ANNUAL EXAMINATION - ANSWER KEY -2018 II PUC - … · Cyclotron is used to accelerate charged particle. 4. State Faraday’s law of electro magnetic induction. Induced emf in a circuit

1

ANNUAL EXAMINATION - ANSWER KEY -2018

II PUC - PHYSICS

I. PART A

1. What is an an equipotential surface?

Surface with same potential at every point on it.

2. Define ‘drift velocity’ of free electrons.

Effective velocity of an electron due to applied external electric field.

3. Give an application of cyclotron.

Cyclotron is used to accelerate charged particle.

4. State Faraday’s law of electro magnetic induction.

Induced emf in a circuit is direcly proportional to rate of exchange of magnetic flux linked

with it.

5. If the peak value of a.c. current is 4.24 A, what is its root mean square value?

0 0.707 4.24 2.99 32

rms

II A

6. Mention one power loss in transformer.

Flux leakage/ hysteresis loss.

7. Two lenses of power +1.5D and -0.5D are kept in contact on their principal axis.

What is the effective power of the combination?

P = P1 + P2 = 1.5 – 0.5 = 1 D

8. The decay of proton to neutron is possible only inside the nucleus. Why?

Mass of proton is less than mass of neutron and hence spontaneous disintegration is not

possible outside the nucleus / Exchange of Mesons is possible for small distance between

nucleons / Nuclear force is exchange force / Nuclear force is short range force.

9. What is ‘depletion region’ in a semi conductor diode?

Region between p type and n-type material with immobile charge carriers/Region between

p and g-type materials, depleted of mobile charge carriers.

10. What is the output of this combination?

Ans : 1

2

II. PART B

11. Mention any two factors on which the capacitance of a parallel plate capacitor

depends.

(a) Area of plates

(b) gap between plates

(c) dielectric constant of medium between plates

12. State Kirchhoff’s laws of electrical network.

(a) Algebraic sum of currents at a node is equal to zero.

(b) Algebraic sum of emf’s and potential drops in closed loop of circuit is zero.

13. Define magnetic ‘declination’ and ‘dip’ at a place.

(a) Angle between geographic meridian and magnetic medium.

(b) Angle between between earth’s total magnetic field at a given place and horizontal

along magnetic meridian.

14. Write the expression for magnetic potential energy of a magnetic dipole kept in a

uniform magnetic field and explain the term.s

Potential energy , U = -M B cos , M = Magnetic moment

B = Magnetic fied

= angle between magnetic axis and external magnetic field

15. Give any two applications of X-rays.

(a) X – ray photography to identify dislocation/fracture etc

(b) Used in the treatment of cancer

(c) Used in X – ray crystallography

16. What is ‘myopia’? How to rectify it?

Inability to see objects at far distance, is called myopia or short sightedness.

Concave lens is used to rectify myopia.

17. Draw the diagram representing the schematic arrangement of Geiger-Marsden

experimental set up for the alpha particle scattering.

Geiger- Marsden experimental set up for the alpha particle scattering.

Lead slit

Lead cavity

Thin gold foil

ZnS screen

Microscope

3

18. Write any two characteristics of nuclear forces.

(a) short range force

(b) Exchange forces

(c) Charge independent force

(d) Mass independent force

PART C

19. Give three properties of electric charge.

(a) Charge is additive

(b) Charge is quantized

(c) Charge is conserved

20. States Ampere’s circuital law and arrive at the expression for the magnetic field

near a straight infinite current carrying wire.

Law state that line integral of magnetic field around a

closed path in vacuum is equal to product of absolute

permittivity and current threading the closed path.

Consider a long straight conductor carrying current I.

Let P be a point at a perpendicular distance r from the

conductor. Consider a circle of radius r around the conductor

passing through P as an Aperian loop. Let d be a small line element of the loop at P. The

magnetic field B

at P due to current carrying conductor is along the tangent to this circule

(From right hand thumb rule) and parallel to d .

Applying Ampere Circuital law to the loop.

oB dl I

cos oBd I Since = 0 and B

is same at all the points on circular loop, 0B dl I

B (2r) = 0I ∵ 2dl r , circumference of the loop.

0

2

IB

r

21. What is hysteresis? Define the terms ‘ coercivity’ and ‘retentivity’ of a

ferromagnetic material.

Lagging of magnetic field in a ferromagnetic substance behind magnetic intensity applied is

called hysteresis.

(a) coercivity is the reverse magnetic field applied to a ferromagnetic substance to destroy

the residual magnetism.

(b) retontivity: is the magnetic field remaining in a ferromagnetic substance, in the

absence of applied external magnetic field.

I

rpdl

B

4

22. Arrive at Snell’s law of refraction, using Huygen’s principle for refraction of a

plane wave.

Snell’s law using Huygens Principle.

Let XY represent the surface separating medium 1 and medium 2. Let v1 and v2 be the

Speed of light in medium 1 and medium 2 respectively. Consider plane wave front AB

incident in Medium 1 at an angle i. Let t be the time taken by the secondary waves to

travel the distance BC. Then BC = v1t. The secondary waves from A will travel a distance

v2t. Draw an arc at D in medium 2with AD as radius. The tangent from C touches the arc at

D. Then CD represent the refracted wavefront. Let r be the angle of refraction.

In ABC, sini =BC

AC……(1)

In ACD, sinrAD

AC …..(2)

Eqn(1)/(2) 1 1

2 2

sin

sin

v t vi BC

r AD v t v …….(3)

If n1 and n2 are the absolute refractive indices of medium – 1 and medium – 2 respectively

and if c represents the speed of light in vacuum, then,

1 21 2

c cn and n

v v

2 1

1 2

n v

n v ……..(4)

From (3) and (4) 2

1

sin

sin

ni

r n n2 sini = n2 sinr

This is the Snell’s law of refraction.

23. Write Bohr’s postulates of hydrogen atom model.

C Y

rr

D

V2tr

A

i iX 1

2

V1-t

B

5

Bohr’s postulates:

1. Electrons revolve round the nucleus only in certain stable stationary orbits without the

emission of radiant energy.

2. Electrons revolve around the nucleus only in those orbits in which the angular

momentum of the electron is integral multiple of2

h

where h is Planck’s constant.

i.e. Angular momentum, mvr2

hn

where n = 1, 2, 3 ……… ,

where n is called principal quantum number.

3. When an electron jumps from orbit of higher energy to another of lower energy it emits

a photon having energy equal to the difference in energy of the electron in outer and the

inner orbits. (Bohr’s frequency condition)

i.e. hv = E2 – E1

Where E2 and E1 are the energies of the electron in outer and inner orbits.

24. Derive the expression for the half-life of a radio active nuclide.

Half life of a radioactive element is the time taken for 50% of the initial number of atoms to

disintegrate.

Number of nuclei in the sample at any instant of time ‘t’ is N = N0e-t ………………(1)

When t = T ( half life), number of nuclear present , 0

2

NN

In equation (1), 002

TNN e 2Te

T = loge2 = 2.303 log102 = 2.303 0.3010 = 0.693

0.693Half life T

25. Write any three distinctions between p-type and n-type semi conductor.

n - type p- type

It is obtained when a pure

semiconductor is doped with

pentavalent impurities.

It is obtained when a pure

semiconductor is doped with trivalent

impurities

Number of free electrons is greater

than number of holes

Number of holes is greater than

number of free electrons

Conductivity is mainly due to

electrons

Conductivity is mainly due to holes

Majority charge carries are in the

conduction band

Majority charge carriers are in the

valence band

6

26. Draw the block diagram of generalized communication system.

27. Define and electrical potential due to a point charge and arrive at the expression

for electric potential at a point due to a point charge.

Electric potential at a given point is defined as work done in moving a unit positive charge

from infinity to that point, against electric field .

Consider a point P at a distance r from a point charge q at O. A unit positive charge is

placed at A, distant x from O. The electric field at A is2

0

1

4

qE

x along OA.

The work done in moving a unit positive charge from A to B through a small distance dx is

dw = - E dx (F = E and -ve sign indicates that F and E are in opposite directions)

dw =2

0

1

4

qdx

x

The total work done in moving the unit positive charge from infinity to P against the electric

field isx r

x

W dw

20

1

4

r qW dx

x

2

0

1

4

rqW dx

x

20

1 1 1( )

4

rq

W dxx x x

0 0

1 1 1

4 4

q qW Potential at P is V

r r

28. Obtain the expression for the effective emf and the effective internal resistance of

two cells connected in parallel such that the currents are flowing in the same

direction.

message Transmitter SignalInformationSource signal

Channel Signal Receiver Message User

Noise

+q

O r P B dx A+ 1C E

x

= I

Eeq , req

V

R

I

V

R

I

E2 , r2

E1 , r1

I2

7

Cells are said to be in parallel if negative terminals of all the cells are connected together to

one terminal and positive terminals are connected to the other terminal. Consider two cells

having emf’s E1, E2 and internal resistances r1, r2 respectively connected in parallel.

Let I1 and I2 are the currents through the cells respectively, then the main current is

I = I1 + I2 …….(1)

If V is the terminal p.d across each cell then for the first cell, V = E1 = I1r1

or 11

1

E VI

r

And for the second cell, V = E2 – I2r2

or 22

2

E VI

r

Substituting I1 and I2 in eqn. (1) 1 2

1 2

E V E VI

r r

1 2

1 2 1 2

1 1(2)

E EI V

r r r r

Let the parallel grouping of cells be replaced with a single cell of emf Eeq and internal

resistance req then, V = Eeq – I req

or ..............(3)eq

eq eq

E VI

r r

Comparing (2) and (3), 1 2 1 2

1 2 1 2 1 2

1 1 1eqeq

eq eq

E E E r rand or r

r r r r r r r r

1 2 1 2 2 1 1 2 1 2 2 1

1 2 1 2 1 2 1 2eq eq

E E E r E r r r E r E rE r

r r r r r r r r

.

29. Derive an expression for magnetic field on the axis of a circular current loop,

using Blot-Savart’s law.

Consider a circle coil having N turns with main radius r. Let P be a point on the axis, at

distance x from the centre O.

The magnetic field at P due to AB is,

02

sin

4

IddB along

a

PM…….(1)

where ‘a’ is the distance between the point

and the element AB.

024

IddB

a

A B

A’ B’

P

dB

dB

db cos

db cos

2 dB sin

xOI

8

( 90 ) . Similarly magnetic field at P due to the element A’B’ is

024

IddB

a

along PN …..(2)

The magnetic field dB is resolved into horizontal and vertical components, dBsin and

dBscos respectively, where is the angle between line joining the point to the element

and axis of the coil.

The vertical components of magnetic fields being equal and opposite cancel each other and

horizontal components add up.

Hence magnetic field at P due to two current elements = 2dBsin

The resultant magnetic field at P due to one turn of the coil = 2dB sin

22 sin

4o Idl

a

from (1)

2

2sin

4o I

da

2

2sin

4o I

ra

Since the field is due to 2 elements ,2

Circumferencedl r

2

3

2

4o Ir

a

From ROP, sinr

a

2

2 2 3/2

2

4 ( )o Ir

r x

From AOP, a = (r2 + x2)1/2 or a3 = (r2 + x2)3/2

For N turns of the coil2

2 2 3/2

2

4 ( )o N I r

Br x

.

The direction of the magnetic field is along the axis passing through the centre and away

from the observer facing the coil if the current in the coil is clockwise and towards the

observer if the current is anticlockwise.

30. Arrive at the expression for the impendance of a series LCR circuit using phasor

diagram method and hence write the expression for the current through the

circuit.

Consider a wire of resistance ‘R’, coil of

inductance ‘L’ and capacitor of capacitance

‘C’ connected in series with a sinusoidal AC

source.

V = V0sin t ……………(1)

Let, I = I0sin( t - ) -----(2) be the current, which is same through the circuit elements at

any time.

Where is the phase difference between the voltage and current in the circuit.

If VR, VL and VC are the p.d across R, L and C respectively, then we have,

v~

CR

r

L

r

vR VL VC

9

VR = IR, VL = IXL and VC = IXC. VR is in phase with current I, VL leads I by /2 and

VC lags behind current I by /2.

If V is the p.d across the combination, (applied voltage) then V is the phasor sum of VR, VL

and VC as shown in the figure. VR, VL and VC are represented by the vectors OA, OB and OC

respectively. If VL > VC, the resultant of these two values is VL – VC represented by the

vector OD.

From the phasor diagram,

V2 = VR2 + (VL – VC)2

V2 = (IR)2 + (IXL – IXC)2

2 2( ) ( )L CV IR IX IX

2 2( )L CV I R X X

V = I Z

Impedance 2 2( )L CZ R X X which is the effective opposition offered by the circuit

for the flow of AC through it. Also tan L C L CL L

R

V V X XIX IX

V IR R

1tan L CX X

R

.

Current I = 00

sinsin

V tVI t

Z Z

31. Deduce the relation between n, u, v, and R for refraction at a spherical surface,

where the symbols have their usual meaning.

In the diagram, XPY represents a spherical surface of radius of curvature R, separating two

media of refractive induces n1 and n2 such that n2 > n1. O is the luminous point object

placed on the principal axis in rarer medium. A ray OP incident along the

principal axis proceeds without deviation.

Another ray OA incident at an angle I is refracted

along AI. Let r be the angle of refraction.

The refracted rays really converge at I,

which is the real image of the object O.

Let AOP = , AIP = and ACP =

In AOC , + = i ……(1)

In ACI , r + = or r = - ……(2)

For small aperture AP can be considered to be

straight and perpendicular to principal axis.

Now, tan = , tan , tanAP AP AP

PO PI PC

N

XAi

O P C

I

r

YRarer (n1) Denser(n2)

10

For small angles, tan , tan , tan

, ,AP AP AP

PO PI PC

According to Snell’s law n1 sini = n2 sinr

For small angles, n1i = n2r

n1 ( + ) = n2 ( - ) From (1) and 2)

1 2

AP AP AP APn n

PO PC PC PI

1 1 2 2 ........(3)n n n n

PO PC PC PI

According to the sign convention, PO = -u, object distance PI = +v, image distance and PC

= + R, radius of curvature.

In eqn(3) 1 1 2 2n n n n

u R R v

1 2 2 1n n n n

u v R

. When n1=na = 1 and n2 = n , we have

1 1n n

u v R

32. What is a rectifier? With suitable circuit describe the action of a full wave rectifier

by drawing input and output waveforms.

Rectification is the process of conversion of AC into DC.

A rectifier is said to be full wave rectifier, if the pulsating dc passes through load resistance

during both the halves of input AC cycles.

The circuit consists of two diodes D1 and D2 with their P-regions connected to the two ends

of secondary of a transformer. The N-regions of the diodes are connected to one end of

load resistance RL whose other end is connected to the centre – tap.

During the positive half cycles D1 conducts as it is forward biased and D2 does not conduct

since it is reverse biased in effect. Thus pulsating de passes through load resistance from

A to B. During the negative half cycles D1 is reverse biased and D2 is forward biased in

effect and hence D2 conducts. Again pulsating dc passes through load resistance from

time

time

11

A to B.

Thus current through load resistance is unidirectional, but fluctuating in each cycle as

diodes D1 and D2 conduct alternatively resulting in full wave rectification.

33. Three charges each equal to +4nC are placed at the three corners of a square of

side 2cm. Find the electric field at the fourth corner.

‘E’ at D due to 4nC at A-9

9 4 -11 2 2

0

1 q 4×10E = =9×10 × =9×10 NC

4πε r (0.02)

‘E’ at D due to 4nC at B-9

9 4 -12 2

4×10E = 9×10 × =9×10 NC

(0.02)

2 2 2 21 2 1 2 1 22 cos90R E E E E E E

4 2 4 2 8(9 10 ) (9 10 ) 162 10R

R = 127279.22 NC-1, bisects the angle be E1 and E2 have along BD

E at D due to electricity at B9

9 13 2

4 109 10 45918.36

(0.028)E NC

Net electric field R + E3 = 127279.22 + 45918.36

= 173197.58 NC-1 = 1.732 105N/S bisectors the angle between E1 and E2

34. 100 mg mass of nichrome metal is drawn into a wire of area of cross-section

0.05 mm2. Calculate the resistance of this wire. Given density of nichrome 8.4

103 kgm-3 and resistivity of the material as 1.2 10-6 m

Given:

Mass = 100 10-3 gmass

Density =volume

A = 0.05 (10-3)2m2

6

3

100 10

8.4 10

massVolume

Density

Density = 8.4 103 kgm-3 V = 11.904 10-9m3

= 1.2 10-6m

lR

A

Volume = ℓ A

R =6

6

1.2 10 0.238

0.05 10

9

6

11.904 10

0.05 10

V

A

R = 5.714 ℓ = 0.23808 m

A B

CDE2

E3

E1

0.02 m + 4nC

+ 4nC

0.02 m

0.02 m

0.02 m

+ 4nC

E4

12

35. A circular coil of radius 10 cm and 25 turns is rotated about its vertical diameter

with an angular speed of 40 rad S-1, in a uniform horizontal magnetic field of

magnitude 5 10-2T. Calculate the emf induced in the coil. Also find the current in

the coil if the resistance of the coil is 15 .

Given:

Radius = 10 cm E = n A B

W = 40 rad s-1 E = n(r2) B

B = 5 10-2 T = 25 3.142 (10 10-2)2 5 10-2 40

E = ? Peak value of emf, E = 1.571 V

R = 15 Peak value of Current1.571

0.10415

EI A

R

36. In Young’s double slit experiment the slits are separated by 0.28 mm and the

screen is placed at a distance of 1.4 m away from the slits. The distance between

the central bright fringe and the fifth dark fringe is measured to be 1.35 cm.

Calculate the wavelength of the light used. Also find the fringe width if the screen

is moved 0.4 m towards the slits, for the same experimental set up.

Given:

d = 0.28 10-3 m

D = 1.4 m

n5 = 1.35 10-2 m

= ? xn = (2n-1)2

D

d

1.35 10-2 = (2 5 -1) 3

1.4

2 0.28 10

2 31.35 10 2 0.28 10

9 1.4

= 6 10-7 m

If screen is moved 0.4 m towards the slits.

D’ = 1.4 – 0.4 = 1m7

3

' 6 10 1

0.28 10

D

d

= 21.42 10-4m = 2.142 mm

37. Light of frequency 8.41 1014 Hz is incident on a metal surface. Electrons with

their maximum speed of 7.5 105 ms-1 are ejected from the surface. Calculate the

threshold frequency for photo emission of electrons. Also find the work function

of the metal in electron volt (eV). Given Planck’s constant h = 0.625 10-34 Js

and mass of the electron 9.1 10-31 Kg.

13

Given:

Frequency of incident radiation V = 8.41 1014 Hz

Vmax = 7.5 105 ms-1

Threshold frequency, 0 = ?

Work function , 0 = ?

h= 6.625 10-34 Js

me = 9.1 10-31 kg

h = 0 + 2max

1

2mv 2

0 max

1

2hv mV

= 6.625 10-34 8.41 1014 -1

231 5 2[9.1 10 (7.5 10 ) ]

= 55.71625 10-20 – 2.559375 10-19 = 10-19 [ 5.5716 – 2.5594]

0 = 3.012 10-19J19

19

3.012 101.88

1.6 10eV

0 = h0 Threshold frequency19

00 34

3.012 10

6.625 10V

h

= 0.455 1015 hz

*****************