enlarge a small object, reduce a large object,or make a far away object seem near. Optical instruments like binoculars,telescopes, and microscopes contain more than just a single lens. Lefs consider asymmetric two-lens system. In Figure 10-16, we see two converging lensesseparated by a distance d. For simplicity, the two lenses have identical radii ofcurvature. The object that we are interested in is at a distance si from Lens 1.

Figure 10-16

Byusing equation (10.12), we can determine where the image si' will be located.The first image is shown in Figure 10-17. Note the image is inverted, because theobject is outside of fi.

Object

Figure 10-17

We now let our inverted image be the object for Lens 2 (see Figure 10-17). Thisrequires that we calculate how far away the new object is from the left side ofLens 2, by using the relationship S2 = d - si'. The final image can be found byusing equation (10.12) again, where o is replaced by S2 and i is replaced by S2'.Keep in mind that the test writers are free to use any of the common designationsfor image position, focal length, and objectposition.

s2' = S2*f2s2-f2

If the value of S2' is positive (+), the image is a real (Figure 10-18). If the value ofS2' is negative (-), the image is virtual. Because the first image (and objectof Lens2) is outside of f2, the final image will be a real image on the right side of Lens2.


Optics



Object

Optics

Final

Image

Figure 10-18

In this example, we chose a symmetric pair oflenses and placed them in such away where their R points were essentially at the same point. When the lensesareof unequal radii of curvature and the system is not symmetrical, it may prove tobe a bit more challenging to visualize. Math may prove to be the tool of choicefor more complicated systems.

A microscope is an example of an asymmetric system, where the first lens(referred to as the objective) is placed in such a fashion that the object is close toor at its focal point Thisallows the image to be extremely magnified before beingmagnified and inverted by the second lens (referred to as the ocularor eyepiece).The magnifying power of a microscope is a product of the magnifications of bothlenses.

Lens AberrationsReal lenses can produce aberrations, owing to their shape and composition, sothey do not always cast perfect images of an object. When lenses are not shapedproperly, many types of aberrations can arise. Three common aberrations arespherical, astigmatic, and chromatic aberrations. Spherical aberrations developwhen a misshapen lens still has cylindrical symmetry with respect to the opticalaxis,but has not been properly rounded; the image is clear when viewed throughthe center of the lens, but it becomes distorted as one moves the image off of theoptical axis. Astigmatic aberrations appear in a lens whose shape is neitherproperly rounded nor cylindrically symmetric. In this case, the image becomesblurred and distorted, regardless of the objecfs orientation. The twoaforementioned aberrations can happen with any type of light passing throughthe lens, even monochromatic light Chromatic aberrations occur because ofdifferences in wavelengths for different colors of light. Recall that in a prism,light refracts differently depending upon its wavelength. Since the focal pointdepends upon refraction through the lens, light of different colors refracts todifferent focal points. This is true for most lenses and can be seen if you lookcarefully at a projected image-it will have a rainbow halo around it (i.e., notevery wavelength oflighthas focused coherently at the viewingscreen).



Interference Phenomena

InterferenceIf you've seen the spectrum of colors in a soap bubble, a stack of microscopeslides, or an oil slick, you've seen examples of interference. The green, forexample, will show brightly only in certain regions of the microscope slide. Thegreen regions are regions of constructive interference for green light; wheregreen is not seen, the green interference is destructive. The same goes for everyother color you see. These everyday interference patterns are seemingly random.In laboratory experiments, the interference is carefully controlled; but it stillarises for the same reasons. When two or more waves overlap in phase (e.g., crestoverlaps crest, trough overlaps trough), the interference is constructive. Theregions of constructive interference are often referred to as intensity maxima.When the waves are completely out of phase (sometimes referred to as "180° outof phase"), the interference is destructive. The regions of destructive interferenceare often referred to as intensity minima. All other types of overlap will produceinterference somewhere between these two extremes. Any wave can interferewith another wave; but in order for the interference to be most noticeable, a fewrequirements must usually be met by the interfering waves:

The waves must be monochromatic—they must be of one and the samefrequency.

The waves must be reasonably coherent-they must retain their periodicshape, without breaks in their phase.

The waves must have the same polarization—waves of different polarizationcan never completely cancel out, since they oscillate along different axes.

These requirements are determined in Young's double-slit experiment. Hisexperiment involved light, but we can see the same qualitative results usingwater waves. Imagine two slits separated by a distance d, from which waterripples emanate (Figure 10-19). The wave crests (black semi-circles) radiate upand hit a screen. Because any overlapping waves can interfere, the resultingwaves at the screen are constructively larger when two crests overlap (or twotroughs overlap). This constructive interference is noted by bright spots on thescreen and by peaks on an intensity-versus-position plot (adjacent to the screen).Where troughs and crests overlap, dark (zero-intensity) waves result.

screen

Figure 10-19




Physics Light and Optics Interference Phenomena

The amplitude of these peaks typically depends upon the intensity of the waves,whereas the spacing depends upon other quantities. For example, what shouldhappen to the peaks if the wavelength increases, but all else remains fixed?Looking at Figure 10-20, the spacing should increase (remember, the sameinterference rules apply, regardless of k, d, or L). What should happen if weincrease d, while keeping all other variables fixed? Figure 10-21 shows that thespacing should decrease. As a quick self-quiz, sketch out what should happen ifL is decreased, while holding k and d constant. You should find that the spacingdecreases if L decreases. This direct proportionality between L and the maximaspacing occurs because the interference pattern has more distance over which tospread, if the screen is farther from the two slits.

screen

Figure 10-20

ii >^

screen

A

Figure 10-21

If we call the space between two adjacent maxima y (as in Figure 10-21), therelation between y and the other experimental variables is expressed as:

y = VLd

(10.13)



This tells you about the interference pattern, once it occurs; but how do youknow whether a passage is askingyou about interference? Look for any or all ofthese features:

waves overlapping in space-regardless of their type, but typically light,sound, or water.

interference-related words-"minima," "maxima," "bright regions," "darkregions," "constructive," or "destructive."

For example, if you shine monochromatic laser light onto a thin film of oil as itfloats on water, you see light and dark patterns. This must be an interferenceeffect; and because it involves a thin film of material, it is called thin-filminterference. It is this type of light interference that you see in bubbles andmicroscope slides. So how do the thickness of the film and its optical propertiesfactor in? Let's answer this by asking what it takes to create constructiveinterference.

Shine laser light on the oil film in Figure 10-22a. Upon hitting the oil, the lightwill partially reflect and partially refract (note that this is a general behavior ofany wave mat hits an interface between two media). The refracted ray willpartially reflect off of the oil-water interface and subsequently return to the air. Ifthis returning wave and the initially reflected wave are in phase, then a bright(constructive) spot results. If they are completely out of phase, then a dark spotresults. Whether the waves are in or out of phase depends upon the path-lengthdifference fi between the two waves (i.e., the extra distance one wave has totravel with respect to the other wave). Here, the refracted and subsequentlyreflected ray is the ray that travels farther. In this case, 6 should be:

5 = 2d = N(X/n) (10.14)

where the 2d comes from the fact that one wave must travel down and back upthrough the oil (note that the laser wave is drawn at an angle only for the sake ofclarity). The index of refraction n (of the thin film) accounts for the shorterwavelength of light within the oil than in air. The variable N is an integer, whichmeans that some whole number of wavelengths must fit into this path-lengthdifference, if the two waves are to be in phase upon traveling back up into theair.

(a) (b)

oil n = 1.25

3^#S^icRa v&j.i&: a .1-- ~l' .tiy^y•

Figure 10-22

This relation does not hold for all examples of thin-film interference. A subtlecomplication can arise that depends upon the refractive indexes of the materialsabove and below the thin film. Referring to Figure 10-22b, suppose that a waterysoap film (of the same thickness as the oil) is surrounded completely by air and ishit with the same light that hit the oil. In this case, the light would interferedestructively. Why?




Physics Light and Optics Interference Phenomena

The effective path-length must be somehow different in this case-specifically, itmust be half a light wavelength different, so that the two interfering waves are180° out of phase. Why does this happen? It turns out that when a wave reflectsfrom an optically denser medium, its phase flips 180° (i.e., a crest turns into atrough). In the case of the oil film on water, the phase flips for both the wavethat reflects off of the air-oil interface and the wave that reflects off of the oil-water interface.In both cases, the light bounces off of a material whose refractiveindex n is higher than that of the material through which the wave moves. Forthe oil film on water, the phases of both waves flip; the effect cancels, and thewaves interfere constructively. For the soap film, only one of the two reflectingwaves flips its phase. This single flip effectively adds half a wavelength to thepath-lengthdifference (note that 180° of phase is akin to a half-wavelength shift)and results in destructive interference. Keep in mind that the constructive anddestructive interference occur in these two problems because of a specificcombination of the film thickness, light wavelength, and the refractive indexes.Generally, light incident normally upon a thin film reflectsconstructively, if:

d=^nthin mm>+ * ^n^ mm>(* of flips) (10.15)(i.e., if the effective path-length difference is an integral number of wavelengths).Identifying a thin-film problem is the same as identifying any interferenceproblem, with the additional requirement that there be some thin film. Thin-filminterference has industrial applications. Thin plastic coatings are often sprayedonto camera lenses to create an anti-reflective lens. No light reflecting meansmore light in through the lens. The wavelength selected to have the leastreflection (e.g., yellow-green, as it is in the middle of the visible spectrum),determines the thickness of the anti-reflective coating of the lens. This coatingactsas the thin film, creating destructive interference of that specific wavelength.

Thin-filminterference is used to size small objects (e.g., objects whose diametersare only a few microns across). If a hair is sandwiched between two microscopeslides, there will be a thin, wedge-shaped film of air between the slides (as inExample 10.12b). Laser light shining on this film interferes constructively ordestructively, as the thickness of the film varies across the slides. The distancebetween the constructive interference zones allows us to find the size of the hair.

Example 10.11aWhich of the following changes to a Young's double-slit experiment will spreadthe interference pattern maxima the LEAST?

A. Decreasing the slit separationby a factor of 4.B. Increasing the distance betweenthe slits and viewingscreenby a factor of4.C. Increasing the wavelength of the light by a factor of 4.D. All changeswill spread the maximaby the same amount.

Solution

According to equation (10.13), the distance between adjacent maxima is directlyproportional to both the wavelength of light, k, and the distance between thescreen and the slits, L. It is inversely proportional to the distance between theslits, d. Decreasing the slit separationby a factor of 4 will increase the separationdistance by a factor of 4. Increasing the distance between the slits and viewingscreen by a factor of 4 will increase the separation distance by a factor of 4.Increasing thewavelength of the lightby a factor of4 will increase the separationdistance by a factor of 4. All of the changes result in an increase in the distancebetween maxima by a factor of 4, so all of the changes are equal.




Example 10.11bWhen laser light is sent through two slits, a periodic pattern of constructive anddestructive spots appears on a distant viewing screen. Suppose the two slits arenow replaced with a diffraction grating (essentially several slits), as shown inthe diagram below. If the same total light intensity is used in both the double- slitand the diffraction grating experiments, then the bright spots of the interferencepattern in the diffraction grating experiment will be:

screen

i—i izzi c=] cz] rzzi czi nzn ezd izzi rzz] czi tzzi czi izzi nn i—r

A. more intense and spaced together more closely than in the double-slitexperiment.

B. more intense and spaced farther apart than in the double-slit experiment.C. less intense and spaced together more closely than in the double-slit

experiment.D. less intense and spaced farther apart than in the double-slit experiment.

Solution

The image may not be drawn to scale, so simply comparing this image with thatof Figure 10-19 will not answer the question; we need to consider the effects onthe intensity and maxima spacing more generally. Without going into depth, weknow that a fixed amount of light should reach the screen, so if the number ofspotsdecreases, the intensity of each remaining spot should increase. This makeschoices C and D highly unlikely answers. The maxima appear at the screen whenlight from all the slits are constructively interfering. In Young's experiment, thereare only two slits whose light must interfere constructively. The maxima,therefore, appear periodically across the screen, whenever the path-lengthdifference between the two light rays is equal to some integral number ofwavelengths. With a diffraction grating, the constructive spots will, again,appear when the light rays from all slits are interfering as constructively as theycan. Since the light rays emanating from several slits must now interfere inphase, the likelihood of a maximum is much smaller than it is in the double-slitexperiment. Therefore, the spacing between maxima must be larger with thediffraction grating; this rules out choices A and C.

The diffraction grating maxima will also be much narrower than the double-slitmaxima. Regarding the intensity, we know that a bright spot at the screen mustoccur because several light paths are constructively interfering. Since "several" isa bigger number than two (as in the double-slit experiment) and because themaxima are now much narrower, the maxima intensity must be greater for thediffraction grating experiment.





PhysiCS Light and Optics Interference Phenomena

Example 10.12aA thin hair is placed between two microscope slides (n = 1.5), as shown. Whenlaser light is shined down onto the slides, periodic intensity maxima are seen tospan the slides. Which of the following changes will decrease the distancebetween the maxima?

top viewhair

side view

IBII^x

I. Decrease the wavelength of the laser light.II. Fill the air gap between the slides with carbon disulfide (n = 1.6).III. Move the hair to the left, while keeping the slides in contact with each other

at their left edges.

A. I onlyB. I and m onlyC. Hand III onlyD. Ill only

Solution

A shorter wavelength will decrease the distance between adjacentmaxima. Thisis shown in equations (10.13) and (10.14). This makes Statement I valid. Fillingthe gap with carbon disulfide, as opposed to air, would lead to a smallerdifference betweenthe indexof refraction of the medium and that of the glass. Asmaller difference in n values results in less bending of light, so the light wouldneed to travel farther to get the same maxima as it got when air filled the gaps.This would spread the bright spots, thereby increasing the distance betweenadjacent maxima. Statement II is invalid. For Statement III, it is perhaps best tovisualize what happens when the hair is moved to either the left or right. If thehair moves to the left, then the glass slides grow farther apart. If the hair movesto the right, then the glassslidesgrow closertogether.Takento an extreme, if thetwo slides touchoneanother, then all of the bright spots are gone (pushed offofthe plate if you will). Opening the gap creates the spots and closing the gapeliminates the spots, so widening the gap must bring adjacent maxima closertogether. This makes Statement III a valid statement.


Example 10.12bA thin layerofplastic (n = 1.46) coats the surface of a glassplate (n = 1.52). Whenmonochromatic light is shined normally onto the plastic from the air above,constructive interference occurs. Destructive interference will occur when thecoated glass is immersed in which of the following liquids?

I. Water (n = 1.33)

II. Carbon tetrachloride (n = 1.46)III. Benzene (n = 1.50)

A. I onlyB. I and III onlyC. Hand III onlyD. nionly



Solution

The system involves light traveling from air (n = 1) to plastic (n = 1.46) to glass (n= 1.52). Since only the medium above the plastic changes (and not the thicknessof the plastic itself)/ a change from constructive to destructive interference mustrelate to phase flips. Phase flips of the light occur when the light bounces off of amedium whose refractive index exceeds that of the medium through which thelight passes. In the original constructive set-up, light bounces off of an air-plasticinterface and a plastic-glass interface. For both reflections, the refractive index ofthe reflecting medium is greater than that of the medium through which the lightpasses. For both reflections, the phase of the light will flip by 180°. To getdestructive interference, we would need to change the experiment such that oneof these two phase flips does not occur. If this is achieved, the single-phase flipwould lead to destructive (i.e., 180° out-of-phase) interference between the twofinal rays. Since the plastic-glass interface is fixed (nochoice allows us to changeit), we must get rid of the phase flip at the air-plastic interface. To do so, weselect a liquid whose refractive index is greater than that of the reflecting plastic(greater than n = 1.46). Benzene is the only liquid with this property, makingchoice D correct. As an aside, notice that water will not change the constructiveinterference, because it is less optically dense than the plastic (i.e., a phase flipwill occur at the water-plastic interface). Carbon tetrachloride will lead to onlyone reflection, at the plastic glass interface. There will be no reflection at thecarbon tetracnloride-plastic interface, because they have the same refractiveindex (i.e., there will be no optical interface for the purposes of reflection andrefraction); this phenomenon is known as index matching, and it is useful inapplications where a limited number ofreflections and refractions aredesired.


DiffractionLight passing through the slits and gratings that we have been discussingspreads out as it makes its way to the screen. You don't see this spreading,however, when you view light passing through your window. The light strikesthe floor, making a pattern of the window. So, when and why does a wavespread? Light waves, orany type ofwave, always bend and distort when passingthrough an opening or around an object. This distortion is known as diffraction.Diffraction is most noticeable in everyday life when watching water waves passby some obstacle. In Figures 10-23a and Figures 10-23b, two different waves(viewed from above) passthrough thesame opening. When the opening is muchlarger than the wavelength, the wave doesn't distort much; when the opening issmaller than the wavelength or is of comparable size, the wave spreads outnoticeably.

(a)

I Intensity *»| Intensity

Figure 10-23




PhysiCS Light and Optics Interference Phenomena

The general relation is this: Diffraction is more significant when the wavelengthis smaller than or comparable to the opening the wave crosses; diffraction is lesssignificant (or negligible) when the wavelength is much smaller than theopening. This applies to all waves. So why do the light waves entering yourroom through the window not spread much? Their wavelength is much smallerthan the window. Why do sound waves spread so effectively when enteringthrough that same window? The wavelength of the typical sound wave is on theorder of a meter-close to the size of the window. In Young's experiment, thewavelengthwas close to that of the slit widths. This is why the light spread afterpassing through the slits.

The light does more than simply spread-it also gives rise to interferencepatterns. Since there is no other wave with which it can diffract, you could saythat diffraction is a form of "self-interference." The patterns at the screens resultfrom the water waves in Figure 10-23a and Figure 10-23b. The pattern isrepresented by a plot of intensity vs. position on the screen. Notice that there is arelatively bright (i.e., high intensity) central maximum, and some much smallermaxima to its sides. These result from self-interference. Notice also that when thewave spreads more, sodoes theresulting diffraction pattern. Diffraction patternsstill occur in the absence of significant spreading-they are just much harder tosee, due to the relatively weak intensities of the side maxima.

How does the width of this central maximum relate to the wavelength of thewave and to the distance separating the slit and screen? As this is an interferenceeffect, the width should relate to these parameters in the same way as tiiemaxima separation y related to them in equation (10-13). For instance, anincreased wavelength would increase the diffractive spreading, and thereforeincrease the width of the diffraction pattern at the screen. Likewise, moving thescreen away from the slits would increase the width of the diffraction pattern,because thediffracted wave has more distance overwhich tospread out.

Example 10.13aIn a diffraction experiment, light diffracts through a single slit and onto aviewing screen. The central diffraction maximum is LEASTnoticeable when:

A. the monochromaticity ofthe light is increased.B. using only one wavelength of light.C. increasing the coherence of the light.D. using full spectrum light.

Solution

Whenever you've seen examples of light diffraction (presumably insome physicslab), you probably used a laser to generate the light. Laser light shining througha small opening forms definite diffraction patterns; such patterns are not seenwhen incandescent light is sent through the same slit. Laser light is fairlymonochromatic (i.e., of one wavelength) and coherent. With these facts in mind,choices A through C are not good choices. Only choice D, which is close to thetype of light an incandescent bulb emits, would make diffraction less noticeable.Remember, diffraction is an interference effect ("self-interference," if you will).Therefore, anything that makes interference more pronounced (e.g.,monochromatic, coherent lightof one polarization) would make diffraction morepronounced. You could have ruled out choices A and B immediately, becausethey say the same thing.



25 Light and Optics Review Questions

I. Screen and Lens

II. Prism Dispersion

III. Interference


(1-7)

(8 -14)

(15 - 21)

(22 - 25)


^^


Students in a physics laboratory conduct a simpleexperiment with a converging lens. An object is placed at adistance d to the left of the lens. Its focused image isprojected onto a screen to the right of the lens. For thisparticularexperiment, the image formed at a distance of 2d tothe right of the lens, as depicted in Figure 1:

Thin lens Screen

Figure 1

For lenses and mirrors, the image distance and objectdistance arerelated by the thin-lens equation:

L + L = L,i o f

where i = image distance, o = object distance, and f = focallength. All distances are measured from the center of thelens.

1. Whatis the focal length of the lens in Figure 1?

A. d2

B. ^3

C. d

D. 2d

2. Which graph BEST represents l/0 versus ty forconverging lens?

A. B.


3. What is the magnitude of the magnification for theobject and image in Figure 1?

A. I2

B. 1

C. V2D. 2

4. Suppose that the object in Figure 1 were moved to theleft, and that d > f throughout the moving process. Tokeep the image on the screen, the screen would have tobe:

A. moved towards the lens.

B. moved away from the lens.

C. left in the same position.

D. moved away from the lens only if the object ismoved infinitely far from the lens.

5. If the lens has a focal length ofd/2, how far away fromthe lens should we place the screen?

A. d/2

B. 2d/3C. d

D. 2d

6. Suppose that the converging lens were replaced with adiverging lens of focal length -2d. The image wouldthen be:

A. real, and located 21tothe leftof the lens.3

B. real, and located 21to the right of the lens.3

C. virtual, and located 2110 the right ofthe lens.3

D. virtual, and located 21 to theleftof the lens.3

7. Suppose that the lens in Figure 1 had a focal length d.How far to the left of the lens should the object beplaced to get an unmagnified, real image?

A. 4d

B. 2d

C. d

D. d2



When white light passes through a prism, light rays ofeach color are refracted (bent) by different amounts toproduce different angles of deviation. Red light is refractedthe least by a prism. The refraction process through a prismis demonstrated in Figure 1.

Red

OrangeYellowGreen

BlueViolet

Figure 1

This dispersion of light is seen in various solids. Therefractive index determines the amount of spreading. As therefractive index increases, the degree of refracting increases.The refractive indexes for six different wavelengths of lightin six different solids are shown in Table 1.

Material

(0°C)Violet

410 nm

Blue

470 nm

Green

530 nm

Yellow

590 nm

Orange610 nm

Red

670 nm

Crown glass 1.538 1.531 1.526 1.523 1.522 1.520

Light flint 1.604 1.596 1.591 1.588 1.587 1.585

Dense flint 1.698 1.684 1.674 1.667 1.665 1.662

Quartz 1.557 1.55 1 1.547 1.544 1.543 1.542

Diamond 2.458 2.444 2.426 2.417 2.415 2.410

Ice 1.317 1.314 1.311 1.309 1.308 1.306

Table 1

In a vacuum, all light waves-regardless of their

frequencies-travel at the same speed of 2.98 x 108 m/s.Through a medium, however, light travels at a slower ratethan through a vacuum, and different frequencies travel atdifferent rates of speed. These differences in speed are notsubstantial in air, so rays of colored light essentially traveltogether as white light in air.

8. The first sign of sunlight on the horizon at sunrise iswhite light, not colored light. How can this beexplained?

A. The different colors of light travel through the airat very different rates of speed.

B. The different colors of light all travel through theair at roughly the same rate of speed.

C. The different colors of light travel through thevacuum of space at different rates of speed.

D. The different colors of light leave the Sun atdifferent times, but their rates of speed changeenough in Earth's atmosphere so that all colorsarrive at our eyes at the same time.

Copyright©by The Berkeley Review15 269

9. What is true of monochromatic lightand the amount bywhich it is refracted by a prism?

A. The greater its frequency, the less it is refracted.B. The longer its wavelength, the more it is refracted.C. The shorter its wavelength, the more it is refracted.

D. The greater its energy, the less it is refracted.

10. Which kind of light will bend the MOST when passingthrough these prisms?

A. Green light passing through dense flint.B. Violet light passing through crown glass.C. Red light passing through ice.

D. Blue light passing through quartz.

11. As a prism is rotated to increase the incident angle oflight striking its surface, what is observed?

A. The colors disperse less, but they stay separate.B. The colors disperse less, and they recombine to

become white light.

C. The colors disperse more, and they stay separate.D. The colors disperse more, and they recombine to

become white light.

12. Two identically shaped prisms are juxtaposed as shownbelow. Violet light exits the system above red light,because the average index of refraction of Prism II is:

13.

14.

White violet

A. less than Prism I; light is dispersed more in Prism I.

B. less than Prism I; light is dispersed less in Prism I.

C. more than Prism I; light is dispersed more in Prism I.

D. more than Prism I; light is dispersed less in Prism I.

Which of the following kinds of light will travel at theGREATEST speed?

A. Violet light passing through air

B. Blue light passing through water

C. Yellow light passing through a vacuum

D. Orange light passing through kerosene

What is true of different colors of light in a medium?

A. As frequency increases, the energy decreases.B. As wavelength increases, the energy increases.

C. As speed decreases, the wavelength decreases.D. As frequency increases, the wavelength decreases.



The fact that light has wave characteristics can bedemonstrated using a double-slit apparatus. Monochromaticlight passes through two narrow slits that are close togetherand falls on a screen, forming a series of bright and darkbands (fringes). A schematic of a double-slit apparatus isshown in Figure 1:

Figure 1

Light rays (e.g., Rays 1 and 2) passing through each slitgenerally travel different distances to reach the same point onthe viewing screen. Because path-length differences yieldphase differences (between the twolightrays striking a pointon the screen), various degrees of constructive or destructiveinterference arise. These constructive and destructive regionsare the brightanddarkfringes, respectively. The criterion forconstructive interference is that the path-length differencemust equal a multiple of thewavelength of theincident light.Quantitatively, this is written as:

m k = d sin 0

where 6 is the angular location of the mm maximum, m issometimes called theorder, mX is thepath-length difference.For destructive interference, the path-length difference mustequal a multiple of halfthe wavelength of the incident light.Thus:

(m +1) k = d sin 62

A diffraction grating consists of several slits, but it worksessentially in the same way as two slits do. The gratings,however, produce fewer bright spots for a given incidentlight intensity than a double-slit set-up does. Each brightfringe also subtends a smaller angular range than its double-slit counterpart The formula for determining the primarybrightfringes in a diffraction grating is:

m k = d sin 9

where 0 is the angular location of the mm maximum, m issometimes called the order, mX is thepath-length difference.For destructive interference, the path-length difference mustequal a multiple of halfthe wavelength of the incident light.Thus:


Figure 2

relativeintensityat screen

Here, we can predict the location of the dark fringes byusing:

m k = a sin 0

where a is the width of the single slit, and 0 is the angularlocationof the m* minimum.

Interference effects arealso seen in the reflection of lightby thin films. Light is reflected from the front and backsurfaces of the film. The light reflected off of the backsurface of the film has a longer path to travel than the lightreflected off of the front of the film. Again, this path-lengthdifference leads to phase differences, depending on thethickness of the film, theangle at which the light falls on thefilm, and also the refractive properties of the spacesurrounding the film.

15. A typical double-slit is actually two single slits placednear one another. The equations describing the locationof the bright and dark fringes that result from two-slitinterference do not take into account the diffractiveeffects of each single slit. This is because we assumethat:

A. single slit effects are masked by the double-slitpattern.

B. d is much greater than k.C. d + a is comparable to k.D. a is comparable to or smaller than k.

16. Certain colors observed in natural materials areproduced by thin-film interference, while most are theresult of absorption. Differentiating one effectfrom theothercan be done MOSTeasily by:

A. observing the light exiting the sample fromdifferent angles with respect to the material'ssurface.

B. observing the light exiting the sample through aprism, set at a fixed angle with respect to thematerial's surface.

C. observing the light exiting the sample through apolarizer, set at a fixed angle with respect to thematerial's surface.

D. The twoeffects are alwaysindistinguishable.


17. If we compare the bright fringes produced by a double-slit apparatus and a diffraction grating, we find that thefringes produced by:

A. both experiments have similar intensity and similarwidth.

B. the double-slit apparatus have a greater intensity;the fringes for both have a similar width.

C. the diffraction grating have a greater intensity; thefringes for the diffraction grating are narrower.

D. the diffraction grating have a greater intensity; thefringes for the double-slit apparatus are narrower.

18.

19.

20.

21.

If the double-slit apparatus is placed under water andthe same light is used, the bright fringes will:

A. now be closer together.

B. now be farther apart.

C. remain at the same locations.

D. become dark fringes.

Soap bubbles change colors when observed over time.This is MOST likely due to:

A. refraction.

B. evaporation.

C. dispersion.

D. selective absorption.

A diffraction grating with 8000 lines/cm is illuminatedwith white light What is the spectrum of the HIGHESTorder (i.e., maximum value of m) that includes anyportion of the visible spectrum?

A. 1

B. 2

C. 3

D. 4

The effects of passing monochromatic light through asingle-slit apparatus are fundamentally different fromthe effects of passing monochromatic light through adouble-slit apparatus. This statement is:

A. true; light from two slits can interfere with itself,whereas light from a single slit has no other waveto interfere with it.

false; both effects are fundamentally the same,

true, but only as long as we use monochromaticlight. It is false, if we use light of severalfrequencies.

true, but only as long as we do not consider adouble-slit apparatus as an apparatus with twosingle slits.

B.

C.

D.



22. Total internal reflection could occur when a light wavetraveling through:

I. a vacuum encounters an air interface.

II. air encounters a glass interface.

III. glass encounters a plastic interface.

A. I and II only

B. II and III only

C. I, II, and III

D. None of the conditions listed above will lead to

total internal reflection.

23. What happens to parallel light rays coming from theleft, and then traveling through a diverging lens?

A. They are directed through the focal point to the leftof the lens.

B. They bend away from the center of the lens,appearing to have come from the focal point to theleft of the lens.

C. They bend towards the center of the lens, travelingthrough the focal point to the right of the lens.

D. They bend away from the center of the lens,appearing to have come from the focal point to theright of the lens.

24. Which of the following equations can be used todetermine the critical angle?

A. sin 0critical = n-/nmaterialB. Sin 0critical = "material/^

* /sin Ocntical ~ '"materialD. Sin ©critical = (nair)(nmaterial)

25. Wave diffraction would be MOST pronounced whensending:

A. 500-nm light waves through a one-meter hole.

B. 500-nm light waves through a one-centimeter hole.C. 5-mm ultrasonic waves through a one-meter hole.

D. 5-mm ultrasonic waves through a one-centimeterhole.

1. B 2. A 3. D 4. A 5. C

6. D 7. B 8 B 9. C 10. A

11. C 12. D 13. C 14. D 15. D

16. A 17. C 18. A 19. B 20. C

21. B 22. D 23. B 24. A 25. D

YOU ARE DONE.

Answers to 25-Question Light and Optics Review

Passage I (Questions 1-7) Screen and Lens

Choice B is the best answer. The quickest way to get the focal length of the lens is to use the thin lens equation given in thepassage. Since the clear image is on the screen, the image distance is the same as the screen distance.

> 1_ + 1 = 1 => _3_ = 1 =* f = 2dof 2ddf 2df 3

1 + 1 = 1

You may also have noted that in order for the image to project onto the screen, it must be a real image. With a convex(converging) lens, the object must be outside of the focal length in order to create a real image, so choices C and D are notpossible. The best answer is choice B.

Choice A is the best answer. The choices are linear or possibly exponential, with either an increasing or decreasingrelationship. As an object gets closer to the lens, the image gets farther away from the lens, so the graph should express adecreasing relationship. This eliminateschoices B and D. The thin-lens equation is:

Rearranging gives:

1 + 1 = 1i o f

1=.1+1

This has the form y=mx + b, with y=-1, x=1, m=-1, and b=1. So the graph of 1 versus 1 is a straight line with ai o f o i

negative slope. The best answer is choice A.

3. Choice Dis the bestanswer. Themagnification for lenses and mirrors isgiven by:

M = ^i- = --L M = - 2d = .2

Since the magnitude ofthe magnification was asked for, all we have to do is take the absolute value ofM. Thus, we get M=2. The best answer is choice D.

Choice A is the best answer. As the object moves farther away from the converging lens, the image gets closer to the focalpoint of the lens. As a limiting case, ifthe object were to be moved infinitely far from the lens, the image would be exactly ata focal point. This means that the image will move closer to the lens as the object moves away from the lens; we musttherefore movethe screen to the left (i.e., closer to the lens). The best answer is choice A.

Choice C is the bestanswer. Using the thin-lens equation:

1 + 1 = 1 1 = -1 + 1 =


1 + 2 = 1d d d

i = d

6. Choice Dis the best answer. The fastest way to the solution is to use the thin-lens equation:

7.

1 + 1 = 1i o f

1 = -1 + 1 = -l.J_ = -J_i of d 2d 2d

= -2d3

The negative sign on i means the image is virtual. For lenses, a virtual image is always on the same side of the lens as theobject This means that the imageis virtual and to the leftof the lens. The best answer is choice D.

Choice B is the best answer. We want the magnification equal to -1 instead of+1, because a real image in this case wouldbe inverted.

M = -i-o

Plugging into the thin-lensequation gives:

1f

1 = -J-o

1 + 1 = 1 1 + 1 = 1


I = o

2=1o d

o = 2d

Copyright ©by TheBerkeley Review® 272 MINI-TEST EXPLANATIONS

Passage II (Questions 8 -14) Prism Dispersion

8. Choice B is the best answer. Choice A should be eliminated, because if the light rays of different colors in sunlight traveledat drastically different speeds, then the fastest constituent color of light would be the first to appear. Choice C should beeliminated, because all light rays travel at the same speed through a vacuum (2.98 x 10^ m/s). Choice D should beeliminated, because rays of white light (comprising all colors) leave the Sun at the same time. Choice D is downright loony!Choice B is the best answer, because after the different colors of light enter the atmosphere, they travel to the surface of theEarth at roughly the same speed, allowing the arrival of all of the colors to appear to be simultaneous. The best answer ischoice B.

9. Choice C is the best answer. Figure 1 shows that violet light is refracted by the greatest amount for light in the visible rangeof the EM spectrum. The data in Table 1 confirm that violet light has the largest index of refraction in all six media, so violetlight is refracted by the greatest amount in all six media. Violet light has the highest frequency, shortest wavelength, andgreatest energy of any color of visible light. This means that greater frequency results in greater refraction, eliminatingchoice A. The shorter the wavelength of visible light, the more it is refracted, which eliminates choice B and makes choice Cthe best answer. The greater the energy of light, the greater its frequency, and the more it is refracted, so choice D iseliminated. The best answer is choice C.

10. Choice A is the best answer. The greatest degree of bending will be experienced by the light-and-medium combinationpossessing the greatest index of refraction. This can be read directly from Table 1. Green light passing through dense flinthas an index of refraction of 1.674. Violet light passing through crown glass has an index of refraction of 1.538. Red lightpassing through ice has an index of refraction of 1.306. Blue light passing through quartz has an index of refraction of 1.551.The greatest index of refraction among the choices given is green light passing through dense flint. The best answer ischoice A.

11. Choice C is the best answer. As the incident angle increases, the amount of refraction at the first interface increases-remember: a bigger incident angle correlates to a bigger refracted angle. This increases the spread of the various light rays asthey pass through the prism and also increases the refracted angles at the second interface in the prism; dispersion is evengreater. If the beamsare dispersed more, they probably will not come back together at any point in space. The best answer ischoice C.

12. Choice D is the best answer. The light must be tracked step by step to grasp what is occurring in the double-prism system.White light passing through the first prism is refracted so that violet light exits at the greatest angle, making it the lowestbeam as it leaves the first prism. Passing through the second prism, violet light is again refracted to the greatest degree, butnow it is refracted in the opposite direction (up rather than down, because the angle of the interface is different than the firstinterface). The degree of refraction must greater in Prism II than in Prism I, to explain why the violet light crossed over thered light, and exited above the red light, as shown in the picture. That means Prism II must have a greater index of refractionthan Prism I, causing light to be more dispersed in Prism II than in Prism I. This makes choice D the best answer. Note thatthe incident angle is close to the exiting angle, meaning that despite the difference in dispersion in the two prisms, thedeviation of each prism is approximately the same and the plane through which the light first enters into Prism I is parallel tothe plane through which the light exits from Prism II. The best answer is choice D.

13. Choice C is the best answer. All light in a vacuum has the same speed, regardless of color (wavelength). In addition, lighttravels more slowly in a medium than in a vacuum. This means that the greatest speed is experienced by the light in avacuum, making choice C the best answer.The fact that the light is yellow is irrelevant in this question. Where the color doesbecome important is when we compare light rays passing through thesame medium. In that case, the light ray with the longerwavelength travels faster. The best answer is choice C.

14. Choice D is the best answer. The energy, wavelength, and frequency of light are related according to the followingequation:

E = h/ = h^-,k

where E is energy, h is Planck's constant, / is frequency, c is speed, and k is wavelength. According to the equation, asfrequency increases, the energy of the light increases, eliminating choice A. As wavelength increases, the energy of the lightdecreases, eliminating choice B.

There is no distinct relationship between speed and wavelength. Within a vacuum, all light has the same speed, regardless ofits wavelength. While this does not eliminate choice C beyond doubt, it does not mandate our selection of it as the bestanswer, either. As frequency increases, the wavelength of the light decreases. The best answer is choice D.

,®Copyright © by The Berkeley Review 273 MINI-TEST EXPLANATIONS

Passage HI (Questions 15 - 21) Interference

15. Choice D is the best answer. Diffraction effects do exist for each of the two single slits-that is why the light spreads outfrom each slit in the first place. Such effects are more noticeable when the slit size is smaller than the wavelength of theincident light (although diffraction is still observed when the slit size iscomparable to the wavelength). Requirements relatingto the slit separation d are irrelevant, because if the aforementioned requirement is not met, then there can be no lightspreading and, therefore, no interference between light from the two slits. This eliminates choices B and C. The best answeris choice D.

16.

17.

18.

19.

Choice A is the best answer. As mentioned in the passage, the path difference isdependent on the angle at which the lightstrikes and exits the film. Observing the exiting light from different angles, with respect to the material's surface, wouldchange this path difference. This is basically saying that as the angle changes, the distance isdifferent, because the diagonalpathway traversed changes. This, in turn, changes which wavelengths of light constructively interfere with each other (i.e., itchanges the observed colors). Choice B is incorrect, because the prism refracts only light exiting the sample; it does notchange the color of this light. Choice C is incorrect: A polarizer polarizes only the exiting light, and it does not change itscolor. The intensity of the lightmay drop, which will impact thecolor's brightness, but it will nonetheless be thesame coloroflight. The best answer is choice A.

Choice C is the best answer. As mentioned in the passage, fewer bright spots are seen when using a diffraction grating thanwhen using the double slit-apparatus, for the same intensity ofincident light. Since the same incident intensity makes up thebright spots in both cases, the bright spots must be brighter for the diffraction grating-after all, there are fewer spots havingthe same total energy as the many more spots in the double-slit setup. This rules out choices Aand B. To see why choice Ciscorrect, recall that the passage states that the diffraction grating spots subtend a smaller angle than those of the double-slits.This means that diffraction grating spots are narrower than those produced by a double-slit setup. This happens because abright spot occurs when light rays from all the slits are constructively interfering at that region of the screen. Since diffractiongratings have many more than two slits, constructive interference is less likely to occur. A bright spot would not be spreadover much of the screen, because any spreading quickly reduces the chance that all slits are interfering constructively. Thedouble-slit apparatus requires only that two slits interfere constructively; this relaxed requirement also relaxes the narrownessof the bright fringes. The best answer is choice C.

Choice A is the best answer. When the double-slit setup is placed under water, the wavelength ofthe light is reduced. This isbecause water has a higher index of refraction than air, so light travels more slowly in water than it travels in air. If the wavespeed of the light is reduced, then the wavelength is also reduced. The frequency of the light remains the same, because thefrequency is determined by the source. The spacing of the bright fringes is determined by the wavelength, according to:m k = d sin0.

According to the relationship, as the wavelength gets smaller, 0also gets smaller, so the bright fringes will be spaced closertogether. This makes choice Athe best answer of the four choices. Choice Dis incorrect: Although the bright and dark fringesshift, they do not necessarily shift in such a way as to replace the bright fringes (observed when the experiment is done in air)with dark fringes (observed when it is done under water). Choice Cshould have been eliminated immediately, because addingthe experiment to water changes the wavelength of the light, so some change in the interference pattern will be observed. Thebest answer is choice A.

Choice Bis the best answer. Constructive interference is based on apath difference traveled by the two waves (the wave thatreflects off of the front of the soap film and the wave that reflects off of the back of the soap film). As the soap film evaporatesover time, the thickness of the soap film changes (perhaps getting thicker or getting thinner). This change in soap filmthickness in turn changes the path difference for the two waves, so as the soap film evaporates, different colors will be theones that are adding constructively. Choices Aand Care incorrect, because choice Cis saying essentially the same thing aschoice A, a refractive effect-it occurs when the index of refraction for a material varies with the wavelength of light. Whentwo answer choices are fundamentally the same from aphysical point of view, those two choices are usually both wrong, andthus should both be eliminated. This can prove useful when speeding through questions or when you aren't quite certain°whata question is asking, but you recognize that two answer choices are saying essentially the same thing. Choice Dis incorrect,because the wavelengths of light that are selectively absorbed by amaterial (assuming absorption is significant) depend uponthe molecular makeup of the material. Since this makeup does not change much with time, the colors absorbed should notchange much with time. The best answer is choice B.

®Copyright © by The Berkeley Review 274 MINI-TEST EXPLANATIONS

20. Choice C is the best answer. To solve for the largest m that includes any portion of the visible spectrum, we must use theequation given in the passage: mk = d sin8. Upon manipulating the equation to isolate the m, we arrive at the followingrelationship:

m _ dsinO

where d = 1/8000 lines/cm. The wavelength of visible light ranges from roughly 400 nm to roughly 700 nm. Since we wantthe largest m, we choose the smallest wavelength and the largest sin 0~these numbers are 400 and 1, respectively. Solvingform:

[(1/8000) x 1Q'Z m/linel

400 x 10"9 m

10' 10'm =

8000 x 400 32 x 105100

32

The value of m has to be an integer, so we must approximate the answer as 3, because it is the closest of the choices to100/32. The best answer is choice C.

21. Choice B is the best answer. Although we speak of single-slit effects and double-slit effects, both effects are the result ofwave superposition-constructive and destructive interference. This means that the effects of passing monochromatic lightthrough a single-slit are not fundamentally different from passing monochromatic light through a double-slit. Choice A isincorrect: Light from a single slit interferes with itself. Choice C is incorrect: Light with several frequencies will still displayinterference effects, although we may not always be able to see them as sharply. Choice D is incorrect: A double-slit is twosingle slits. It is possible to speak of single-slit and double-slit effects in the same experiment; for certain experimentalconditions, it is possible to see single-slit diffraction patterns within the double-slit interference pattern. This will occur whenthe slit width and the wavelength of monochromatic light are chosen in such way where fringe interference occurs atdifferent points than the bright spots of the double-slit interference. The best answer is choice B.


22. Choice D is the best answer. Total internal reflection can occur only when a wave is moving from a region with a higherindex of refraction to one with a lower index of refraction, because the ray will bend away from the normal. This occurswhen light is traveling from a slower medium into a faster medium (as we see when light travels from water into air). Noneof the Options I, II, or III corresponds to a situation where light is traveling from a slower medium to a faster medium. Lightis faster in a vacuum than it is in air, light is faster in air than in glass, and light is faster in glass than plastic. Unnerving as itmight be to choose, the best answer is none of the above, choice D. Although "none of the above" is rarely an answer choice,if it does happen to be a choice on your actual MCAT, treat it as just another answer and squelch any bias or preconceivednotions you may have about choosing such an answer. The best answer is choice D.

23. Choice B is the best answer. First off, a converging lens focuses parallel light rays that pass through it toward the focalpoint of the lens. This is how the lenses in cameras, microscopes, and reading glasses work. However, light rays comingtoward and then passing through a diverging lens diverge away from the center of the lens. This fact immediately eliminateschoice C. This is something you should have committed to memory as both a informational fact as well as a mentalvisualization. Choice A is also obviously incorrect, because if light rays were coming in parallel to each other, why wouldthey all bend and go through the first focal point before ever even coming into contact with the lens? If they haven't yetinteracted with the lens, then they would not undergo any sort of bending due to the lens. That leaves us with choices B andD. The only difference between the two is that in choice B the light rays appear to have come from the first focal point, andin choice D, they appear to have come from the secondfocal point. Imaginea ray diagram for a typical diverging lens: If theraysappearto comefrom the second focal point, then they would have to be refracted by more than ninety degrees. Since weknow that this is physically impossible, the light rays must appear to come from the first focal point This is why we extendour dashed line in ray diagrams back to the first focal point, using the straight lines associated with the diverging rays leavingthe lens. The best answer is choice B.

24. Choice A is the best answer. When dealing with a critical angle, vve are considering a refracted angle of 90°. If you look atthe answer choices, you'll notice that this question requires that you manipulate Snell's law for a refracted angle of 90°.Because sin 90° is 1, the sine 90° term drops out of the calculation, leaving just n on that side of Snell's relationship. Theresult is that the sine of the critical angle equals the ratio of the medium with the lower index of refraction to the mediumwith the higher index of refraction. The math is as follows:

nmateriaisin 8criticai =n^-sin 90° .-. sin 8criticai =nair/nmaterialsin 90° =nair/nmatcriai



25. Choice D is the best answer. Wave diffraction is most pronounced when thewavelength of thewaves passing through a slitare comparable to the diameterof the slit Of the answerchoices, 5 millimeters is relativelyclose to 1 centimeter(it is half aslarge), so that is the scenario with the greatest observed amount of diffraction. The best answer is choice D.

Copyright ©by TheBerkeley Review® 276 MINI-TEST EXPLANATIONS

52-Question Light and Optics Practice Exam

I. Critical Angle Experiment

II. Minors

III. Telescopes


IV. Functioning of the Eye

V. Different Lenses


VI. Thin Films

VII. Double Lens


Light and Optics Exam Scoring Scale


42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3

(1-6)

(7-12)

(13 -17)

(18 - 21)

(22 - 27)

(28 - 32)

(33 - 36)

(37 - 42)

(43 - 48)

(49 - 52)


To study the refractive properties of a series of materials,light of 589 nm (from a yellow sodium lamp) was used. Thelight was transmitted through a half-cylinder disk of thematerial at gradually increasing incident angles until no lightleft the material from the flat surface to the air. For light thattransmitted from flat face of the half-disk to the air, the beamwas allowed to strike a semi-circular screen. For each of the

light rays that left the middle of the flat face of the half-disk,the point of highest intensity against the screen was recorded.The path of the light was traced back to determine theincident and refracted angles. Figure 1 shows the apparatus.

Incident rayof589nm,<

Point a

Half-diskof Material

Screen

Figure 1.Apparatus for the Refraction Experiment using aHalf-disk of Material and Light from a 589 nm Laser

Air has an index of refraction of 1.003, which vveassume to be 1 for the purposes of this experiment. Thestudents measured the threshold angle for six solid materials.

Polystyrene 42.2° Crown Glass 41.1°

Rock Salt 40.4° Flint Glass 38.1°

Zirconia 31.3° Rutile 22.4°

The index of refraction, n, can be calculated from thecritical angle associated each solid using Snell's Law:

"material'Sin 0critical = najr-sin 90°

What is the refracted angle for yellow light passingfrom air into a rectangular block of flint glass with anincident angle of 60°?

1.

2.

A.

B.

C.

D.

0°

32.1

38.1'

90°

For four lenses of identical dimensions, which materialwould generate the focal point closest to the lens?

A. Polystyrene

B. Rock salt

C. Rutile

D. Zirconia

®Copyright ©by The Berkeley Review 278

3.

4.

5.

6.

To be useful in the experiment, the material must be allof the following EXCEPT:

A. transparent to visible light.B. of uniform composition.

C. at a uniform temperature.D. absorbing of light at 589 nm.

Which of the following explanations supports thenotion that monochromatic light is essential for theexperiment to work?

I. Monochromatic light avoids complications causedby diffraction.

II. Monochromatic light avoids complications causedby dispersion.

III. Monochromatic light avoids complications causedby reflection.

A. I only

B. II only

C. I and II only

D. II and III only

Light that initiatesfrom point a in Figure 1:

A. will always strike the middle of the screen.

B. does not leave the half-cylinder disk.C. splits to strike the screen at multiple points.D. slows when leaving the disk and entering air.

If this experiment were carried out in water instead ofair, what would be expected?

A. The critical angle would be smaller in water thanair; the angle of refraction would be greater inwater than air.

B. The critical angle would be smaller in water thanair; the angle of refraction would be smaller inwater than air.

C. The critical angle would be greater in water thanair; the angle of refraction would be greater inwater than air.

D. The critical angle would be greater in water thanair; the angle of refraction would be smaller inwater than air.



A mirroris a device that focuses light rays by reflectingthem to a point Mirrors can be planar (such as the mirror thathangs in a bathroom) or spherical (such as the front and backof a metallic spoon). Images in planar mirrors are always thesame size as the object whose light they reflect. Sphericalmirrors can be either convex or concave, depending on thecurvature of the mirror. A convex spherical mirror is alsoknown as a diverging mirror. Light rays striking the surfaceof a diverging mirror are reflected away from its axis. Theselight rays never actually meet, but they appear to meet Aconcave spherical mirror is also known as a convergingmirror. Light rays striking the surface of a converging mirrorare reflected toward its axis, so the rays actually do meet.Images in spherical mirrors may be enlarged or reduced aswell as inverted or upright

The object distance (s), image distance (s'), and focallength (f) of a spherical mirror are governed by the thin lensequation:

I-I +lf s s'

7.

8.

Using a plane mirror that is 1 meter tall would preventwhich standing person from seeing approximately theirentire body length?

A. A person standing 1 m from the mirror whoseheight is 2 meters.

B. A person standing 2 m from the mirror whoseheight is 1 meter.

C. A person standing 4 m from the mirror whoseheight is 0.5 meters.

D. A person standing 0.5 m from the mirror whoseheight is 4 meters.

What do spherically convex and plane mirrors have incommon?

I. Both mirrors form only virtual images of an object.

II. Both mirrors can form images of an object that aresmaller than the object.

III. Both mirrors form images of an object that arecloser to the mirror than the object.

A. I only

B. I and II only

C. I and III only

D. II and III only


9. If an object is placed in front of a concave mirror at adistance from its surface that is three times the mirror'sfocal length, the image will be:

A. 2.f from the mirror, in front of it2

B. ^- f from themirror, inback of it.2

C. 4-ffrom the mirror, infront of it.3

D. 4-ffrom themirror, in back of it.3

10. The focal length of a spherical mirrordependson:

A. the radius of curvature of the mirror.

B. the index of refraction of the mirror.

C. the density of the reflective material.

D. the compressibility of the reflective material.

11. For parallel rays of white light incident on a sphericalmirror, the light's color components have focal lengthsthat:

A. are independent of the component frequency.

B. increase as the component frequency increases.C. decrease as the component frequency increases.D. increase as the component frequency increases, for

concave mirrors only; convex mirrors would yieldthe inverse relationship between focal length andfrequency.

12. When a lens and a mirror are immersed in water, wefind that the focal length of the lens:

A. and mirror both increase.

B. decreases, while the mirror remains the same.

C. increases, while the mirror remains the same.

D. and mirror both remain the same.



A telescope allows one to view large objects that are faraway. Refractive telescopes function by having the objectiveform a reduced real image (I) of the object while theeyepiece forms a virtual, enlarged image of the object. Theobjects astronomers observe through refracting telescopes areat such great distances that the first image, I, is very nearly ata focal point of the objective. The final image, I', is formed atinfinity, when the first image is at a focal point of theeyepiece. The length of the telescope, defined as the distancebetween the objective and the eyepiece, is the sum of thefocal lengths of the objective and the eyepiece, f\ + /2-

Objective Eyepiece

Figure 1

The magnitude of the telescope's angular magnification,M, is equal to the ratio of the focal length of the objective tothe focal length of the eyepiece. The calculation of M isshown below:

M = »0

where 8 = - y'/fl and 8 = y'//2. After substituting for 8 and0':

M =*/•h f\

y'/r, '*7i

In reflecting telescopes, the other broad category oftelescopes,a large concave mirror replaces the comparativelysmaller objective of the refracting telescope. The mirrorreflects the image to a smaller plane mirror, which reflectsthe image to the eyepiece. There are three typesof focus usedin reflecting telescopes: prime focus, Newtonian focus, andCassegrainian focus. In telescopes with prime focus, there isno second plane mirror, and the eyepiece actually sits in thepath of the light. The eyepiece is arranged differently withrespect to the mirrors.

13. What is NOT an advantage of a Newtonian reflectingtelescope over a refracting telescope?

A. Mirrors are free of chromatic aberrations.

B. Spherical aberrations are easier to correct.

C. The shape of the objective is not as variable as aconcave mirror.

D. The second mirror blocks incident light.


14. What must be true of the relative sizes of refracting andreflecting telescopes?

A. Refracting telescopes are larger, because theobjective must be small.

B. Reflecting telescopes are larger, because theconcave mirror must be small.

C. Refracting telescopes are larger, because theobjective must be large.

D. Reflecting telescopes are larger, because theconcave mirror must be large.

15. The focal point of the ocular (eyepiece) in a refractingtelescope is:

A. at the positionof the image I of the objective.B. at the position of the objective.

C. necessarily exactly between the objective andocular.

D. necessarily farther from the ocular than theobjective.

16.

17.

When comparing the focal lengths of the two lenses in arefracting telescope to the telescope's length, we findthat:

A.

B.

C.

D.

bothfocal lengthsare shorter than the telescope,bothfocal lengthsare longer than the telescope,the objective's focal length is longer than thetelescope; the ocular's focal length is shorter thanthe telescope.

the objective's focal length is shorter than thetelescope; the ocular's focal length is longer thanthe telescope.

Replacing the concave mirror in a Newtonian reflectingtelescope with a plane mirror would limit thefunctionalityof the telescope, because:

A. starlight would no longer be magnified.B. no light from the new mirror could reflect onto the

secondary plane mirror.

C. the new mirrorcould form only virtual images.D. a star's image would be half as far from the new

mirror as the new mirror is from the star itself.



18.

19.

20.

In any compound lens, the image of the first lens willserve as the object of the second lens. Which of thefollowing statements) is/are TRUE regarding theimage created by the first lens?

I. The orientation of a virtual image can be uprightwith respect to the object

II. A real object will always form a real image.

III. Light must travel through the first lens and formthe image before it travels through the second lens.

A. I only

B. I and II only

C. I and HI only

D. I, II, and III

Where will the image form when an object is placed 15cm to the left of a concave mirror that has a focal lengthof 12 cm?

A. 18 cm to the left of the mirror

B. 24 cm to the left of the mirror

C. 60 cm to the left of the mirror

D. 6.67 cm to the right of the mirror

In the absence of an atmosphere, the color of the skywould be:

A. blue.

B. black.

C. white.

D. red.


21. Placing a glass lens in oil will:

A. increase the focal lengthof the lens relative to air.B. decrease the focal length of the lens relative to air.

C. have no effect on the focal length; the focal lengthis a property of the lens material.

D. increase the focal length because the wavelength ofthe light is increased.



The human eye is an excellent example of how physicsprinciples may be applied to biology. The human eye acts asa camera, taking light from an object and focusing it to makean image. Like a camera lens, the focal length or power ofthe lens of the eye is determined in part by the radii ofcurvature of the lens. A typical human eye is spherical inshape, approximately 2 cm long. The focusing mechanism ofthe eye is the cornea-lens system.

Light enters the eye through a fairly inflexiblemembrane called the cornea. The cornea has a small radius

of curvature and the highest index of refraction of all of thetransparent constituents of the eye. The cornea covers atransparent bulge on the eye. The amount of light thatactually enters the eye is regulated by the iris, a colored ringbehind the cornea. The iris can adjust its size, controlling theamount of light that enters the pupil. Behind the iris is thelens, whose shape is controlled by the ciliary muscles. Whenthe ciliary muscles contract, the lens becomes more round,thus bending light rays more.

Between the cornea and the lens is a fluid called the

aqueous humor, and behind the lens is a second fluid calledthe vitreous humor. The two humors have an index of

refraction close to that of water. The light is finally broughtto focus on the retina, a membrane that contains the rods andthe cones, nerve cells that are sensitive to light and respondby generating electrical impulses.

Defects in vision, such as nearsightedness andfarsightedness are caused by aberrations in the eye system.Nearsightedness means that a person is able to focus clearlyon objects that are close to the eye, while a farsighted personcan only focus clearly on objects at a great distance from theeye.

22. If a normal human eye were focusing on an object 2meters away, what would be the effective focal lengthof the eye?

A. 0.002 meters

B. 0.04 meters

C. 0.01 meters

D. 0.02 meters

23. Farsightedness could be attributed to which of thefollowing defects within the eye?

I. Cornea too curved.

II. Cornea not curved enough.

III. Ciliary muscles not able to contract.

IV. Ciliary muscles not able to relax.

A. I and III only

B. II and III only

C. I and IV only

D. II and IV only


24. When the ciliary muscles of the eye contract, which ofthe following statements are true?

I. The eye assumes a more rounded shape.

II. The eye assumes a less rounded shape.

III. The focal length of the eye decreases.

IV. The focal length of the eye increases.

A. I and III onlyB. II and IV onlyC. I and IV onlyD. I, III, and IV only

25. Which type of lens might be used to correct fornearsightedness?

A. Convex

B. Concave

C. Either convex or concave, depending on the focallength

D. Either convex or concave, depending on hownearsighted the person is

26. Most of the bending of light through the eye occursbecause of the:

A. lens.

B. iris.

C. cornea.

D. humors.

27. The index of refraction of most materials is inverselyproportional to the wavelength of light Given that theeye has evolved so that yellow light focuses on theretina, where would you expect blue light to focus?

A. In front of the retina

B. Behind the retina

C. On the retina

D. On the cornea



A lens is a piece of transparent material that focuseslight to produce an image. This focusing process makes useof the fact that light is refracted twice when it passes througha lens. The focal length of a lens is a measure of how muchthe light rays are bent. A lens with short focal length bendsthe light rays to a greater extent than a lens with a long focallength.

A lens can either be converging or diverging, dependinghow it bends light rays that pass through it A converging (orconvex) lens bends light rays toward the axis of the lens.The light rays passing through a converging lens will actuallymeet at a point. Converging lenses can produce either realimages (the light rays really do meet at some point) or virtualimages (the light rays only appear to meet at a particularpoint). A diverging (or concave) lens is one that bends lightrays away from the axis of the lens. The light rays will neveractually meet any place; diverging lenses can only producevirtual images.

The thin lens equation governs the relationship betweenthe focal length of a lens, object distance, and imagedistance:

1 = 1 + 1f o i

where f is the focal length, o is the distance from the lens tothe object and i is the distance from the lens to the image.

Lenses can magnify an object's size. Magnificationrelates the dimensions of the image to the dimensions of theobject and can be written as:

M = iL=.lh o

where h' is the height of the image and h is the height of theobject

A compound lens is two or more lenses acting togetherto produce a final image from an object. Microscopes andtelescopes are examples of compound lenses. The simplestmicroscopes and telescopes consist of two convex lenses.The lens that first focuses the object is called the objective.The lens that you actually look through is called the ocular.

28. To measure the focal length of a converging lens,parallel light rays are passed through the lens. Theserays will meet at the focal point; the distance from thelens to this point can be measured. To measure thefocal length for a diverging lens:

A. the same procedure can be carried out.

B. a second diverging lens of known focal length mustbe employed.

C. a second converging lens of known focal lengthmust be employed.

D. diverging lenses do not have a focal length, sincelight rays are diverged away from the lens axis.


29. Fora microscope, whereshould theobjectbe placed?

A. Inside the focal lengthof the objectiveB. Just outside the focal pointof the objectiveC. Faraway from the focal pointof the objectiveD. As close to the objective as possible

30. When white light passes from air to a prism, what istrue?

A. Light with the greatest speed is refracted by thegreatest amount.

B. Lightwith the highestfrequency is refracted by thegreatest amount.

C. Light with the lowest frequency is refracted by thegreatest amount.

D. All light is refracted by the same amount.

31. In a compound lens, the image of the first lens willserve as the object of the second lens. Which of thefollowing statements are true?

I. A virtual image cannot serve as a real object

A real object will always form a real image.

Light must travel through lens 1 and form an imagebefore it travels through the second lens.

A. I only

B. II and III only

C. I, II, and HI only

D. None of the above statements are true.

II.

III.

32. For a telescope with a convex lens with a short focallength as its ocular, the objective is a:

A. concave lens with a focal length that is very long.B. concave lens with a focal length that is very short.

C. convex lens with a focal length that is very long.D. convex lens with a focal length that is very short



33. For a convergent mirror with f = 0.30 cm, at whatposition should an object be placed to generate animage that is three times the object is magnitude ofsize?

A. At the 0.20 cm mark

B. At the 0.40 cm mark

C. At the 0.50 cm mark

D. At the 0.60 cm mark

34. Which of these images are consistent with Snell's Law?

35.

A. I and V only

B. I, III and IV only

C. I and IV only

D. I, III, IV, and V only

What is NOT true for visible light?

A. It is emitted when an electron falls from an excited

state to a lower excited or ground state.B. As the wavelength increases, the frequency of the

photon decreases.C. It is of higher energy than microwave EM

radiation.

D. It has a range of wavelengths that spans from 200nm to 400 nm.


36. When parallel rays of white light pass through a glasslens:

A. the rays all meet at the same focal point.

B. the index of refraction of the glass is the same forall the colors.

C. the focal length for violet light is smaller than thefocal length for red light

D. whether the rays all have the same focal pointdepends on whether the lens is converging ordiverging.



A scientist wants to measure the precise thickness of apiece of glass. In order to do so, he bombards the glass withelectromagnetic waves, measuring the phase differencebetween the wave that bounces off of the glass' near surfaceand the wave that bounces off of the glass' far surface.Knowing that when a wave is traveling from a medium oflower index of refraction to one of a higher index ofrefraction and is reflected, the wave changes phase by xradians, the scientist tries the following two experimental setups in order to obtain two data sets:

Set-up I

Set-up 2

Air Glass Plastic

The relationship between phase difference (<j>) and pathdifference (Ar) used to determine the thickness (t) of the glassis given by the following expression, where (k) is thewavelength:

Equation 1 Ar = -?M>2k

For all of his calculations, the scientist used nAh- = LOT,

nGlass = 1-50, and npiastic = 1.60, along with radio waves ofwavelength one meter. The glass was uniformly thick.

37. In Set-up 1, a light ray traveling in air and incident onthe glass could:

I. pass directly through the glass, entering and exitingthe glass at the same angle.

II. achieve total internal reflection on this air-to-glassinterface.

III. achieve total internal reflection on the far glass-to-air interface.

A. II only

B. HI only

C. I and II only

D. I and HI only


38. If in using Set-up 2, the phase difference is measured tobe x radians, then which of the following is a possiblevaluefor the thickness of the glass, if the incident angleis assumed to be negligibly small?

A. 0.250 m

B. 0.375 m

C. 0.500 m

D. 0.625 m

39.

40.

41.

42.

If the interference between the two reflected waves in

Set-up 1 is constructive, then for the same piece ofglass the reflected waves in Set-up 2 would:

A. constructively interfere.

B. destructively interfere.

C. constructively interfere, but with less intensity thanin Set-up 1.

D. not interfere at all.

Assuming a light ray is passing from glass to plastic asin Set-up 2, which of the following would BESTdescribe the relationship between the incident angle(8j), the reflected angle (0R), and the refracted angle(0F)? (All angles are taken with respect to the normal.)

A. 81>8R>0F

B.

C.

D.

8l = 9R> 9F

1 ^ R ^ F

e, = eR < eF

For Set-up 1 and Set-up 2, respectively, if the light'spath-length within the glass is an integer multiple of thereflected light rays' wavelength, then the reflected raysin the two set-ups will be, respectively:

A. in phase and out of phase.

B. in phase and in phase.

C. out of phase and out of phase.

D. out of phase and in phase.

In order to measure the precise thickness of the piece ofglass, the scientist must first know its approximatethickness. Why is this the case?

A. These measurements need not be very precise.

B. Since waves are periodic, the scientist can obtain aset of possible solutions only from the data.

C. The measurement is hurt by the fact that only partof the wave reflects to its source, while the rest is

transmitted.

D. Each set-up gives the scientist a different thickness,and he must know which one is correct.



Lenses can be used to focus or disperse light rays. Inorder to produce a desired effect, it is often necessary to use acombinations of lenses. For a single lens, the focal point (f),the object distance (p), the image distance (q), and the lensmagnification (m) are related by the following equations:

1 = 1f P

Equation 1 + 1

Equation 2 m = - —

A physicist is curious about the amount of magnificationand where the final image will appear, if he puts a diverginglens 10 cm in front of a converging lens and changes theobject's distance from the first lens. For a diverging lens witha focal length of -10 cm and a converging lens with a focallength of 10 cm, the results of the physicist's study are givenin Table 1. It should be noted that in the representation of thedata the diverging lens is referred to as Lens 1, while theconverging lens is Lens 2.

IfObject

Lens 1

Figure 1

Lens 2

10 cm

Trial Pi(cm) Ql(cm) P2(cm) Q2(cm) m

A 20 -6.67 16.67 25 -0.5

B 10 -5 15 30 -1.0

C 5 -3.33 13.33 40 -2.0

D -3.33 5 5 -10 3.0

E -20 -20 30 15 0.5

Table 1

Table 1 shows object distances, image distances, andmagnifications.

43. As an object moves towards a converging lens andcrosses the focal point, what happens to its image?

A. The image goes from being real and upright to realand inverted.

B. The image goes from being real and inverted toreal and upright

C. The image goes from being real and upright tovirtual and inverted.

D. The image goes from being real and inverted tovirtual and upright


44. In moving the object from the position in Trial C to theposition in Trial D, what happens to the final image?

A. It becomes larger and inverts.

B. It becomes smaller and inverts.

C. It becomes larger and stays upright.

D. It becomes smaller and stays upright

45. For which trial is the final image formed on Lens 1?

A. Trial A

B. Trial B

C. Trial C

D. Trial D

46. What is the magnification of Lens 2 in Trial B?

A. -2.0

B. -0.5

C. 0.5

D. 2.0

47. For a converging lens, when is the image formed atinfinity?

A. When the object is at half the focal length from thelens.

B. When the object is at one focal point

C. When the object is at twice the focal length fromthe lens.

D. When the object is infinitely far from the lens.

48. When an object is placed inside the focal length on theright side of a diverging lens, what type of image isformed?

A. A virtual, upright image.B. A virtual, inverted image.

C. A real, upright image.

D. A real, inverted image.


49. An object is placed in front of a concave mirror, justinside its focal length. The image can best be describedas:

A. real and inverted.

B. real and upright.

C. virtual and inverted.

D. virtual and upright.

50. How can a mirage, the result of refraction of light, bedistinguished from a real object?

A. A mirage cannot be distinguished from a realobject.

B. Look through a polarizer.

C. Look through a prism.

D. Look through a light-scattering material.

51. A polarizer was observed to filter 50% of ambientlightemitted from a bulb, independent of the orientation ofthe polarizer. How can it be explained that when thepolarizer was used to filter ambient light that hadreflected of a plastic surface that the amount of lighttransmitted varied with the polarizer's alignment?

A. The plastic absorbed light, reducing the amount ofreflected light

The plastic reflected only one frequency of light.

The plastic acted as a semi-polarizer, reflectingand refracting light differently according to the EMwave's orientation.

The plastic emitted light in selected frequencyranges.

B.

C.

D.


52. When a swimmer opens his eyes under water, objectsdo not appear in focus. This is due to:

A. an increase in thefocal length of theeye.B. a decrease in the focal length of theeye.C. the scatteringof light by the water.

D. a decrease in the effectiveness of the iris atabsorbing light.

1. B 2. C 3. D 4. B 5. A 6. D

7. D 8. A 9. A 10. A 11. A 12. C

13. D 14. D 15. A 16. A 17. A 18. A

19. C 20. B 21. A 22. D 23. B 24. A

25. B 26. C 27. A 28. C 29. B 30. B

31. D 32. C 33. B 34. A 35. D 36. C

37. D 38. A 39. B 40. B 41. D 42. B

43. D 44. A 45. D 46. A 47. B 48. A

49. D 50. B 51. C 52. A

YOU ARE DONE.

Answers to 52-Question Light and Optics Practice Exam

Passage I (Questions 1 - 6) Critical Angle Experiment

Choice B is the best answer. This question is best solved by first eliminating what we can using common sense and thenfinding the best answer from the remaining choices using mathematics. We know that flint glass is denser than air, so thelight slows down upon entering the flint glass from air. This means that with an incident angle of 60°, the refracted anglemust be less than 60°. This eliminates choice D. We also know the refracted angle cannot be 0° unless the incident raystrikes the boundary in a perpendicular fashion. Because the incident angle is 60°, the refracted angle cannot be 0°, so choiceA is eliminated. We are down to two choices. The passage tells us that light striking the boundary between flint glass andthe surrounding air at an incidentangle of 38.1° results in a refracted angle of 90°. If vve consider the reverseof this processwhere light travels from air and strikes a boundary made of flint glass, then the angles should be reversed. So, if the lightcoming from air strikes the boundary at an angle just slightly less than 90°, then the refracted rays would travel a path that isslightly less than 38.1° from the normal axis. This means that an incident angle of 60° should result in a refracted angle lessthan 38.1°, so choice C is eliminated. All that remains is choice B, so it must be the best answer. To solve this questionusing math is arduous and not the most time efficient option, but we shall address the process nonetheless. From theinformation in the passage and Snell's law, we know:

nfiint giasssin 38.1° =nairsin 90° =1x1=1 .*. nfiintgiass =1/s|n 33 1°

The question asks for the refracted angle when travels from air and strikes a boundary of flint glass with an incident angle of60°. To solve this, we must employ Snell's law.

nair-sin60° = nfjintgiasssin 0rcfracted

1xsin 60° =(V^ 38.r)'sin ^refracted •'• sin 6refractcd =sin 60° xsin 38.1°Because sin 60° is a number less than 1, sin 9refractcd < sin 38.1° .\ ©refracted < 38.1°

Determination of the exact number is challenging without a calculator, but luckily for us, it is a multiple-choice question.This is enough math to support our selection of choice B. The best answer is choice B.

Choice C is the best answer. The focal point of a lens depends on the angle of curvature of the lens and the index ofrefraction of the lens material. In each case, the surrounding medium is air and the lenses are of identical dimensions, andtherefore of identical curvature. The difference in focal points between the four lenses depends strictly on the materials. Thematerial with the greatest index of refraction will bend the light the most and in doing so create a focal point (where the raysintersect)closer to the lens. This question is lookingfor the material with the greatest index of refraction. The passage givesus critical angles, which decrease as n-value for the material increases. This means that we are looking for the material withthe smallest critical angle. This describes rutile, so choice C is the best answer. Even if we had no idea how to answer thequestion, the question is asking for an extreme value (closest to the lens). Without any clue, we should at the very leasteliminate rock salt and zirconia, because their critical angles are less than the critical angle for polystyrene. The best answermust be either choice A or choice C. The best answer is choice C.

Choice D is the best answer. The experiment requires that yellow light (k = 589 nm) pass through the two mediums (one ofwhich is air) without being reflected and refracted more than once (only at the boundary), and without being absorbed. If thelight undergoes multiple points of refraction, then the ultimate angle cannot be attributed solely to the difference between thematerial and air. If the material absorbs light then there is a chance that no light will pass, which would result in no intensityat the screen for a reason other than exceeding the critical angle. This means the material must be transparent and uniform,making choices A, B, and C all valid. The light should not be absorbed, so choice D is the best answer. You should notethat choices A and D are opposite answers, so it is probable that one of the two is the best choice. The best answer is choiceD.

Choice B is the best answer. Diffraction occurs when light interacts with a slit in a material, which is not the case in thisexperiment. While monochromatic light may be necessary to study diffraction, it does not impact this experiment in terms ofdiffraction. Statement I is invalid, which eliminates choices A and C. We know Statement II must be true based on theremaining choices, but just to verify, different frequencies of light have different indexes of refraction, so if polychromaticlight were used, the different wavelengths would exhibit a different degree of bending, resulting in dispersion of the incidentlight. As such, each color would result in a different critical angle. Statement II is in fact valid. Reflection occursindependent of the frequency of light so whether the light is monochromatic or polychromatic, reflection is still involved.Statement III is invalid, making choice B the best answer. The best answer is choice B.


Choice A is the best answer. Light initiating from Point a will strike the boundary at a right angle. For light that strikes aninterface in a perpendicular fashion, it's speed will change, but there will be no change in the orientation of the rays. Toobserve refraction, light must strike the boundary at some angle other than perpendicular. This makes choice A the bestanswer. Choice D should have been eliminated, because the light speeds up when transmitted to the air. Choice B shouldhave been eliminated, because the incident angle is 0°, which is less than the critical angle for every material in the passage.Choice C should be eliminated, because light doesn't exhibit such behavior. The best answer is choice A.

Choice Dis the best answer. The index of refraction for water is greater than the index of refraction for air, so light passingfrom the disk to water will be bent less than light passing from the disk to air. This means that the angle of refraction issmaller in water than air, eliminating choices A and C. In order to get a refracted angle of 90° in water, a greater incidentangle is required than what is required to get a refracted angle of90° in air. Thus, the critical angle between the disk and airis less than it is between the disk and water. This makes choice Dthe best answer. The math to support the critical anglesconclusion is as follows.

The

nmateriarSin ©critical = nSUrrounding medium"Sin 90° .'. sin 8critica| ="surrounding medium/n

The surrounding medium with the larger index of refraction, in this case water, has the greater sin 0criticai and therefore hasthe larger critical angle, 0criticai- The relative refracted angles ofwater and aircan also bedetermined using Snell's law.

nmateria!"Sin ©incident = nair*s'n ©refracted in air and nmalcrial'Sin ©incident = nwater*s»n ©refracted in water

So: sin erefractcdinair =nmaterial/^-sin eincident and sin 9refracted in water ="material^^-sin eincidentBecause nvvatcr isgreater than nair, sin 8refracted in air isgreater than sin 9refractcd in watcr .-. 9rcfracled in air > 0refracted in water

The math supports the selection of choice D. The best answer is choice D.

Passage II (Questions 7 -12) Mirrors

Choice D is the best answer. First and foremost, the distance from a plane mirror plays no role in the limits of the imagebesides intensity. We need consider only the height of the person in front of the mirror, especially the person who is 4 m(nearly thirteen feet) tall, as they are probably from outer space. To understand this better, draw a picture. Let the person bestanding a distance d in front of the mirror. Then the image must appear to be a distance d behind the mirror. In order forlight to reach the person's eyes from the eyes of the image, a light ray must appear to come from the eyes of the image. Inactuality, the lightfrom the eyes of the person leaves them, strikes the mirror's surface, and then reflects into their eyes. Forlight to reach the person's eyes from the feet of their own reflected image, a light ray must appear to come from the feet ofthe image and enter the person's eyes. In actuality, light from the person's feet strikes the mirror and reflects into their eyes. Ifwe draw out all the rays, we see that there are two similar triangles:

Plane Mirror

T

h;__d_ j_h " 2d ~ 2

Even a person standing 2 meters tall could see approximately their entire body length in a mirror 1 meter long (rememberthat the body extends a bit above the eyes). The best answer is choice D.

Choice A is the best answer. Recall some general observations about plane mirrors: They always form virtual images;images formed in plane mirrors are of the same sizeas theoriginal object; images formed in plane mirrors are as far from themirror as the original object is. The last two observations invalidate statements II and III, leaving choice A as the bestanswer. To confirm this selection, we need ask only whether spherically convex mirrors always form virtual images (i.e., weneed test only the validity of statement I). Yes, spherically convex mirrors can form only virtual images-the fact that youalways see an upright image in any convex mirror is proof of this. Thus, statement I is true. The best answer is choice A.


9. Choice A is the best answer. Let's use the equation given in the passage to solve this one:

10.

11.

12.

1 -

f1 + 1s s'

where s = +3f (a plus sign is used, because the object is in front of the mirror). Solving for s' yields: s' = 3f/2. This rules outchoices C and D. If you stick with the convention that positive lengths represent objects and images in front of a mirror, thenyou can cross out choice B in favor of choice A. Another way to see that choice A is better than choice B is to realize thatwhen any object beyond a concave mirror's (or even a convex lens') focal point, the image is always real (i.e., it has real lightrays meeting in real space). Since real space is in front of a mirror, choice A must be correct. Yet another method for gettingchoice A would be to trace out the light rays from the object. The best answer is choice A.

Choice A is the best answer. The radius of curvature of the mirror determines where the normal lines to the surface are.

This, in turn, determines where the reflected rays go and means that the radius of curvature affects the focal length. Choice Ais correct. The index of refraction applies to refraction, not reflection, so while it affects the focal length of a lens, it isimmaterial the focal length of a mirror. The density and compressibility of the reflective material may affect the ease withwhich the mirror can be shaped, but they do not affect the focal length. The best answer is choice A.

Choice A is the best answer. The law of reflection states that the angle of incidence is equal to the angle of reflection. Thislaw is independent of the frequency (and wavelength) of the incident light and independent of the refractive index of thematerial of which the mirror is made. The best answer is choice A.

You may notice that this is the fourth straight question where choice A is the best answer. This was done to work on yourpsyche and see if such a pattern influences your decision.

Choice C is the best answer. Mirrors work by reflection, so the focal length is determined by the radius of curvature of themirror and by nothing else. Although the wavelength of the light changes under water, the rays still meet at the same point.This eliminates choice A. A lens on the other hand, works by refraction, which depends on both the radius of curvature of thetwo sides of the lens and the difference in refractive index between the lens and the surrounding medium. If a lens isimmersed in water, then the medium surrounding the lens has a higher index of refraction, which results in less bending ofthe light rays. By bending less, the rays intersect farther from the lens, meaning that the focal length increases. This makeschoice C the best answer. The best answer is choice C.

Passage III (Questions 13 -17) Telescopes

13. Choice D is the best answer. Chromatic aberrations are caused by refraction, so mirrors are in fact free of chromaticaberrations, while lenses are not. Because the reflecting telescopes employ concave mirrors as the objective, while refractingtelescopes employ lenses, choice A is an advantage of the reflecting telescope. With a concave mirror, spherical aberrationscan easily be corrected. With a lens, the correcting is not done as easily, owing to the curvature of both sides.

This points out another advantage of the reflecting telescope, and it eliminates choice B. In a refracting telescope, theobjective must be a convex lens. In a reflecting telescope, the mirror can be parabolic or spherical, making choice C anadvantage of the reflecting telescope. Choice C can be eliminated. In a reflecting telescope, the plane mirror sits at the focalpoint, which is in front of the concave mirror. This means that some incident light is blocked by the plane mirror (the secondmirror), so this is a disadvantage of the reflecting telescope. Choice D is the best answer.

Cassegrainian Focus


ReflectingMirror

Ocular

14. Choice D is the best answer. It is stated in the passage that reflecting telescopes have large reflecting mirrors. Thiseliminates choice B. Because a reflecting telescope has a large mirror, it must be considerably larger than a refractingtelescope. This eliminates choices A and C and makes choice D the best answer. Most telescopes found in observatoriestoday are reflecting telescopes, which accounts for observatories being so large. Portable telescopes, like the kind you wouldtake on a field trip, are typically refracting telescopes, and they are generally small. The best answer is choice D.


15. Choice A is the best answer. The passage tells us that in a refracting telescope, the rays that pass through the eyepiece mustleave in a parallel fashion. This occurs when the object lies at the focal point. Given that the image of the objective is theobject of the eyepiece, we need the first image to form at the focal point of theocular (eyepiece). This makes choice A thebest answer. In Figure 1, vve see that the image created by the objective is at the focal point of the objective. This isremarkable, because the object (be ita star, planet, or galaxy) is ineffect infinitely far from the objective-remember the thin-lens equation? The best answer is choice A.

16. Choice A is the best answer. Looking at Figure 1 in the passage, we see that both focal lengths are shorter than thetelescope. The passage defines the telescope length as the distance between the objective and the eyepiece; this distance isalso defined as the sum of the focal lengths of the two lenses. The best answer is choice A.

17. Choice A is the best answer. Let's recall some basic facts about plane mirrors: Forone thing, they reflect light. This rulesoutchoice B, because lightcould possibly reflect onto the secondary mirror. Plane mirrors form a virtual image of the object;this image isequal in size to the object and as far from the mirror as the mirror is from the object. The main disadvantage ofthese properties, in using the plane mirror to build a telescope, is that the plane mirror does not magnify the image of theobject. What looks (to the naked eye) like a speck of starlight reflects onto the secondary mirror as a speck of starlight. Thebest answer is choice A.


18. Choice A is the best answer. Remembering the UV-IRrule for objects and images: An upright image is always virtual, withrespect to the object. This makes Statement I valid. A real object may or may not form a real image. It depends upon thecombination of lenses being used. For example, diverging lenses form virtual images from real objects. This makesStatement II invalid. Statements with absolute language, such as "always," should be treated as invalid until you havecompleted an exhaustive search for contradictory answers. One exception disproves an extreme statement. Finally, light maytravel through two lenses before forming the final image. If the second lens is within the focal length of the first lens, thenlight will travel through the first and second lens before forming the image. This makes Statement III invalid. Because onlyStatement I is valid, choice A is the best answer. The best answer is choice A

19. Choice C is the best answer. This question can be solved systematically, by first noting that vve are dealing with a single,converging lens. Our first approach is to visualize the positions of the object, mirror, and image (in essence to create aminimalistic ray diagram in our mind). For a converging system involving a single mirror or single lens, if the object isbetween the radius of curvature and focal point, then the image will be an inverted, real image formed beyond the radius ofcurvature. Given that the focal length is 12 cm, the radius of curvature must be 24 cm, which means that the image mustform at a distance greater than 24 cm from the mirror. Only choice C fits describes a point beyond 24 cm. The image is onthe left side of the mirror, because reflected rays can only form a real image on the original side of the mirror (where theobject exists).

If you don't like the visualization approach to lens and mirror questions, then you are always welcome to employ math. Youneed to keep sign conventions in mind, so for a converging mirror, the focal length is a positive value, Had you used anegative value for f, which you would do for a diverging mirror, then you would have errantly come up with a value of -6.67,which is one of the answer choices. Let's take advantage of the shortcut version of the thin lens equation given in the text ofthis chapter:

' =(° X°/(o -0=°5 X12)/(15 -12) =°5 X12)/3 =15x4 =60 cmA +60 means it's a real image that is 60 cm from the mirror on the "real" side, which in this case is the left side. The bestanswer is choice C.

20. Choice B is the best answer. The sky has color because the Sun's light is scattered by the particles in the atmosphere. In theabsence of any atmosphere, there can be no scattering, so vve would see no color. (Ever see any pictures of the Moon? TheMoon has no atmosphere; the sky as seen from the lunar surface is black.) The best answer is choice B.

21. Choice A is the best answer. The focal length of a lens is a measure of how much the lens can bend incoming light rays.The index of refraction of oil is higher than that for air, and thus is closer to the index of refraction of the lens. This meansthat light going from air to glass will be refracted more than light going from oil to glass. The focal length of the lens istherefore greater (farther from the lens) in oil than it is in air. This eliminates choices B and C and makes choice A the bestanswer. The wavelength of light is shorter in oil than air, so choice D is also eliminated. The best answer is choice A.


Passage IV (Questions 22 - 27) Functioning of the Eyes

22. Choice D is the best answer. To solve for the focal length of the eye, vve use the thin lens formula:

1=1+ 1f s S'

where f is the focal length, s is the object distance, and s' is the image distance. Choosing s' to be 0.02 meters, the length of

23.

24.

25.

26.

27.

the eye as mentioned in the passage, and s to be 2 meters, from the question, vve solve for f:

1 1 .*. f is 0.02 meters.f 2 meters 0.02 meters

The 1/2 part of the equation above is negligible compared to 1/0.02, so we can essentially ignore it. This leads to theconclusion that f = 0.02 meters. The best answer is choice D.

Choice B is the best answer. Farsightedness, as defined in the passage, is the inability to focus on nearby objects.Farsightedness occurs because the lens of the eye is not powerful enough to focus on light rays from an object close to theeye. Remember that focusing on objects close up requires a small focal length and therefore, a more powerful lens thanfocusing on objects that are far away. Looking at the thin lens equation, one way this could happen would be if the eye wastoo short (if s' was too small), but that is not an option. The cornea needs to have a rounded shape to bend light rays from anobject close to the eye. This means that an inability to focus on nearby objects is because the curvature of the lens is notgreat enough, which can be attributed to either the cornea not being round enough, or the ciliary muscles inability to contractthe lens. Statements II and III go together, so choice B is the best answer. The best answer is choice B.

Choice A is the best answer. When the ciliary muscles of the eye contract, the eye becomes more round. The effect of thisis to cause incoming light rays to get bent more as they enter the eye, which means that the focal length had to decrease, soStatements I and III are valid. You should notice that Statements I and III are complementary and that Statements II and IVare complementary. This eliminates choices C and D. The ciliary muscles of the eye contract when vve are focusing onobjects closer to the eye. Remember-the smaller the focal length, the more powerful the lens and the more it is able to bend

light rays. Lens power, P, is defined as Vf- A small focal length is required to focus on objects close to the eye. If you arenot sure about this last part, try plugging some numbers into the thin lens equation:

1 =1 + 1f s s-

Upon focusing on an object 2 meters away means that the eye has a focal length of 0.02 meters, or a powerof 50 diopters. Ifvve use 0.25 meters for the object distance:

1 = ! + 1f 0.25 meters 0.02 meters

and solve for f, vve get f = 0.019 meters, or a power of 54 diopters. The best answer is choice A.

Choice B is the best answer. A nearsighted person can see objects thatare near, butcannotsee objects thatare far away. Tocorrect for such a condition, the ideal lens should take a far away object and create an image that is near. With a concave(divergent) lens, the image in an upright, virtual image that is located inside of the focal length. This means that a far awayobject will generate a nearby image, which means that the divergent lens has taken an object outside the field of vision andgenerated an image within the field of vision. A nearsighted person benefits from a concave lens. You may also note that anearsighted person has a retina that is abnormally longer, so the image focuses before the retina. A divergent lens will spreadthe rays before they enter the eye, so that the image can now focus at the retina. The best answer is B.

Choice C is the best answer. The passage tells us that the cornea has the highest index of refraction of all of the transparentparts of the eye, making choice C the best answer. A substance with a high index of refraction bends light rays more than asubstance with a small index of refraction. So whatjob does the lens do? The passage tells us that the cornea is inflexible-itcannot bend and change its radius of curvature. However, the lens is able to bend and change its radius of curvature, therebychanging the focal length of the cornea-lens system. The lens makes it possible to focus on objects at variable distances fromthe eye. The best answer is C.

Choice A is the best answer. Blue light has a smaller wavelength than yellow light. Therefore, it sees a higher index ofrefraction than does yellow light. This means that the blue light will be bent more than the yellow light upon passing throughthe eye, and therefore the blue light will focus in front of the retina. This eliminates choices B and C and makes choice A thebest answer. Light will not be focused on the cornea, so choice D is eliminated. The best answer is A.

Copyright© by The Berkeley Review0 292 REVIEW EXAM EXPLANATIONS

Passage V (Questions 28 - 32) Different Lenses

28. Choice C is the best answer. A diverging lens does not have a point where all the light rays actually meet. In order tomeasure the focal length of a diverging lens, a converging lens mustbe used. The two lenses will form a compound lens-theimage of the first lensacts as an object for the second lens. By knowing the location of the final real image, and by knowingthe focal length of the converging lens, vve can work backwards to determine the focal length of the diverging lens. Thismakes choice C the bestoption of the choices. Choice A is incorrect, because parallel light rays do not meet any place whenthey travel througha diverging lens. Choice B is incorrect, because a real image is needed, so that vve may actually measurean imagedistance. Diverging lenses only produce virtual images. Choice D is incorrect. Although the light rays are indeeddiverged away from the lens, vve can still speakof how much these light rays are refracted, and this is all vve mean by focallength. The best answer is C.

29.

30.

31.

32.

Choice B is the best answer. The purpose of a microscope is to enlargeobjects. By placing the object just outside the focallength of the objective, vve insure a large imagedistance. This in turn provides a large magnification, according to:

M = -io

ChoiceA is incorrect. By placing the object just inside the focal length of the objective, we create a virtual image. Choice Cis incorrect. By placing the object far away from the focal point, vve insure a small image distance, which means a smallmagnification. Choice D is incorrect, for the same reason that choice A is incorrect. This will put the object inside the focallength again, which means the production of a virtual image. The best answer is choice B.

Choice B is the best answer. This question focuses on recall. Light traveling through a medium will exhibit the same wavespeed regardless of the frequency, so choice A is eliminated. Because vve see that white light is dispersed (refracted), choiceD can be eliminated. The light with the highest frequency (violet light) is "bent" to the greatest extent, which means thatlight with the greatest frequency is most refracted. Choice C is eliminated. The best answer is choice B.

Choice D is the best answer. A virtual image may indeed serve as a real object. This is the principle behind glasses andcontacts. This makes Statement I invalid, which eliminates choices A and C. A real object may or may not form a realimage. It depends upon the combination of lenses being used. For example, diverging lenses form virtual images from realobjects. This makes Statement II invalid, which eliminates choice B and leaves only choice D. Finally, light may travelthrough two lenses before forming the final image. If the second lens is inside the focal length of the first lens, then light willtravel through the first and second lens before forming the image. This confirms that Statement III is invalid. When solvingthe problem quantitatively, vve assume the formation of something called a "virtual object." This is the principle behind thetelephoto lens. The best answer is choice D.

Choice C is the best answer. Because the ocular for this particular telescope is a convex lens, the objective needs to be aconvex lens. This will doubly invert the image, resulting in a final image that is upright. This eliminates choices A and B.The focal length of the ocular is dictated in part by the physical size of the telescope and in part by what vve want thetelescope to do. A telescope focuses on objects that are very far away and tries to magnify them. In order to enlarge them,vve need an objective with a long focal length, to make s1 as large as possible. The best answer is choice C.


33. Choice B is the best answer. For a convergent mirror to generate a real image, the object must be beyond the focal point, sochoice A can be eliminated immediately. With f = 0.3 cm, R must be 0.6 cm. If the object is placed at the radius of curvature,then the image will also form at R, resulting in an image with the same magnitude of size as the object. Choice D iseliminated. Only math can make the final distinction between choices B and C. If the object is at 0.4 cm, then the image willform at (0.4 x 0.3)/(0.4 - 0.3) = 0.12/0.10 = 1.2 cm. Magnification is found using m = -(dj/d0) = -(1.2/0.4) = -3. Because m is

-3, the image is three times as large as the object and it is inverted. The best answer is choice B.

34. Choice A is the best answer. Snell's Law tells us that light will change direction upon entering a new medium, as long as theincident ray is not perpendicular to the boundary. Light bends toward the normal if the light enters a denser medium from aless dense medium, and light bends away from the normal if light enters a less dense medium from a denser medium. PictureI is consistent with light entering a denser material from a less dense material; Picture V is consistent with light entering aless dense material from a denser material. Picture II can be eliminated; the light does not show a direction change uponentering the new medium. Although the light in Pictures III and IV do show a direction change upon entering the newmaterial, it ends up traveling along the normal. This could only happen if the incident angle was 0°. Only Pictures I and Vare valid. The best answer is choice A.


35. Choice D is the best answer. We are told here that ultraviolet radiation and visible light are associated with transitionsbetween different electronic energy levels, so it is true that visible light can be emitted when an electron falls from an excitedstate to a ground state. This eliminates choice A. For all electromagnetic radiation, including visible light, the wavelength ofa photon increases as its frequency decreases. This obeys the equation v = f-k (where v is the speed of the photon, / is thefrequency of the photon, and k is the wavelength of the photon), so choice B is eliminated. Visible light has more energythan microwave radiation. Even if you were unaware of this, the passage states that electronic transitions are due to visiblephotons, while vibrational transitions can be caused by microwaves. Because electronic transitions are of a higher energythan vibrational transitions, it can be concluded that visible light is of a higher energy than microwaves. This makes choice Cvalid, which eliminates it from being "NOT true." Visible light ranges from about 400 nm to about 700 nm, so choice D isnot true-so it is the best answer. The best answer is choice D.

36. Choice C is the best answer. The indexof refraction for a material is somewhat wavelength dependent. When two colors oflight travel through the same lens, they are subsequently refracted different amounts, and thus will not meet at the same focalpoint. Choices A and B are incorrect for the reasons just mentioned. Whether the lens is converging or diverging, differentcolors are still refracted different amounts, so for converging or diverging lenses, different colors will have different focalpoints. This eliminates choice D. Because violet light is refracted more than red light, it has a shorter focal length. The bestanswer is choice C.

Passage VI (Questions 37 - 42) Thin Films

37.

38.

39.

40.

41.

Choice D is the best answer. If vve think about the path a light ray takes in Setup 1, vve see it will first bend toward thenormal to the surface when entering the glass, because it is moving from a region of a lower index to oneof a higher index.Upon exiting the glass, the light ray will refract away from the normal, because it is moving from a region of higher index ofrefraction to one of a lower index. The net result is the light ray entering and leaving the glass at the same angle, so vve cansee that Statement I is correct. Since total internal reflection can occur only when the light is moving from a region with ahigher index (nGLAss = 1-50) to one of a lower index (nAjR = 1.01), Statement II is incorrect, while Statement III is correct.The answer that corresponds to choosing both Statements I and III is answer choice D. The best answer is choice D.

Choice A is the best answer. If the light ray is coming in approximately perpendicular to the surface of the glass, then Ar =2t, where t represents the thickness of the glass, because the light ray that reflected off of the second surface must travel intothe glass and back out. Thequestion states that the waves are re out of phase, so this problem becomes a simple calculation:

Ar = 2t = A^j)2re

and t = A<j, =4re

-im_(jirad) = 0.25m4rerad

Thus, vve can see that the best answer is choice A, or the pieceof glass is 0.25 m thick. The best answer is choice A.

Choice B is the bestanswer. If in Setup 1 the interference isconstructive, then vve know that the phase difference betweenreflected waves is some multiple of 2rc. Since in Setup 1 vve have a phase change in going from air to glass (but not fromglass to air), while in Setup 2 there is a phase change in going from air to glass (as well as going from glass to plastic), Setup1 and Setup 2 aren outof phase with each other. This means that in Setup 2, the phase difference between reflected waves issome odd multiple of jc. Odd multiples of x correspond to destructive interference. The best answer is choice B.

Choice B is the bestanswer. The incident and reflected beams are always equal, so vve can eliminate choices A and C. Also,when light is moving from a region with a lower index of refraction (nGLAss = 1-50) to a region with a higher index ofrefraction (nPLASTic = 1-60), the light ray bends towards the normal to the surface. Angles of refraction are measured withrespect to the normal, so the initial angle is greater than thefinal angle, or 6j = Or > 0p. The best answer is choice B.

Choice D is the best answer. In order to answer this question, vve must look at the phase change of the wave at eachboundary. It is stated in the passage that when light goes from a region with a lower index of refraction to a region with ahigher index of refraction, the reflected wave changes phase by re. In this case, there is a phase change of x in the reflectedwave when light passes from air to glass, as well as from glass to plastic. Applying this to Setup 1, only the first reflectedwave gathers a phase change of re, while the wave returning from the second surface obtains no phase change uponreflection. If the thickness of the glass is some multiple of the wavelength, then the two reflected waves for Setup 1 willarrive outof phase by re, so they interfere destructively. Similarly, in Setup 2, the first reflected wave gathers a phase changeofre, as well as the wave returning from the second surface. This happens because the wave is traveling from a region with alower index to one with a higher index in both cases. Assuming again that the thickness of the glass is some multiple of thewavelength, the two reflected waves for Setup 2 both arrive jt out of phase from the incident wave, or in phase with eachother. This combination corresponds to constructive interference. Destructive interference for Setup 1, and constructiveinterference for Setup 2 corresponds to out of phase for 1and in phasefor 2. The best answer is choice D.


42. Choice B is the best answer. Unless the answer is obvious, vve should solve this one by the process of elimination. Choice Ais incorrect, because the whole reason the scientist was doing this experiment in the first place was to obtain a precisemeasurement. In disproving choice C, as long as some of the wave is reflected, it is the only part of the wave vve areinterested in for our measurements. Furthermore, choice D is not good, because the results of any experiment should bequestioned, if they do not agree with another experiment meant to measure the same thing. This leaves us with the onlychoice that makes sense, choice B. In taking data, it is possible to measure a phase difference only between 0 and 2re, becausethe waves are periodic. In trying to apply the data to the equation, Ar =^/g^, an experimental phase difference of, say, recouldcorrespond to an actual phase difference of re, 3re, 5re, etc....Therefore, it would be useful to have an approximate valueof the thickness first, in order to know which "actual phase" the experimental phase corresponds to. The best answer ischoice B.

Passage VII (Questions 43 - 48) Double Lens

43. Choice D is the best answer. We can find out what happens to an object's image as vve cross the focal point by looking atthe lens equation, where p represents the position of the object and q represents the position of the image:

1 = 1 + 1

If the object is outside the focal point, then p > f, so:

1>1f P

This is to say that the denominator on the left being a smaller number makes the overall fraction larger. When this is true, qmust be positive in order to satisfy the lens equation. Thus, when vve are outside the focal point of a converging lens, theimage is real. Looking at the magnification equation:

m-flP

we also notice that since q and p are both positive, the image is inverted. Only choices B and D satisfy this condition forwhen the object is outside the focal point. When we move the object inside the focal point of a converging lens, then f > p,so:

1>1P f

When this is true, q must be negative in order to satisfy the lens equation. Thus, when vve are inside the focal point of aconverging lens, the image is virtual. Also, since q is negative and p is positive, upon looking at the magnification equationvve see that the image must then be upright. In conclusion, when moving an object from outside the focal point of aconverging lens to inside the focal point, the image goes from being real and inverted to virtual and upright. This is one ofthose random facts that will make future questions dealing with a single mirror or a single lens easier to solve. The bestanswer is choice D.

44. Choice A is the best answer. Looking at the data in Table 1, one notices by comparing Trials C and D that the magnificationgoes from -2.0 to 3.0. Since the magnitude of the magnification in Trial D is larger than in Trial C, the final image getsbigger. Also, the image goes from being inverted in Trial C to being upright in Trial D. The best answer is choice A.

45. Choice D is the best answer. To find out where the final image is formed, look at the data for q2- If the final image is

formed on Lens 1, q2 should be 10 cm to the left of Lens 2, or -10 cm. Table 1 shows that this corresponds to choice D. The

best answer is choice D.

46. Choice A is the best answer. If vve want to find the magnification of Lens 2, vve must use both p2 and q2 and themagnification formula, as follows:

m = -5= _SL = .30 = .2P P2 15

We calculate a value of -2 for the magnification of Lens 2 in Trial B. The best answer is choice A.

47. Choice B is the best answer. For any lens, the image is formed at infinity when Vn = 0. Since the focal length of a

converging lens is positive, then Vn = 0 when p = f. Thus, the image of a converging lens is formed at infinity when the

object is at the focal length of the lens. The best answer is choice B.


48. Choice A is the best answer. We must look at the lens equation to figure outthis question. Since the object is to the right ofthe lens and the lens isdiverging, both p and q are negative. If the object is placed inside the focal length, the magnitude offis greater than the magnitude of p, so the magnitude of Vp is greater than the magnitude of Vf. Since Vp is a larger negativenumber, then in order to satisfy the lens equation, q must be positive. If q is positive, the magnification equation indicatesthat the image isupright. This rules outchoices Band D. The image will be virtual; this can be shown by a bit of ray tracing(see the diagram below). Since actual light rays do not come from the point inspace occupied by the image, the image is saidto bevirtual. This makes choice A correct. Asa general reminder, diverging lenses always form virtual images.

Lens 1


Questions 49 - 52 Not Based on a Descriptive Passage49.

50.

51.

52.

Choice D is the best answer. When an object is placed just inside the focal length of a concave (converging) mirror, theresulting image will be virtual, upright, and enlarged. You can verify this using the lens equation:

1 = 1 + 1

Ifs is less than f, then s'must be a negative number for the equation to work out. Anegative value for the image implies thatit is a virtual image. This eliminates choices Aand B. For a single mirror or single lens, a virtual image is always upright.This confirms that choice D is the best answer. The best answer is choice D.

Choice B is the best answer. Refracted light can be polarized, at least partially so. By looking through a polarizer androtating it, vve can see whether the intensity ofthe light changes. If the intensity does not change, then the light was comingstraight from the real object. If the intensity does change, then the light was refracted-the apparent object was a mirage. Thebest answer is choice B.

Choice C is the best answer. As a good polarizer should do, the one in the question filters 50% ofambient light as long asthe light is a random assortment of EM waves. Light reflected off of the plastic doesn't show the same behavior as therandom light, being that the amount of light reflected off the plastic and passing through the polarizer varies with theorientation of the polarizer. Because the orientation ofthe polarizer impacts the amount of light that passes, the best answermust be involve some form of polarization. While the plastic may absorb light, that would explain a reduction in intensity,but not a dependence on the orientation of the polarizer. Choice A is eliminated. If the plastic were to reflect only onefrequency oflight, then the reflected light would have adistinct color, but that wouldn't explain the polarization. Choice Biseliminated. The plastic does not emit light, because it is not an energy source. This eliminates choice D. The best explanationis that light reflects off the plastic differently, depending on the orientation of its electric wave relative to the plastic surface.This means that the reflected light off of the plastic (the glare) is semi-polarized. This means that the polarizing filter'simpact will depend on its orientation relative to the plastic surface. When the polarizer is perpendicular to the plastic surface,it will do its greatest filtering. The best answer is choice C.

Choice A is the best answer. Water has ahigher index ofrefraction than air, the medium that the eye normally receives lightfrom. Light rays are bent and focused as they pass through the eye because the rays go from air to a medium with a higherindex of refraction. Under water, the medium and the components of the eye now have similar indexes of refraction. Thismeans that the light rays do not bend as much when they pass from water into the eye. Less bending means a less powerfullens, which means an increase in the focal length of theeye. This eliminates choice B. The best answer is choiceA.


Topic

Absolute Potential

Air Foil

Alternating CurrentAmpereArchimedes' PrincipleBatteries

Beat FrequencyBernoulli's PrincipleBlood Pressure

Blue Shift

Brittle

Bulk Modulus

BuoyancyCapacitanceCapacitorCapacitors in ParallelCapacitors in SeriesCapacitor ChargingCathode-ray TubeClosed PipeCombination of Lenses

Concave (Converging) MirrorConcave (Diverging) LensConductivityConductor

Continuity EquationConvex (Converging) LensConvex (Diverging) MirrorCoulomb's Law

Critical AngleCyclotronDecibel

DensityDielectric Constant

Diffraction

Diffraction GratingDispersionDoppler EffectDuctile

Echolocation

Elastic Fields

Elastic Limit

Electric Circuit

Electric Motor

Electrical Potential EnergyElectrical Power

Electrical Resistance


Physics Book 2 Index

Page

124

75

199

131,17360

176

39

73-74, 9896

14

79

81,9460

184

184

193

192

185-186

154

20-21,41257-258

247-250

251

177

173

71,98251-252

247

116

242

160

9

55

126,184265,271

263

244

12-15

79

4,26, 36120

78

173

138

123

181

178

233-234

Topic Page

Electromagnetic Spectrum 233-234

Electromotive Force (emf) 176-177

Electrophoresis 143

Electrostatics 115

Equipotential Line 123

Faraday's Law 137,142,157FarsightednessFiber Optic Cable 243

Field Lines 121

Floating Object 62

Fluid 55

Focal Length 246

Fundamental Frequency 19

Harmonics 16,41

Hydraulic Press 84

Image Distance 246

Index of Refraction 238

Inductance 213

Insulator 173

Intensity (Brightness of Light) 234

Intensity (Loudness of Sound) 9-10

Interference Phenomena 259

Joule Heating Experiment 202

Kirchhoff's Junction Rule 188-190

Kirchhoff's Loop Rule 188-190

Laminar (Streamline) Flow 76

Lens 251

Lens Aberrations 258

Lenz's Law 136

Longitudinal Wave 7

Magnetic Field 131

Magnification of Image 252

Mass Spectrometer 139,140,159

Microscope 258,283

Millikan Oil Drop Experiment 152

Mirror 247,279

Modulus of Elasticity 78

Molecular Inertia 7

Nearsightedness 282

Object Distance 246

Ohm's Law 179-180

Open Pipe 18-19,41

Optical Activity 236

Pacemaker 217

Pascal's Principle 57-58

Pitch (Sound Frequency) 9

Plane Polarized Light 235

Plastically Deformed 78

Topic Page

Poiseuille's Principle 66,69,101Prism 269

Radius of Curvature 247

Rarefaction 6

Real Image 246Red Shift 14

Reflection 238

Refraction 238

Relative Density 57Resistivity 177Resistor 191

Resistors in Parallel 192

Resistors in Series 191

Resonant Frequency 16,42Reynold's Number 76Right-hand Rule 132-135Shear 78,81Solenoid 138

Sonar 26, 30,36Snell's Law 240

Specific Gravity (Relative Density) 55,57,63Speed of Sound 6

Topic

Standing WaveStrain

Stress

Sunken ObjectSurface Tension

TelescopeThin Film Interference

Thin Lens EquationTotal Internal Reflection

Tuning ForkTurbulence

Turbulent How

Ultimate Tension StrengthUltrasound

Velocity SelectorVirtual ImageViscosityViscous Retarding ForceVisible SpectrumWheatstone BridgeYoung's Double Slit ExperimentYoung's Modulus (of Elasticity)

Page

17

78, 91, 9478, 91, 94

62

64

280,283261

246

242, 275, 27838

76

76

78

35

139,140,156246

66

67

233-234

216

259

79,86

Topic

Physics Book 2 Test-taking SkillsPage Topic Page

Basic Relationship 127Equivalent Resistance/Capacitance Shortcut 197Energy Conservation 180, 181Formula Identification 239

Floating Object Calculation Trick 62-63,112Graph Identification 68

Limiting Cases 183Multiple Concepts 61, 130,185Quantitative Change 70Sunken Object Calculation Trick 62-63,112Thin Lens Shortcut Equation 252-256Unit Analysis 183

Tracking your ProgressIt is important to be organized in your review and to recognize where you are strong and whereyou need to spend more time. On these pages, you should keep a scoring profile of yourperformance and a list of items you wish to return to at a future date and review again. Makinga checklist and keeping a log can prove very helpful in staying on task during your reviewperiod. It will also prove useful to write down any key terms and equations from each chapter,so you have a quick source to reference during your final days of reviewing. Keeping all of thisin your review books will allow you to have one source for all of your review.

Sound and Doppler EffectDate Read: Date Reviewed: Text Questions to Repeat:

Score on Review Questions: 1\ - Score on Practice Test: /CO

Review Questions to Repeat: Practice Test Questions to Repeat:

Key Equations for Sound and Doppler Effect: Key Terms for Sound and Doppler Effect:

Fluids and SolidsDate Read: Date Reviewed: Text Questions to Repeat:

Score on Review Questions: 1* ^ Score on Practice Test: /co


Key Equations for Ruids and Solids: Key Terms for Fluids and Solids:

Electrostatics and ElectromagnetismDate Read: Date Reviewed: Text Questions to Repeat:

Score on Review Questions: /- - Score on Practice Test: / cO


Key Equations for Electrostatics andElectromagnetism:

Key Terms for Electrostatics andElectromagnetism:

Electricity and Electric CircuitsDate Read: Date Reviewed: Text Questions to Repeat:

Score on Review Questions: /\ ^ Score on Practice Test: / co


Key Equations for Electricity and ElectricCircuits:

Key Terms for Electricity and ElectricCircuits:

Light and OpticsDate Read: Date Reviewed: Text Questions to Repeat:

Score on Review Questions: * c Score on Practice Test: / CO


Key Equations for Light and Optics: Key Terms for Light and Optics:

IfeERKELEY

B R-E'V'I'E-W®

PERIODIC TABLE OF THE ELEMENTS

1

H

2

He

1.0

3

4.0

4 5 6 7 8 9 10

Li Be B C N o F Ne

6.9 9.0 10.8 12.0 14.0 16.0 19.0 20.2

11 12 13 14 15 16 17 18

Na Mr AI Si P s CI Ar

23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

39.1

37

40.1 45.0 47.9

40

50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8

38 39 41 42 43 44 45 46 47 48 49 50 51 52 53 54

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

85.5 87.6 88.9 91.2 92.9 95.9 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3

55 56 571LaT72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

Cs Ba Hf Ta W Re Os Ir Pt Au Hg Ti Pb Bi Po At Rn

132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222)

87 88 89 RAc§

104 105 106 107 108 109 110 111 112 113 114 115 116 117 118

Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Uut Uuq Uup Uuh Uus Uno

(223) (226) (227) (261) (262) (266) (264) (277) (268) (271) (272) (277) (287) (289) (291) (292) (292) (293)

58 59 I 60 61 1 62 63 64 65 1 66 67 68 69 70 1 71

t Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

140.1

90

140.9 1144.2 (145) 150.4 152.0 157.3 158.9 | 162.5 164.9 167.3 168.9 173.0 175.0

91 92 93 94 95 96 97 98 99 100 101 102 103

§ Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

232.0 (231) 238.0 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)

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