march 2014 physics complete answers

HIGHER SECONDARY PUBLIC EXAMINATION – MARCH – 2014

+ 2 PHYSICS ANSWER KEY

Part - I

Q. NO. Answer Q. NO. Answer

1. (a) 21

x 16. (b) Absorbs green light

2. (c) 0

17. (a) A stream of electrons

3. (b) kq

4 18. (c) Phase and amplitude

4. (a) N9102 19. (b) 10.2 eV

5. (b) 1:2n 20. (a) mE

h2

6. (a) T510 21. (a) Zero

7. (c) High specific resistance 22. (d) 24

11Na

8. (b) 0.25H 23. (c) isotones

9. (a) AC - only 24. (b) Mean life

10. (b) Room heater 25. (c) 931 MeV

11. (a) the average value of current is zero 26. (d) Type of semiconductor

12. (b) Power is transmitted in a direction

perpendicular to both the fields 27. (b) 2.0V

13 (a) Pure line spectrum 28. (a) A

14. (a) Contracts 29. (c) Ionospheric propagation

15. (d) topaz 30. (b) Twice the signal frequency

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Part - II

31. Define electric dipole moment. Give its unit.

The magnitude of the dipole moment is given by the product of the magnitude of the one of the

charges and the distance between them.

Electric dipole moment, p = q2d (or) 2qd.

It is a vector quantity and acts from –q to +q. The unit of dipole moment is Cm.

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32. Calculate the potential at a point due to a charge of 4 x 10-7

C. located at 0.09m away.

Given Data: q= 4 x 10-7

C, r = 0.09m, 04

1

V

r

q = V4

2

79 104

109

104109

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33. What are the changes observed at the transition temperature in superconductors?

At the transition temperature the following changes are observed;

(1) The electrical resistivity drops to zero. (2) The conductivity becomes infinity (3) The magnetic flux lines

are excluded from the material.

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34. State Kirchoff’s voltage law.

Kirchoff’s Second law: (voltage law) The algebraic sum of the product of resistance and current in each

part of any closed circuit is equal to the algebraic sum of the emf’s in that closed circuit. This law is a

consequence of conservation of energy.

---------------------------------------------------------------------------------------------------------------------------------- J. Mahesh, M.Sc., B.Ed., M.Phil., PG Teacher in PHYSICS, GBHSS, POIGAI, VLR-632114

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35. An incandescent lamp is operated at 240V and the current is 0.5A. What is the resistance of the lamp?

Given Data: V= 240V, I = 0.5A, V = IR, 4805.0

240

I

VR

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36. Define Peltier Coefficient.

The amount of the heat energy absorbed or evolved at one of the junctions of a thermocouple when

1 A of current flows for one second (H = It). Its unit is volt.

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37. State Len’z law in electromagnetic induction.

The induced current produced in a circuit always flows in such a direction that it opposes the

changes or cause that produces it. e = - dt

d

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38. Magnetic field through a coil having 200 turns and cross sectional area 0.04m2 changes from

0.1 wbm-2

to 0.04 wbm-2

in 0.02sec. Find the induced emf.

Given Data: N = 200, A = 0.04m-2, B1 = 0.1wbm

-2, B2 = 0.04wbm

-2, t = 0.02s.

NBAdt

de =

dt

dBNA =

dt

BBNA 12 = -200 x 4 x 10

-2 x

2102

1.004.0

= 24V.

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39. Give the uses of ultraviolet radiation.

(1) To destroy the bacteria (2) In detection of forged documents, in finger prints in forensic laboratories. (3)

To preserve the food items (4) To find the structure of atoms.

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40. What are the necessary condition for total internal reflection to take place?

Condition: (1) Light must travel from denser to rarer medium.

(2) Angle of incidence inside the denser medium should be greater than critical angle.

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41. Distinguish between soft x-rays and hard x-rays.

Soft X-rays Hard X-rays.

1. wavelength of 4 Å or above.

2. lesser frequency

3. lesser energy

4. low penetrating power

5. produced by low potential difference

1. wavelength of 1Å

2. higher frequency

3. higher energy

4. high penetrating power

5. produced by high potential difference

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42. An electric beam passes through a transverse magnetic field of 2 x 10-3

tesla and electric field E of 3.4

x 104 V/m acting simultaneously. If the path of the electrons remain un deviated calculate the speed of the

electrons.

Given Data: E = 3.4 x 104V/m, B = 2 x 10

-3T, 17

3

4

107.1102

104.3

ms

B

EV

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43. Write any three applications of photoelectric cells.

1) To reproducing sound in cinematography. 2) Used in burglar alarm and fire alarm.

3) In controlling the temperature of furnaces. 4) To automatic switching on and off the street lights.

5) In opening and closing of door automatically.

---------------------------------------------------------------------------------------------------------------------------------- J. Mahesh, M.Sc., B.Ed., M.Phil., PG Teacher in PHYSICS, GBHSS, POIGAI, VLR-632114

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44. Write any three properties of neutrons.

1) Neutrons are the constituent particles of all nuclei except hydrogen. (2) It has no charge, hence they are

not deflected by electric and magnetic fields. (3) They are stable inside the nucleus. But they are unstable

outside the nucleus and decays an emission proton, electron and antineutrino, with half life of 13 minutes.

(4) They are neutral hence it easily penetrate any nucleus (5) Based on kinetic energy it classified as slow

neutrons (0 to 1000eV) and fast neutrons (0.5MeV to 10MeV)

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45. What is pair production and annihilation of matter?

The conversion of a photon into an electron- positron pair on its interaction with the strong electric

field surrounding a nucleus is called as pair production.

The converse of pair production in which an electron and positron combine to produce a photon is

known as annihilation of matter.

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46. What are the essential components of an oscillator?

Draw its block diagram.

Its essential components are

(1) tank circuit

(2) amplifier and (3) feedback circuit

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47. What are the advantages of integrated circuits(IC)?

(1) extremely small in size (2) low power consumption (3) reliability

(4) reduced cost (5) very small weight (6) easy replacement

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48. What are the important characteristics of an operational amplifier (Op-amp)?

(1) It as very high input impedance or even infinity which produces negligible current at the inputs (2) very

high gain (3) very low output impedance or even zero.

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49. Find the output F of the logic circuit given below.

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50. What is ‘Skip distance’?

In the skywave propagation for a fixed frequency the shortest distance between the point of transmission

and the point of reception along the surface is known as the skip distance.

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Part – III

51. Deduce an expression for the effective capacitance of capacitors of capacitance C1,C2 and C3

connected in series.

BCAY

CABAY

CA

BA

.

J. Mahesh, M.Sc., B.Ed., M.Phil., PG Teacher in PHYSICS, GBHSS, POIGAI, VLR-632114

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1. Three capacitors of capacitance C1C2 and C3 connected in series with ‘V’ potential difference.

2. Each capacitor carries the same amount of charge q.

3. Potential difference across C1 is V1.

Potential difference across C2 is V2.

Potential difference across C3 is V3.

4. V = V1 +V2 + V3

5. Potential difference across each capacitors is ,,,3

32

21

1 Cq

VC

qV

Cq

V

6. 321 C

q

C

q

C

qV =

321

111

CCCq

7. 7. If Cs be effective capacitance then V = sC

q

8.

321

111

CCCq

C

q

s

321

1111

cCCCs

The reciprocal of effective capacitance is equal to sum of reciprocal of the capacitance of individual

capacitance.

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52. Explain the working of Daniel cell with a neat diagram.

Daniel Cell:

1. Its is a primary cell.

2. Electrolyte – strong solutions of copper sulphate.

3. Anode – copper vessel.

4. Cathode – zinc rod

5. Dil. Sulphuric acid contained in porous pot.

6. zinc rod reacting with dil H2So4 produces Zn++

ions and

2 electrons. And reacts with copper sulphate solutions

producing Cu++

ions which deposit on anode copper vessel.

7. Two electrons on zinc rod pass through the external circuit and

reach copper vessel thus neutralizing the copper ions, this

constitutes an electron current from copper to zinc.

8. Daniel cell produces an emf of 1.08 volt.

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53. How will you compare the e.m.f of two cells using a potentiometer?

Comparison of emf of two given cells using potentiometer:

Principle: emf of the cell is directly proportional to its

balancing length of potentiometer .

Construction: (i) The potentiometer wire AB is connected

in series with a battery (Bt), key(K), rheostat(Rh).

(ii) The end A of potentiometer is connected to the terminal C

of a DPDT, switch. The terminal D is connected to jockey (J)

through a galvanometer (G) and high resistance.

(iii) E1 and E2 are two cells to be compared.

Theory: (i) Let I be the current flowing through the primary

circuit and r be the resistance of the potentiometer wire per

meter length. (ii) E1 is connected to circuit and adjusted for

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zero deflection in galvanometer. (iii) By principle of

potentiometer, the potential difference across the balancing

length 11 IrlE (1) (iv)E2 is connected to circuit and adjusted for zero deflection in galvanometer. (v)

By principle of potentiometer, the potential difference across the balancing length 22 IrlE (2)

(vi) Dividing equ (1) by (2) 2

1

2

1

2

1

l

l

Irl

Irl

E

E

(vii) If emf of one cell (E1) is known the emf of the other cell can be calculated using relation 1

212

l

lEE

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54. A current of 4 A flows through 5 turn coil of a tangent galvanometer having a diameter of 30cm. if

the horizontal component of Earth’s magnetic induction is 4 x 10-5

T. Find the deflection produced in the

coil. [given µ0 = 4π x 10-7

Hm-1

]

Given Data: I = 4A, N = 5, a = 30cm, Bh = 4 x 10-5

T, µ0 = 4 x 10-7

H/m

tan2

0n

aBI h ;

60

20

10410302

451014.34

2tan

52

7

0

haB

nI; 0466.1tan 1 = 64º28´

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55. Obtain an expression for the self inductance of a long solenoid.

1) Let us consider of N-turns with length ‘L’ and area of cross-section A.

2) It carries a current I and B is the magnetic field at any point inside the solenoid.

3) Magnetic flux per turn = B x area of cross section of one turn l

NIB 0

4) Magnetic flux per turn = l

NIA0

5) Total flux l

NIA0 x N = l

IAN 2

0 (1)

6) But = LI (2) where ‘L’ is the coefficient of self induction of long solenoid.

7) comparing (1) and (2) l

IANLI

2

0

8) from above l

ANL

2

0

9) If the core is filled with a material of permeability () then l

ANL

2

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56. In a Newton rings experiment the diameter of the 20th

dark ring was found to be 5.582mm and that of

the 10th

ring 3.36mm. If the radius of the plano convex lens is 1m. Calculate the wavelength of light used.

Given Data : D20 = 5.82mm = 5.82 x 10-3

m, D10 = 3.36mm = 3.36 x 10-3

m, R = 1m, λ = ?

mR

rr nnm

22 (or)

40

1036.382.536.382.5

1104

1036.31082.5

4

6232322

mR

DD nnm

Å5645107.564540

1046.218.9 106

m

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57. Write any five properties of canal rays.


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(1) They are the stream of positive ions (2)They are deflected by magnetic and electric field. (3)They are

travel in straight line.(4) The velocity of canal rays are much smaller than the velocity of cathode rays (5)

They affect photographic plates (6)They can produce fluorescence. (7)They ionize the gas.

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58. Derive Einstein’s photoelectric equation.

According to Einstein’s when a photon of energy h is incident on a metal surface its energy is used up

in two ways.The emission of photo electron is the interaction between a photon and an electron.

1) A part of the energy W spent in releasing the photoelectron is known as photoelectric work function

of the metal.

2) The remaining energy of the photon is used to impart kinetic energy, to the liberated electron.

Energy of incident photon = Work function + kinetic energy of the electron

12

1 2 mvWh

If max is the maximum velocity with which the photoelectron can be ejected, then

22

1max

2 mvWh

This equation is known as Einstein’s photoelectric equation. When the frequency of the incident radiation

is equal to threshold frequency 0 of metal surface kinetic energy of electron is zero. Then 30 Wh

max2

02

1mvhh

(or) max2

02

1mvhh

42

1max

2

0 mvh

This is another form of Einstein’s photoelectric equation.

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59. List uses and limitations of an electron microscope.

Uses of electron microscope: (1) To study the structure of textile fibres, surface of metals, composition of

paints. (2) In medicine and biology to study virus and bacteria (3) To study atomic structure and structure of

crystals.

Limitations of Electron Microscope: (1) It is operated only in high vacuum. (2) So we cannot use to study

living organisms. (3) But when we till use it would evaporated and disintegrate under such conditions.

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60. A reactor is developing energy at the rate of 32MW. Calculate the required number of fissions per

second of 92U235

. Assume that energy per fission is 200MeV.

Given Data: Energy required = 32 x 106J/s

Energy per fission = 200MeV = 200 x 106eV=200 x10

6 xx1.6 x 10

-19J

Required number of fission per second N = 18

196

6

101106.110200

1032

issionenergyperf

yliberatedtotalenerg

(or)

A piece of none from an archaeological site is found to give a count rate of 15counts per minute.

A similar sample of fresh bone gives a count rate of 19 counts per minute. Caloiculate the age of the

specimen. Given T1/2 = 5570 years.

Given Data: Count rate of fresh sample N0 = 19 counts per minute

Count rate of bone N = 15 counts per minute, 55702

1 T years,

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Age of the sample t = ?

teNN 0 ; 5570

6931.0 ; 15 = 19 te ;

15

19te ;

15

19log3096.2

6931.0

557010t

t = 1899 years.

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61. Obtain the relation between the current amplification factors α and β of a transistor.

Common emitter configurations. Since 95% of the injected electrons reach the collector the collector current

is almost equal to the emitter current. Almost all transistors have α in the range 0.95 to 0.99.

We know that cB

c

E

C

II

I

I

I

; 1

1

c

B

c

cB

I

I

I

II

;

11

1 ;

1

Usually β lies between 50 and 300. Some transistors have β as high as 1000.

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62. Draw the block diagram of amplitude modulated (AM) radio transmitter.

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Part – IV

63. Explain the principle construction and working of Van-de-Graff generator.

Van- de-Graff Generator:

1. It is an electrostatic machine produces large electrostatic

potential difference of order of 107V.

2. Principle: (a) electrostatic Induction (b) action of points.

3. Construction: (i) A hollow metallic sphere ‘A’ is mounted on

insulating pillars.(ii) It has two pulleys such as B and C.

(i) A belt made of silk moves over the pulleys.(iv) ‘D’ or ‘E’

comb shaped conductor. (v) comb D is maintained at 104 volt positive supply.

The current amplification factor or current gain of

a transistor is the ratio of output current to the

input current. If the transistor is connected in

common base mode the current gain E

c

II

and

if the transistor is connected in common emitter

mode the current gain B

C

II

. A NPN

transistor connected in the common base and

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4. Working: (i) By electrostatic induction and action of points from comb D positive charges are repelled

on the belt moves up and reaches near the comb E acquires negative charges and sphere occurs positive

charge at the outer surface of the sphere.(ii) After reaching limiting value (maximum) no more charge can

be placed on the sphere it starts leaking to the surrounding due to ionization of air.

5. Minimizing leaking of charge: By enclosing it in a gas filled steel chamber at a very high pressure.

6. Uses: It is used to accelerate positive ions(protons, deuterons) for purpose of nuclear disintegration.

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64. Deduce an expression for the force on a current carrying conductor placed in a magnetic field. Find

the magnitude of the force.

Force on current carrying conductor placed in magnetic field:

(1) let PQ be the conductor of length ‘L’ area f cross section A, placed in a uniform magnetic field ‘B’

(2) The current I = nAeVd(1) where I current, n – number of electrons per unit volume.

(3) x –plying equ (1) by ‘l’ on the both sides, Il = nAel Vd (2)

(4) The current element lnAeVIl d (3)

(5) negative sign indicates the direction of current is

in opposite to Vd.

(6) Magnetic lorentz force )( BxVef d (4)

(7) number of free electrons N = nAl (5)

(8) magnetic lorentz force for all electrons fNF (6)

(9) sub (4) & (5) in above equ (6) BxVenAlF d

(10) BxVnAleF d (7)

(11) sub (2) in (7) BxIlF

(12) The magnitudes of the force F = BIlsin,

(a) If = 90, F = BIl conductor experiences maximum force,

(b) If = 0, F = 0, conductor experiences no force.

(13) The direction of the force is given by fleming’s left hand rule.

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65. Describe the principle construction and working of a single phase AC generator.

A.C. Generator: Its converting mechanical energy into electrical energy .

1. Principle: Electromagnetic induction

2. The important parts of an a.c generators are (i) armature (ii) field magnetic (iii) slip rings (iv) brushes.

3. Armature: It consists of a large number of loops or coils insulated wire wound on soft iron core or ring.

Soft iron core is used to (i) supports the coils (ii) increases the magnetic field.

4) Field magnets: (i) permanent magnet for low power dynamos.

(ii) electromagnet for high power dynamos

5) slip rings : (i) Two metal rings to which the ends of the armature coil is connected (ii) the slip rings rotate

with the armature.

6) Carbon brushes: (i) constantly touch the revolving slip rings.

(ii) used to carry the current generated in armature to the external circuit.

7) Working: (a) whenever there is a change in orientation of the coil magnetic flux changes producing

induced emf.(electromagnetic induction)

(b) the direction of induced current is given by fleming’s right hand rule.

(c) If the coil is rotated in anticlockwise direction AB moves downwards DC moves upwards.


(d) The current flows along DCBA and in external circuit flows from B1 to B2.

(e) On further rotation AB moves upwards. DC moves downwards then

(f) The current flows along ABCD and in external circuit flows from B2 to B1

(g) Alternating current changes its direction for every half- revolution of the armature coil.

(h) frequency

2 , where is angular velocity.

(i) Peak value of emf E0=NBA and induced emf at any instant is given by

e = E0sint

where N- number of turns, B – magnetic field, A – area of coil.

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66. Derive an expression for the bandwidth of interference fringe’s in young’s double silt.

Expression for Bandwidth : (1) The two slits A and B serve as the coherent sources.

(2) Screen is kept at distance of one meter from the slits. Draw AM lar BP.

(3) Path differences between waves from A and B

BMMPBP

(4) ABM, BM=d sin , sin = ,When ‘’ is small,.

(5) = .d (1)

(6) COP, tan = D

x

CO

OP ,When ‘’ is small, tan = ,

D

x (2)

(7) (2) (1), D

xd (3)

8) Bright fringe: path difference nD

xd , Where n = 0,1,2,3,……. n

d

Dx

(9) Dark fringe: path difference 2

)12(

nD

xd Where n = 1,2,3 …..

2)12(

nd

Dx

(10) The distance between any two consecutive bright or dark bands is called bandwidth.

(11) The distance between (n+1)th

and nth

order consecutive bright fringes from ‘O’ is given by bandwidth =

nd

Dn

d

Dxx nn )1(1 =

d

D

(12) Condition: (1) Screen should be as far away from the source as possible .

(2) The wavelength of light used must be larger.

(3) Two coherent sources must be as close as possible.

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67. State the postulates of Bohr’s atom model. Obtain an expression for the radius of nth orbit of an

electron of hydrogen atom based on Bohr’s theory.

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Bohr Postulates: (1) An electron cannot revolve round the nucleus in all possible orbits. The electrons can

revolve round the nucleus only in those allowed or permissible orbits for which the angular momentum of

the electron is an integral multiple of h/2.( where h is planck’s constant 6.626 x 10-34

Js).These orbits are

called stationary orbits or non-radiating orbits and an electron revolving in these orbit does not radiate any

energy (2) An atom radiates energy only when an electron jumps from stationary orbit of higher energy to a

orbit of lower energy. If the electron jumps from an orbit of energy E2 to an orbit of energy E1 a photon of

energy h = E2-E1, is emitted. This condition is called Bohr’s frequency condition.

Consider an atom whose nucleus has a positive charge Ze where Z is the atomic number that gives the

number of protons in the nucleus and e the charge of the electron which is numerically equal to that of

proton. Let an electron revolve around the nucleus in the nth

orbit of radius rn.

By coulomb’s law the electrostatic force of attraction between the nucleus and the electron =

)1(4

12

0

nr

eZe

Where 0 is the permittivity of the free space.

Since the electron revolves in a circular orbit it experiences a centripetal force 22

2

nn

n

n mrr

mv

Where m is the mass of the electron vn and ωn are the linear velocity and angular velocity of the electron in

the nth

orbit respectively.

The necessary centripetal force is provided by the electrostatic force of attraction.

For equilibrium from equation (1) and (2)

)4(4

1

)3(4

1

2

2

2

0

2

2

2

0

nn

n

n

n

n

mrr

Ze

r

mv

r

Ze

From equation (4) )5(4 2

0

22

n

nmr

Ze

The angular momentum of an electron in nth

orbit is L = mvnrn = mrn )6(n

By Bohr’s first postulate the angular momentum of the electron )7(2

nhL

Where n is an integer and is called as th principal quantum number. From equations (6) and (7)

mrn2 n =

2

nh (or)

22 n

nmr

nh

Squaring on both sides

222

22

4 n

nrm

hn

)8(

from equation (5) and (8) 422

22

3

0

2

44 nn rm

hn

mr

Ze

(or) )9(

2

0

22

mZe

hnrn

from equation (9) it is seen that the radius of the nth orbit is proportional to the square of the principal

quantum number. Therefore the radii of the orbits are in the ration 1:4:9….

For hydrogen atom Z = 1. From equation (9) )10(2

0

22

me

hnrn

Substituting the known values in the above equation we get rn = n2 x 0.53Aº. If n =1, r1 = 0.53Aº.

This is called Bohr radius.

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68. Discuss the principle and action of a Bainbridge mass spectrometer to determine the isotopic masses.

Uses : To accurate determination of atomic masses.


Atoms with one or more electrons removed have a net positive

charge and they become positive ions. A beam of positive ions produced

in a discharge tube is collimated into a fine beam by two narrow slits S1

and S2. This fine beam enters into a velocity selector. The velocity

selector allows the ions of a particular velocity to come out of it by

combined action of an electric and magnetic field. The velocity selector

consists of two plane parallel plates P1 and P2, which produce a uniform

electric field E and an electromagnet to produce uniform magnetic

field B (represented by the dotted circle). These two field are at

right angles to each other and to the direction of the beam.

The electric field and magnetic field are so adjusted that the deflection produced by one field is

nullified by the other, so that the ions do not suffer any deflection within the velocity selector. Let E and B

be the electric field intensity and magnetic induction respectively and q be the charge of the positive ion.

The force exerted by the electric field is equal to qE and the force exerted by the magnetic field is equal to

Bqv where v is the velocity of the positive ion.

qE = Bqv

v = B

E

only those ions having this velocity v pass out of the velocity selector and then through the slit S3 to

enter the evacuation chamber D. These positive ions having the same velocity are subjected to another

strong uniform magnetic field of induction B´ at right angles to the plane or the paper acting inwards. These

ions are deflected along circular path of radius R and strikes the photographic plates. The force due to

magnetic field B´qv provides the centripetal force.

R

mvqvB

21 (or)

v

qRBm

1

; SubstitutingB

Ev then E

qRBBm

1

Ions with different masses trace semi-circular paths or different radii and produce dark lines

on the plate. The distance between the opening of the chamber and the position of the dark line gives the

diameter 2R from which radius R can be calculated.

Since q, B, B´, E and R are known the mass of the positive ions and hence isotopic masses can be

calculated.

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69. What is rectification? Explain the working of a bridge rectifier. Draw the input and output signals.

The process in which alternating voltage or alternating current is converted into direct voltage or

direct voltage or direct current is known as rectification.

A bridge rectifier is shown in the diagram. There are four diodes D1 D2 D3 and D4 used in the circuit,

which are connected to form a network. The input ends A and C of the network are connected to the

secondary ends S1 and S2 of the transformer. The output ends B and D are connected to the load resistance

RL.


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During positive input half cycle of the a.c voltage the point A is positive with respect to C. The

diodes D1 and D3 are forward biased and conduct whereas the diodes D2 and D4 are reverse biased and do

not conduct. Hence current flows along S1ABDCS2 through RL. During negative half cycle , the point C is

positive with respect to A. The diodes D2 and D4 are forward biased and conduct, whereas the diodes D1 and

D3 are reverse biased and they do not conduct. Hence current flows along S2CBDAS1 through RL. The same

process is repeated for subsequent half cycles. It can be seen that current flows through RL in the same

direction, during both half cycles of the input a.c signals. The output signal corresponding to the input signal

is shown in diagram. The efficiency of the bridge rectifier is approximately 81.2%. Because the full

secondary voltage is applied to the conducting diodes in series with the load resistance, the load resistance

the load voltage is twice that of the full-wave rectifier discussed earlier.

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70. Explain the function of a vidicon camera tube.

VIDICON CAMERA TUBE

Vidicon camera is a television camera which converts the light energy into electrical energy. It

functions on the principle of photo conductivity, where the resistance of target material decreases when

exposed to light.

CONSTRUCTION

The vidicon consists of a glass envelope with an optically flat face plate. A photosensitive, target

plate is available on the inner side of the face plate. The target plate has two layers. To the front, facing the

face plate, is a thin layer of tin oxide. This is transparent to light but electrically conductive. The other side

of the target plate is coated with a semiconductor, photosensitive antimony trisulphide. The tin oxide layer is

connected to a power supply of 50V.

Grid-1 is the electron gun, consisting a cathode and a control grid. The emitted electrons are

accelerated by Grid-2. The accelerated electron are focused on the photo conductive layer by Grid-3.

Vertical and Horizontal deflecting coils, placed around the tube are used to deflect the electron beam for

scanning the target.

WORKING

The light from a scene is focused on the target. Light passes through the face plate and tin oxide,

incident on the photo conductive layer. Due to the variations in the light intensity of the scene, the resistance

of the photo conductive layer varies. The emitted electrons from antimony trisulphide reach the positive tin

oxide layer. So, each point on the photo conductive layer acquires positive charge. Hence, a charge image

that corresponds to the incident optical image is produced. As the electron beam from the gun is incident on

the charge image, drop in voltage takes place. As a result, a varying current is produced. This current

produces the video-signal output of the camera.

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march 2014 physics complete answers

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