PHYSICS

(Solved Paper of CEE)

Assam Electronics Development Corporation Limited

(AMTRON) (A Government of Assam Undertaking)

1) Which of the following substance (A, B, C) has the highest specific heat?

A) A

B) B

C) C

D) All has equal specific heat.

Answer [C]

Hints:

Q= MsdT dQ/dt=MsdT/dt Pdt=MsdT

dT=(P/Ms)dt ------ (1)

From the graph we can observe that rise in temperature in graph-A is more

than in B & C

From -1

Rise in temperature dT ∝1/S

dT is maximum for A and minimum for C

S is maximum for C and minimum for A

2)A barometer tube of length 90 cm contains some air over mercury.The reading of the

mercury level is 74.8 cm when true atmospheric pressure is 76 cm and temperature is 30 . If

the reading is observed to be75.4 cm on some another day when temperature is 10 , then

what will be the true pressure?

A) 74.25 cm

B) 75.65 cm

C) 76.57 cm

D) 77.26 cm

Answer [C]

Hints:

2

22

1

11

/

1 4.75

2.1

768.74

T

vp

T

vp

pp

p

p

As no. of moles of gas is constant in the barometer tube

56.76166.14.75

,

166.1

283

6.14

303

2.152.1

/

1

1

p

therefore

p

p

3) For the following circuit, the potential drop across 3 Mf capacitor when switch S is open and

switch S is closed will be

A) 12 V, 10 V

B) B) 8V, 9 V

C) C) 9 V, 8 V

D) D) 12 V, 8 V

Answer: [D]

Hints:

Case(1),

When switch is opened

Potential drop across VCeqv 18

In series q is same

VV

VVV

V

V

V

V

C

C

V

V

CV

CqV

12

18

2

36

/1

/

1

21

2

1

2

1

1

2

2

1

Case (2),

If switch is closed,

Capacitors are at steady state

So no current will flow through there

Ai

i

1

1818

Potential drop across F3 is same as to resiostor8

Potential drop across resiostor8 is V818

Potential drop across F3 capacitor is 8 V

So, answer is 12V, 8V

4) What is the current passing through the 1 ohm resistor in the circuit given?

A) 1/19 Ampere

B) 2/19 Ampere

C) 3/19 Ampere

D) 4/19 Ampere

Answer: [B]

Hints:

Applying Kirchoff’s law in loop-1

1234

0234

233

1

1

1

ii

ii

ViiiV AA

Apllying Kirchoff’s lawin loop-2

Applying Kirchoff’s law in loop-3

Applying junction law

5

4

432

1 2

iii

iii

3

22

34

34

ii

ViiV cc

2232

0223

1233

14

41

41

ii

ii

ViiV BB

Solving these equations we get

Ai19

2

5) Two wires AB and AC carry 1 Ampere current each as shown in the figure. One end wires

extend to infinity and the 60BAC . The magnetic field at the point at a distance 1 m from

A and lies on the bisector of the BAC will be

A) 3244

B) 3222

C) 3228

D) 3242

Answer [A]

Hints:

Direction of current is wrong. But assuming that is correct

2]2

32[

]90sin60[sin

90&60

]sin[sin

,

1]2

32[

]90sin60[sin

90&60

]sin[sin

2

2

21

212

1

1

21

211

rkiB

rkiB

rkiB

also

rkiB

rkiB

rkiB

21 BBBnet

)324(4

)32(2

]32[2

]2

32[4

2/1

]2

32[2

]2

32[]

2

32[

iB

iB

kiB

rkiB

rkiB

rki

rkiB

net

net

net

net

net

net

6) The radiation of a wavelength 332 nm is incident on a metal of work function 1.07 eV. The

value of the stopping potential will be

A) 0.70 V

B) 1.14 V

C) 1.68 V

D) 2.06 V

. Answer [D]

Hints:

Energy incident = 7.3332

1240

hc

ev07.1

Stopping potential must be able to stop the most energetic electron

vv

evev

evev

66.2

66.2

734.3

max

max

max

7) The de-Broglie wavelength of a neutron at C27 is . What will be its de Broglie

wavelength at 927 .C

A) . /2

B) . /3

C) . /4

D) 3 . /2

. Answer [A]

Hints:

T

Tv

now

v

mv

h

1

,

1

22

2

277.92

7.9227

27

7.92

7.92

27

T

T

8) When U238

92 decays to final product Pb206

82 , the number of particle and particle

emitted are:

A) 8 particle , 6 particle

B) 6 particle , 8 particle

C) 8 particle , 8 particle

D) 6 particle , 6 particle

Answer [A]

Hints:

0

1

4

2

206

82

238

92 nnPbU

Balance mass as particle has no mass no.

238-206=32

So,

No. of particle = 32/4=8

By charge balance

92=82+28+ n (-1)

92=82+16- n

n =6 & n =8

9) A radioactive substance of half-life 6 minutes is placed near a Geiger counter which is found

to register 1024 particles per minute. How many particles per minute it will register after 42

minutes?

A) 8

B) 16

C) 24

D) 32

Answer [A]

Hints:

Half life= 6 min =

2ln

6

692.0

6

2ln

At t= 0, 1024 particles per minute

After 42 minutes, 7 half-life is complete

Therefore, no. of particles = 72

1024

No. of particle = 8

10) For the given logic circuit, the output will be 1 under which of the following conditions?

A) A=0, B=0, C= 1

B) A=1, B=0, C=1, D=0

C) A=0, B=1, C=0, D=1

D) A=0, B=1, C=1, D=1

Answer [D]

11) A particle is moving with constant acceleration and 32 ,,1

VVV are the average velocity of

the particle in three successive intervals 321 ,, ttt . Which of the following relation will be

correct?

A) 32

21

32

31

tt

tt

VV

VV

B)31

21

32

21

tt

tt

VV

VV

C) 32

21

32

21

tt

tt

VV

VV

D)32

21

32

21

tt

tt

VV

VV

Answer [C]

Hints:

If constant acceleration

2

fi

avg

VVV

32

21

32

21

33

22

11

11

:

322

222

,

122

2

tt

tt

VV

VV

sveequationSolvingabo

atUV

atUV

Similarly

atUV

atUUV

=

12)The maximum height attained by a projectile is increased by 10% by increasing its speed of

projection, without changing the angle of projection. The percentage of increase in the time of

flight of the projectile will be:

A) 5%

B) 10%

B) 15%

D) 20%

Answer: [A]

Hints:

Hg

U

2

sin 22

%5

20/1

sin2

20/1/

210

1

2

2

T

T

T

T

U

U

T

T

UT

g

UT

UDelU

U

delU

U

delU

H

delH

UH

13) The resultant of the three vectors shown in the figure and the angle made by the resultant

with X-axis:

A) 37&10m

B) 5.35&6.8 m

C) 37&)35( m

D) None of these

Answer [B]

Hints:

CBAR

A= 4i+3J

B= 3i

C=2j

From the above equations:

5.35&6.877

57

2334

mR

jiR

jijiR

14) A clock has its minute hand 4.0 cm long. The average velocities of the tip of the minute

hand between 6:00 a.m. and 6:30 a.m., and 6:00 a.m. and 6:30 p.m. respectively are:

A) sec/104.4 3 cm , sec/108..1 4 cm

B) sec/104.4 3 cm , sec/104.4 3 cm

C) sec/108..1 4 cm , sec/108.1 4 cm

D) sec/106.9sec,/104.5 33 cmcm

Answer [A]

Hints:

C-1

4

3

3

108.1

10177.0

60750

8

sec1044.4

6030

8

t

rV

t

rV

avg

15) A body weighs 98 Non a spring balance at the North Pole, Its weight recorded on the same

scale if it is shifted to the equator is:

A) 98 N

B) 0

C) 98.34 N

D) 97.66 N

Answer [D]

Hints:

we know:

766.9

0,

90,

cos

2

22

eff

eff

eff

g

rgg

Atequator

golesgravityatp

Atpoles

Rgg

So weight at equator:

W=mg effect=97.66

= 10*9.766

=97.66

16) A lift weighing 250 Kg is to be lifted up at constant velocity of 0.20 m/sec. What would be

the minimum horse power of the motor to be used?

A) 1.3 HP

B) 0.65 HP

C) 1.5 HP

D) 0.75 HP

Answer [B]

Hints:

P= VF.

F=mg

=250*10

=2500

V=0.2 m/sec

From (1):

P=2500*0.2

=500 watt

P=0.65 HP

17) Starting from rest, a fan takes 16 second to attain the maximum speed of 400 rpm.

Assuming constant acceleration, the time taken by the fan in attaining half the maximum speed

is:

A) 2 Seconds

B) 4Seconds

C) 8 Seconds

D) None of these

Answer: [C]

Hints:

.,

sec8

)(6/503/20

sec/3

20200,

:

6

5

48

40

1603

40

3

40

60

2*400

400

:

tan

CHence

t

t

radrpmt

Again

ngSubstituti

rpmW

Given

t

tcons

fif

f

f

if

18) The value of acceleration due to gravity in )/( 2sm at a point 5-km above the earth’s surface

and 5 km below the earth surface are respectively:

A) 9.78,9.79

B)9.78,0

C) 9.79,0

D) 9.79, 9.78

Answer [A]

Hints:

We know:

For a height h above the ground:

2sec/78.96400

5*218.9

)2

1(

mg

R

hgg

h

h

At Depth,

.,

sec/79.9

)6400/51(8.9

)/1(

2

AHence

m

rdggd

19) The three vessels shown in figure have some base area. Equal volumes of a liquid are

poured in the three vessel. The force on the base will be:

A) Maximum at Vessel A

B) Maximum at Vessel B

C) Maximum at vessel C

D) Equal in all vessels

Answer[C]

Hints:

hP

ghP

Height is maximum in “C”.

Hence pressure at base will be maximum in Vessel C.

Therefore force is maximum.

Hence, “C” is correct.

20) A water drop of radius 1 cm is broken into 1000 equal droplets. If the surface tension of

water is 0.075 N/m, then the gain in surface energy will be

A) 0

B) J4105.8

C) J4105.7

D) Infinite

Answer [B]

Hints:

Initial Volume= Final Volume (Applying conservation of Volume)

mr

nr

nr

rnR

001.0

)1000/(1/1

1

3

4

3

4

3

1

3

1

33

33

Increase in surface energy= J

ST

4

22

105.8

01.0001.010004075.0

21) A compound microscope consists of an objective of focal length 1.0 cm and an eyepiece of

focal length 5.0 cm separated by 12.2 cm. At what distance from the objective should an object

is placed to focus it properly so that the final image be formed at the least distance of distinct

vision 25 cm?

A) -1.1 cm

B) -2.1 cm

C) -1.5 cm

D) -2.5 cm

Answer [A]

Hints:

For Objective:

V=

1/

1)(

)1)((

xx

x

x

fu

uf

So, distance of image formed by objective from eyepiece is=12.2-1x

x

=1

2.122.12

x

xx

For eyepiece:

U= -1

2.122.11

x

x

V=-25 f=5

So, -25= 51

)2.112.12(/5)

1

2.112.12(

x

x

x

x

Solving and getting:

X=1.14

22) The distance between two parallel planes of a crystal is 3 A

. What should be the

wavelength of X-ray used for getting the first order Bragg reflection at a glancing angle of

7.14 ?

A) 1 A

B) 1.3 A

C) 2.5 A

D) 1.5 A

Answer [D]

Hints:

A/Q:

2dsin n

."",

5.1

)7.14sin(1032 10

iscorrectDHence

A

23) The density of water and steam are 1000 3/ mkg and 0.6

3/ mkg respectively. The latent

heat of vaporisation of water= 2.25 kgj /10 6 . The increase in internal energy of 1 kg of water

at 100 degree celcius, when it is converted into steam at the same temperature and 1 atm(100

kPa) is

A) 2 J610

B) -2 J810

C) - 2 J1010

D) -2 J1210

Answer [A]

Hints:

Given:

.,

102

)mod1(

1067.1

)106/10(10

6/10

/

10

/6.0

1

/1000

6

5

35

3

33

3

3

AHence

U

ynamicsrstlawoftheWUQ

W

W

VPW

mV

mV

mV

mkg

kgm

mkg

steam

steamsteamsteam

w

steam

w

w

24) A liquid cools from 70 c to 60 c in 5 minutes. If the temperature of the surrounding is

constant at 30 c , then the time taken by the liquid to cool from 60 c to 50 c is:

A) 5 min

B) 10 min

C) 7 min

D) 8 min

Answer [C]

Hints:

A/Q, Newton’s law of cooling:

.,

min7

:&

)2/(

,

35/2

,

)2

(/

2112

2112

CHence

t

gcalculatinEquating

TTTKt

TT

Now

K

wegetngSubstituti

TTT

KtTT

25) Figure below shows a triangular loop PQR carrying a current I, The triangle is equilateral

with edge length l. A uniform magnetic field B exists in a direction parallel to PQ. The forces

acting on PQ, QR, and RP respectively are

A) 0, ilB2/3 , ilB2/3

B) 0, 1/2ilB, ilB2/3

C) 0, ilB2/3 ,1/2ilB

D) ilB2/3 , 0, 0

Answer [A]

Hints:

As we know that:

BilBliF effeqB )(

For PQ:

l

ilB

BilF

ll

ForPR

F

Bparallell

eff

pR

PQB

eq

B

2

3

2

3

:

0)(

Similarly for QR:

ilBFB2

3

26) A positively charged particle projected towards east is deflected towards north by a

magnetic field. The magnetic field may be

A) Towards west

B) Towards south

C) Upward

D) Downward

Answer [D]

Hints:

".",

ˆˆ

)ˆˆ(

ˆ

ˆˆ

DHence

KB

BiJ

jF

iV

BVQF

B

B

27) Two long straight wires each carrying a current of 5.0 A, are kept parallel to each other at a

separation of 2.5 cm. The magnitude of magnetic force experienced by 5.0 cm of wire is

A) 8 N510

B) 4.0 N510

C) 2.0 N510

D) 1.0 N510

Answer [D]

Hints:

.,

101

5.22/55104

2/

5

7

21

DHence

N

diiFB

28) A parachutist drops freely from an aeroplane for 10s before the parachute opens out. Then

he descends with a net retardation of 2.5 2/ sm . If he jumps out of the plane at a height of 2495

m and g= 10 2/ sm , then his velocity on reaching the ground will be

A) 2.5 1/ sm

B) 7.5 1/ sm

C) 5 1/ sm

D) 10 1/ sm

Answer [C]

Hints:

Using:

.,

2

2

1

22

2

CHence

asUV

atutS

29) If

kji ,, are unit vectors along X, Y an Z axes respectively, then the angle between

the vectors

KJi and

i is given by

A) = 3/1cos 1

B) = 3/1sin 1

C) = 2/3cos 1

D) =

2

31Sin

Answer [A]

Hints:

Let,

.,

)3

1(

:

.

1

AHence

Cos

Solvingfor

ABCosBA

i

A

B

kji

30) A projectile has same range R when the maximum height attained by it is either 21orhh ,

Then 211, andhhR will be related as

A) R= 21hh

B) R= 2 21hh

C) R= 3 21hh

D) R=4 21hh

Answer [D]

Hints:

Rangeg

U 2sin2

Hg

U

2

sin 22

max

Range is same for complementary angles.

Max. height/Range= tan RH /4

R=4 21hh

Hence “D”.

31) A ball of mass m is moving towards a batsman at a speed “V”. The batsman strikes the ball

and deflects it by an angle without changing its speed. The impulse imparted to the ball is

given by

A) mvcos

B) mvsin

C) 2mvcos /2

D) 2mvsin /2

Answer [D]

Hints:

Impulse= Change in momentum

R= 2ACos 2/

=2mvcos2

)180(

=2mvsin 2/ .

Hence,”D”.

32) A block of mass m is lying on a horizontal surface of coefficient of friction . A force F is

applied to the block at an angle of with the horizontal. The block will move with a minimum

force F, if

A) tan

B) cot

C) sin

D) cos

Answer [A]

Hints:

tan , by solving horizontal and vertical equation., and equating

dy/dx=0.

33) The potential energy ( in joule) of a body of mass 2 kg moving in X-Y plane is given by

U=6x+8y, where the position coordinates x and y are measured in metres. If the body is at rest

at point (6m, 4m) at time t=0, then it will cross the y-axis at time tequal to:

A) 1s

B) 2s

C) 3s

D) 4s

Answer [B]

Hints:

Potential Energy (U) =6x+8y

Mass=2 kg

Body is at rest at (6,4) at t = 0

93

86

)//(/

4

ji

j

ji

onacceleratimassi

dydudxdudxduf

Particle will cross the Y-axis, when x=0. At x=6, v=0; distance=-6, U=o, a=-3

sec2

36

3

/6

,

222

t

t

tV

atUV

smV

Weget

asUV

ngSubstituti

34) A wind-powered generator converts wind energy into electrical energy. Assume that the

generator converts a fixed fraction of the wind energy intercepted by its blades into electrical

energy. For wind speed v, the electrical power output will be proportional to:

A) v

B) v2

C) v3

D) v4

Answer [C] Hints: Bernouli’s Theorem

35) n small balls, each of mass m, impinge elastically each second on a surface with velocity u.

The force experienced by the surface will be:

A) mnu

B) 2mnu

C) 4mnu

D) mnu2

1

Answer [B]

Hints:

Let there be “n” ball striking per second.

For 1 ball-

t

pF

mnvP

mvP

2

,2

=2mnv

Hence, (B).

36) A particle falls from a height h on a fixed horizontal surface and rebounds. If e is the

coefficient of restitution, then the total distance travelled by the particle before it stops

rebounding is:

A) h 2

2

1

1

e

e

B) h 2

2

1

1

e

e

C) h/2 2

2

1

1

e

e

D) h 2

2

1

1

e

e

Answer [A]

Hints:

Total Distance travelled

.,

)1

1(

.....)1(.2

...22

2

2

22

42

AHence

e

eh

heheh

heheh

37) When W joule of work is done by a flywheel, its frequency of rotation increases from v1Hz

to v2Hz. The moment of inertia of the flywheel about its axis of rotation is given by:

A) )(2 2

1

2

2

2 vv

W

B) )(2 2

1

2

2

2 vv

W

C) )(4 2

1

2

2

2 vv

W

D) )(2 2

1

2

2

2 vv

W

Answer [A]

Hints:

Work done

.,

)(2/

2/)(2

2,

2

1

2

2

2

2

1

2

2

2

22

AHence

VVWI

VV

Now

I

if

38) A disc is rotating with angular velocity w . A force F acts at a point whose position vector

with respect to the axis of rotation is r . The power associated with the torque due to the force is

given by:

A) wFr

).*(

B) wFr

*)*(

C) )*.( wFr

D) )*(* wFr

Answer [A]

Hints:

.,

).(

.

AHence

Fr

Power

39) An extremely small and dense neutron star of mass M and radius R is rotating with angular

velocity w. If an object is placed at its equator, then it will stuck to it due to gravity, if

A)M>G

Rw

B)M>G

wR 22

C)M>G

wR 23

D)M>G

wR 32

Answer [C]

Hints:

.,

/

0/

0

,,

23

22

2

2

CHence

GRM

RRGM

Rg

ifedstayattachObject

rgg eff

40) A body of mass m is dropped from a height nR above the surface of the earth (here Ris the

radius of the earth). The speed at which the body hits the surface is:

A)1

2

n

gR

B)1

2

n

gR

C)1

2

n

gRn

D)1

2

n

gRn

Answer [D]

Hints:

By Applying conservation of energy:

.,

1/2

/2/1)/(0 2

DHence

ngnRV

RGMmmvRnRGMm

UU fi

41) The volume of a wire remains unchanged when the wire is subjected to a certain tension.

Then Poisson’s ratio of the material of the wire is:

A) 0.25

B) 0.4

C) 0.5

D) 0.75

Answer [C]

Hints:

As the length and radius is changed, Volume remains same:

.,

'5.02/1

21)1(

)1()1(

)1()1(

))((

:,'

2

222

22

CHence

sratiopoisson

l

lr

r

r

r

l

l

r

r

l

l

r

rr

l

lllr

rrlllr

rrNewR

llNewLsaysLet

42) Equal masses of two substances of density 1 and

2 are mixed together. The density of the

mixture would be:

A) 212

1

B) 21

C)21

21

D)21

212

Answer [D]

Hints:

Let us say mass of each substance is”M”.

.,

/2

/

2121

21

DHence

MMMM

etotalvolum

totalmassmix

43) A vessel contains oil (density 0.8 g cm-3) over mercury (density 13.6 g cm-3). A

homogeneous sphere floats with half volume immersed in mercury and other half immersed in

oil. The density of the material of the sphere (in g cm-3) is:

A) 3.3

B) 6.4

C) 7.2

D) 12.8

Answer [C]

Hints:

F (Body) = mg.

Let the total volume be “V”.

.,

2.7

2/8.02/6.13

CHence

gVgVgV

body

body

44) A cylindrical tank having cross-sectional area A is filled with water to a height of 2m. A

circular hole of cross sectional area a is opened at a height of 75cm from the bottom. If a/A =

2.0 then the velocity with which the water emerges from the hole is:

A) 4.9 ms-1

B) 4.95 ms-1

C) 5.0 ms-1

D) 5.5 ms-1

Answer [D]

Hints:

Velocity is given by the formula

2)(1

2

A

a

gh

Substituting h=12.5 and a/A=0.2,

We get, V=5.5m/sec

Hence, D.

45) If the displacement (x) and velocity (v) of a particle executing simple harmonic motion are

related through the expression 4v2=25-x2, then its time period is given by:

A)

B) 2

C) 4

D) 6

Answer [C]

Hints:

Comparing the equation

.""

4/2,2/1,

)(,)25(4/1 222222

iscorrectC

THence

xAVwithxV

46) A simple pendulum attatched to the ceiling of a stationary lift has a time period T. The

distance y covered by the lift moving upwards varies with time t as y=t2, where y is in metre

and t in second. If g=10ms-2, then the time period of the pendulum will be:

A) T5

4

B) T6

5

C) T4

5

D) T5

6

Answer [B]

Hints:

T= 2/2&/2 glgl , in case the lift is stationary and accelerating

upward respectively.

Since,

.2

2/

2

y

y

a

tdtdyV

ty

.,

6/5

10/12/

BHence

TT

TT

47) The extension in a string, obeying Hooke’s law is x. The speed of the wave in the stretched

string is v. If the extension of the string is increased to 1.5x, then the speed of the wave in the

string will be:

A) 1.22 v

B) 0.61 v

C) 1.50 v

D) 0.75 v

Answer [A]

Hints:

As we know that,

.,

.22.1

5.1/

5.1

,

,

AHence

VV

x

xVV

xV

xV

extensionxV

ExtensionT

TV

48) When the temperature of a rod increases from t to t+ t , the moment of inertia of the rod

increases from I to II . If the coefficient of linear expansion of the rod is , then the ratio

of I

Iis:

A)t

t

B)t

t2

C) t

D) 2 t

Answer [D]

Hints:

.,

2/

/

/2/

2

DHence

tII

tRR

RRII

rI

49) The ratio of adiabatic bulk modulus and isothermal bulk modulus of a gas is

v

p

C

C

A) 1

B)

C)1

D)

1

Answer [B]

Hints:

We know that:

Adiabatic bulk modulus= ϒρ

Isothermal bulk modulus=ρ

Ratio=ϒ

Hence, B

50) An ideal gas is expanded isothermally from volume v1to v2 and then compressed

adiabatically to original volume v1. If the initial pressure is P1, the final pressure P3, and net

work done is W then:

A) P3 >P1, W > 0

B) P3 <P1, W < 0

C) P3 >P1, W < 0

D) P3 =P1, W = 0

Answer [C]

Hints:

Work is negative, 13 PP

Option C is correct.

51) A vessel of volume V contains an ideal gas at temperature T and pressure P. The gas is

allowed to leak till its pressure falls to P’. Assuming that the temperature remains constant

during leakage, the number of moles of the gas that have leaked is:

A) )( PPRT

V

B) )(2

PPRT

V

C) )( PPRT

V

D) )(2

PPRT

V

Answer [C]

Hints:

From Ideal gas equation:

PV= nRT

Initially:

.,

/

:,.

),(

21

2

1

CHence

PPRTVnn

leakedofmolesNO

leakageAfterRTnVP

RTnPV

52) Two spheres A and B having radii 3 cm and 5 cm respectively are coated with carbon black

on their outer surfaces. The wavelengths of maximum intensity of emitted radiations are 300

nm and 500 nm respectively. If the powers radiated are QA and QB respectively then QA/QB is:

A)3

5

B)3

5

C)

2

3

5

D)

4

3

5

Answer: [C]

Hints:

Wein’s displacement law

Now,As we know that:

Power Ratio,

".",3/5/

:&,4,

///

2

2

424244

CHenceQQ

RTgSustitutinrAWhere

TrTreTAeTAQQ

BA

BBAAAAba BB

53) A metallic sphere of radius R is charged to a potential V. The magnitude of electric field at

a distance r (r>R) from the centre of the sphere is:

A)r

V

B)2R

Vr

C)2r

VR

D) 0

Answer [C]

Hints:

At a distance “r” from the Centre, where r>R:

3/5300/500//, maxmax

max

ABBA TTHence

bT

.,

/

:""

/

/

/&/

2

2

CHence

rVRE

inEQngSubstituti

KVRQ

RKQV

rKQERKQV

54) A particle of mass m and charge +q is midway between two charged particles each having

a charge +q, and at a distance 2L apart. The middle charge is displaced slightly along the line

joining the fixed charges and released. The time period of oscillation is proportional to:

A) L1/2

B) L

C) L3/2

D) L2

Answer [C]

Hints:

As the particle is placed mid-way between two charge +q. Force on the

central charge by both the charge particle will be equal and away from it.

If the central particle is displaced by “x”,then the net force acting on it will

be.

Fnet= F2-F1

.,

:,

/2

)..(,

/4

,

/4

:,

)/()/(

2/3

2

32

32

2222

CHence

LT

wegetByequating

T

MHforSxaNow

Lxkqa

maFnetNow

LxKqFnet

wegetSolving

xLKqxLKq

55) If n drops each of capacitance C and charged to a potential V, coalesce to form a big drop,

the ratio of the energy stored in the big drop to that in each small drop will be:

A) n:1

B) n4/3 : 1

C) n5/3 : 1

D) n2 : 1

Answer [C]

Hints:

Let the Charge on each drop is q.

C=q/V

Charge on final drop=Q= nq

Let radius of each small drop is “r” and big drop is “R”.

V=nv (Volume)

.,

1/2/1/'2/1/'

:

//'

)(

3/4.3/4

3/5

3/2

3/1

33

CHence

nqvQVuU

rgyRatioofene

rKqnRKQV

rnR

rnR

56) Two identical capacitors have the same capacitance C. One of them is charged to potential

v1and the other to v2. The negative ends of the capacitors are connected together. When the

positive ends are also connected the decrease in energy of the combined system is:

A) 2

2

2

14

1vvC

B) 2

2

2

14

1vvC

C) 22

2

2

14

1vvC

D) 22

2

2

14

1vvC

Answer [C]

Hints:

From charge conservation:

.,

)(4/1,

)(

2/

2

2

21

21

2121

1

CHence

VVCWeget

UUSystemU

VVV

CVCVCVCV

QQqq

fi

57) What is the total current supplied by the battery to the circuit shown in the figure below?

A) 1 A

B) 4 A

C) 6 A

D) 8 A

Answer [B]

Hints:

We know that:

:

6

eq

eqb

eqbb

gforRCalculatin

Ri

RiV

2 ohm is parallel to 6 ohm and in series with 1.5 ohm. The combination is

parallel to 3 ohm resistance.

R (Effective)= 3/2 Ω

I (Current in the circuit)=6/1.5=4 Ampere

Hence “B”.

58) An electric bulb has a rating of 500 W, 100 V. It is used in a circuit having a 200 V supply.

What resistance must be connected in series with the bulb so that it delivers 500 W?

A) 10 ohm

B) 20 ohm

C) 30 ohm

D) 40 ohm

Answer [B]

Hints:

Let us say resistance of bulb is “R”,

.,

20

200)(5

200)(

)(5

20500

:

20

/)100(500

/

,

2

2

2

2

BHence

R

RR

VVoltageSupply

currentcircuitAi

i

RiP

Now

R

R

RVP

ext

bulbext

59) A charged particle is released from rest in a region of steady and uniform electric and

magnetic fields which are parallel to each other. The particle will move in a

A) Straight line

B) Circle

C) Helix

D) Cycloid

Answer [A]

Hints:

As the charge particle is released from rest:

enet

b

b

e

benet

FF

FV

BVqF

EqF

FFF

BE

0,0

)(

Since Electric field is acting along the straight line, and force on particle

will be always in the direction of electric field. Hence the particle will show

straight-line motion.

Hence, A.

60) A particle of charge q moves with a velocity v =ai in a magnetic field B=bj +ck where a, b

and c are constants. The magnitude of the force experienced by the particle is:

A) 0

B) qa(b+c)

C) qa(b2-c2)1/2

D) qa(b2+c2)1/2

Answer [D]

Hints:

.,

)ˆˆ(

)ˆˆ(ˆ(

)(

22

DHence

cbqaF

jackabq

kcjbiaq

BVqF

B

B

61) An L-C-R series circuit with R =100 is connected to a 200 V, 50Hz a.c. source When

only the capacitance is removed, the voltage leads the current by 60 C and when only the

inductance is removed, the current leads the voltage by 60 C . The current in the circuit is:

A) 3

2A

B) 2

3A

C) 1 A

D) 2 A

Answer [D]

Hints:

Consider cases like, When capacitance is removed the circuit is L-R Circuit,

and when Inductor is removed the circuit is R-C Circuit, and calculate :

3100,3100.& clcl XXXX

Current in L-C-R circuit is

:

.,

2

2

DHence

Ai

XXRZ

XXRZ

ziV

cl

cl

62) When a glass prism of refracting angle 60 C is immersed in a liquid, its angle of minimum

deviation is 30 C . The critical angle of glass prism with respect to the liquid medium (in

degree)

A) 42

B) 45

C) 50

D) 52

Answer [B]

Hints:

For minimum deviation:

Angle of incidence(i)= angle of emergence(e)

Angle of refraction(r)= prism angle(A)/2

Now as we know that:

"."

45

)2/1(sin

2/1/

:,

)2/sin()2/sin(

)sin(sin

1

BHence

gleCriticalan

nn

valuesabovengSubstituti

AnAn

rnin

prismmedium

prismmmedium

prismmedium

63) A convex lens and a concave lens are placed in contact. The ratio of magnitude of the

power of the convex lens to that of the concave lens is 4 : 3. If the focal length of the convex

lens is 12 cm, then the focal length of the combination will be:

A) 16 cm

B) 24 cm

C) 32 cm

D) 48 cm

Answer [D]

Hints:

As we know that:

Power (P)=1/focal length(f)

Here 21 & ff are focal lengths of convex and concave lens respectively.

cmf

ff

pP

16123/4

3/4/

3/4/

2

12

21

Hence focal length of concave lens=16 cm

Focal length of Combination can be calculated by the formula:

21 /1/1/1 fff ncombinatio

Keeping f2 negative, as for concave lens focal length is negative.

We get focal length of combination= 48 cm.

Hence,D.

64) Monochromatic light is refracted from air into glass of refractive index . The ratio of the

wavelength of the incident and refracted waves is:

A) 1 : 1

B) 1 : µ

C) µ : 1

D) µ2 : 1

Answer [C]

Hints:

Speed of light in any medium is given by:

/c Where “c” is the speed of light in vacuum, and is the refractive

index of the medium.

Velocity of light in air = c

Velocity of light in glass =c/

We know that:

".",

/

)(,

/

)(/)()(

CHence

c

cVV

sameremainsfrequecy

f

V

f

V

frequencyfspeedvwavelength

ri

r

r

i

i

refractedincident

65) Two coherent sources of light of intensity ratio n are employed in an interference

experiment. The ratio of the intensities of the maxima and minima in the interference pattern is:

A)

1

1

n

n

B)

2

1

1

n

n

C)

1

1

n

n

D)

2

1

1

n

n

Answer [D]

Hints:

Given:

221min

2

21max )(

/

III

III

nII BA

After calculating their ratio, and substituting “n”, we get

2)1

1(

n

n

Hence,D.

66) The wavelength of K line from an element of atomic number 41 is . Then the

wavelength of K line of an element of atomic number 21 is:

A) 4 λ

B) λ/4

C) 3.08 λ

D) 0.26 λ

Answer [A]

Hints:

Apply Moseley’s law.

67) If a hydron atom at rest emits a photon of wavelength , then the recoil speed of the atom

of mass m is given by:

A) h/mλ

B) mh/λ

C) mhλ

D) none of these

Answer: (A), self explanatory.

68) A nucleus ruptures into two parts which have their velocities in the ratio of 2: 1. What will

be the ratio of their nuclear sizes (radii)?

A) 21/3 : 1

B) 1 : 21/3

C) 31/2 : 1

D) 1 : 31/2

Answer [B]

Hints:

CBA

momentum of B & C are equal.

)(2// givenVVmm

VmVm

bccb

ccbb

Hence there area will also be in the same ratio.

Calculating the ratio, we get:

BHence

RR bc

,

)2

1(/ 3/1

69) The half-life of a radioactive sample is 6.93 days. After how many days will only 20

1 of

the sample be left over? [Take loge (20) =3]

A) 20 days

B) 27 days

C) 30 days

D) 35 days

Answer [C]

Hints:

Use teNN

, Hence(c) is the correct option.

70) Three forces start acting simultaneously on a particle moving with velocity v. These forces

are represented in magnitude and direction by three sides of a triangle taken in the same order.

The particle will now move with a velocity:

A) less than ν

B) more than ν

C) ν only

D) cannot say

Answer [C]

Hints:

As the three forces are acting along the three sides of the triangle in the

same order, hence their net forces will cancel out each other and will be

zero, Hence acceleration=0, thus velocity will remain constant.

71) Two small satellites move in circular orbits around the earth respectively, at distances

r and (r + r) from the centre of the earth. If their time periods are T and T+ T

( r >>r, T>>T), then

A) r

rTT

2

3

B) r

rTT

2

3

C) r

rTT

3

2

D) r

rTT

Answer [A]

Hints:

According to kepler’s law:

32 RT

Then, take log on both sides and differentiate with respect to “x”.

r

rTT

2/3

Hence,(A).

72) If n1, n2 and n3 are the fundamental frequencies of three segments into which a string

is divided, then the original fundamental frequency n of the string is given by:

A) 321

1111

nnnn

B)

321

1111

nnnn

C) 321 nnnn

D) 321 nnnn

Answer [A]

Hints:

Let the length of three segments be

3,21 &, xxx Respectively.

).(,

/1/1/1/1,

2/2/2/2/

2/

2/

2/

:

2/

:

,

321

321

33

22

11

321

AThus

nnnnHence

nvnvnvnv

xvn

xvn

xvn

Similarly

xvn

hatNowweknowt

xxxx

73) If the electric flux entering and leaving an enclosed surface are 1 and 2 respectively,

then the electric charge inside the surface will be:

A) (ɸ2-ɸ1)/ε0

B) (ɸ1+ɸ2)/ε0

C) (ɸ1-ɸ2)/ε0

D) ε0(ɸ1+ɸ2)

Answer: (D)

74) What is the reading of ammeter shown in the figure below?

A) 3 A

B) 4 A

C) 1.5A

D) 6 A

Answer [A]

Hints:

Current through capacitor is zero in steady state. So current will not flow

through capacitor. 2Ω &2Ω Re in parallel, and their equivalent will be 1Ω.

Now, I=V/R=3/1=3 Ampere’

Hence(A).

75) An astronaut is looking down on earth’s surface from a space shuttle at an altitude of

500 km. Assuming that the astronaut’s pupil’s diameter is 5 mm and the wavelength of visible

light is 500 nm, the astronaut will be able to resolve linear objects of the size of about:

A) 0.5 m

B) 5 m

C) 50 m

D) 500 m

Answer [C]

Hints:

50 m is the correct answer.

76) The velocity of the most energetic electrons emitted from metallic surface is double when

the frequency v of the incident radiation is doubled. The work function of the metal is

A) 0

B) hν/3

C) hν/2

D) 2hν/3

Answer [D]

Hints:

We know that,

Kinetic energy= hmv2

2/1

Using this equation, we get:

.,

3/2

dHence

hd

77) Two identical photocathodes receive light of frequencies f1 and f2. If the velocities of the

photoelectrons (of mass m) coming out are v1 and v2 respectively, then

A) )(2

21

2

2

2

1 ffm

hvv

B)

2/1

2121 )(2

ff

m

hvv

C) )(2

21

2

2

2

1 ffm

hvv

D)

2/1

2121 )(2

ff

m

hvv

Answer [A]

Hints:

Same as above question, take reference.

78) If the time period of a simple pendulum is glT /2 , then the fractional error in

acceleration due to gravity is:

A) 2

24

T

l

B) T

T

l

l

2

C) T

T

l

l

2

D) none of these.

Answer [C]

Hints:

Given:

"."

/2//

)/(4

)/(4

:),(

/2

22

22

cHence

TTllgg

Tlg

glT

sidesbothSquaring

glT

79) A ball moving with a momentum 5 kg ms-1 strikes a wall. If the initial and final

momenta make equal angles of 45o, then the magnitude in change in momentum is

A) 5.05 kg ms-1

B) 7.07 kg ms-1

C) 10 kg ms-1

D) 0 kg ms-1

Answer [B]

Hints:

Change in momentum= if pp

As both makes 45 degree each, hence both will be perpendicular to each

other.

22 PPP

Hence “B” is the correct option.

80) A force Fy (3x + 2)N is acting on a body. The work done by this force if it tends to

displace the body from x= 0 m to x= 4 m will be

A) 32 J

B) 16 J

C) 0 J

D) (12x + 8) J

Answer [C]

Hints:

As here yF is the acting force .Therefore the force is acting along y-

direction. Thus no work will be done as displacement is along x-direction.

It’s a tricky Question.

“C” is the correct option.

81) A motorcyclist is moving along a circular path of radius 500 m. If his speed is 30 ms-1

then the type(s) of acceleration and the resultant acceleration will be

A) one and 18 ms-2

B) two and 2 ms-2

C) two and 2.7 ms-1

D) none of these

Answer [A,D]

Hints:

In circular motion, there are two types of acceleration.

Tangential and centripetal.

Tangential acceleration is responsible for change in speed, as here the

speed is constant. Hence tangential acceleration =0.

Centripetal acceleration is given by formula= Rv /2

By calculating we get “a”= 1.82sec

m

Here in option “A” there is a printing mistake, as its 1.8 not 18. So “A” and

“D” are correct.

82) Two heavy spheres of mass m are kept separated by a distance 2r. The gravitational

field and potential at the midpoint of the line joining the centres of the spheres are:

A) 2

2

r

Gm and 0

B) 0 and r

Gm2

C) 0 and r

Gm2

D) none of these

Answer [B]

Hints:

As we know that, gravitational forces are always attractive, and hence will

act in opposite direction, thus will be zero.

Hence gravitational field will not be present at the midpoint of the line

joining.

Vnet= -Gm/R + (-Gm/R)

=-2Gm/R.

Hence “B” is the correct answer.

83) The transverse displacement of a vibrating string is

)120cos(3

2sin06.0 txy

If mass per unit length of the string is 2x10-2 kg m-1, the tension in the string will be

A) 648 N

B) 864 N

C) 684 N

D) 468 N

Answer [A]

Hints:

Use kVTV /&/

84) A and B are two charged bodies with charges q1 and q2, and with a total charge q. If the

force between A and B has to be maximum, then the ratio of charge on A to the total charge

will be

A) 1

B) 0.25

C) 2

D) 0.5

Answer [B]

85) From the diagram given below, it can be calculated that the resistivity of the material of the

rod is

A) 3 x 10-6 Ω cm

B) 2.8 x 10-5 Ω m

C) 1.5 x 10-3 Ω cm

D) 12 x 10-4 Ω m-1

Answer [B]

Hints:

We know that:

V=iR

R=V/i

=40/3Ω

lRA

A

lR

/

By calculating, we get:

“B” as the correct option.

86) If an electron has to be accelerated by a cyclotron to a kinetic energy of 5 MeV with a

magnetic field of strength 2T, which one of the following will be correct?

A) Cyclotron frequency, f=5.6 x 1014 Hz

B) Speed, v= 1.32 x 109 ms-1

C) r can be calculated using r=(mv/qB)

D) None of the above

Answer [C]

Hints:

“C” is the correct option as we know that, r=mV/qB.

87) The inductance of two coils are 10 H and 20 H respectively. If the mutual inductance

between the coils is 5 H and the mutual fluxes aid, each other, then the total inductance of the

circuit will be:

A) 15 H

B) 25 H

C) 9.37 H

D) none of these

Answer [D]

Hints:

MLLLeq 221

Substituting the values, we get:

L (equivelant) = 20H, 40H

Hence ”D”.

88) In Young’s double-slit experiment, if the superimposing waves have amplitude a0 and

intensity I0, then the average intensity of the light in the fringe pattern formed on a screen will

be

A) 6 I0

B) 4 I0

C) 2 I0

D) I0 /2

Answer [C]

Hints:

As I is intensity of superimposing waves.

."",

2

2/

0

4

minmax

min

max

iscorrectCHence

I

IIIavg

I

II

89) Sodium light has two wavelengths and they are 5890 A and 5896 A . This light falls on a

slit of width 2 m. The diffraction pattern formed is observed on a screen. The spacing

between the first maxima for the two wavelengths is:

A) 9 x 10-4 m

B) 3 x 10-8 m

C) 6 x 10-4 m

D) 8 x 10-6 m

90) A transistor is connected in CE configuration. In the collector circuit VCC = 8 V, RC= 800

. If the voltage drop across RC is 0.5 V and =0.96, then IC, VCE and IB will be:

A) 0.625 mA, 7.5 V and 0.026 mA

B) 0.625 mA, 0.026 V and 7.5 mA

C) 1 mA, 8.5 V and 0.028 mA

D) 6.91 mA, 9.6 V and 0.29 mA

Answer [A]

Hints:

.,

026.0

651.0

96.0/

625.0

800/5.0

5.0

AHence

mA

III

mAI

II

III

mA

I

RI

CEB

E

Ec

EBc

c

c

91) The output (Y) of the combination of gates is:

A) Y= A . B

B) Y = A + B

C) Y= BA.

D) Y= BA

Answer [A]

92) From the diagram given below, it can be calculated that the modulation factor is:

A) 4

B) 2

C) 0.6

D) 8

Answer [C]

Hints:

As we know:

Modulation factor= Amplitude change of carrier wave/Amplitude of normal

carrier wave

Amplitude change of carrier wave= 8-5= 3

Amplitude of Normal Carrier wave=5

m-factor = 3/5=0.6

Hence, “C”.

93) If the velocity of surface wave (v) depends upon surface tension (T), coefficient of

Viscosity ( ) and density ( ), then the expression for v will be:

A) n

T

2

B) n

T

C) 2T

n

D) n

Answer [B]

Hints:

By Dimensional Analysis,

94) Two balls are thrown horizontally in opposite directions from the same point from a height

h with velocities 4 m/s and 3 m/s. The separation between the two balls when their velocities

are perpendicular will be

A) 6.5 m

B) 5.25 m

C) 2.45 m

D) none of these

Answer [C]

Hints:

".",

422.2

10/327

10/32

:,

,0.

ˆˆ3

ˆˆ4

2/1

CHence

m

tUS

t

WegetSolving

VV

jgtiV

jgtiV

BA

B

A

95) Three blocks of mass 3 kg, 2kg and 1 kg are placed side-by-side on a smooth surface as

shown in figure. If a horizontal force of 24 N is applied on 3 kg block, then the netforce on 2 kg

block will be

A) 2 N

B) 4 N

C) 8 N

D) 12 N

Answer [C]

Hints:

Net Acceleration= Net Force/Net Mass=24/6=4

Fnet on 2 kg block i= 2* 4 N

=8 N .

96) A block of mass 10 kg is moving in x-axis direction with constant speed of 10 m/sis

subjected to a retarding force F = -0.1 x J/m during its travel from x=20 m to x=30m. What will

be the final KE of the block?

A) 500 J

B) 475 J

C) 450 J

D) 400 J

Answer [B]

Hints:

As we know that:

W

joulek

k

kk

KWnet

N

xdx

dxFW

f

f

if

xnet

475

50025

25

25

1.0

30

20

97) A satellite is orbiting close to the earth above the equator with a period of rotation of 1.5

hours. 1f it is above a point x on the equator at some time, then after how much time will it be

above x again?

A) 1.5 hours if it is rotating from east to west

B) 1.6 hours if it is rotating from west to east

C) 1.6 hours if it is rotating from east to west

D) 1.8 hours if it is rotating from west to east

(B)

98) The speed of the wind passing over the wings of a small aeroplane is 70 m/s and below the

wing is 60 m/s. If the mass of the plane is 1000 kg and the area of the wing is 14 m2, then what

will be the net vertical force on the aeroplane?

A) 620 N upward

B) 920 N upward

C) 620 N downward

D) 920 N downward

Answer [B]

Hints:

Here pressure below aeroplane and above aeroplane will be termed as

ab PP & respectively.

Bernoulli’s equation states that:

"",

920

)()(

780

)(2/

2/12/1

22

22

BHence

Newton

weightAreaPPupwardFnet

UUPP

zUPZUP

ab

baab

aaabbb

99) A liquid is contained in a vertical tube of semi-circular cross-section as shown in figure. If

the angle of contact is zero, then the ratio of the force of surface tension on flat side to curved

side of the tube will be

A) 1/π

B) 3/2π

C) 1/4π

D) 2/π

Answer [D]

Hints:

Force on flat side/Force on curved side= rTrT /2 = /2

Hence,D.

100) A railway engine whistling at a constant frequency moves with a constant speed and it

goes past a stationary observer standing beside the railway track. Then the frequency of (n’) of

the sound heard by the observer with respect to time (t) can be best represented by which of the

following curve?

A)

B)

C)

D)

Answer [D]

Hints:

Using Doppler’s Effect:

fVVVVf s )/(

“D” is the correct option.