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COUNSELLING CODE1222P.B. College of Engineering

Campus Address : Administration Office :Irungattukottai, Chennai – Bangalore National Highway 3/117, Brahmin Street, Karambakkam,Sriperumbudur Taluk, Kanchipuram Dist., Chennai – 602 117. Porur, Chennai – 600 116.Tel. : 044-27198033, 9940026861, 9600112733 Tel. : 044-24766386, Telefax : 044-24766941E-mail : [email protected] Website : www.pbce.in Cell : 9600113031, 9962206098

MODEL QUESTION PAPER – IPHYSICS

Time: 3 Hrs. Max. Marks: 150

Note: i) Answer all the questionsii) Choose and write the correct answeriii) Each question carries one mark.

PART – IAnswer all the questions: (30 x 1 = 30)

1. The Electric field outside the two oppositely charged plane sheets, each of chargedensity is ___________

a) b) c) d) Zero

Answer: d) Zero

2. The value of permittivity of free space is ____________

a) 8.854 x 10-12 C2N-1m-2 b) 9 x 108C2N-1m-2

c) C2N-1m-2 d) C2N-1m-2

Answer: a) 8.854 x 10-12 C2N-1m-2

3. The law that governs the force between electric charge isa) Ampere’s law b) Faraday’s law c) Coulomb’s law d) Ohm’s law

Answer: c) Coulomb’s law

4. A hollow insulated conduction sphere is given a positive charge of 10c. What will bethe electric field at the centre of sphere, if its radius is 2 metres ?(a) 20 NC-1 (b) 5 NC-1 (c) Zero (d) 8 NC-1

Answer: (c) Zero

5. How many electrons constitute current of one ampere(a) 6.25 x109 (b) 6.25x1018 (c) 6.25x105 (d) 6.25x106

Answer: (b) 6.25x1018

6. Of the following devices, which has small resistance ?(a) Voltmeter (b) Ammeter of range 0-10A(c) Moving coil galvanometer (d) Ammeter of range 0-1A

Answer: (b) Ammeter of range 0-10A

-

19x109

14x9x109

2


7. Which of the following equation represents Biot-Savart law ?

a) dB = (b) dB =

(c) dB = (d) dB =

Answer: (d) dB =

8. In an a.c circuit with an inductor(a) voltage lags current by /2 (b) Voltage and current are in phase(c) Voltage leads current by (d) Current lags voltage by /2

Answer: (d) Current lags voltage by /2

9. The core used in audio frequency chokes is(a) iron (b) Carbon (c) lead (d) steel

Answer: (a) iron

10. The Q factor (quality factor) of an A.C. circuit containing a resistance R, inductance Land capacitance C is

(a) Q = (b) Q = (c) Q = (d) Q =

Answer: (c) Q =

11. In an A.C. circuit, the voltage leads the current by a phase of /2, then the circuithas(a) only an inductor (L) (b) only a capacitor (C)(c) only a resistor (R) (d) L,C and R is series

Answer: (a) only an inductor (L)

12. In an electromagnetic wave, the phase difference between the electric field E andmagnetic field B is(a) /4 (b) /2 (c) (d) 0

Answer: (d) 0

13. Of the following, which one is uniaxial crystal?(a) Mica (b) Aragonite (c) Topaz (d) Quartz

Answer: (d) Quartz

14. Angle between the electric component and magnetic component of electromagneticwave is.(a) 0 (b) /4 (c) /2 (d)

Answer: (c) /2

o

4Idlr2

o

4Idl sin

r2

o

4o

4Idl x r

r2

Idl x rr3

LC

1R

CL

1R

LC

LR

o

4Idl x r

r3

1R

LC

3


15. In diffraction pattern, the fringes are(a) equally spaced (b) of same intensity(c) unequally spaced (d) halving the full dark.

Answer: (c) unequally spaced

16. The first excitation potential energy or the minimum energy required to excite theatom from ground state of hydrogen atom is(a) 13.6 eV (b) 10.2 eV (c) 3.4 eV (d) 1.89 eV

Answer: (b) 10.2 eV

17. The minimum wavelength of X-rays produced in an X-ray tube at 1000KV is(a) 0.0124Å (b) 0.124Å (c) 1.24Å (d) 0.00124Å

Answer: (a) 0.0124Å

18. The direction of viscous force in Millikan oil drop experiment is(a) always downwards (b) always upwards(c) Opposite to the direction of the motion of oil drop(d) either upwards or downwards

Answer: (c) Opposite to the direction of the motion of oil drop

19. of cathode ray particle __________

(a) depends upon the nature of the cathode(b) depends upon the nature of the anode(c) depends upon the nature of the gas atoms present inside the discharge tube(d) is independent of all these

Answer: (d) is independent of all these

20. A clock moving with velocity c/2 measures the time interval between two events in itsframe as 1 second for an observer at rest, the time interval between the same twoevents would be?(a) ¾ second (b) 3/2 second (c) 4/3 second (d) 2/ 3 Second

Answer: (d) 2/ 3 Second

21. In the photo electric phenomenon, if the ratio of the frequency of incident radiationincident on a photosensitive surface is 1 : 2 : 3, the ratio of the photo electric currentis _______(a) 1:2:3 (b) 1 : 2 : 3 (c) 1 : 4: 9 (d) 1 : 1: 1

Answer: (d) 1 : 1: 1

22. One amu is equal to(a) 931eV (b) mass of carbon atom. (c) 1.66x10-27kg (d) mass of oxygen atom

Answer : (c) 1.66x10-27kg

23. The binding energy of 26Fe56 nucleus is _________(a) 8.8 MeV (b) 88 MeV (c) 493 MeV (d) 41.3 MeV

Answer: (c) 493 MeV

em

4


24. The radius of nucleus is 5.2 F. The number of nucleons in the nucleus is __(a) 52 (b) 104 (c) 64 (d) 128

Answer: (c) 64

25. A radioactive substance has a half life period of 30 days .The disintegration constantis _________(a) 0.023/day (b) 0.231 / day (c) 02.31 / day (d) 23.1 /day

Answer: (a) 0.023/day

26. In CE single stage amplifier , the voltage gain at mid – frequency is 10. The voltagegain at upper cut-off frequency is(a) 10 (b) 14.14 (c) 7.07 (d) 20

Answer: (c) 7.07

27. Find the voltage across the resistor as shown in the figure (silicon diode is Used)

+2.7V

D

180

(a) 2.4 V (b) 2.0 V (c) 1.8 V (d) 0.7 V

Answer: (b) 2.0 V

28. The following arrangement performs the logic function of gate(a) AND (b) OR (c) NAND (d) EXOR

Y

Answer: (b) OR

29. In an AM super heterodyne receiver, the local oscillator frequency is 1.245 MHZ. Thetuned station frequency is(a) ground wave (b) Sky waves (c) Space waves (d) Micro waves

Answer: (b) Sky waves

30. Printed documents to be transmitted by fax are converted into electrical signals bythe process of(a) reflection (b) scanning (c) modulation (d) light variation.

Answer: (b) scanning

5


PART – IIAnswer all the questions: (30 x 1 = 30)

31. State Gauss’s law of electrostatics.Gauss’s law : The law relates the flux through any closed surface and the net chargeenclosed within the surface. The law states that the total flux of the electric field E

over any closed surface is equal to times the net charge enclosed by the surface.

=

32. Calculate the effective capacitance of the combination shown in the figure.

C2 C3

C1

10 Fµ 10 Fµ

A B

5 Fµ

= + = + =

=

Effective Capacitance (Cp) = C1 + Cs

= 5F + 5F= 10F

33. Write any three applications of super conductors.Super conductors form the basis energy saving power system, namely the super

conducting generators, which are smaller in size and weight, in comparison withconventional generators.

Super conducting magnets have been used to levitate trains above its rails. Theycan be driven at high speed with minimal expenditure of energy.

34. State Kirchhoff’s second law for electrical network.

Kirchoff’s second law (or) kirchoff’s voltage law states that the algebraic sum of theproducts of resistance and current in each part of any closed circuit is equal to thealgebraic sum of the emf’s in that closed circuit. This law is a consequence ofconservation of energy.

35. Distinguish between electromotive force and potential difference.

Emf Potential difference1.It is the difference of potential betweentwo terminals of a cell in an open circuit.2.It is independent of external resistanceof the circuit.

1.It is the difference in potentials betweenany two points in a closed circuit.2. It is proportional to the resistancebetween any two points.

1Cs

1C2

1C3

1Cs

110

110

15

1Cs

15

Cs = 5

1oq

o

6


. 36. What is Ampere’s Circuital law?

The line integral .B dl

for a closed curve is equal to o times the net current Io

through the area bounded by the curve.

37. Why a d.c ammeter cannot read a.c?For direct current XL = 0 and Xc = . Alternating current varies in magnitude anddirection periodically. But direct current is a unidirectional current and does notvary periodically. Hence a d.c. ammeter cannot read a.c

38. A capacitor of capacitance 2F is in an a.c. circuit of frequency 1000 Hz. If the rmsvalue of the applied e.m.f. is 10V, find the effective current flowing in the circuit andcapacitive reactance.

C = 2F, V = 1000 Hz, Eeff =10V

Formula : XC = =

Working : XC = = 79.6

Irms = = = 0.126 A

Irms = 0.126A

39. In Young’s experiment, the width of the fringe obtained with light of wavelength6000Å is 2mm. Calculate the fringe width, if the entire apparatus is immersed in aliquid of refractive index 1.33.

Given : = 6000 Å = 6x10-7m = 2mm = 2x10-3m

= 1.33’= ?

Working : ’ = = =

’ = = 1.5 x10-3m (or) 1.5 mm

Bandwidth in a liquid = 1.5 x 10-3m

40. Define optic axis of a crystal

Inside the crystal there is a particular direction in which the ordinary andextraordinary rays travel with same velocity. This direction is called optic axis of thecrystal.

41. Define Ionisation potential and Ionisation Potential energy.

Ionization potential is that accelerating potential which makes the impinging electronacquire sufficient energy to knock out an electron from the atom and thereby ionizethe atom. The energy required to remove an electron from the atom is calledionization potential.

12 x 10-6 x 2 x 103

1C

1C x 2

Eeff

XC

1079.6

D’d

Dd

2x10

7.33

7


42. Write the application of Moseley’s law.Applications:(i) Moseley’s law has led to the discovery of new elements like hafnium (72),technetium (43), rhenium (75) etc.(ii) Any discrepancy in the order of the elements in the periodic table can be removedby Moseley’s law by arranging the element according to the atomic numbers and notaccording to the atomic weights.

43. Define stopping Potential.Stopping potential:The minimum negative (retarding) potential given to the anode for which the photoelectric current becomes zero is called the cut – off or stopping potential.

43. Define curie.

Curie is defined as the quantity of a radio active substance which gives 3.7 x 1010

disintegrations per second or 3.7 x 1010 becquerel. This is equal to the activity of onegram of radium.

45. Tritium has a half life of 12.5 years. What fraction of the sample will be left over after25 years ?Working: Let the amount of sample be 1g in the beginning.After 12.5 years fraction of sample undecayed = ½After 25 years fraction of sample undecayed = ½ x ½ = ¼

46. Prove the following logic expression. (A +B ) (A+B) = B

(Ā + B) (A + B) = B

(Ā + B) (A + B) = ĀA + ĀB + AB + BB

= O + Ā B + AB + B ( ĀA=0)= (Ā+ A) B + B ( Ā+ A = 1)= B + B (B + B =B)= B

(Ā+ B) (A+ B) = B is proved.

47. Draw NOT gate using transistor.

y = A

RC

RBA VCC

NOT GATE USING TRANSISTOR

+

_

8


48. Define output impedance of a transistor.

The output impedance ro is defined as the ratio of variation in the collector emittervoltage to the corresponding variation in the collector current at a constant basecurrent in the active region of transistor characteristic curves.

Output impedance, r0 =

49. What is Zener breakdown?

When both sides of the PN junction are heavily doped, then the depletion layer isnarrow. Zener breakdown takes place in such a thin narrow junction.

When a small reverse bias is applied, a very strong electric field is producedacross the thin depletion layer. This field breaks the covalent bonds, due to whichextremely large number of electrons and holes are produced, which gives rise to thereverse saturation current (Zener current).

50. What is interlaced scanning ?

In the interlaced scanning, the total numberlines are divided into two groups calledfields. During the presentation of the first field, only odd numbered lines are scannedand during second field all even numbered lines are scanned. Half way along thebottom of the first field, the vertical retrace returns the scanning beam to the top ofthe image and completes the unfinished lines.

Part – III

Note: (i) Answer question No. 56 Compulsorily. (7X5 = 35)(ii) Answer any six of the remaining 11 questions(iii) Draw diagrams wherever necessary.

51. Three charges -2x10-9C, +3 x10-9 C, -4x10-9 C are placed at the vertices of anequilateral triangle ABC of side 20cm. Calculate the work done in shifting the chargesfrom A, B and C to A1, B1, C1 respectively which are midpoints of the sides of thetriangle.

Working:

The P.E of the system of charges, U = + +

A-2x10 C-9

-4x10 C-9+3x10 C-9

CBB1

C1A1

Work done in displaying the charges from A,B and C to A1, B1, C1, respectively.

W = Uf – Ui

Ui and Uf are the initial and final potential energy of the system.

VCE

IC IB

14π

q1q2

rq2q3

rq3q1

r

9


Ui = [-6x10-18 – 12x10-18 + 8 x 10-18]

= -4.5 x10-7 J

Uf = [ -6 x 10-18 – 12 x10 –18 + 8 x 10-18]

= - 9 x 10-7 J

Work done = - 9x 10-7 – (-4.5 x 10-7)

W = -4.5 x10-7 J

52. State and verify Faraday’s second law of electrolysis.

Faraday’s second Law of electrolysis states that mass of a substance liberated at anelectrode by a given amount of charge is proportional to the chemical equivalent ofthe substance. So if E is the chemical equivalent of a substance, then from thesecond law

m E

Verification: Two electrolytic cells containing different electrolytes, CuSO4 solutionand AgNO3 solution are connected in series with a battery, a rheostat and anammeter. Copper electrodes are inserted in CuSO4 and silver electrodes are insertedin AgNO3

The cathodes are cleaned, dried, weighed and then inserted in the respective cells.The current is passed for some time. Then the cathodes are taken out, washed, driedand weighed. Hence the masses of copper and silver deposited are found as m1 andm2.

It is found that = , where E1 and E2 are the chemical equivalents of

copper and silver respectively.m E

Thus, the second law is verified.

A

CuSO4 AgNO3

Rh

+ +_ _

Bt

VERIFICATION OF FARADAY’S SECOND LAW

9 x 109

0.20

9 x 109

0.100

m1

m2

E1

E 2

10


53. Explain the working of a Leclanche cell with a diagram.Working of Lechanche cell:

At the cathode due to oxidation reaction, Zn atom is converted into Zn++ ions and2 electrons. Zn++ ions reacting with ammonium chloride produces zinc chloride andammonia gas.i.e Zn++ + 2NH4Cl 2 NH3 + ZnCl2 + 2H+ + 2e-

Carbon Rod

Mixture of MnOand Charcoal

3

Porous Pot

Zinc Rod

AmmoniumChloride Solution

Glass Vessel

LECLANCHE CELL

The ammonia gas escapes. The hydrogen ions diffuse through the pores of theporous pot and react with manganese dioxide. In this process the positive charge ofhydrogen ion is transferred to the carbon rod.

When Zinc rod and carbon rod are connected externally, the two electrons fromthe zinc rod move towards carbon and neutralize the positive charge. This currentflows from carbon to zinc.

Leclanche cell is useful for supplying intermittent current. The emf of the cell isabout 1.5V and it can supply a current of 0.25 A.

54. A rectangular coil of 500 turns and of area 6 x 10-4m2 is suspended inside a radialmagnetic field of induction 10-4 T by a suspension wire of torsion constant 5x10-10 Nmper degree. Calculate the current required to produce deflection of 10°.

Given: No. of turns of the coil (n) = 500Area of the coil (A) = 6x10-4 m2

Magnetic field (B) = 10-4TTorsional Constant (C) = 5 x10-10 Nm per degreeAngle of deflection = 10˚

= x 10 = rad

Formula: nBIA = C

Current I =

Working: Current = =

Current required = 0.1666 mA

180

18

CnBA

5 x 10-10 x 10500 x 10-4 x 6 x 10-4

50 x 10-10

3000 x 10-8

11


55. Explain how an e.m.f. can be induced by changing the area enclosed by the coil.

PQRS is a conductor bent in the shape as shown in the figure. L1M1 is a slidingconductor of length l resting on the arms PQ and RS. A uniform magnetic field ‘B’acts perpendicular to the plane of the conductor. The closed area of the conductor isL1QRM1. When L1M1 is moved through a distance dx in time dt, the new area isL2QRM2. Due to the change in area L2L1M1M2, there is a change in the flux linkedwith the conductor. Therefore, an induced emf is produced.

change in area dA = Area L2L1M1M2

dA= l dx

B

QP

R

S

L1 L2

M1 M2

dx

EMF INDUCED BY CHANGING THE AREA

Change in the magnetic flux, d = B.dA= Bldx

e = -

e = - = - Blv

Where v is the velocity with which the sliding conductor is moved.

56. (i) A parallel beam of monochromatic light is allowed to incident normally on aplane transmission grating having 5000 lines per cm. A second order spectralline is found to be diffracted at an angle of 30°. Calculate the wavelength of thelight.

Given: N = 5000 lines / cm

= 5000 x 102 lines/mm = 2 ; = 30° ; = ?

Formula : sin = Nm => =

Working : = =

wavelength () = 5 x 10-7m = 5000Å

ddt

Bldxdt

sin

Nmsin 30°

5 x 105 x 2

0.5

5 x 105 x 2

12


(ii) A Soap film of refractive index 4/3 and thickness 1.5 x10-4cm is illuminated bylight at an angle 60°. The reflected light is examined by a spectroscope in whichdark band corresponds to a wave length of 5000 Å. Calculate the order of thedark band.

Formula : For dark band , 2 t cosr = n

working : For dark band, =

sin r = =

=¾ sin60° =¾ x =

cos r = 1-sin2r

= 1 - = 1 - =

2t cos r = n

n =

= 2 x x x

= 6.09 = 6

The order of the dark band = 6

57. Explain the spectral series of hydrogen atom.

Whenever an electron in a hydrogen atom jumps from higher energy level to the lowerenergy level, the difference in energies of the two levels is emitted as a radiation ofparticular wavelength. The different wavelengths constitute spectral series which arethe characteristic of the atoms emitting them. The following are the spectral series ofhydrogen atom.

(i) Lyman series : when the electron jumps from any of the outer orbits to thefirst orbit, the spectral lines emitted are in the ultraviolet region of thespectrum and they are said to form a series called lyman series.Here n1= 1, n2 = 3, 4, 5, …………..

The wave number of the lyman series is given by

= R 1 -

(ii) Balmer series : when the electron jumps from any of the outer orbits to thesecond orbit, we get a spectral series called the Balmer series. All the lines ofthis series in hydrogen have their wavelength in the visible region.

sin isin r

sin i

32

sin 60°

3 38

2t cosr

2764

378

9 x 364

43

15 x 10-7

378

1n22

13


Here n1 = 2, n2 = 3, 4,5,……

Lyman Series

Balmer Series

BrackettSeries

PfundSeries

Paschan Series

nucleus

n = 1

n = 2

n = 3

n = 4n = 5

n = 6

SPECTRAL SERIES OF HYDROGEN ATOM

The wave number of the Balmer series is

= R - = R -

The first line in this series (n2 = 3) is called the H - line, the second (n2 = 4),the H – line and so on.

(iii) Paschen series: This series consists of all wavelengths which are emittedwhen the electron jumps from outer most orbits to the third orbit.Here n2 = 4,5,6, ……and n1 = 3. This series is in the infrared region with thewave number given by

= R - = R -

iv) Brackett series:

The series obtained by the transition of the electron from n2 = 5,6,….to n1 = 4is called Brackett series. The wavelengths of these lines are in the infraredregion. The wave number is

= R - = R -

v) Pfund series : The lines of the series are obtained when the electron jumpsfrom any state n2 = 6,7,…..to n1 = 5. This series also lies in the infraredregion. The wave number is

= R - = R -

122

1n22

14

1n22

132

1n22

19

1n22

142

1n22

116

1n22

152

1n22

125

1n22

14


58. Explain Lorentz – Fitzgerald contraction with an example.

Consider two frames of references S and S′ to be initially at rest. A rod isplaced in the frame of reference S′ and an observer O is in S.

The length of the rod in S′ is measured by the observer in S as l0 .

Now the frame of reference S′ moves with a velocity v along the positive x-axis. Now, the length of the rod is measured as l by the observer in S. Then,

l = l0 1 –

i.e., l < l0

Thus the length of the rod moving with a velocity V relative to the observer at

rest is contracted by a factor 1 – in the direction of motion. This is known as

Lorentz – Fitzgerald Contraction.

Object seen when observer at rest Object seen when observer is moving

A circular object will appear as an ellipse for a fast moving observer.

59. Calculate the kinetic energy of a proton moving with velocity 0.900 times the velocityof light.

Working :

= 0.9 x c = 0.9 x3 x 108 m/s

= 2.7 x 108 m/s

Mass of proton = 1.007 x 1.67 x 10-27 kg

= 1.68169 x10-27 kg

K.E = x (1.68169 x 10-27) x (2.7x108)212

v2

c2

v2

c2

15


K.E in Joule = x (1.68169 x 10-27) x (2.7x108)2

= 6.1297 x 10-11 J

K.E. in MeV =

= 3.826 x 102 MeV

60. Explain Soddy – Fajan’s radioactive displacement laws.In 1913, Soddy and Fajan framed the displacement laws governing radioactivity. Theyare - decay : when a radioactive nucleus disintegrates by emitting - particle, theatomic number decreases by two and mass number decreases by four The - decaycan be expressed as

zXA Z-2 YA-4 + 2He4

Example : Radium 88Ra226 is converted to radon( 86Rn222) due to - decay88Ra226 86Ra222+ 2He4

Gamma decay

– decay: when a radioactive nucleus disintegrates by emitting a - particle, theatomic number increases by one and the mass number remains the same. - decaycan be expressed as

zXA z+1yA + -1eo

Example:Thorium (90Th234) is converted to Protoactinium (91Pa234 ) due to - decay

90Th 234 91Pa234+ -1eo

At a time, either or - particle is emitted. Both or particles are not emittedduring a single decay.

-decay : When a radioactive nucleus emits - rays, only the energy level of thenucleus changes and the atomic number and mass number remain the same.During or - decay, the daughter nucleus is mostly in the excited state. It comes toground state with the emission of – rays.

12

6.1297 x 10-11

1.602 x 10-19

16


Example : During the radioactive disintegration of radium (86Ra226) into radon(86Ra222), gamma ray of energy 0.187 MeV is emitted, when radon returns from theexcited state to the ground state.

61. Explain the working of a half – wave diode rectifier.

A circuit which rectifies half of the a.c. wave is called half wave rectifier.The a.c. voltage (Vs) to be rectified is obtained across the secondary ends S1, S2 of thetransformer. The P-end of the diode D is connected to S1 of the secondary coil of thetransformer. The N-end of the diode in connected to the other end S2 of the secondarycoil of the transformer, through a load resistance RL. The rectified output voltage Vdc

appears across the load resistance RL.

Step downtransformer

P1

RL

D

N

P2

S2

Line Voltagefrom an AC Power

P

S1

Vdc

HALF WAVE RECTIFIER

During the positive half cycle of the input a.c voltage Vs, S1 will be positive and

the diode is forward biased and hence it conducts. Therefore, current flows throughthe circuit and there is a voltage drop across RL. This gives the output voltage.

During the negative half cycle of the input a.c voltage (Vs), S1 will be negative and

diode D is reverse biased. Hence the diode does not conduct. No Current flowsthrough the circuit and the voltage drop across RL will be zero. Hence no output

voltage is obtained. Thus Corresponding to an alternate input signal, unidirectionalpulsating output is obtained.

V8

Vdc

0 π 2π3π 5π ωt

ωt

4π

0 π 2π 3π 5π4π

(a) input

(b) input

HALF WAVE RECTIFIER SIGNALS

17


62. Mention the principle of RADAR and Write its applications.

Principle of RADAR :

Radar works on the principle of ‘radio echoes’. The transmitter in radar,radiates the high power electrical pulses into space. When these pulses are incidenton any distant target such as a mountain, ship or aircraft, they get scattered in alldirections. The transmitter antenna receives a part of the scattered energy. Thistransmitter antenna also acts as receiving antenna for the receiving pulse. The pulsetravels with the speed of light 3 x 108 ms-1. In other words, these pulses cover adistance of 300 metres for every micro second. Hence by measuring the time taken bythe pulse to reach the target and back to the transmitter, the range or distance of thetarget can be easily determined. To locate the direction of the target, directionalantennas are used.

Applications of radar :

i) Air and sea navigation is made entirely safe, with radar installations. Highflying planes and ship at sea, can get detailed reports of mountains, ice bergs,rivers, lakes, shore lines etc., which they can avoid.

ii) Radar systems are used for the safe landing of air crafts. On approaching theair field, the pilot is guided by signals from radar set, so that it flies along theline of the runway and lands safely, whatever be the visibility.

PART - IV

Note: (i) Answer any four questions in detail (4x10=40)(ii) Draw diagrams wherever necessary.

63. State Gauss’s law. Applying this, calculate electric field due to (i) an infinitely longstraight charged wire with uniform density.

Gauss’s Law: The law relates the flux through any closed surface and the net change

enclosed within the surface. The law states that the total flux of the electric field E

over any closed surface is equal to times the net charge enclosed by the surface.

=

Electric field due to an infinite long straight charged wire: Consider a uniformlycharged wire of infinite length having a constant linear charge density (charge perunit length). Let P be a Point at a distance r from the wire and E be the electric fieldat the point ‘P’. A cylinder of length ‘l’, radius ‘r’, closed at each end by plane capsnormal to the axis is chosen as Gaussian surface. Consider a very small area ds onthe Gaussian surface. By Symmetry, the magnitude of the electric field will be thesame at all points on the curved surface of the cylinder and directed radially outward.E and ds are along the same direction.

1o

qo

18


INFINITELY LONG STRAIGHT CHARGED WIRE

The electric flux () through curved surface = E ds cos

= E ds ( = ; Cos = 1)

= E (2rl)

( The surface area of the curved part is 2 rl )

Since E and ds are right angles to each other, the electric flux through the plane caps= 0 Total flux through the Gaussian Surface, = E. (2rl)The net charge enclosed by Gaussian surface is , q = l

By Gauss’s law, E (2rl) = (or) E =

The direction of electric field E is radially outward if line charge is positive and inwardif the line charge is negative.

64. Explain the motion of a charged particle in a uniform magnetic field.

Let us consider a uniform magnetic field of induction B acting along the Z –axis. Aparticle of charge q and mass m moves in XY plane. At a point P, the velocity of theparticle is .The magnetic Lorentz force on the particle in F = q ( x B). Hence F acts along POperpendicular to the plane containing and B. since the force acts perpendicular toits velocity, the force does not do any work. So, the magnitude of the velocity remainsconstant and only its direction changes. The force F acting towards the point O actsas the centripetal force and makes the particle to move along a circular path. Atpoints Q and R, the particle experiences force along QO and RO respectively

o

0r)

19


X

Y

Z

B

O

Q

B

B

B

Rν

ν

ν

FF

F

P

MOTION OF A CHARGED PARTICLE

Since and are at right angles to each other,F = Bqv sin90° = BqvThis magnetic Lorentz force provides the necessary centripetal force

B q =

r = …………….. (1)

It is evident from this equation, that the radius of the circular path is proportional to(i) mass of the particle and (ii) Velocity of the particle

From equation (1), =

= ………………. (2)

This equation gives the angular frequency of the particle inside the magnetic field.Period of rotation of the particle,

T =

T = …………………(3)

From equations (2) and (3), it is evident that the angular frequency and period ofrotation of the particle in the magnetic field do not depend upon (i) the velocity of theparticle (ii) radius of the circular path.

65. Obtain the phase relation between current and voltage in an a.c circuit with inductoronly.( Draw necessary graph)

Let an alternating source of emf be applied to a pure inductor of inductance L. Theinductor has a negligible resistance and is wound on a laminated iron core. Due to analternating emf that is applied to the inductive coil, a self induced emf is generatedwhich opposes the applied voltage (eg) choke coil.

m

rmBq

r

Bqm

Bqm

22mBq

20


The instantaneous value of applied emf is given by e = Eo sin t …………….. (1)

Induced emf e= -L

e = E sin to ω

o

e=E sin to ω

P = eii =I sin to

ω

eL

i

e,i&

p

π 2π t

(a) (b) (c)

where L is the self inductance of the coil. In an ideal inductor circuit induced emf is

equal and opposite to the applied voltage . Therefore e = - e

Eo sint = -[-L ]

Eo sint = L

di = sint dt

Integrating both the sides

i = sin dt

=

= -

i = sin (t – )

i = Io sin t – ………….. (2)

where Io = . Here, L is the resistance offered by the coil.

It is called inductive reactance. Its unit is ohm.From eqns (1) and (2), it is clear that in an a.c. circuit containing a pure inductor thecurrent ‘i’ lags behind the voltage ‘e’ by the phase angle of /2. It is shown graphicallyin Fig (b).The diagram (fig c) represents the phasor diagram of a.c. circuit containing only L.

66. Explain emission and absorption spectra.

(i) Emission Spectra :

When the light emitted directly from a source is examined with a spectrometer, theemission spectrum is obtained. Every source has its own characteristic emissionspectrum.

didt

didtdidtEo

L

EoL

EoL

-cos t

Eo cos tL

EoL

2

2

EoL

21


Types of the emission spectra are1. Continuous spectrum2. Line spectrum3. Band Spectrum

1. Continuous Spectrum :It consists of unbroken luminous bands of all wavelengths containing all the

colours from violet to red. These spectra depend only on the temperature of thesource and is independent of the characteristic of the source. Incandescentsolids, liquids, electric filament lamps etc., give continuous spectra

2. Line spectrum :

Line spectra are sharp lines of definite wavelengths. It is the Characteristic ofthe emitting substance. It is used to identify the gas.

H H H H

Line Spectrum of Hydrogen

Atoms in the gaseous state i.e. free excited atoms emit line spectrum. Thesubstance in atomic state such as sodium in sodium vapour lamp etc., give linespectra.

3. Band Spectrum :

It consists of a number of bright bands with a sharp edge at one end butfading out at the other end . Band spectra are obtained from molecules. It is thecharacteristic of the molecule. Calcium or Barium Salts in a Bunsen flame etc.give band spectra.

ii) ABSORPTION SPECTRA :

When the light emitted from a source is made to pass through an absorbingmaterial and then examined with a spectrometer, the obtained spectrum is calledabsorption spectrum. It is the characteristic of the absorbing substance.

Types of the absorption spectra are :

1) Continuous absorption Spectrum

2) Line absorption spectrum

3) Band absorption spectrum

1. Continuous absorption Spectrum:

A pure green glass plate when placed in the path of white light, absorbseverything except green and gives continuous absorption spectrum.

22


2. Line absorption spectrum :

When light from the carbon are is made to pass through sodium vapour andthen examined by a spectrometer, a continuous spectrum of carbon arc with twodark lines in the yellow region is obtained as shown in the figure.

5896Å 5890Å 5896Å 5890Å

3. Band absorption spectrum :

If white light is allowed to pass through iodine vapour or dilute solution of bloodor chlorophyll or through certain solutions of organic and inorganic compounds,dark bands on a continuous bright background are obtained. The bandabsorption spectra are used for making dyes.

67. Draw a neat diagram of He- Ne laser and explain its working with the help of energylevel diagram.

He-Ne Laser :A continuous and intense laser beam can be produced with the help of gas

lasers. He –Ne laser system consists of a quartz discharge tube containing helium andneon in the ratio of 1 : 4 at a total pressure of about 1mm of Hg. One end of the tubeis fitted with a perfectly reflecting mirror and the other end with a partially reflectingmirror.

A Powerful radio frequency generator is used to produce a discharge in the gas,so that the helium atoms are excited to a higher energy level.

HE – NE LASER

When an electric discharge passes through the gas, the electron in the dischargetube collide with the He and Ne atoms and excites them to metastable states ofenergy 20.61 eV respectively above the ground state. Some of the excited heliumatoms transfer their energy to unexcited Ne atoms by collision. Thus, He atoms helpin achieving a population inversion in Ne atoms. When an excited Ne atom drops

Emission & Absorption spectrum of sodium

23


down spontaneously from the metastable state at 20.66 eV to lower energy state at18.70eV, it emits a 6328Å Photon in the visible region. This photon travelling throughthe mixture of the gases is reflected back and forth by the reflector ends, until itstimulates an excited neon atom and causes it to emit a fresh 6328 Å photon inphase with the stimulating photon. This stimulated transition from 20.66 eV level to18.70 eV level is the laser transition. The output radiations escape from the partiallyreflecting mirror. The neon atoms drop down from the 18.70 eV level to lower state E,through spontaneous emission emitting incoherent light. From this level E, the Neatoms are brought to the ground state through collision with the walls of the tube.Hence the final transition is radiationless.

68. Explain the construction and working of a GM Counter (Geiger – Muller).

Geiger - Muller Counter :

Principle: Geiger - Muller counter is used to measure the intensity of theradioactive radiation. When nuclear radiations pass through gas, ionisation isproduced. This is the principle of this device.

R

C

Toamplifier

W

_ +

H.T.

E E

G.M. COUNTER

Construction : The G.M. tube consists of a metal tube with glass envelope (C)acting as the cathode and a fine tungsten wire (W) along the axis of the tube,which acts as anode. The tube is well insulated from the anode wire.

The tube is filled with an inert gas like argon at a low pressure. One end isfitted with a thin mica sheet and this end acts as a window through whichradiations enter the table. A high potential difference of about 1000 V is appliedbetween the electrodes through a high resistance R of about 100 mega ohm.

Working : When an ionizing radiation enters the counter, primary ionization takesplace and a few ions are produced. These ions are accelerated with greater energydue to the high potential difference and they cause further ionization and these ionsare multiplied by further collisions. Thus an avalanche of electrons is produced in ashort interval of time. This avalanche of electrons on reaching the anode generates acurrent pulse, which when passing through R develops a potential difference. Thisis amplified by electronic circuits and is used to operate an electronic counter. Thecount in the counter is directly proportional to the intensity of the ionizingradiation.

24


Limitation: The ionization of the gas is independent of the type of the incidentradiation. Hence, G.M. counter does not distinguish the type of radiation that entersthe chamber.

69. What is an operational amplifier? Explain its action as (i) inverting amplifier, (ii) non– inverting amplifier.

Op –amp is a solid state device capable of sensing & amplifying dc and acinput signals.

Circuit symbol and Pin-out configuration of an OP-AMP

positively voltage supply terminal

Other leads for null adjustments

_

Output Terminal

Negatively Voltagesupply terminal

Invertinginput

Non-Invertinginput +

SYMBOL FOR OP – AMP

The OP-AMP is represented by a triangular symbol. It has two inputterminals and one output terminal. The terminal with negative sign is called as theinverting input and the terminal with positive sign is called as the non – invertinginput. The input terminals are at the base of the triangle. The output terminal isshown at the apex of the triangle.

(i) Inverting amplifier : The basic OP – AMP inverting amplifier is shown in thefigure. The input voltage Vin is applied to the inverting input through the inputresistor Rin. The non-inverting input is grounded. The feedback resistor R isconnected between the output and the inverting input.

_

+

V2 Iin

If

Rin

P

Rf

Vo ut

Virtualground

INVERTING AMPLIFIER

Since the input impedance of an op – amp is considered very high, no currentcan flow into or out of the input terminals. Therefore Iin must flow through Rf and isindicated by If (the feedback current). Since Rin and Rf are in series, then Iin = If. Thevoltage between inverting and non– inverting inputs in an OP – AMP is essentiallyequal to zero volt. Therefore, the inverting input terminal is also at 0 volt. For thisreason the inverting input is said to be at virtual ground. The output voltage (Vout)is taken across Rf.

25


It can be proved thatIf = –

Since Iin = If, then

=

Rearranging the equation, we obtain

=

The voltage gain of an inverting amplifier can be expressed as

Av =

The amplifier gain is the ratio of R to Rin

Finally, the output voltage, can be found by

Vout = x V in

The output voltage is out of phase with input voltage.

(ii) Non – inverting amplifier : The basic OP – AMP non-inverting amplifier isshown in the figure. The input signal Vin is applied to the non – inverting inputterminal. The resistor Rin connected from the inverting input to ground. Thefeedback resistor Rf is connected between the output and the inverting input.

_

+

Ii

If

R in

P

Rf

Vout

NON - INVERTING AMPLIFIER

Resistors R and Rin form a resistive ratio network to produce the feedbackvoltage (VA) needed at the inverting input. Feedback voltage (VA) is developed acrossRin. Since the potential at the inverting input tends to be the same as the non-inverting input (as pointed with the description of virtual ground), Vin = VA.

Since VA = Vin the gain of the amplifier can be expressed as

AV =

However, VA is determined by the resistance ratio of Rin and R ; thus,

VA = Vout

Vout

Rf

Vin

Rin

–Vout

Rf

–Vout

Vin

Rf

Rin

–Rf

Rin

–Rf

Rin

Vout

VA

Rin

Rf + Rin

26


=

= 1 +

Av = 1 +

Finally, the output voltage can be found by, Vout = 1 + Vin.

It is seen that the input and output voltages are in phase.

69. With the help of a block diagram, explain the function of a monochrome TV receiver.

The receiving antenna intercepts radiated RF signals and the tuner selectsthe desired channel frequency band. The antenna provide RF picture and soundsignals for the RF amplifier stage. The RF amplifier stage is then coupled into themixer stage. The mixture is connected to the local oscillator. The RF audio and videosignals are heterodyned into intermediate frequency by the mixer and local oscillator.The RF amplifier, mixer and local oscillator stages are combinedly called as the RFtuner. The output signal from the tuner circuit is amplified by using a common IFamplifier. Then the video and audio components are separated by a detector. Thesound signals are detected from FM waves, amplified and then fed into the loudspeaker, which reproduces the sound.

Videoamplifier CRT

Picturetube

Ligh

t

Videodetector

Scanning &Synchronising

circuits

Common IFamplifier

FM Sounddemodulator

Mixer

RF Turner

RFamplifier

Sount IFamplifier

Audioamplifier

LocalOscillator

Receivingantenna

Loud Speaker

ELEMENTARY BLOCK DIAGRAM OF A MONOCHROME TV RECEIVER

The video components are first passed into a detector which separates thepicture signal from the synchronising pulses. The line synchronising pulses and thesynchronising pulses are fed into the horizontal and vertical deflector plates of thepicture tube. The blanking pulses are given to the control grid of the electron gun ofthe picture tube. The picture signals are applied to the filament of the electron gun ofthe picture tube. According to the variations of potential in the picture, electrons areemitted from the electron gun. Thus, the intensity of the fluorescent screen of thepicture tube is in accordance with the variation of potential in the picture and thepicture is reproduced.

Vout

VA

Rf + Rin

Rin

Vout

VA

RfRin

RfRin

RfRin

27


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GOLD MEDALIST - J. SUJATHA, M.E. (Aero) – UNIVERSITY RANK I

ANNA UNIVERSITY RANK HOLDERS

B. SUBASHREE YUVARAJ P.S. PRABHAJA

V. THULASI STEPHEN B S. ANITHAD. PRASHANTH R. VISHNU PRIYA M. DEVAMBIKA

B. NANCY PUSHPARANI U. SARASWATHI G. SARGUNARUBI

G. SAI DIVYA G. GAJALAKSHMI OLIVER PRASANNA

S. SHATHEESH KUMAR J. REVATHI JYOTHI LAKSHMIN. SIVARAM A. ANTO PREETHA

Campus Address : Administration Office :Irungattukottai, Chennai – Bangalore National Highway 3/117, Brahmin Street, Karambakkam,Sriperumbudur Taluk, Kanchipuram Dist., Chennai – 602 117. Porur, Chennai – 600 116.Tel. : 044-27198033, 9940026861, 9600112733 Tel. : 044-24766386, Telefax : 044-24766941E-mail : [email protected] Website : www.pbce.in Cell : 9600113031, 9962206098

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