TOPPER SAMPLE PAPER 4

Class XII- Physics

Solutions

Time: Three hours Max. Marks: 70 General Instructions

(a) All questions are compulsory.

(b) There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry two marks each, questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each.

(c) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions.

(d) Use of calculations is not permitted.

(e) You may use the following physical constants wherever necessary.

19

8 1

34

7 1

23 1

23

27

1.6 10

3 10

6.6 10

4 10

1.38 10

6.023 10 /

1.6 10

µ λ

−

−

−

− −

−

−

= ×

= ×

= ×

= ×

= ×

= ×

= ×

o

B

A

n

k

e C

c ms

h Js

TmA

JK

N mole

m kg

1. Suppose one proton A and one electron B are placed between two parallel plates having a potential difference V as shown in the figure.

Will A and B experience equal or unequal force?

2. An electron and a proton moving with the same speed enter the same magnetic field at right angles to the direction of the field. For which of the two particles will the radius of circular path be smaller?

3. For which frequency of light is the human eye most sensitive? 4. The electric current in a wire in the direction from B to A is decreasing.

What is the direction of induced current in the metallic loop kept above the wire as shown in the figure?

5. A lens of focal length 30 cm is cut as shown in figure. What will be the new focal length?

6. Are matter waves electromagnetic wave? Write de – Broglie wave equation.

7. Why a transistor cannot be used as a rectifier?

8. What will happen if energy of the electron orbiting around nucleus

becomes positive?

9. Name the dielectric whose molecules have (i) non- zero (ii) zero dipole moment. Define the term ‘dielectric constant’ for a medium.

10. Two hollow spheres of radius r, and 2r are given. The space between

them is filled with material of resistivity ( )ρ as shown. Calculate its

resistance.

11. A current carrying solenoid contracts in length. Why?

12. An A.C. voltage sino

E E wt= is applied across an inductor L. Obtain an

expression for the current I.

13. How much work is required to be done to reduce the separation between two like charges of magnitude 100 Cµ each from 20 cm to 10 cm?

14. Why is spark produced in the switch, when the light is put off?

OR

An iron bar falling through a hollow region of a thick cylindrical shell made of copper experiences a retarding force. What can you conclude about the nature of the iron bar?

15. Show that only an accelerated charge can produce an electromagnetic

wave. 16. An eye- piece of a telescope consists of two Plano – convex lenses

1 2andL L each of focal length f separated by a distance of 2 3.f Where

should 1L be placed relative to the focus of the objective lens of the

telescope so that the final image through 2L is seen at infinity?

17. The half life of 238

92U against α - decay is 94.5 10× years. What is the

activity of 1 g sample of238

92U ?

18. What is an antenna? What role does it play in communication system?

What should be the length of a dipole antenna?

19. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the beaker is measured by a microscope to be 9.4 cm. What is the refractive index of water?

If water is replaced by a liquid of refractive index 1.63 up to the same height. By what distance would the microscope have to be moved to focus on the needle again?

20. Calculate capacitance of capacitor as shown below.

OR

A conducting slab of thickness‘t’ is introduced without touching between the plates of a parallel plate capacitor, separated by a distance ‘d’ (t < d). Derive an expression for the capacitance of the capacitor.

21. In the following circuit, the transistor used has a 100.β = Find

, , and for 2cBECE BCV V V I mA= .

22. Draw a graph to show the variation of the angle of deviation δ with that of

angle of incidence ‘i’ for a monochromatic ray of light passing through a glass prism of refracting angle ‘A’. Hence deduce the relation.

( )

sin2

sin 2

mA

A

δ

µ

+ =

23. An electron and a photon each have a wavelength of 1 nm.

Find i) Their momenta

ii) Energy of photon

iii) Kinetic energy of electron

24. A potential difference of V volts is applied to a conductor of length L and diameter D. How are the electric field and resistance of the conductor affected when in turn

iv) V is halved

v) L is doubled

vi) D is halved

25. The energy levels of an atom of element are shown in the following diagram. Which one of the level transitions will result in the emission of photons of wavelength 620 nm?

26. Explain the terms (i) pulse – amplitude modulation (PAM) and (ii) pulse –

code modulations (PCM). Which modulation is preferred in transmitting signals and why?

27. Show how the following gates can be obtained by using NAND gates only

(i) OR gate (ii) AND gate (iii) Not gate

28. (a) Estimate the average drift speed of conduction electrons in a copper

wire of cross – sectional area 7 21.0 10 m

−× carrying a current of 1. 5 A. Assume that each copper atom contributes roughly one conduction

electron. The density of copper is 3 39.0 10 ,kg m× and its atomic mass is

63. 5 u.

(b) Compare drift speed obtained above with

vii) Thermal speed of copper atoms at ordinary temperatures.

viii) Speed of propagation of electric field along the conductor which causes the drift motion.

OR Determine the current in each branch of the network shown in figure.

29. Explain the principle and working of a cyclotron with the help of a labeled diagram. For a cyclotron having oscillator frequency as10MHz , what should be the operating magnetic field for accelerating protons? If the radius of its ‘dees’ is 60 cm, what is the kinetic energy of the proton beam produced by accelerator? Express your answer in units of

( )19 27 13. 1.6 10 , 1.67 10 , 1 1.602 10 .pMeV e C m kg MeV J− − −= × = × = ×

OR

State Biot – Savart law. Using Biot – Savart law, derive an expression for the magnetic field at the centre of circular coil of numbers of turns ‘N’, radius ‘r’ and carrying a current ‘ i’. A semicircular arc of radius 20 cm carries a current of 10A. Calculate the magnitude of the magnetic field at the center of the arc.

30. Explain the phenomenon of total internal reflection. State two conditions

that must be satisfied for total internal reflection to take place. Derive the relation between the critical angle and the refractive index of the medium. Draw ray diagram to show how a right angled isosceles prism can be used

to (i) deviate ray through o180 (ii) to invert it.

OR

Prove that 1 2 2 1

u v R

µ µ µ µ− −+ =

When refraction occurs from rarer to denser media at a convex refracting spherical surface.

TOPPER SAMPLE PAPER 4

Class XII- Physics

Solutions

1. The proton A and the electron B will experience the electrostatics force of

equal magnitude but in opposite direction. 1

2. Asmv

rqB

= , for same V and B, .m

rq

α Since m

q is smaller for an electron,

the radius of the circular path followed by the electron will be smaller. 1

3. 145.405 10 .Hz× 1

4. Induced current in the loop will be in the clockwise direction 1

to oppose its origin

5. Since ( )1 2

1 1 11n

f R R

= − −

Taking ( )

1 2

2 11, ,

nR R R R

f R

−= = − = …………………. (1)

1

2

When lens is cut, then 1 2

1 1, ,

'

nR R R

f R

−= = ∞ = ………..(2)

1

2

Using eq (1) and eq (2), ' 2 60f f cm= =

6. No, Matter waves are not electromagnetic waves. 1

2

The de – Broglie equation is .h

mvλ =

1

2

7. A rectifier device conducts only in forward biased and not in

reverse biased. But a transistor conducts either way. 1 8. As negative energy of the electron orbiting around nucleus keeps the

electron bounded with the nucleus. If the energy of the electron becomes positive, the electron will no longer be bounded with the nucleus.

1

9. (i) water ( )2H O 1

2

(ii) oxygen ( )2O 1

2

The ratio of permittivity of a medium to the permittivity of vacuum is called dielectric constant for the medium. 1

10. Small resistance = A

ρ�

24

drdR

r

ρ

π=

1

2

1

2

∴ Total resistance 2

22

11

1

4 4

rr

rr

dr

r r

ρ ρ

π π

= = − ∫

1

2

1 2

1 1

4R

r r

ρ

π

= −

1

2

11. When current through a solenoid, it flows in same direction as shown. As

current through each turn is in same direction, they attract each other. Hence, the solenoid contracts in length. 1

1

12. Due to the applied emf E, the current ‘I’ in the inductor will be such that:

ordI E

E L dI dtdt L

= = 1

2

0 0sin

or sinE t dt E

dI I t dtL L

ωω= = ∫

1

2

0 cosE

I tL

ωω

= − 1

2

0or sin2

I I t πω = −

1

2

when 00

EI

Lω=

13. Here 1 2 2100 , 10 , 20q q C r cm r cmµ= = = = 1

2

Work done

1 2 2 1

0 1 24

q q r rW

r rπε

−∴ =

1

2

( )2

6 9

2

10 1100 10 9 10

200 10

−

−

= × × × ×

1

2

450 .J= 1

2

14. When light is put off, a large emf is produced to oppose the decay of current

in the circuit. This large induced emf across the gap causes sparking in the switch. 2

OR

Iron bar experiences a retarding force. It means eddy currents are there, which are produced whenever there is a change in magnetic flux. So we conclude that iron bar is a magnet. 2

15. A stationary charge produces only an electric field around it. When a charge

moves with a constant velocity it produces a constant magnetic field in addition to the electric field. As the charge is accelerated, both electric and magnetic fields change with time and space. One becomes the source of the other, thus giving electromagnetic waves. 2

16. Image of the object is formed at focus of objective. This image acts like an

object for virtual image of 1L . Now virtual image produced by 1L should be

located at the focus of 2.L

1

2

Since 1 2 2 / 3L L f=

Image distance from 1 is / 3L v f= − 1

2

1 1 1

f v u∴ = −

1

2

1 3 1

f f u= − −

/ 4u f⇒ = − 1

2

17. Given that

9

27

12

12

4.5 10 years

1.42 10 sec

T

T

= × ×

= ×

One k mol of an isotope has a mass equal to the atomic weight of that

isotope expressed in kg. Hence, 1 g of 238

92U contains 3

6104.2 10

238 /

kgk mol

kg k mol

−−= ×

1

2

One k mol of any isotope contains Avogadro’s number of atoms.

So 1g of 238

92U contains atoms equal to

6 26

20

4.20 10 6.025 10

25.3 10

N k mol

N atoms

−= × × ×

= ×

1

2

1

2

20

17

0.693Activity

0.693 25.3 10

1.42 10

R NT

R

λ= =

× ×=

×

∵

1

2

4 1 41.23 10 1.23 10 .R s Bq

−= × = × 1

2

18. An antenna is a length of conductor which radiates or picks up

electromagnetic waves carrying signals. Basic function of antenna is to convert high frequency signals into electromagnetic waves and vice – versa.

1

In most cases, the length of an antenna is / 2λ , whereλ is the wavelength of radio signal applied. 1

19. (i) Actual depth t =12.5 cm 1

2

Apparent depth a = 9.4 cm

12.51.33

9.4a w

t

aµ = = = 1

(ii) 1.63a lµ =

1

2

12.5

7.71.63

l

a l

ta cm

µ= = =

1

2

Therefore, the microscope has to be moved by ( )9.4 7.7 ,cm−

i.e., 1.7cm. 1

2

20. The capacitor can be considered as split into two capacitors in series as

shown below

1

2

Here, 1 2 01 2

2 2o

K A k Ac and c

d d

ε ε= =

1

2

The capacitance of capacitor

1 2

1 1 1

c c c= +

1

2

1 2

1 2

c cc

c c⇒ =

+

1

2

1 2

1 2

2 o A k kc

d k k

ε =

+ 1

OR

Let σ be the surface charge density of capacitor plates of area A. Electric field between the plates in the air space is

o

o

Eσ

ε=

1

2

1

2

As in case of conducting slab .p o

E E= Net electric field inside the

conducting slab is zero. Now potential difference between the plates of capacitor is

1

2

( ) ( )o

o

V E d t d tσ

ε= − = −

1

2

Q Aσ=

1 /

o oA CQ

CV d t t d

ε= = =

− −

1

2

where oo

E AC

d=

1

2

21. Since 100β =

32 10

?

?

?

c

CE

BE

BC

I A

V

V

V

−= ×

=

=

=

3

3 3

' 4 10

20 2 10 4 10 20 8

CE BE CV V I

− +

= − × ×

= − × × × = − 1

12CEV volt=

c

b

I

Iβ =

52 10c

b

II A

β−= = ×

52 10

20 4 16

BE EE b

BE

V V I

V valt

−= − × ×

= − = 1

and

16 12

4

BC BE CE

BC

V V V

V volt

= −

= −

=

1

22.

1

2

1

2

At minimum deviation position, there is only one angle of incidence. In the minimum deviation position

1 2( ) ( )i i e ii r r r= = = 1

2

(iii) Refracted ray LM on is parallel to the base We know that

1 2r r A+ =

and ( )m i e Aδ = + − 1

2

At minimum deviation position, we have

2

2

r A

r A

=

=

and 2

mA

iδ +

= 1

2

It µ is the refractive index of the material of the prism, then according to

Snell’s law

sin

sin

i

rµ =

Thus,

sin2

sin 2

mA

A

δ

µ

+ =

1

2

23. Sinceh

pλ =

34

9

25

6.63 10( ) omentum,

10

6.63 10 /

hi M p

p kg m s

λ

−

−

−

×= =

= ×

1

34 8

9 19

( ) Energyof photon

6.63 10 3 101.24

10 1.6 10

hcii E hv

E keV

λ−

− −

= =

× × ×= =

× ×

1

(iii) Kinetic energy of election 2

22

h

meλ=

( )( )

234

29 31 19

6.63 10

2 10 9.11 10 1.6 10

1.51eV

−

− − −

×=

× × × × ×

=

1 24. Effect on electric field

( )V

i EL

=

When V is halved

1

2 2

V EE

L= =

Electric field gets halved. 1

2

(ii) When L is doubled

'2 2

V EE

L= =

Electric field gets halved. 1

2

(iii) When D is halved. No effect on electric field. 1

2

Effect on resistance -

(i) When potential is halved, current also gets reduced in the same proportion. Thus, resistance does not change.

i. e., constantV

RI

= = 1

2

(ii) l

RA

ρ=∵

As length is doubled, resistance also gets doubled. 1

2

(iii) When D is halved, area reduces to one fourth. Thus, resistance becomes

four times. 1

2

25. The energy of a photon of wavelength ishc

Eλλ

= 1

2

For 620nmλ =

34 8

9 19

6.62 10 3 102.0

620 10 1.6 10E eV

−

− −

× × ×= =

× × × 1

1

2

Thus transition D for which E is 2.0 eV will take place. 1

26. Pulse amplitude modulation (PAM)

Here carrier wave is in the form of pulses 1

2

And information signal is continuous wave.

(i) Amplitude of the pulse varies in accordance with the modulating signal.

It could be single polarity or double polarity PAM. 1

2

1

2

(ii) Pulse – code modulation (PCM)

It is modulation technique employed in digital communication. In PCM carrier wave is a continuous wave and information signal is in the form of

coded pulse. 1

2

Common modulating techniques are

i) Amplitude shift keying (ASK)

ii) Frequency shift keying (FSK) 1

2

iii) Phase shift keying (PSK)

Since is more error and noise free communication so it is preferred.

1

2

27. (i) Realization of OR gate using NAND gate

1

(ii) Realization of AND gate using NAND gate

1

(iii) Realization of NOT gate using NAND gate

1

28. (a) We know that d

I V ne A=

d

IV

enA= (1)

1

2

Given that 1.5I A=

7 21.0 10A m

−= ×

191.6 10e C−= ×

1

2

Density of copper 3 39.0 10 /kg m= ×

Atomic mass of copper = 63.59u Therefore, the number of atoms or number of free electrons

per unit volume of copper ( )23

66.0 109.0 10

63.59n

×= × ×

1

2

28 38.5 10n m

−= ×

Thus, from equation (1), we get

28 19 7

1.5drift velocity

8.5 10 1.6 10 1.0 10d

V− −

=× × × × ×

1

3 11.1 10 / 1.1d

V m s m s− −= × =

1

2

(b) (i) Thermal speed of copper atoms at temperature T is obtained from

formula 21 3 3

2 2

BB

K TMV K T V

M= ⇒ =

1

2

at 300 ,T k=

22 10V m s= ×

Thus, drift speed of electrons is much smaller about 510− times the typical

thermal speed at ordinary temperatures. 1

2

(ii) Electric field travels along a conductor with a speed of electro magnetic

wave ( )83.0 10 .m s× Drift speed in comparison is 1110−

times the

speed of electric field. 1 OR We take closed loop ADCA and apply Kirchhoff’s second rule

( ) ( )2 2 3 1 14 , 2 10 0I I I I I I− − + + − − + = 1

2

1 2 37 6 2 10I I I− − = (1)

For closed loop ABCA, we get

( )2 2 3 14 2 10 0I I I I− − + − + =

1 2 36 2 10I I I− − = (2) 1

2

For closed loop BCDEB, we get

( ) ( )2 3 2 3 1

1 2 3

2 , 2 5 0

2 4 4 5

I I I I I

I I I

− + − + − + =

− − = − (3)

1

2

On solving equation (1), (2) and (3) we get

1

2

3

2.5

0.63

1.88

I A

I A

I A

=

=

=

1

2

Thus, the currents in various branches of the network are

2 0.63AB I A= = 1

2

1 2 1.87AD I I A= − = 1

2

1 2.5CA I A= = 1

2

2 3 1 0.01CD I I I A= + − = 1

2

3 1.88DEB I A= = 1

2

2 3 2.51BC I I A= + = 1

2

29. Principle of Cyclotron

A positive ion can be accelerated to a very high energy by making it to pass through a moderate electric field again and again by making use of magnetic field. 1 Working High frequency oscillator maintains a modest alternating potential difference

between the dees. Suppose positive ion produced at any instant finds 1D at

negative potential. It gets accelerated towards it. 1

2

1

2

Perpendicular magnetic field makes it to move in a circular path. Particle traces a semicircular path of radius r such that

2mv

Bqvr

mvr

Bq

=

=

1

2

The time taken ‘t’ is equal to the half time period of electric oscillator. Hence, as soon as the ion arrives in the gap between the dees, the polarity of the

dees is reverse and positive ion gets accelerated towards 2D . This way ion

keeps on accelerating until it is removed out of the dees by applying

deflecting electric across a window. 1

2

Given that 610 10v Hz= ×

0.6R m= Kinetic energy =? At resonance

2

Bqv

mπ=

( )2 mv

B q ee

π= =∵

1

2

27 7

19

2 3.14 1.67 10 10

1.6 10B

−

−

× × × ×=

×

0.66B T= 1

2

Kinetic energy 2 2 2

2

B e R

m=

1

2

( ) ( ) ( )

22 219

27

0.66 1.6 10 0.6

2 1.67 10

−

−

× × ×=

× ×

7.5MeV= 1

2

OR Bio- Savart’s Law -The magnetic field induction at a point P due to a current element is given as.

2

sin

4

oIdl

dBr

µ θ

π= 1

Magnetic field at the center of a circular coil carrying current. According to Biot – Savart’ s law magnetic field at the centre of the coil carrying I due to

current element Idl is

( )290

4

ooIdl

dBr

µθ

π= =∵

1

2

Magnetic field due to whole loop is

24

oI

B dB dlr

µ

π= =∫ ∫

1

2

2

24

oI

B rr

µπ

π= ×

1

2

2

oI

Br

µ=

1

2

When there are N number of turns

2

oNI

Br

µ=

1

2

Numerical:- Magnetic field at the center of semicircular are of radius ‘r’ carrying current I is

4

oI

Br

µ=

1

2

7 1

7

5

Given 20 0.2

10

4 10

4 10 10

4 0.2

1.57 10 .

o

r cm m

I A

Tm A

B

B tesla

µ π

π

− −

−

−

= =

=

= ×

× ×=

×

= ×

It is perpendicular to the plane of the paper directed in words. 1

30.Total Internal Reflection:-

Total Internal reflection is the phenomenon of reflection of light rays back to the denser medium when they are incident on the interference of a denser and a rarer medium at an angle of incidence greater than the critical angle. 1

Conditions for total internal reflection:-

(a) Light rays should go from denser medium to rarer medium.

(b) The angle of incidence should be greater than the critical angle ci where

1cSin i

µ= 1

Then the rays are totally internally reflected.

0

2 1 0

1 2

for angle

90

sinsin

sin sin 90

1

sin

c

c

c

i i

r

ii

r

i

µ

µ

=

=

= =

⇒ =

1

(i) To deviate ray through o180 . ( )fig a

1

(ii) To invert object. fig (b)

1

OR

Let the 1µ be refracting index of rarer medium and 2µ be the refracting index of

spherical convex refracting surface XY of small aperture.

From A draw AM such that AM OI⊥ 1

2

( )

In

r γ Ext. angle property

IAC

B

∆

+ =

γr β∴ = − 1

2

Similarly in OAC∆

γi α= +

1 According to snell’s law

22 1

1

sin

sin

i ir i

r r

µµ µ

µ= ≈ ⇒ =

1

2

So, ( ) ( )1 2 γµ α γ µ β+ = − (i)

Let tanAM AM

OM POα α≈ = =

1

2

tanAM AM

MI PCβ β= = =

As spherical surface has small aperture.

tanAM AM

yMC PC

β= = = 1

2

Substituting the value in eq. (I), We have

1 2 2 1

PO PI PC

µ µ µ µ−+ =

1

2

by sign convention put , ,PO u PI v PC R= − =+ = +

We get 1 2 2 1

u v R

µ µ µ µ−+ =

−

Which is the required relation. 1