PHYSICS PRACTICAL MANUAL CUM RECORD I & II...

PHYSICS PRACTICAL MANUAL CUM RECORD I & II Sem

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MEASURING INSTRUMENTS

1. MEASUREMENT OF LENGTH

For the measurement of length usually meter scales are used with an accuracy upto millimeters. To measure length say 1/100th of a centimeter (or) 1/100th of a millimeter, it is not possible to measure accurately using meter scale. Hence the following instruments are used for more accuracy. 1. Vernier Calipers (Accuracy upto 1/100th of cm) 2. Screw gauge (Accuracy upto 1/100th of mm)

1. VERNIER CALIPERS

Design of the Instrument

It consists of two scales viz, main scale and vernier scale. The main scale is a long steel bar graduated in inches on the top and in centimeters on the bottom. 10 main scale divisions are equal to 1cm. The vernier scale has 10 divisions which are equal to 9 main scales divisions. There are two metal

jaws A & B. A is fixed and B is movable which can be fixed at any position using the screw(s) as shown in figure 1.

Fig 1. Vernier Caliper

The projections P1 and P2 of the jaws in the upward direction are used to measure the internal diameter of the calorimeter, cylinder, etc.

Least Count Least count of vernier calipers is defined as the smallest length that can be accurately measured and is equal to the difference between the main scale division (MSD) and the vernier scale division (VSD).

PHYSICS PRACTICAL MANUAL CUM RECORD 2010-2011

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Derivation

LC = 1 MSD – VSD 10 MSD = 1 cm 1 MSD = 1 /10 cm 10 VSD = 9 MSD 1 VSD = 9/10 MSD (or) 9/10 X 1/10 = 9/100 cm LC = 1/10- 9/100 cm

= 1/100 cm = 0.01 cm

No Zero Error To find the zero error of the vernier calipers, the two jaws A and B brought in contact with each other. If the zero of the vernier scale coincides with the zero of the main scale then there is no zero error (Z.E = 0) (fig. 1.1). Hence there is no need to apply zero correction (Z.C = 0).

Fig. 1.1 No zero error.

Positive Zero Error If the zero of the vernier scale lies on the right side the zero of the main scale, then the instrument has an error called positive zero error (fig. 1.2). This error should be subtracted from the final reading. Thus, the zero correction is negative.

Example If the 4th vernier division coincides with any of the main scale division, then zero error = Vernier Scale coincidence x Least count Fig. 1.2 positive zero error. = VSC x LC = 4 x 0.01 = 0.04 cm Zero Correction = - 0.04 cm

Negative Zero Error If the zero of the vernier scale lies on the left side of the zero of main scale, then the instrument is said to have an error called negative zero error (fig. 3). This error should be added to the final reading. Thus, the zero correction is positive.

Fig. 1.3 negative zero error.


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Example If the 6th division of the vernier coincides with any of the main scale division, then Zero Error = - (no. of vernier scale divisions – Vernier scale coincidence) x Least count

= - (10-6) x 0.01 cm

= - 4 x 0.01 = - 0.04 cm

Zero Correction = 0.04 cm

To find the length of the specimen

The specimen whose length (or) outer diameter is to be determined is held between the two jaws. The position of zero of the vernier on the main scale gives the main scale reading (MSR). The vernier division which coincides with any one of the main scale division gives the vernier scale coincidence (VSC), then Observed reading (OR) = MSR+ (VSC X LC) in cms Actual (or) correct Reading = OR ±Zero Correcti0n in cms

Example OR = 1.2 + (3 x 0.01) cm (Zero Error = Nil) CR = 1.23 cm

2. SCREW GAUGE

Fig 2. Screw gauge


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Design of the Instrument It consists of two scales namely pitch scale and head scale. Pitch scale is a millimeter scale engraved on the cylinder (C), which is rigidly attached with the frame (f). The head scale carrier 100 equal divisions. The specimen can be held in-between the two edges A (fixed) and B (movable) shown in fig. 2.

Derivation of Least count LC = Pitch/No. of head scale divisions Pitch = Distance moved on the pitch scale / No. of head scale rotations To find pitch the head is given say, two rotations and the distance moved by the head on the pitch scale is noted Pitch = 2mm/2 = 1 mm Number of head scale division = 100 LC = 1/100 mm LC = 0.01 mm

To find zero error To find zero error the stud A and the tip B are kept in contact. If the zero of the head scale coincides with zero of the pitch scale on the reference line, the instrument is said to have no zero error (Fig. 2.1). Fig. 2.1 No zero error.

Positive zero error If the zero of the head scale lies below the reference line of the pitch scale, the zero error is positive and the correction is negative (Fig. 2.2).

Example If the 2nd division of the head scale coincides with the reference line of the pitch scale then Zero error = head scale coincidence x Least count = 2 x 0.01 mm = 0.02 mm Zero Correction = -0.02 mm Fig. 2.2 positive zero error.

Negative zero error If the zero of the head scale lies above the reference line of the pitch scale the zero error is negative and the correction is positive (Fig. 2.3).

Example If the 96th division of the head scale coincides with reference line of the pitch scale then, Zero error = - (100 – head scale coincidence) x least count = - (100 – 96) x 0.01 mm = - 0.04 mm Zero correction = 0.04 mm Fig. 2.3 negative zero error.


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To find the diameter of the given wire The wire is gripped gently between the faces A and B. The number of pitch scale divisions just in front of the head scale gives the pitch scale reading. The division on the head scale that coincides with the reference line gives the head scale coincidence. The readings are noted and are tabulated.

Example Suppose if the PSR = 5mm H.S.C = 35 div. HSR = HSC x L.C. = 35 x0.01 = 0.35 mm Observed reading = PSR + HSR = 5 + 0.35 = 5.35 mm If the zero error is 0.04 mm (Positive zero error), then the zero correction is – 0.04 mm, Correct reading = OR±ZC = (5.35-0.04) mm = 5.31 mm

3 . TRAVELLING MICROSCOPE

Design of the Instrument Travelling Microscope consists of an ordinary compound microscope which slides along a graduated vertical pillar, attached to horizontal base resting on the leveling screws. The main scales divisions along with vernier scale divisions are marked on the horizontal base and the vertical pillar as shown in fig.3. The microscope can be moved up and down in the vertical pillar and can be moved in to and fro direction over the horizontal base. Thus the microscope can be moved both in the vertical and horizontal directions. Two fine adjustment screws are provided for the horizontal and vertical movements respectively. The image of the object can be focused by adjusting the side screw (S) attached to the microscope. The eye piece of the microscope is provided with a cross wire. Fig. 3 Travelling Microscope.


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The main scale is divided into millimeter and half a millimeter. Therefore the value of one main scale division (MSD) is 0.5 mm. The vernier scale of the travelling microscope is divided into 50 divisions which are equivalent to 49 main scale divisions.

Least count Derivation LC = 1 MSD – 1 VSD 20 MSD = 1 cm 1 MSD = 1/20 cm Here 50 VSD = 49 MSD 1 VSD = 49/50 MSD = (49/50) x (1/20) cm 1 VSD = 49/1000 cm LC = (1/20 – 49/1000) cm = 1/1000 cm LC = 0.001 cm

4. MEASUREMENT OF ANGLES

Fig. 4 Spectrometer.

Spectrometer – Design of the instrument Spectrometer consists of collimator (c) telescope (T), prism table (p) and vernier table as shown in fig 4. The collimator consists of a convex lens fixed at one end and a slit of adjustable width on the other end. The telescope consists of an objective (O) at one end and an eye piece fixed with the cross wires on the other end. The prism table consists of two circular disc connected by three leveling screws. The vernier tables have two verniers VA and VB each having main scale and vernier scale.


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Here one main scale division is equal to half a degree. Each vernier scale has 30 divisions, which is equal to 29 main scale divisions.

Initial adjustments

(i) Eye piece adjustment: The eye piece (E) is adjusted until the cross wires are clearly seen when viewed through the telescope.

(ii) Distant object method: A clear, well defined inverted and diminished image of a distance object is seen by adjusting the telescope.

(iii) Slit adjustment: The slit is made narrow with the help of the screw, provided aside of the slit.

(iv) Collimator adjustment: The telescope is brought on line with the collimator. The slit is illuminated by a source of light. If the image of the slit appears blurred, then the screw of the collimator is adjusted until a clear image is seen when viewed through the telescope. Now the rays of light emerging from the collimator will be rendered parallel.

(v) Prism table adjustment: The spirit level is placed on the prism table, parallel to the line joining any two leveling screws. The air bubble in the spirit level is brought to the centre by adjusting the two screws. It is then placed in a perpendicular direction and the air bubble is brought at the centre by adjusting the third screw. Now the prism table will be horizontal.

(vi) Spectrometer base: The base of the spectrometer is adjusted to the horizontal with the help of the three leveling screws.

Least count LC = 1 MSD – 1 VSD 1 MSD = ½ degree 30 VSD = 29 MSD 1 VSD = 29/30 MSD 1 VSD = 29/30 x ½ degree 1 VSD = 29/60 degree LC = (1/2-29/60) degree = (30-29)/60 degree = 1/60 degree Since 60’ = 1° 1’ = 1°/60 LC = 1’ Measurement Before doing any experiment using spectrometer, the above mentioned initial adjustments have to be made. While performing the experiment, the main scale and vernier scale readings are noted from both the verniers in VA and VB in degree and minutes.


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Fig. 5

Fig.6. Different order of spectrum


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Expt. No.: Date:

1. a. PARTICLE SIZE DETERMINAION USING DIODE LASER

Aim

To determine the size of the given particle using the laser source.

Apparatus required

A laser source, laser grating, given particle, screen, scale, etc.

Formula

Explanation of symbols

Symbol Explanation Unit

λ Wavelength of the laser source Å

m Order of spectrum Unit

Ym Distance of mth order from Zeroth order metre

D1 Distance between the particle and the screen metre

Theory `

A device for producing spectra by diffraction is known as diffraction grating. While producing diffraction spectra using the particle, the size of the particle should be comparably equal to the wavelength of the source. The diffracted wave undergoes constructive and destructive interference

effect. The intensities of the spectra depend up to the diffraction angle. By measuring the diffraction angle interms of orders of spectra. The wavelength of the given laser source and the size of the

particle can be determined.


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To fine the size of the given particle

Wavelength of the given laser source = ……………………… Å

Ym S.No. Order (m)

LHS RHS

Mean Ym

Y2m D2

1 d

Unit No. X10-

2m X10-

2m X10-2m X10-4m2 X10-4m2 X10-2m Metre

1

2

3

4

5

6

7

8

9

10


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Procedure

To find the size of the given particle

Now the laser grating is removed and the size of the particle of be found is introduced. The

laser source is switched ON and the light is made of fall on the particle. The screen is moved back and forth until the clear image of the spectrum is seen and the distance between the screen and the

particle (D1) is noted. Due to diffraction of laser light by the particle, different orders of spectrum are obtained as shown in fig.5. The positions Y1, Y2, Y3…… of the spots belonging to the first order, second order, third orders etc. on either side of the central maximum are noted in similar way as

noted above. Then by using the given formula the size of the particle can be determined.

Result:

The size of the given particle = ………………………….. meter.

VIVA – VOCE

1. Define the term “Diffraction”, with its conditions.

2. What is meant by grating element?

3. Is the diffraction possible if laser light is replaced by ordinary light for the same particle? Explain.

4. What happens to the order of spectrum, if the distance between the particle and the screen is increased?

5. Will the laser undergo diffraction through ordinary grating? Explain.

6. What will happen to the order of spectrum if the size of the particle is decreased?


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WORK SHEET

CALCULATIONS:


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WORK SHEET

CALCULATIONS:


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Fig. 7 Diffraction Pattern

Fig. 8 Angle of divergence.


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Expt. No.: Date:

1. b. DETERMINATION OF WAVELENGTH OF LASER USING GRATING

Aim

To determine Wavelength of the given laser source, using a laser grating.

Apparatus required

A laser source, laser grating, screen, scale, etc.

Formula



λ Wavelength of the laser source Å

m Order of spectrum Unit

Xm Distance of mth order from Zeroth order metre

D Distance between the particle and the screen metre

N Number of rulings in the grating Lines/metre

Theory

A device for producing spectra by diffraction is known as diffraction grating. While producing diffraction spectra using the particle, the size of the particle should be comparably equal to the

wavelength of the source. The diffracted wave undergoes constructive and destructive interference effect. The intensities of the spectra depend upto the diffraction angle. By measuring the diffraction angle interms of orders of spectra. The wavelength of the given laser source and the size of the

particle can be determined.


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(i) To find the wavelength of the laser source

Number of rulings in the grating = ………………………

Xm S.No.

Order (m) LHS RHS

Mean Xm

X2m D2 λ

Unit No. X10-

2m X10-

2m X10-2m X10-4m2 X10-4m2 X10-2m Aº

1

2

3

4

5

6

7

8

9

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(ii) To find the Angle of Divergence (Ф) using Laser source

S.No. Distance between laser and screen (X 10-2 m )

Radius of circular image (X 10-2 m )

Angle of divergence Ф = r2-r1/d2-d1 (degree)

1 d 1 = r 1 =

2 d 2 = r 2 =


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Procedure

(i) To find wavelength of the laser source

The laser source and the laser grating are mounted on separate stands as shown in fig.7. A fixed distance (D) is kept between the laser grating and the screen. The laser source is switched ON

and the beam of laser is allowed to fall on the laser grating. The diffracted beams are collected on the screen. The diffracted beams are in the form of spots as shown in fig. 7.

In the figure 7, the intensity of the irradiance is found to decrease, from zeroth order to higher

orders, i.e. the first order is brighter than the second order and so on. The positions X1, X2, X3… of the spots belonging to the first order, second order, third order etc., on either side of the central

maximum are marked on the screen and is noted.

The experiment is repeated for various values of D and the positions of the spots are noted. Then by using the given formula the wavelength of the source can be calculated and the mean is taken.

(ii) To find the angle of a divergence (Ф)

Angle of divergence given the angular spread of the laser beam. A simple diagrammatic explanation of finding the angle of divergence show in fig. 8.


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WORK SHEET

CALCULATIONS:


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Result

(a) The wavelength of the given laser source =……………………… Å

(b) Angle of Divergence of the Laser beam= ………………….. Degrees.

VIVA – VOCE

1. Define the term “Diffraction”, with its conditions.

2. What is meant by grating element?

3. Is the diffraction possible if laser light is replaced by ordinary light for the same particle? Explain.

4. What happens to the order of spectrum, if the distance between the particle and the screen is increased?

5. Will the laser undergo diffraction through ordinary grating? Explain.

6. What will happen to the order of spectrum if the size of the particle is decreased?

7. What is meant by “Numerical aperture” and “Acceptance angle”?

8. What is the principle used in the propagation of light through optical fibres?


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9. What are the parts of an optical fibre?

Fig 9. Numerical Aperture


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Expt. No.: Date:

1. c. DETERMINATION OF ACCEPTANCE ANGLE OF OPTICAL FIBRE

Aim

To measure the numerical aperture and acceptance angle of the given optical fibre.

Apparatus required

Optical fibre, numerical aperture measurement Jig., scale, etc.

Formula

Numerical aperture of the given optical fibre NA = (No unit)

Acceptance angle ‘θ’max = sin-1 (NA) degrees



d Distance between the tip of the optical fibre and the aperture of the Numerical Aperture (NA) Jig.

metre

r Radius of the circular opening in NA jig metre

Theory

Numerical Aperture (NA) and Acceptance Angle:

It is the light collecting efficiency of the fibre and is the measure of the maximum amount of light that can be accepted by the fibre. Using Snell’s law mathematically we can say

NA = sin θmax = √n12-n2

2 Where, n1 = refractive Index of core,

n2 = refractive index of cladding

θmax = Acceptance angle of the fibre

= sin-1(Numerical aperture)

i.e. θmax = sin-1 (NA)


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Measurement of Numerical aperture

S.No.

Length of

the given

fibre

Distance between NA

Jig opening and the fibre (d)

Radius of the circular

opening in Numerical aperture Jig (r)

Numerical

aperture=

Unit meter X10-3m X10-2m (No Unit)

1.

2.

3.

4.

5.

6.

Acceptance angle (θ)


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θmax = sin-1 (NA)

θmax = ……………………. degrees

Procedure

Measurement of Numerical aperture and Acceptance angle

A known length of fibre is taken. One end of the fibre is connected to the laser source and the other end is connected to the numerical aperture (NA) Jig as shown in fig. 9. The source is switched ON. The opening in the NA jig is completely opened so that a circular red patch of laser light is

observed on the screen. Now the opening in the NA Jig is slowly closed with the knob provided, so that at a particular points the circular light patch in the screen just cuts. The radius of the circular

opening (r) of NA Jig at which the circular patch of light just cuts is measured.

The distance between the NA jig opening and the fibre can be measured directly with the help of the calibration in NA jig. By substituting the values in the given formula the numerical aperture can

be calculated.

The same procedure can be adopted for various distances between the fibre and the opening of NA jig. The same procedure can be adopted for various length of fibre cable.

By finding the mean of numerical aperture (NA) and substituting it in the given formula the acceptance angle can be determined.

Result

(i) The Numerical aperture of the given optical fibre = …………. (No unit)

(ii) The acceptance angle of the given optical fibre = …………… Degrees

VIVA – VOCE

1. What is meant by “Numerical aperture” and “Acceptance angle”?

2. What is the principle used in the propagation of light through optical fibres?


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3. What are the parts of an optical fibre?

4. What is the type of the laser beam used in the experiment?

WORK SHEET

CALCULATIONS:


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WORK SHEET

CALCULATIONS:


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Fig. 10. Air Wedge


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Fig.11. Fringe pattern.

Expt. No.: Date:

2. DETERMINATION OF THICKNESS OF A THIN WIRE - AIR WEDGE METHOD

Aim To determine the thickness of a given thin wire (or) thin sheet of paper by forming interference fringes due to an air-wedge.

Apparatus required

Traveling microscope, two optically plane glass plates, 45° angle glass plate, given wire (or) thin paper, sodium vapour lamp, Transformer etc.

Formula

Thickness of the given wire (or) thin sheet of paper is given by t =



l Distance between edge of contact and the wire (or) paper metre

λ Wavelength of the monochromatic source of light (5893 Å) metre

β Mean fringe width metre

Theory

Two plane glass plates are inclined at an angle by introducing thin materials. (e.g.. hair), forming a


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wedge shaped air film. This film is illuminated by sodium light, interference occurs between the two rays, one reflected from the front surface and the other obtained by internal reflection at the back surface. Since in the case of a wedge shaped film, thickness of the material remains constant only in direction parallel to the thin edge of the wedge, straight line fringes parallel to the edge of the wedge are obtained.

(i) To find the band width (β) LC = 0.001 cm TR = MSR + (VSC X LC)

Microscope reading

MSR VSC TR

Width of 15 fringes

Fringes width (β) Order of

the fringes

X10-2m (Div) X10-2m X10-2m X10-2m

n

n+5

n+10

n+15

n+20

n+25

n+30

n+35

n+40

n+45

n+50

Mean = ……………. X 10-2 m


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(ii) To find the distance between the edge of contact and the material of wire (or) paper.

LC = 0.001 cm TR = MSR + (VSC x LC)

Microscope reading

Position of the Microscope MSR x 10-2m

VSC (Div)

TR x 10-2m

l = R 1~R2 x 10-2m

At the edge of contact (R1)

At the edge of material of wire (or) paper (R2)

Procedure

Two optically plane glass plates are placed one over the other and tied by means of a rubber band at one end. The given material of wire (or) paper is introduced on the other end, so that an airwedge is formed between the plates as shown in fig.10. This set up is placed on the horizontal bed

plate of the traveling microscope.

Light from the sodium vapour lamp (S) is rendered parallel by means of a condensing lens (L). The parallel beam of light is incident on a plane glass plate (G) inclined at an angle of 45º and gets

reflected. The reflected light is incident normally on the glass plates in contact Pc. Interference taken place between the light reflected from the top and bottom surfaces of the glass plates and is viewed

through the traveling microscope (M). Hence large number of equally spaced dark and bright fringes are formed which are parallel to the edge of contact (fig. 11).

The microscope is adjusted so that the bright (or) dark fringe near the edge of contact is made to coincide with the vertical cross wire and this is taken as the nth fringe. The reading from the

horizontal scale of the traveling microscope is noted. The microscope is moved across the fringes using the horizontal traverse screw and the readings are taken when the vertical cross wire coincides

with every successive 5 fringes (5, 10, 15….). The width of every 15 fringes is calculated and the width of one fringe is calculated. The mean of this gives the fringe width (β).

The cross wire is fixed at the inner edge of the rubber band and the reading from the microscope is noted. Similarly reading from the microscope is noted keeping the cross wire at the

edge of the material. The difference between these two values gives the value of ‘l’. Substituting ‘β’ and ‘l’ in the given formula, the thickness of the given material can be determined.


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WORK SHEET

CALCULATIONS:


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Result

The thickness of the given material = ………………….. meter.

VIVA – VOCE

1. What is the principle behind the formation of fringes in Air wedge?

2. What is meant by an “Air Wedge” and explain how it can be formed?

3. What is meant by fringe width?

4. What is the theory of formation of thin films? Give examples.


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5. What is the use of inserting 45˚ angled glass plate?

6. Why do we get straight line fringes in an airwedge?

7. Explain the reason for color formation in soap bubbles, when white light falls on it.

8. What happens to the fringe width, if the thickness of the material is increased?

9. What is the condition for the formation of bright fringes?

10. Why do we get bright and dark fringes alternatively?


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Fig.12 Ultrasonic Interferometer

Expt. No.: Date:

3. DETERMINATION OF VELOCITY OF SOUND AND COMPRESSIBILITY OF THE LIQUID -

ULTRASONIC INTERFEROMETER

Aim

(i) To determine the velocity of Ultrasonic waves in the given liquid using Ultrasonic Interferometer. (ii) To determine the compressibility of the given liquid.

Apparatus required

Ultrasonic Interferometer, measuring cell, frequency generator, given liquid, etc.

Formula

(i) Velocity of Ultrasonic waves in the given liquid v = nλ ms-1

Where, Wavelength metre (or) Å

(ii)Compressibility of the given liquid m2 / N


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n Frequency of the generator which excites the crystal Hertz

λ Wavelength of the Ultrasonic metre

ρ Density of the given liquid Kg/m2

d Distance moved by the micrometer screw metre

x Number of oscillations Unit

Fig. 13 Distance moved by the reflector Vs Crystal current


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Theory

An Ultrasonic Interferometer is a simple and direct to determine the velocity of Ultrasonic waves in liquid with a high degree of accuracy. Here the high frequency generator generates variable frequency, which excited the Quartz Crystal placed at the bottom of the measuring cell (Fig. 12). The excited Quartz Crystal generates Ultrasonic waves in the experimental liquid. The liquid will now serve as an acoustical grating element. Hence when Ultrasonic waves passes through the rulings of grating, successive maxima and minima occurs, satisfying the condition for diffraction.

Initial adjustments: In high frequency generator two knobs are provided for initial adjustments. One is marked with ‘Adj’ (set) and the other with ‘Gain’ (Sensitivity). With knob marked ‘Adj’ the position of the needle on the ammeter is adjusted and with the knob marked ‘Gain’, the sensitivity of the instrument can be increased for greater deflection, if desired.

Procedure

The measuring cell is connected to the output terminal of the high frequency generator through a shielded cable. The cell is filled with the experimental liquid before switching ON the generator is switched ON, the Ultrasonic waves move normal form the Quartz Crystal till they are reflected back by the movable reflector plate. Hence, standing waves are formed in the liquid in-between the reflector plate and the quartz Crystal. The distance between the reflector and crystal is varied using the micrometer screw such that the anode current of the generator increases to a maximum and then decreases to a minimum and again increases to a maximum in the anode current is equal to half the wave length of the Ultrasonic waves in the liquid (Fig.


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13.). Therefore, by nothing the initial and final position of the micrometer screw for one complete oscillation (maxima – minima- maxima) the distance moved by the reflector can be determined. To minimize the error, the distance (d) moved by the micrometer screw is noted for ‘x’ number of oscillations (successive maxima), by noting the initial and final reading in the micrometer screw and is tabulated. From the total distance (d) moved by the micrometer screw and the number of oscillations(x), the wavelength of ultrasonic waves can be determined using the formula λ = 2d/x. From the value of λ and by noting the frequency of the generator (n), the velocity of the Ultrasonic waves can be calculated using the given formula. After determining the velocity of the Ultrasonic waves in liquids. The compressibility of the liquid is calculated using the given formula.


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Result

(i)The velocity of Ultrasonic waves in the given liquid =……………. ms-1

(ii)Compressibility of the given liquid = ……………. m2N-1

VIVA – VOCE

1. What is meant by Compressibility?

2. Explain the terms stress and strain.

3. What is the frequency range of Ultrasonics?

4. What is mean by acoustical grating?

5. Explain the principle of determining the compressibility of liquids.

6. What is meant by nodes and Antinodes?

7. What do you understand by the term “Over tones”?

8. Is Ultrasonic wave, an Electromagnetic wave? Explain.

9. What are the various liquids that can be used for finding the compressibility using Ultrasonic

interferometer?

10. What type crystal is thrown into vibrations in the case of Ultrasonic interferometer?


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WORK SHEET

CALCULATIONS:


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WORK SHEET

CALCULATIONS:


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N =

si

n θ

/nλ

Line

s / m

Mea

n θ

degr

ees

VB

degr

ees

Mea

n θ

= 2θ

/2

VA

degr

ees

VB

degr

ees

Ord

er o

f the

spec

trum

(n) =

Tot

al R

eadi

ng =

MSR

+ (V

SCX

LC)

2 θ

= R

1-R

2

VA

degr

ees

TR

degr

ees

VSC

divi

sion

s

Ver

nier

B (V

B)

MSR

degr

ees

TR

degr

ees

VSC

divi

sion

s

Ver

nier

A (V

A)

MSR

degr

ees

(i) T

o fin

d th

e nu

mbe

r of l

ines

per

met

re o

n th

e gr

atin

g (N

)

LC =

1´

λ=

5893

A°

Diff

ract

ed ra

y re

adin

gs

Left

(R

1)

Rig

ht (

R2)


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Expt. No.: Date:

4. DETERMINATION OF WAVELENGTH OF MERCURY SPECTRUM -

SPECTROMETER GRATING

Aim

To determine the wavelength of the mercury (Hg) spectrum by standardizing the plane transmission grating.

Apparatus required

Spectrometer, plane transmission grating, mercury vapour lamp, sodium vapour lamp, spirit

level, lens, torch light etc.

Formula

(i) The number of lines drawn on the grating per meter is given by

lines /meter

(ii)The wavelength of the prominent lines of the mercury spectrum is given by

Å



n Order of spectrum Unit

λ Wavelength of the Sodium vapour lamp (5893 Å) Metre (or) Å

θ Angle of diffraction Degree


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Fig. 14. Position of the Grating Fig.15. Spectrometer Grating First to Get Reflected Ray order Diffraction


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Procedure

(i) Adjustment of grating for normal incidence

Preliminary adjustments of the spectrometer are made. The grating (G) is mounted on the grating table with its ruled surface facing the collimator. The slit is illuminated by a source of light (either

sodium (or) mercury vapour lamp) and is made to coincide with the vertical cross wire. (The vernier scales are adjusted to read 0˚ and 180˚ for the direct ray. The telescope is rotated through an angle

90˚ and is fixed. The grating table is adjusted until the image coincides with the vertical cross wire. Both the grating table and the telescope are fixed at this position (fig.14). Now rotate the vernier tables through 45˚in the same direction in which the telescope has been previously rotated. The light

from the collimator incidents normally (perpendicularly) on the grating. The telescope is released and is brought on line with the direct image of the slit. Now the grating is said to be in the normal

incidence position (fig. 15).

(ii) Standardization of grating (To find the number of lines drawn in the grating per meter)

The slit is illuminated by sodium vapour lamp. The telescope is released to get the diffracted

image of the first order on the left side of the central direct image. The readings are tabulated from the two verniers VA and VB. Similarly readings are taken for the right side of the central direct image

(fig. 15). The difference between the two readings gives 2θ, where θ is the angle of first order

diffraction. The number of lines per meter (N) of the grating is calculated using the given formula.

The experiment is repeated for the second order and the readings are tabulated.

(iii) Determination of wavelength of the mercury spectrum

The sodium vapour lamp is replaced by mercury vapour lamp. The diffracted images of the first order are seen on either side of the central direct image (fig.15). As before the readings are tabulated

by coinciding the vertical cross wire with the prominent lines namely violet, blue, blueish green,

green, yellow, red of the mercury spectrum. The difference between the readings give 2θ, from this

θ can be found. The wavelength of each spectral line is calculated using the given formula.


45


46

Result

(i) Number of lines drawn in the grating per meter = ………….. Lines/ meter.

(ii) Wavelength of various spectral lines of the mercury spectrum are

λV = ……….…………. Aº λB = ……….…………. Aº

λBG = ……….…………. Aº λG = ……….…………. Aº

λY = ……….…………. Aº λR = ……….…………. Aº

VIVA - VOCE

1. How many rulings are there in the grating element given?

2. What is the condition for diffraction?

3. What is the difference between reflection and scattering?

4. What do you understand by the term least count?

5. What is the difference between transmission and reflection grating?

6. Why the skies appear red during sunset?

7. What is meant by dispersion?

8. Define wave packet.

9. What is the use of collimator and telescope?

10. How the astronomical telescopes differ from ordinary telescope?


47

WORK SHEET

CALCULATIONS:


48

WORK SHEET

CALCULATIONS:


49

Fig. 16 Lee’s Disc apparatus arrangement

(i) To find (dθ/dt)θ2

θ 1 = ………..°C θ 2 = ………..°C

Temperature Time Taken S.No

°C Kelvin Sec.

1 θ 2 + 5

2 θ 2 + 4

3 θ 2 + 3

4 θ 2 + 2

5 θ 2 + 1

6 θ 2

7 θ 2 - 1

8 θ 2 - 2

9 θ 2 - 3

10 θ 2 - 4

11 θ 2 - 5


50

Expt. No.: Date:

5. DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD

CONDUCTOR - LEE’S DISC METHOD.

Aim

To determine the thermal conductivity of a bad conductor using Lee’s disc apparatus.

Apparatus required

Lee’s disc apparatus, two thermometers, bad conductor, stop watch, steam boiler, vernier caliper, screw gauge, biscuit balance etc.

Formula

Thermal conductivity of the given band conductor



M Mass of the metallic disc Kg

S Specific heat capacity of the material of the disc J/Kg/K

(dθ/dt)θ2 Rate of cooling at steady temperature θ2 Kelvin

θ 1 Steady temperature of the steam chamber Kelvin

θ 2 Steady temperature of the metallic disc Kelvin

R Radius of the metallic disc metre

h Thickness of the metallic disc metre

x Thickness of the bad conductor metre


51

Fig. 17 Model Graph.

(ii) To find the diameter of the metallic disc

LC = 0.01 cm ZE = ………….

ZC = …………. x 10-2m

S.No. Main Scale

Reading

(MSR)

Vernier Scale Coincidence (VSC)

Observed Reading OR=MSR+(VSCXLC)

Correct Reading CR = OR±ZC

Unit X 10-2m Div. X 10-2m X 10-2m

1

2

3

4

5

6

Mean = ……….. x 10-2m


52

Description

Lee’s disc apparatus consists of a brass metal disc (B) suspended horizontally by three strings form a stand. A hollow steam chamber (A) with inlet and outlet for steam is placed above. The given bad conductor is placed between them. Two thermometers T1 & T2 are inserted to measure the

temperatures of A and B respectively.

Procedure

The experimental arrangement is as shown in fig 16. Steam is allowed to pass through the steam chamber. The temperature indicated by two thermometers start rising. After 30 minutes a steady state is reached (i.e) the temperature of the lower disc will no longer rises. At this stage,

steady temperatures θ1 and θ2 are recorded from the thermometers T1 and T2. Now the cardboard is removed and the lower disc is heated directly by keeping it in contact

with the steam chamber. When the temperature of the lower disc attains a value of about 10°C more than its steady state temperature, the chamber is removed and the lower disc is allowed to cool down on its own accord.

When the temperature of the disc reaches 5°C above the steady temperature (θ2) of the disc i.e., (θ2+5°C), a stop watch is started and the time is noted for every 1°C fall of temperature until

the metallic disc attains temperature (θ2-5°C).

The thickness and radius of the metallic disc is found using screw gauge and vernier caliper respectively. The thickness of the bad conductor is found using screw gauge. The mass of the metallic disc is found by using biscuit balance. The readings are tabulated in the tabular column.

Graph

A graph is drawn by taking time along the x-axis and the temperature along y-axis (fig. 17)

Cooling curve is drawn. From the cooling curve dθ/dt is calculated by drawing a triangle by taking 0.5°C above and 0.5°C below the steady temperature θ 2. Substituting this dθ/dt in the

given formula, thermal conductivity of the card board can be calculated.


53

(iii) To find the thickness of the metallic disc LC = 0.01 mm ZE = ………….

ZC = …………. x 10-3m

S.No. Pitch Scale

Reading (PSR) Head Scale

Coincidence (HSC) Observed Reading

OR=PSR+(HSCXLC) Correct Reading

CR = OR±ZC

Unit X 10-3m Div. X 10-3m X 10-3m

1

2

3

4

5

Mean = ……….. x 10-3m

(i) To find the thickness of the Bad Conductor LC = 0.01 mm ZE = ………….

ZC = …………. x 10-3m

S.No. Pitch Scale



OR=PSR+(HSCXLC) Correct Reading

CR = OR±ZC

Unit X 10-3m Div. X 10-3m X 10-3m

1

2

3

4

5

6

Mean = ……….. x 10-3m


54

Result

Thermal Conductivity of given Bad Conductor = ……………………. W/m/K

VIVA - VOCE

1. What do you understand by the term conduction, convection and radiation?

2. What is meant by thermal conductivity?

3. Name any four bad conductors?

4. What do you understand by the term steady state?

5. What is meant by “Rate of Cooling”?

6. What is the use of cooling the slab and noting the time?

7. What happen to the thermal conductivity if the thickness of the given bad conductor is

increased?

8. Is the diameter of the bad conductor should match diameter of the disc (D) and the steam chamber (S)? Why?

9. Will there be any change in thermal conductivity if the area of cross section of bad conductor is

decreased: Justify.

10. State some of the applications of bad conductors.


55

WORK SHEET

CALCULATIONS:


56

WORK SHEET

CALCULATIONS:


57

Fig 18 a. Top view of the B-H Curve Unit

Fig 18b. Hysteresis loop


58

Expt. No.: Date:

6. DETERMINATION OF HYSTERESIS LOSS IN A FERROMAGNETIC

MATERIAL.

Aim

To determine the hysteresis loss in the transformer core using B-H curve unit.

Apparatus required

CRO, B-H curve unit, patch cords

Formula

Energy loss =



Number of turns in the primary coil -

Number of turns in the secondary coil -

A Area of cross section of core m2

L Length of the core m

SV Steady temperature of the metallic disc -

SH Horizontal sensitivity of the CRO -

DESCRIPTION The experimental arrangement is shown in the figure 18a. One of the specimens used in

the unit is made using transformer stampings. There are two windings on the specimen (primary and secondary). The primary is fed to low A.C voltage (50Hz). This produces a magnetic field H in the

specimen. The voltage across R1 (resistance connected in series with primary) is proportional to the magnetic field. It is given to the imput of the CRO. The A.C magnetic field induces a voltage in the secondary coil. The voltage induced is proportional to dB/dt. This voltage is applied to passive

integrating circuit. The output of the integrator is proportional to B and fed to the vertical input of the


59

CRO. As a result of the application of voltage proportional to H the horizontal axis and a voltage proportional to B is the vertical axis, the loop is formed as shown in fig 18b.

S.No. N1 N2 R1 ohm R2 ohm SV SH Area of the

loop

Energy

loss

1

2

3

4

5

6

7

8

9

10

Observations

Number of turns in the primary N1 = Number of terms in the secondary N2 =

R1 = ohm R2 = ohm

C2 = µF SV = Vm-1 SH = Vm-1

Calculation

Area o f the loop = m2


60

Energy loss =

PROCEDURE

The unit one to force the B-H loop (hysteresis) of a ferromagnetic specimen using a CRO is shown in figure. A measurement of the area of the loop leads to the evaluation of the energy loss in

the specimen. The top view of the unit is shown in the figure. There are 12 terminals on the panel; sin patch cords are supplied with the kit.

The value of R1 can be selected connecting terminal D to B or C (A-D = 50 ohm; B-D = 150

ohm; C-D = 50 ohm)

A is connected to D. The primary terminals of the specimen is connected to p, p secondary to s, s terminals. The CRO is calibrated as per the instructions given the Instruction Manual of CRO. CRO is

adjusted to work on external mode (the time base is switched off). The horizontal and vertical position controls are adjusted such that the spot is at the centre of the CRO screen.

The terminal marked GND is connected to the ground of the CRO. The H is connected to the Horizontal input to the ground of the CRO. The H is connected to the Horizontal input of the CRO.

The terminals V are switched on. The hysteresis loop is formal. The horizontal and vertical gains are adjusted such that the loop occupies maximum area of the screen of the CRO. Once this adjustment is

made, the gain controls should not be disturbed. The loop is traced on a translucent graph paper. The area of the loop is estimated.

The connections from CRO are removed without disturbing the horizontal and vertical gain

controls. The vertical sensitivity of the CRO is determined by applying a known A.C voltage say 1 volt (peak to peak).

If the spot deflects by X cms for 1 volt, the vertical sensitivity is 1/X x 10-2 (volt/m). Let it be dV. The horizontal sensibility of CRO is determined by applying a known A.C. voltage say 1 volt

(peak to peak). Let the horizontal sensitivity be SH (volt/m).

The transformer core may be replaced by ferrite ring and hysteresis loss in ferrite core can be found.


61

WORK SHEET

CALCULATIONS:


62

Result:

Energy loss = ……………………………. Joules cycle-1 m-3.


63

Fig.19 Non – Uniform Bending Model Graph

To find M/y LC = 0.001 cm TR = MSR +(VSCXLC)

Mean M/y = ……..Kg/m

Microscope Reading

Increasing Load Decreasing Load S.No

Distance between

Knife edges (l)

Load

(M) MSR VSC TR MSR VSC TR

Mean Depression (y)

M/y

Unit X 10-2 m X 10-3 Kg X 10-2 m Div X 10-2 m X 10-2m Div X 10-2m X 10-2 m X 10-2 m X 10-1

Kg m-1

1 W

2 W+50

3 W+100

4 W+150

5 W+200

6

W+250


64

Expt. No.: Date:

7. DETERNINATION OF YOUNG’S MODULUS OF THE MATERIAL –

NON UNIFORM BENDING

Aim

To determine the young’s modulus of the given material of the beam by Non Uniform bending.

Apparatus requited

A long uniform beam usually a meter scale, travelling microscope, pin, weight hanger with

slotted weights, vernier calipers, Screw gauge, knife edges etc., Formula

The Young’s Modulus of the given material of the beam

By Calculation method Nm-2

By Graphical method Nm-2



g Acceleration due to gravity m/s2

l Distance between Knife edge meter

b Breadth of the beam meter

d Thickness of the beam meter

y Depression produced for ‘M’ kg of load meter

M Load applied Kg

k Slope 1/k from graph Kg m-1


65

By Graphical method to find h/M

AXIS W-(W+50) W-(W+100) W-(W+150) W-(W+200) W-(W+250)

X (M)

Y ( h)

To find Breadth (b) of the beam using vernier calipers LC = 0.01cm ZE = …………. ZC = …………. x 10-2m

S.No. Main Scale

Reading (MSR) Vernier Scale

Coincidence (VSC) Observed Reading

OR=MSR+(VSCXLC) Correct Reading CR

= OR±ZC

Unit X 10-2m Div. X 10-2m X 10-2m

1

2

3

4

5

Mean ‘b’ = ……….. x 10-2m

To find the thickness (d) of the beam using screw gauge. LC = 0.01 mm ZE = …………. ZC = …………. x 10-3m

S.No. Pitch Scale



OR=PSR+(HSCXLC) Correct Reading CR

= OR±ZC

Unit X 10-3m Div. X 10-3m X 10-3m

1

2

3

4


66

5

Mean‘d’ = ……….. x 10-3m

Procedure The given beam is placed on the two knife edges (A and B) at a distance say 70 cm or 80 cm. A weight hanger is suspended at the centre(C) of the beam and a pin is fixed vertically on the frame of the hanger as shown in fig.19. Taking the weight hanger alone as the dead load the tip o the pin is focused by the microscope, and is adjusted in such a way that the tip of the pin just touches the horizontal cross wire. The reading on the vertical scale is noted. Now the weight is added in steps of 50 grams. Each time the tip of the pin is made to touch the horizontal cross wire and the readings are noted from the vertical scale of the microscope. The procedure is followed until the maximum load is reached. The same procedure is repeated by unloading the weight in steps of same 50 grams and the readings are tabulated in the tabular column. From the readings the mean of (M/y) is calculated. The thickness and the breadth of the beam are measured using screw gauge and vernier calipers respectively and are tabulated. By substituting the values in the given formula. The young’s modulus of the material of the beam can be calculated. Graph A Graph is drawn taking load (M) along x axis and depression ‘y’ along y axis as shown in figure. The slope of the graph gives the value K = y/M. Substituting the value of the slope in the given formula, the Young’s modulus can be calculated.

Result The Young’s modulus of the given material of the beam by calculation Y = …………………. Nm-2

by Graph Y = …………….…… Nm-2

VIVA - VOCE

1. What is meant by non – uniform bending?

2. Define neutral axis.

3. Name any two methods used to determine the young’s modulus of the beam.

4. Define elasticity.

5. Will there be any change in young’s modulus of the material, if its thickness is increased? Justify.

6. What are the basic assumptions made from the theory of bending?

7. Why iron girders used in buildings are made in the form of I section?


67

8. What is the use of keeping an optimum of 0.7 to 0.8 meter distance between the knife edges?

9. Define elastic limit.

10. What is meant by elastic constants?

WORK SHEET

CALCULATIONS:


68

WORK SHEET

CALCULATIONS:


69

MODEL GRAPH Fig.20 Band gap


70

Fig.21 Model graph

Expt. No.: Date:

8. DETERMINATION OF BAND GAP OF A SEMICONDUCTOR MATERIAL

Aim

To determine the band gap energy of a semiconductor by studying the variation of reverse

saturation current of a point contact diode at any temperature.

Apparatus required

Point contact diode, heating arrangement to heat the diode, ammeter, voltmeter,

thermometer etc. Formula

Band Gap Energy Eg = 0.198 x Slope eV

Where



71


Is Saturation Current µA

T Absolute Temperature Kelvin

Theory

For a semiconductor diode at 0 K the valence band is completely filled and the conduction band is completely filled and the conduction band is empty and it behaves as an insulator. If the

temperature is increased, some of the valence electrons gains thermal energy greater than the forbidden band gap energy (Eg) and it moves to conduction band, which constitutes some current to flow through the semiconductor diode.

(i) Measurement of current for various temperatures

Power supply = …………….. Volts

S.No. Temperature in

centigrade Temperature in

Kelvin 1000/T Is Log Is

Unit °C K K-1 X 10-6 Amp Amp

1

2

3

4

5

6

7

8

9

10


72

11

12

13

14

15

16

17

18

19

20

Band Gap energy

Eg = 0.198 x ( ) eV

Eg = 0.198 x Slope eV

Eg = ……………… eV

Procedure

The circuit is given as shown in fig.20. The point contact diode and the thermometer are

immersed in a water (or) oil bath, in such a way that the thermometer is kept nearby the diode. The power supply is kept constant (say 3 volts). The heating mantle is switched ON and the oil bath is

heated up to 70°C. Now the heating mantle is switched OFF and the oil bath is allowed to cool slowly. For every one degree fall of temperature the microammeter reading (Is) is noted. A graph is plotted taking 1000/T along X axis and log Is along negative Y axis, (Since Is is in the

order of micro-amperes, log Is value will come in negative). A straight line is obtained as shown in model graph (fig.21). By finding the slope of the straight line, the band gap energy can be calculated

using the given formula. The same procedure can be repeated for various constant power supplies (4volts, 5 volts)

Result

The band gap energy of the given diode is = ……………….. eV


73

VIVA - VOCE

1. What is meant by conduction band and valence band?

2. Explain the term Band gap?

3. What are the types of semiconductors?

4. What do you understand by the term “Fermi Level?

5. What is the band gap energy for germanium and silicon?

6. What is the effect of temperature over a semiconductor?

7. What is the principle used in finding the band gap energy of a semiconductor?

8. What is the use of keeping the diode inside the oil bath?

9. Shall the oil bath be replaced by water bath? State the reason.

10. What is the biasing of the semiconductor diode in the band gap experiment?

WORK SHEET

CALCULATIONS:


74

WORK SHEET

CALCULATIONS:


75

Circuit diagram

Fig 22.(a) Carey Foster’s Bridge


76

Fig. 22(b) Bridge Equivalent

Expt. No.: Date:

9. DETERMINATION OF SPECIFIC RESISTANCE OF GIVEN COIL OF WIRE –

CAREY FOSTERS BRIDGE

Aim

To determination of the specific resistance or resistivity of the material of the coil.

Apparatus requited

Bridge wire resistance coil, Battery, 4 & 2 dial resistance boxes, copper strips, galvanometer, High resistance, key, Jockey, etc.,

Formula

Resistance per unit length of the bridge wire

Specific resistance or resistivity = SA/l ohm meter


77



Resistance per unit length of the bridge wire Ohm / Meter

R Resistance introduced in the variable box R Ohm

l1, l2 The balancing lengths Meter

S Resistance of the coil (to be determined) Ohm

l Length of the coil Meter

A Area of the cross section of the coil Meter2

Table No: 1

To determine the resistance per unit length of the bridge wire.

Balancing Length

S.No. Fractional

Resistance ‘R’

ohm l’1 m l’2m

= R/ (l2’-l1’)

ohm/ meter.

1

2

3

4


78

5

6

7

8

9

10

Mean =……………..

Procedure

Carey Foster’s bridge is more useful for comparing two nearly equal resistances or to determine unknown resistance. It is the modified form of the meter bridge and works on the

principle of Wheatstone’s bridge. Since the end resistances are eliminated, the unknown resistances can be determined very

accurately. It consists of a uniform resistance wire AB, of one meter long. Normally the wire is made from managing which has zero temperature coefficient of resistance. The ends of the wire are soldered

to two thick L-Shaped cooper strips of negligible resistance. There are three more copper strips which are fixed on the wooden board between the first two L-shaped strips so as to form four gaps

1, 2, 3 and as shown in fig22(a). The copper strips are provided with binding screws for electrical connection. Two known equal standard resistances P & Q are connected in gaps 2 and 3 respectively and a variable resistance box R and the given unknown resistance S are connected in

gaps 1 and 4 respectively. Battery with a key and a galvanometer with a safety high resistance and jockey are connected as shown in fig 22a. Contact can be made at any point on the bridge wire by


79

means of jockey. A meter scale is fixed on the board parallel to the length of the wire so as to take readings directly.

Experiment to determine the resistance of the given coil of wire:

Having introduced a value R in the resistance box, the circuit is closed and the balancing point J on the bridge wire where there is no deflection in the galvanometer is determined. The balancing length l1 is measured from the point ‘A’. The experiment is repeated by interchanging R and S and

the balancing length l2 is noted. The experiment is repeated by varying the known resistance R and in each case l1 and l2 are measured.

Determination of :

To determine the resistance per unit length of the bridge wire, the resistance S is replaced by a thick copper strip (i.e. S = 0) and the balancing length ‘l1’ is determined using fractional resistance

in R. Now by interchanging R and copper strip (S = 0, the balancing length l2’ is determined.

0 = R+ (l1’-l2’)

= R/ (l2’-l1’) ohm / meter.

Table: 2

To find the specific resistance or resistivity of the material of the coil

Balancing Length

S.No. Resistance ‘R’

ohm l1 m l2m

S= = R+ (l1-l2)

Ohm

1

2

3


80

4

5

6

7

8

9

10

Mean ‘S’ =……………. Ohm

Now the unknown resistance is inserted in the bridge circuit instead of copper stip. The bridge will have the maximum sensitivity when S = R and hence the error in measuring the value of S is also

minimum in that condition. So, after inserting the unknown resistance in the bridge arm, the jockey is pressed at the 50th cm of the bridge wire and the value of R is varied. The value of R at which there is

null deflecting in the galvanometer is noted. This will give the approximate resistance of the unknown resistance. Let it be X ohm. Now, the value of R is varied is steps of 0.2 ohm between (X+1) ohm to (X-1) ohm and correspondingly the balancing lengths l1 and l2 are determined and tabulated in

the table no 2. Mean of the last column in the tabular column number 1 indicated the resistance of the bridge

wire per unit length (p) and that value is substituted in equation in the tabular column number 2. The mean of the last column in the tabular column number 2 gives the value of the unknown resistance of the given coil of wire.

Special cases:


81

(i)Determination of the specific resistance or resistivity of the material of the coil:

Specific resistance or resistivity = SA/l ohm meter where S is the resistance of coil, A and l are the area of cross section and the length of the wire in the coil respectively. Using a screw gauge, the

radius of the wire can be determined. Hence the area of cross section a = πr2 can be determined. Using the above formula the resistivity of the material of the coil wire can be determined.

Result

1. Resistance per unit length of the bridge wire = …………………. Ohm / Meter

2. Specific resistance of resistivity of the material of the coil = ………….. Ohm - Meter

VIVA - VOCE

1. State the Principle of wheatstone’s Bridge

2. What do you mean by the term balancing length?

WORK SHEET

CALCULATIONS:


82

WORK SHEET

CALCULATIONS:


83

Fig. 23. Coefficient of viscosity - poiseuilles flow method


84

(i) Measurement of time for liquid flow

h0 = ………….. x 10-2 metre

S.No. Burette reading

Time note

while crossing

level

Range

Time for

flow of 5 cc

liquid

Height of

initial reading

h1

Height of final reading

h2

Pressure head

h=(h1+h2/2)-h0

ht

Unit cc Seconds X10-

2m seconds X10-2m X10-2m X10-2m

X10-

2m 1 0 0-5

2 5 5-10

3 10 10-15

4 15 15-20

5 20 20-25

6 25 25-30

7 30 30-35

8 35 35-40

9 40 40-45

10 45 45-50

11 50 - Mean ht = ………… X10-2m -sec

Expt. No.: Date:

10. DETERMINATION OF VISCOSITY OF LIQUID - POISEUILLES

METHOD

Aim

To determine the coefficient of viscosity of the given liquid by poiseuilles flow method.

Apparatus required

A graduated burette, rubber tube, capillary tube, pinch cork, etc., Formula

Coefficient of viscosity (Nsm-2)


85




ρ Density of the liquid Kg/m3

r Radius of the capillary tube metre

v Volume of the liquid collected metre3

h metre

h1 Height from table to initial level of water in the burette

metre

h2 Height from table to final level of water in the burette

metre

ho Height from table to mid portion of capillary tube

metre

t Time taken for the liquid flow Second

l Length of the capillary tube metre

Fig 24. Radius of Capillary Tube

To find the radius of the bore of the capillary tube

LC = 0.001 cm TR = MSR+ (VSCXLC)


86

Microscope Reading MSR VSC TR

Radius (r) Position

X10-

2m Div X10-2m X10-2m Left V1

r = V1~V2

Right V2

Top H1

r = H1~H2

Bottom H2

Procedure

The burette is held vertically on a retort stand. A capillary tube is attached to the lower end of

the burette using a rubber tube as shown in fig 23. The burette is filled with the given liquid. The capillary tube is made horizontal and the liquid is allowed to flow freely through it. When the liquid

comes to a known height (h1), which is the height measured from the axis of the capillary tube, the stop watch is started. The stop watch is stopped when the liquid comes to another level which is of

height h2 from the axis of the capillary tube. Then the driving height is given by . The

driving height, volume of the liquid and the time taken for flow of liquid are noted in the tabular

column. The experiment is repeated for various known heights of the liquid and the time taken is noted. The mean of (ht/V) is taken. The radius of the bore of the capillary tube figure. Can be found

by using a travelling microscope by mounting the capillary tube over a stand. Substituting the above data in the given formula, the coefficient of viscosity can be calculated.


87

Result:

The coefficient of viscosity of the given liquid = ………………………….. Nsm-2.

VIVA – VOCE

1. Define coefficient of viscosity

2. What is meant by capillary rise, give some practical applications?

3. What will happen to the viscosity if the density of the liquid is increased?

4. Name any two highly viscous liquid.

5. Name any two methods used to determine the viscosity.

6. What will happen to the viscosity of the liquid if the temperature of the liquid is increased? Justify your answer.

7. Define stream line motion and turbulent motion.

8. State the relation between pressure and viscosity.

9. Define critical velocity.

10. What is meant by velocity gradient?

WORK SHEET

CALCULATIONS:


88

WORK SHEET

CALCULATIONS:


89


90

Fig 25. Angle of Prism (A)

(i) To find the angle of Prism (A)

LC = 1’ Total Reading (TR) = MSR + (VSCXLC)

VERNIER - A VERNIER –B 2A = R1~ R2 A

MSR VSC TR MSR VSC TR VA VB VA VB Reflected Image (deg) (div) (deg) (deg) (div) (deg) (deg) (deg) (deg) (deg)

Face –I (Left) (R1) (R1)

Face – 2 (Right) (R2) (R2)

Mean A =

Expt. No.: Date:


91

11. SPECTROMETER - DISPERSIVE POWER OF A PRISM

Aim

To determine the refractive index and hence the dispersive power of the given prism using spectrometer.

Apparatus required

Spectrometer, glass prism, sodium vapour lamp, transformer, mercury vapour lamp, pencil torch, lens, spirit level etc.,

Formula

Refractive index of the given prism is given by (No unit)

Dispersive power of the prism in the wavelength region of λ1 and λ2 is = (No unit)



A Angle of the prism degrees

D Angle of minimum deviation degrees

µ1 Refractive index of the prism for 1st colour No unit

µ2 Refractive index of the prism for 2nd colour No unit


92

Fig 26. Angle of Minimum Deviation (D)


93

Procedure

To find the angle of Prism (A)

To initial adjustments of the spectrometer is made. The silt is illuminated by yellow light from the sodium vapour lamp. The given prism is mounted on the prism table such that light emerging from the collimator should be made to incident on both the refracting faces of the prism as shown in

fig 25. The telescope is rotated (left or right) to catch the image of the slit reflected by one of the

refracting face of the prism. The telescope is fixed. By adjusting the tangential screw, the image is made to coincide with the vertical cross wire. The main scale and the vernier scale readings are noted from both the vernier A (VA) and vernier B (VB). Similarly readings are taken for the image reflected

by other refracting face of the prism. The difference between the two readings gives 2A, where ‘A’ is the angle of prism.

To find the angle of minimum deviation (D) and dispersive power

The prism is mounted such that light emerging from the collimator is incident on one of the

refracting face of the prism. The telescope is slowly rotated to catch the refracted image of any one of the colour (say colour -1 (violet)) which emerges from other refracting face of the prism. Now by viewing through the telescope the prism table is slightly rotated in such a way that the

violet image moves towards the direct ray and at a particular position it retraces its original path. This position is called MINIMUM DEVIATION POSITION (fig 26). The prism table is fixed and hence

now all the colours in the adjusted to coincide with the image of each and every colour with the vertical cross wire and the readings are noted. The prism is removed and the direct ray reading is noted.

The difference between the direct ray and the refracted ray readings for each and every colour gives the angle of minimum deviation (D) for that respective colour.

Then, by substituting the values of D and ‘A’ in the given formula, the refractive indices (µ) for each and every colour can be calculated. Finally by choosing any one of the colour refractive index as µ1 and the any other as µ2 the

dispersive power of the prism is calculated using the given formula. Similarly for various values of µ1

and µ2 the dispersive powers are calculated and the mean of all the dispersive powers is calculated.


94

(ii)To find the angle of minimum deviation (D) and refractive index (µ)

LC = 1’ TR = MSR+ (VSCXLC) Angle of Prism (A) = …………………..

VERNIER - A VERNIER –B D = R1~ R2

MSR VSC TR MSR VSC TR VA VB

Mean

(D) Position

(deg) (div) (deg) (deg) (div) (deg) (deg) (deg) (deg)

µ =

sin[(A+D)/2]/sin

[A/2]

Direct

Ray

(R1) (R1)

Violet – I (R2) (R2)

Violet - II

(R2) (R2)

Blue (R2) (R2)

Blueish Green

(R2) (R2)

Green (R2) (R2)

Yellow –

I

(R2) (R2)

Yellow - II

(R2) (R2)

Red (R2) (R2)


95

Result:

(i) Angle of the given prism (A) = ………………………. degrees

(ii) Angle of the minimum deviation (D) = ……….………………. degrees

(iii) The refractive index of the material

of the given prism (µ) = …………..…………….(No unit)

(iv) Mean dispersive power of the prism = ……………………….. (No unit)

VIVA – VOCE

1. What is meant by a spectrum?

2. Define dispersive power.

3. What are the initial adjustments to be made before using the spectrometer?

4. Define angle of prism.

5. What is meant by minimum deviation?

6. Define reflection, refraction and transmission of light.

7. Define refractive index of the prism.

8. Is sodium lamp a monochromatic source? Why?

9. What are the applications of a spectrometer?

10. What is meant by ‘Least Count’ give its significance?


96

WORK SHEET

CALCULATIONS:


97

WORK SHEET

CALCULATIONS:


98

Fig 27. Uniform Bending

(i) To find M/h LC = 0.001 cm

TR = MSR + (VSCXLC)

Load Microscope Reading

(M) Increasing Load Decreasing Load S.No

Distance between

Knife edges (l) MSR VSC TR MSR VSC TR

Mean

Elevation (h)

M/h

X 10-1 Unit X 10-2 m X 10-3 Kg X 10-2

m Div X 10-2

m X 10-2

m Div X 10-2

m X 10-2

m X 10-2 m

Kg m-1

1 W

2 W+50

3 W+100

4 W+150

5 W+200

6

W+250

Mean =


99

Expt. No.: Date:

12. DETERNINATION OF YOUNG’S MODULUS OF THE MATERIAL –

UNIFORM BENDING

Aim

To determine the young’s modulus of the given material of the beam by uniform bending.

Apparatus required

A long uniform beam usually a meter scale, travelling microscope, pin, weight hanger with slotted weights (2 sets), vernier calipers, Screw gauge, knife edges etc.,

Formula

The Young’s Modulus of the given material of the beam

By Calculation method Nm-2

By Graphical method Nm-2




l Distance between Knife edge metre

b Breadth of the beam metre

d Thickness of the beam metre

h Elevation of the midpoint of the beam above the knife-edge metre

M Load applied Kg

k Slope h/M from graph m Kg -1

a Distance between knife edge and slotted weight hanger in the beam

metre


100

By Graphical method to find h/M

AXIS W-(W+50) W-(W+100) W-(W+150) W-(W+200) W-(W+250)

X (M)

Y ( h)

(i) To find Breadth (b) of the beam using vernier calipers ZE = …………. LC = 0.01 cm ZC = …………. x 10-2m

S.No. Main Scale



OR=MSR+(VSCXLC) Correct Reading CR

= OR±ZC

Unit X 10-2m Div. X 10-2m X 10-2m

1

2

3

4

5

Mean ‘b’ = ……….. x 10-2m

To find the thickness (d) of the beam using screw gauge. ZE = …………. LC = 0.01 mm ZC = …………. x 10-3m

S.No. Pitch Scale



OR=PSR+(HSCXLC) Correct Reading CR

= OR±ZC

Unit X 10-3m Div. X 10-3m X 10-3m

1

2

3

4

5

Mean‘d’ = ……….. x 10-3m


101

Procedure The given beam is supported on the two knife edges (A and B) in the same horizontal level, equal lengths projecting beyond each knife- edge. The weight hanger are attached at two points equally distant (a) beyond the knife-edges. A pin is fixed exactly at the mid-point of the beam. A travelling microscope capable of vertical motion is arranged in front of the pin and focused on its tip. The tip is made to coincide with the horizontal cross-wire. Loads are added to both the hangers in steps of M (50 gms) and the microscope is moved up so that the tip of the image of the pin just coincides with the horizontal cross-wire in each case and the MSR, VSC readings are noted. The same procedure is repeated by unloading the weight in steps of same 50 grams and the readings are tabulated in the tabular column. From the readings the mean of (M/h) is calculated. The thickness and the breadth of the beam are measured using screw gauge and vernier calipers respectively and are tabulated. By substituting the values in the given formula the Young’s modulus of the material of the beam can be calculated. Graph A graph is drawn taking load (M) along x axis and elevation h along y axis as shown in figure. The slope of the graph gives the value K=h/M. Substituting the values of the slope in the given formula, the Young’s modulus can be calculated. Result The Young’s modulus of the given material of the beam

By calculation Y = …………………. Nm-2

By Graph Y = …………….…… Nm-2

VIVA - VOCE

1. What is meant by uniform bending?

2. Define neutral axis.

3. Name any two methods used to determine the young’s modulus of the beam.

4. Define elasticity.

5. Will there be any change in young’s modulus of the material, if its thickness is increased? Justify.

6. What are the basic assumptions made from the theory of bending?

7. Why iron girders used in buildings are made in the form of I section?

8. What is the use of keeping an optimum of 0.7 to 0.8 meter distance between the knife edges?

9. Define elastic limit.

10. What is meant by elastic constants?


102

WORK SHEET

CALCULATIONS:


103

WORK SHEET

CALCULATIONS:


104

Fig 28. Torsional Pendulum

(i)To find L/T2

Time for 10 oscillations S.No

Length of the

suspend wire (L) Trail - 1 Trail - 2 Mean

Time period

T T2 L/T2

Unit X 10-2 m sec sec sec Sec s2 Ms-2

1

2

3

4

5

Mean L/T2=………..…ms-2


105

Expt. No.: Date:

13. TORSIONAL PENDULUM - DETERMINATION OF RIGIDITY

MODULUS

Aim

To determine the moment of inertia of the disc and rigidity modulus of the given material of a wire by torsional oscillations.

Apparatus required

Circular metal disc, suspended wire, biscuit balance, metre scale, stop watch, screw gauge,

vernier caliper, etc.,

Formula

Moment of inertia of the disc Kg m2

Rigidity modulus of the material of the given wire Nm-2



M Mass of circular disc Kg

R Radius of the circular disc meter

r Radius of the given wire meter

L Length of the suspension wire meter

T Time period for various lengths Second


106

To find diameter of the circular disc using vernier caliper ZE = ………….

LC = 0.01 cm ZC = …………. x 10-2m

S.No. Main Scale



OR=MSR+(VSCXLC) Correct Reading CR =

OR±ZC

Unit X 10-2m Div. X 10-2m X 10-2m

1

2

3

4

5

6

Mean = ……….. x 10-2m

Radius of the circular disc =……….. x 10-2m

(iii)To find the diameter of the suspension wire using screw gauge. ZE = …………. LC = 0.01 mm ZC = …………. x 10-3m

S.No. Pitch Scale

Reading (PSR) Head Scale Coincidence

(HSC) Observed Reading

OR=PSR+(HSCXLC) Correct Reading CR =

OR±ZC

Unit X 10-3m Div. X 10-3m X 10-3m

1

2

3

4

5

6

Mean = ……….. x 10-3m

Radius of the suspension wire = ……….. x 10-3m


107

Theory

The circular disc is rotated in a horizontal plane so that a twist is given to the wire which holds the disc. Hence the various elements of the wire undergo shearing strains. The restoring couples,

which tend to restore the unstrained conditions, are called into action. Now when the disc is released it starts executing torsional oscillations. The couple which acts on the disc produces in it an angular acceleration which is proportional to the angular displacement and is always directed

towards its mean position. Hence the motion of the disc is a simple harmonic motion. Procedure

A uniform thin wire whose rigidity modulus has to be found is suspended from a rigid support. The other end of the wire is attached to a circular disc using an adjustable chuck. The length of the

suspension wire (L) between the point of suspension and the metal disc is the length of the torsional pendulum as shown in fig 28.

Initially, the length of the pendulum is adjusted; say 50 cm. Torsional oscillations are set up by giving a small twist to the disc. The time taken for 10 oscillations are found and hence time period, which is nothing but the time taken of 5 cm and periods of oscillations are found for each length.

The readings are tabulated in the tabular column and the mean of L/T2 is calculated. The radius of the wire (r) is found using screw gauge and the radius of the disc (R) is found

using vernier calipers. The mass (M) of the metal disc is found using the biscuit balance. Substituting the values, in the given formula the moment of inertia can be calculated. Now by substituting the value of moment of inertia and L/T2 in the given formula rigidity

modulus of the given wire can be calculated.

Graph

The value of L/T2 can also be found by drawing a graph taking ‘L’ along x-axis and T2 along y-axis as shown in figure. The slope of the graph will give the value of L/T2. Substituting the

graphical value of L/T2 in the given formula, the rigidity modulus can be determined by graphical method.


108

WORK SHEET

CALCULATIONS:


109

Result

(i) The moment of inertia of the disc (I) = ………………….. Kgm2

(ii) The rigidity modulus of the suspension wire (n)

(a) By calculation = ……………..……. Nm-2

(b) By Graph = …………………… Nm-2

VIVA - VOCE

1. What is meant by torsional oscillations?

2. What do you understand by the term shearing strain?

3. What is meant by restoring couple?

4. What is meant by simple harmonic oscillation?

5. Define angular acceleration and angular displacement.

6. If the radius of the disc is altered, what will happened to the rigidity modulus of the wire. Explain.

7. What is the use of varying the length of suspension while doing the experiment?

8. Why the disc is circular? Are there any other shapes possible to find the moment of inertia?

9. Explain rigidity modulus.

10. Will the moment of inertia of the disc vary with the material of the wire? If so explain.


110

STANDARD VALUES

Base quantity Symbol Description SI unit Symbol for

dimension

Length l The one dimensional extent of an object. metre (m) L

Mass m The amount of matter in an object. kilogram (kg) M

Time t The duration of an event. second (s) T

Electric current I Rate of flow of electrical charge. ampere (A) I

Temperature T Average energy per degree of freedom of a system. kelvin (K) Θ

Amount of

substance n Number of particles compared to the number of atoms in 0.012 kg of

12C. mole (mol) N

Luminous

intensity L Amount of energy emitted by a light source in a particular direction. candela (cd) J

Plane angle θ Measure of a change in direction or orientation. radian (rad) 1

Solid angle Ω Measure of the size of an object as projected on a sphere. steradian (sr) 1

Derived quantity Sym

bol Description SI units Dimension

Acceleration a Rate of change of the speed or velocity of an object. m s−2

L T−2

Angular acceleration α Rate of change in angular speed or velocity. rad s−2

T−2

Angular speed (or

angular velocity)

ω or

ω The angle incremented in a plane by a segment connecting an object and a reference point. rad s

−1 T

−1

Angular momentum L Measure of the extent and direction and object rotates about a reference point. kg m2 s

−1 M L

2 T

−1

Area A The two dimensional extent of an object. m2 L

2

Area density ρA The amount of mass per unit area of a two dimensional object. kg m−2

M L−2

Capacitance C Measure for the amount of stored charge for a given potential. farad (F = A

2 s

4

kg−1

m−2

) I

2 T

4 M

−1 L

−2

Catalytic activity Change in reaction rate due to presence of a catalyst. katal (kat = mol

s−1

) N T

−1

Catalytic activity

concentration Change in reaction rate due to presence of a catalyst per unit volume of the system. kat m

−3 N L

−3 T

−1

Chemical potential µ The amount of energy needed to add a particle to a system. J mol−1

M L2 T

−2 N

−1

Molar concentration C Amount of substance per unit volume. mol m−3

N L−3

Current density J Amount of electric current flowing through a surface. A m−2

I L−2

Dose equivalent H Measure for the received amount of radiation adjusted for the effect of different types of

radiant on biological tissue.

sievert (Sv = m2

s−2

) L

2 T

−2

Dynamic Viscosity η Measure for the resistance of an incompressible fluid to stress. Pa s M L−1

T−1

Electric Charge Q Amount of electric charge. coulomb (C = A

s) I T

Electric charge density ρQ Amount of electric charge per unit volume. C m−3

I T L−3

Electric displacement D Strength of the electric displacement. C m−2

I T L−2

Electric field strength E Strength of the electric field. V m−1

M I−1

L2 T

−3

Electrical conductance G Meausure for how easily current flows through a material. siemens (S = A

2

s3 kg

−1 m

−2)

L−2

M−1

T3 I

2

Electric potential V The amount of work required to bring a unit charge into an electric field from infinity. volt (V = kg m

2

A−1

s−3

) L

2 M T

−3 I

−1

Electrical resistance R The degree to which an object opposes the passage of an electric current. ohm (Ω = kg m

2

A−2

s−3

) L

2 M T

−3 I

−2

Energy E The capacity of a body or system to do work. joule (J = kg m2 M L

2 T

−2


111

s−2

)

Energy density ρE Amount of energy per unit volume. J m−3

M L−1

T−2

Entropy S Measure for the amount of available states for a system. J K−1

M L2 T

−2 Θ

−1

Force F The external cause of acceleration, acting on an object. newton (N = kg

m s−2

) M L T

−2

Frequency f The number of times something happens in a period of time. hertz (Hz =s−1

) T−1

Half-life t1/2 The time needed for a quantity to decay to half its original value. s T

Heat Q Amount of energy transferred between systems due to temperature difference. J M L2 T

−2

Heat capacity Cp Amount of energy needed to raise the temperature of a system by one degree. J K−1

M L2 T

−2 Θ

−1

Heat flux density ϕQ Amount of heat flowing through a surface per unit area. W m−2

M T−3

Illuminance Ev Total luminous flux incident to a surface per unit area. lux (lx = cd sr

m−2

) J L

−2

Impedance Z Measure for the resistance of an electrical circuit against an alternating current. ohm (Ω = kg m

2

A−2

s−3

) L

2 M T

−3 I

−2

Index of refraction n The factor by which the speed of light is reduce in a medium. 1

Inductance L Measure for the amount of magnetic flux generated for a certain current run through a

circuit.

henry (H = kg

m2 A

−2 s

−2)

M L2 T

−2 I

−2

Irradiance E Power of electromagnetic radiation flowing through a surface per unit area. W m−2

M T−2

Linear density ρl Amount of mass per unit length of a one dimensional object. M L−1

Luminous flux (or

luminous power) F Perceived power of a light source.

lumen (lm = cd

sr) J

Magnetic field strength H Strength of a magnetic field in a material. A m−1

I L−1

Magnetic flux Φ Measure of quantity of magnetism, taking account of the strength and the extent of a

magnetic field.

weber (Wb = kg

m2 A

−1 s

−2)

M L2 T

−2 I

−1

Magnetic flux density B Measure for the strength of the magnetic field. tesla (T = kg

A−1

s−2

) M T

−2 I

−1

Magnetization M Amount of magnetic moment per unit volume. A m−1

I L−1

Mass fraction x Mass of a substance as a fraction of the total mass. kg/kg 1

(Mass) Density (volume

density) ρ The amount of mass per unit volume of a three dimensional object. kg m

−3 M L

−3

Mean lifetime τ Average time needed for a particle to decay. s T

Molar energy Amount of energy present is a system per unit amount of substance. J mol−1

M L2 T

−2 N

−1

Molar entropy Amount of entropy present in a system per unit amount of substance. J K−1

mol−1

M L

2 T

−2 Θ

−1

N−1

Molar heat capacity c Heat capacity of a material per unit amount of substance. J K−1

mol−1

M L2 T

−2 N

−1

Moment of inertia I Inertia of an object with respect to angular acceleration. kg m2 M L

2

Momentum p Product of an object's mass and velocity. N s M L T−1

Permeability µ Measure for how the magnetization of material is affected by the application of an external

magnetic field. H m

−1 M L

−1 I

−2

Permittivity ε Measure for how the polarization of a material is affected by the application of an external

electric field. F m

−1 I

2 M

−1 L

−2 T

4

Power P The rate of change in energy over time. watt (W) M L2 T

−3

Pressure p Amount of force per unit area. pascal (Pa = kg

m−1

s−2

) M L

−1 T

−2

(Radioactive) Activity A Number of particles decaying per unit time. becquerel (Bq =

s−1

) T

−1


112

(Radioactive) Dose D Amount of energy absorbed by biological tissue from ionizing radiation per unit mass. gray (unit) (Gy

= m2 s

−2)

L2 T

−2

Radiance L Power of emitted electromagnetic radiation per solid angle and per projected source area. W m−2

sr−1

M T−3

Radiant intensity I Power of emitted electromagnetic radiation per solid angle. W sr−1

M L2 T

−3

Reaction rate r Measure for speed of a chemical reaction. mol m−3

s−1

N L−3

T−1

Speed v Rate of change of the position of an object. m s−1

L T−1

Specific energy Amount of energy present per unit mass. J kg−1

L2 T

−2

Specific heat capacity c Heat capacity per unit mass. J kg−1

K−1

L2 T

−2 Θ

−1

Specific volume v The volume occupied by a unit mass of material (reciprocal of density). m3 kg

−1 L

3 M

−1

Spin S Intrinsic property of particles, roughly to be interpreted as the intrinsic angular momentum

of the particle. kg m

2 s

−1 M L

2 T

−1

Stress σ Amount of force exerted per surface area. Pa M L−1

T−2

Surface tension γ Amount of work needed to change the surface of a liquid by a unit surface area. N m−1

or J m−2

M T−2

Thermal conductivity k Measure for the ease with which a material conducts heat. W m−1

K−1

M L−1

T−3

Θ−1

Torque (moment of force) T Product of a force and the perpendicular distance of the force from the point about which it

is exerted. N m M L

2 T

−2

Velocity v Speed of an object in a chosen direction. m s−1

L T−1

Volume V The three dimensional extent of an object. m3 L

3

Wavelength λ Distance between repeating units of a propagating wave. m L

Wavenumber k Reciprocal of the wavelength. m−1

L−1

Weight w Amount of gravitation force exerted on an object. newton (N = kg

m s−2

) M L T

−2

Work W Energy dissipated by a force moving over a distance, scalar product of the force and the

movement vector.

joule (J = kg m2

s−2

) M L

2 T

−2

SI Unit Prefixes

Factors Prefix Symbol Factors Prefix Symbol Factors Prefix Symbol Factors Prefix Symbol Factors Prefix Symbol

1012 tera T 103

kilo k 10-1 deci d 10-6

micro µ 10-15 femto f

109 giga G 102

hecto h 10-2 centi c 10-9

nano n 10-18 atto a

106 mega M 101

deca da 10-3 milli m 10-12

pico p 10-15 femto f

S.NO. PHYSICAL CONSTANT

SUBSTANCE STANDARD VALUE

UNIT

Water 1000 Kerosene 830 Castor oil 970 Glycerine 1260 Copper 8900 Brass 8600 Steel 7800 Aluminium 2700

1 Density

Iron 7500

Kgm3

Aluminium 2.5 x 1010 Brass 3.5 x 1010 Cast Iron 5.0 x 1010 Copper 3.4 to 3.6 x 1010 Wrought Iron 8.0 x 1010 Steel (Cast) 7.6 x 1010

2 Rigidity Modulus

Steel (Mild) 8.9 x 1010

Nm-2


113

Brass 0.001 to 0.002 Copper 0.0039

S.NO. PHYSICAL CONSTANT

SUBSTANCE STANDARD VALUE

UNIT

Box wood 1 x 1010 Teak wood 1.7 x 1010 Aluminium 7.1 x 1010 Brass 9.8x 1010 Cast Iron 10 to 12 x 1010 Copper 11.7 x 1010 Wrought Iron and steel 20x 1010

3 Youngs Modulus

Glass 6 x 1010

Nm-2

Water 0.0008 Kerosene 0.002 Coconut oil 0.0154

4

Coefficient of Viscosity (at room temperature) Glycerine 0.3094

Nsm-2

Card board(varies with specimen also) 0.04 Ebonite 0.7 Glass 1

5 Thermal Conductivity

Wood & Rubber 0.15

Wm-1K-1

Crown Glass 1.5 Flint Glass 1.56 Dense Crown Glass 1.62 Dense Flint Glass 1.65

6 Refractive Index

Water 1.333

(No Unit)

Sodium Vapour Lamp 5893 Mercury Vapour Lamp Red 6234 Yellow I 5791 Yellow II 5770 Green 5461 Blueish Green 4916 Blue 4358 Violet I 4078

7 Standard Wavelength

Violet II 4047

A°

Aluminium 0.0043 Brass 0.001 to 0.002 8

Temperature Co-efficient of resistance Copper 0.0039

per °C

PHYSICS PRACTICAL MANUAL CUM RECORD I & II...

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