question paper with solution of aipmt- 2015in a double slit experiment, the two slits are 1mm apart...

$: QUESTION PAPER WITH SOLUTION OF AIPMT- 2015In a double slit experiment, the two slits are 1mm apart and the screen is placed 1m away. ... The refracting angle of a prism is A, and$
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91. A radiation of energy ‘E’ falls normally on aperfectly reflecting surface. The momentum

transferred to the surface is (C = Velocity of

light) :

(1) 2

E

C(2)

E

C

(3) 2E

C(4) 2

2E

C

Sol.(3)

/E C /E C

Momentum of light E

pC

=

So momentum transferred to the surface

2f i

Ep p

C= − =

92. A ship A is moving Westwards with a speed of

110kmh− and a ship B 100 km South of A ,

is moving Northwards with a speed of 110km h− .

The time after which the distance between thembecomes shortest, is :

(1) 10 2 h (2) 0 h

(3) 5 h (4) 5 2 h

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QUESTION PAPER WITH SOLUTION OF AIPMT- 2015

(HELD ON 3 MAY SUNDAY 2015)rd

Sol.(3)

45°

O

A

B

10kmph

10kmph

10

0k

m

2 210 10 10 2BA

V kmph= + =

distance 100cos 45 50 2OB km= ° =

Time taken to reach the shortest distancebetween

50 2 50 2&

10 2BA

A BV

= =

5sn

t hrs=

93. Three blocks ,A B and C , of masses 4 kg ,

2 kg and 1kg respectively, are in contact on a

frictionless surface, as shown. If a force of 14 N

is applied on the 4 kg block, then the contact

force between A and B is :

A B C

(1) 18N (2) 2N

(3) 6N (4) 8N

2.



Sol.(3) Acceleration of system 214

2 /7

m s= =

contact (2 1) 3(2) 6F a N= + = =

94. The electric field in a certain region is acting

radially outward and is given by E Ar= . AA

charge contained in a sphere of radius ' 'acentred at the origin of the field, will be given by:

(1) 3

0Aaε (2)

2

04 Aaπε

(3) 2

0A aε (4)

3

04 Aaπε

Sol.(4) Flux linked with sphere E ds= ⋅

since electric field is radial. It is alwaysperpendicular to the surface.

So 2(4 )Ar rφ π= ⋅

2( )(4 )A a aφ π= (as r a= )

34A aφ =

Now according to gauss law

0

inq

φε

= ⇒ 0inq φ= ⋅∈

so3

04inq A aπ= ⋅∈

95. ,A B and C are voltmeters of resistance R ,

1.5 R and 3R respectively as shown in the

figure. When some potential difference is applied

between X and Y , the voltmeter readings are

,A B

V V and C

V respectively. Then :

X YA

B

C

(1) A B CV V V≠ ≠ (2) A B CV V V= =

(3) A B C

V V V≠ = (4) A B C

V V V= ≠

Sol.(2) Effective resistance of

(1.5 )(3 )&

1.5 3

R RB C R

R R= =

+

In series sequence V R∝

so voltage across ' 'A = voltage across &B C

Now &B C are parallel so B CV V=

⇒ A B CV V V= =

96. In a double slit experiment, the two slits are

1mm apart and the screen is placed 1m away..

A monochromatic light of wavelength 500 nm

is used. What will be the width of each slit forobtaining ten maxima of double slit within thecentral maxima of single slit pattern ?

(1) 0.02 mm (2) 0.2 mm

(3) 0.1mm (4) 0.5mm

Sol.(2) Angular width of central maxima in double slit

experiment d

λ=

Angular width of central maxima in single slit

experiment 2

'd

λ=

According to the question

10 2

'd d

λ λ=

⇒ ' 0.2 0.2d d mm= =

97. One mole of an ideal diatomic gas undergoes a

transition from A to B along a path AB asshown in the figure,

5 A

B

4 6

2P kPa (in )

3(in )V m

The change in internal energy of the gas duringthe transition is :

3.



(1) 12 kJ− (2) 20 kJ

(3) 20 kJ− (4) 20 J

Sol.(3) VU nC T∆ = ∆ & PV

TnR

=

so 2 2 1 1

2 1

PV PVT T T

nR

−∆ = − =

so 2 2 1 1 2 2 1 1

1 1

PV PV PV PVnRU

nRγ γ

− − ∆ = =

− −

⇒

38 1020

2 / 5U kJ

− ×∆ = = −

98. A rod of weight W is supported by two parallel

knife edges A and B and is in equilibrium in a

horizontal position. The knives are at a distance

d from each other. The centre of mass of the

rod is at distance x from A . The normal

reaction on A is :

(1) ( )W d x

d

−(2)

Wx

d

(3) Wd

x(4)

( )W d x

x

−

Sol.(1) By torque balancing about B

A

AN

BN

W

B

( ) ( )AN d W d x= −

⇒ ( )

A

W d xN

d

−=

99. Kepler’s third law states that square of period of

revolution (T ) of a planet around the sun, is

proportional to third power of average distance

r between sun and planet

i.e. 2 3T Kr=

here K is constant.

If the masses of sun and planet are M and mrespectively then as per Newton’s law ofgravitation force of attraction between them is :

2

GMmF

r= , here G is gravitational constgant

The relation between G and K is described

as :

(1) 1

KG

= (2) 24GK π=

(3) 24GMK π= (4) K G=

Sol.(3)

22 34

T rGM

π= ⋅

Comparing

24K

GM

π=

100. If in a p n− junction, a square input signal of

10V is applied, as shown,

5V+

LR

5V-

then the output across LR will be :

(1)

5V

(2) 10V-

4.



(3)

10V

(4) 5V-

Sol.(1) Half wave rectifier

101. Two particles of masses 1 2,m m move with initial

velocities 1

u and 2

u . On collision, one of the

particles get excited to higher level, after

absorbing energy ε . If final velocities of particles

be 1

v and 2

v then we must have :

(1) 2 2 2 2 2 2 2 2

1 1 2 2 1 1 2 2

1 1 1 1

2 2 2 2m u m u m v m vε+ + = +

(2) 2 2 2 2

1 1 2 2 1 1 2 2m u m u m v m vε+ − = +

(3) 2 2 2 2

1 1 2 2 1 1 2 2

1 1 1 1

2 2 2 2m u m u m v m v ε+ = + −

(4) 2 2 2 2

1 1 2 2 1 1 2 2

1 1 1 1

2 2 2 2m u m u m v m vε+ − = +

Sol.(4) Energy will always be conserved so

initial final. . . .K E K E= + Excitation energy

2 2 2 2

1 1 2 2 1 1 2 2

1 1 1 1

2 2 2 2m u m u m v m v ε+ = + +

102. Which of the following figures represent thevariation of particle momentum and theassociated de-Broglie wavelength ?

(1)

P

λ

(2)

P

λ

(3)

P

λ

(4)

P

λ

Sol.(3)hc

Pλ

= ⇒ 1

Pλ

∝ (Rectangular hyperbola)

103. The approximate depth of an ocean is 2700 m .

The compressibi l i ty of water is

11 145.4 10 Pa

− −× and density of water is

3 310 /kg m . What fractional compression of

water will be obtained at the bottom of the ocean?

(1) 2

1.4 10−× (2)

20.8 10

−×

(3) 21.0 10−× (4) 21.2 10−×

Sol.(4) As we know

PB

V

V

=∆

SoV P

V B

∆=

Now P ghρ= & compressibility 1

' 'KB

=

so ( )V

gh KV

ρ∆

=

3 1110 9.8 2700 45.4 10−= × × × ×21.201 10−= ×

104. The two ends of a metal rod are maintained at

temperatures 100 C° and 110 C° . The rate of

heat flow in the rod is found to be 4.0 /J s . If

the ends are maintained at temperatures

210 C° , the rate of heat flow will be :

(1) 4.0 /J s (2) 44.0 /J s

(3) 16.8 /J s (4) 8.0 /J s

5.



Sol.(1) Rate of heat flow ∝ temperature differencebetween two ends

⇒ 2 1( )dQ

T Tdt

∝ −

Here temperature difference in both case is

same (i.e. 10 C° )

So, rate of heat flow will also be same

So, 4 /dQ

J sdt

=

105. A particle of unit mass undergoes one-dimensional motion such that its velocity varies

according to 2( ) nv x xβ −= , where β and n

are constants and x is the position of the

particle. The acceleration of the particle as a

foundation of x , is given by :

(1) 2 4 12 nn eβ − +− (2) 2 2 12 n

n xβ − −−

(3) 2 4 12 nn xβ − −− (4) 2 2 12 n

xβ − +−

Sol.(3) 2nv xβ −=

so2 12 ndv

n xdx

β − −= −

Now2 2 1( )( 2 )n ndv

a v x n xdx

β β− − −= = −

⇒ 2 4 12 na n xβ − −= −

106. The refracting angle of a prism is A , and

refractive index of the material of the prism is

cot ( / 2)A . The angle of minimum deviation is:

(1) 180 2A° + (2) 180 3A° −

(3) 180 2A° − (4) 90 A° −

Sol.(3)

sin2

sin2

m A

A

δ

µ

+ =

∴

sin2

cot ( / 2)

sin2

m A

AA

δ + =

⇒ cos ( / 2) sin2

mA

Aδ +

=

⇒ 90 / 22

mA

Aδ +

° − =

⇒ 180 2m Aδ = ° −

107. A particle is executing SHM along a straight

line. Its velocities at distances 1x and

2x from

the mean position are 1

V and 2

V , respectively..

Its time period is :

(1)

2 2

1 2

2 2

1 2

2V V

x xπ

−

−(2)

2 2

1 2

2 2

1 2

2x x

V Vπ

+

+

(3)

2 2

2 1

2 2

1 2

2x x

V Vπ

−

−(4)

2 2

1 2

2 2

1 2

2V V

x xπ

+

+

Sol.(3) 2 2

1 1V A xω= − & 2 2

2 2V A xω= −

solving these two equations we get

2 2

1 2

2 2

2 1

V V

x xω

−=

−

⇒2 2

2 1

2 2

1 2

2x x

TV V

π−

=−

108. Two similar springs P and Q have spring

constants PK and QK , such that P Q

K K> .

They are stretched, first by the same amount

(case a ) , then by the same force (case b ).

The work done by the springs PW and QW are

related as, in case (a) and case (b); respectively:

(1) ;P Q Q PW W W W< <

(2) ;P Q P QW W W W= >

(3) ;P Q P QW W W W= =

(4) ;P Q Q PW W W W> >

6.



Sol.(1) Case (a)

21

2P PW K x= − ,

21

2Q QW K x= −

P QW W<

Case (b)

21

2P

P

FW

K= − ,

21

2Q

Q

FW

K= −

( )P QW W>

109. Consider 3rd orbit of He

+ (Helium), using non-

relativistic approach, the speed of electron in this

orbit will be [given 99 10K = × constant, 2Z =

and h (Planck’s Constant) 346.6 10 Js

−= × ]

(1) 83.0 10 /m s× (2) 62.92 10 /m s×

(3) 61.46 10 /m s× (4) 60.73 10 /m s×

Sol.(3) For H − like atoms

62.188 10 /Z

v m sn

= × ×

here 2Z = , 3n =

61.46 10 /v m s= ×

110. A wire carrying current I has the shape asshown in adjoining figure. Linear parts of the wire

are very long and parallel to X − axis while

semicircular portion of radius R is is lying in

Y Z− plane. Magnetic field at point O is :

R

I

O Y

Z

X

II

(1) ( )0 ˆˆ 24

IB i k

R

µπ

π= −

(2) ( )0 ˆˆ 24

IB i k

R

µπ

π= +

(3) ( )0 ˆˆ 24

IB i k

R

µπ

π= − −

(4) ( )0 ˆˆ 24

IB i k

R

µπ

π= − +

Sol.(4)

2

31 II

' 'B due to segment ‘1’

01

ˆ[sin 90 sin 0] ( )4

IB k

R

µ

π= ° + −

01 3

ˆ( )4

IB k B

R

µ

π= − =

B due to segment ' 2 '

02

ˆ( )4

IB i

R

µ= −

so ' 'B at center 1 2 3c

B B B B= + +

⇒ ( )0 ˆˆ 24

C

IB i k

R

µπ

π= − +

111. A particle of mass m is driven by a machine

that delivers a constant power k watts. If the

particle starts from rest the force on the particle

at time t is :

(1) 1/ 21

2mk t−

(2) 1/ 2

2

mkt

−

(3) 1/ 2mk t

− (4) 1/ 22mk t−

7.



Sol.(2) P Fv mav= =

⇒dv

K mvdt

=

By integrating the equation

⇒k

vdv dtm

=∫ ∫

⇒2

2

v kt

m= ⇒

2kv t

m=

1/ 22 1

2

dv ka t

dt m

− = =

1 2

2

kF ma m

mt

= =

⇒

2

mkF

t=

112. The fundamental frequency of a closed organ

pipe of length 20cm is equal to the second

overtone of an organ pipe open at both the ends.The length of organ pipe open at both the endsis :

(1) 140cm (2) 80cm

(3) 100cm (4) 120cm

Sol.(4) Fundamental frequency of closed organ pipe

4c

v

l=

2nd overtone frequency of open organ pipe

0

3

2

v

l=

Now0

3

4 2c

v v

l l=

⇒ 0 6 6(20 ) 120cl l cm cm= = =

113. An electron moving in a circular orbit of radius

r makes n rotations per second. The magnetic

field produced at the centre has magnitude :

(1) 0

2

ne

r

µ(2)

0

2

ne

r

µ

π

(3) Zero (4)

2

0n e

r

µ

Sol.(1) Magnetic field due to a circular loop

0 / 2B I rµ=

I ne=

0

2

neB

r

µ=

114. Two identical thin plano-convex glass lenses

(refractive index 1.5 ) each having radius of

curvature of 20cm are placed with their convex

sourfaces in contact at the centre. Theintervening space is filled with oil of refractive

index 1.7 . The focal length of the combination

is :

(1) 50cm (2) 20cm−

(3) 25cm− (4) 50cm−

Sol.(4)

2 1.7n =

1 1.5n =

1 2

1 1 1 1(1.5 1)

20 40f f

= − = =

& 3

1 2 7(1.7 1)

20 100f

− − = − =

Now 1 2 3

1 1 1 1

f f f f= + +

⇒ 1 1 1 7

40 40 100f= + −

⇒ 50f cm= −

8.



115. On observing light from three different starts P ,

Q and R , it was found that intensity of violet

colour is maximum in the spectrum of P , the

intensity of green colour is maximum in the

spectrum of R and the intensity of red colour is

maximum in the spectrum of Q . If PT , QT and

RT are the respective absolute temperatures of

,P Q and R , then it can be concluded from the

above observations that :

(1) P Q RT T T< < (2) P Q RT T T> >

(3) P R QT T T> > (4) P R QT T T< <

Sol.(3) From Wein’s displacement law

1m

Tλ ∝

Now from sequence ' 'VIBGYOR

( ) ( ) ( )m P m R m Qλ λ λ< <

So P R QT T T> >

116. If energy ( )E , velocity ( )V and time ( )T are

chosen as the fundamental quantities, thedimensional formula of surface tension will be :

(1) 2 1 3[ ]E V T− − − (2) 2 1[ ]EV T

− −

(3) 1 2[ ]EV T− − (4) 2 2[ ]EV T

− −

Sol.(4) 2 2( )

F E ES

l l VT= = =

2 2

EV T− − =

117. A Carnot engine, having an efficiency of 1

10η =

as heat engine, is used as a refrigerator. If the

work done on the system is 10 J , the amount

of energy absorbed from the reservoir at lowertemperature is :

(1) 1J (2) 100 J

(3) 99 J (4) 90 J

Sol.(4) For Engine & refrigerators operating betweentwo same temperatures

1

1η

β=

+ ⇒

1 1

10 1 β=

+ ⇒ 9β =

2Q

Wβ = (From the principle of refrigerator)

2910

Q= ⇒ 2 90Q = Joule

118. A mass m moves in a circle on a smooth

horizontal plane with velocity 0v at a radius 0R .

The mass is attached to a string which passesthrough a smooth hole in the plane as shown.

0V

m

0R

The tension in the string is increased gradually

and finally m moves in a circle of radius 0

2

R.

The final value of the kinetic energy is :

(1) 2

0

1

2mv (2)

2

0mv

(3) 2

0

1

4mv (4)

2

02 mv

Sol.(4) Angular momentum remains Constant becauseof the torque of tension is zero.

⇒ i fL L=

⇒ 02

Rmv R mv=

⇒ 02v v=

2 2

0 0

1(2 ) 2

2f

KE m v mv= =

9.



119. For a parallel beam of monochromatic light of

wavelength ' 'λ diffraction is produced by a

single slit whose width ' 'a is of the order of the

wavelength of the light. If ' 'D is the distance of

the screen from the slit, the width of the centralmaxima will be :

(1) 2Da

λ(2)

2D

a

λ

(3) D

a

λ(4)

Da

λ

Sol.(2) Linear width of central maxima

(2 ) 2 2D D Da

λθ θ= = =

θθ

120. A wind with speed 40 m/s blows parallel to the

roof of a house. The area of the roof is 2250m .

Assuming that the pressure inside the house isatmospheric pressure, the force exerted by thewind on the roof and the direction of the force

will be : 3( 1.2 / )

airP kg m=

(1) 52.4 10 N× , downwards

(2) 54.8 10 N× , downwards

(3) 54.8 10 N× , upwards

(4) 52.4 10 N× , upwards

Sol.(4) By Bernaulli’s equation

in 0P P=

in 0V =2

0

10

2P v Pρ+ = +

2

0

1

2P P vρ− = ,

21

2F v Aρ=

52.4 10F = × upward

121. The ratio of the specific heats P

V

C

Cγ= in terms

of degrees of freedom ( n ) is given by :

(1) 12

n +

(2)

11

n

+

(3) 13

n +

(4)

21

n

+

Sol.(4) Fact

122. If radius of the 27

13Al nucleus is taken to be

' 'Al

R then the radius of 125

53 Te nucleus is nearly:

(1)

1/313

53Al

R

(2)

1/353

13Al

R

(3) 5

3Al

R (4) 3

5Al

R

Sol.(3) 1/3R A∝

1/327

125

Al

Te

R

R

=

⇒ 5

3Te AlR R=

123. Figure below shows two paths that may be taken

by a gas to go from a state A to a state C .

46 x10 Pa

P

V

42 x10 Pa

3 32 x10 m- 3 34 x10 m-

A

B C

In process AB , 400 J of heat is added to the

system and in process BC , 100 J of heat is

added to the system. The heat absorbed by the

system in the process AC will be :

10.



(1) 300 J (2) 380 J

(3) 500 J (4) 460 J

Sol.(4) In cyclic process ABCA ,

cyclic0U∆ =

cyclic cyclicQ W=

AB BC CAQ Q Q+ + = closed loop area

3 41400 100 (2 10 ) 4 10

2CA

Q −+ + = × × × ×

400 100 40AC

Q+ − =

460AC

Q J=

124. A block of mass 10 kg , moving in x direction

with a constant speed of 110ms− , is subjected

to a retarding force 0.1 /F x J m= during itss

travel from 20x m= to 30 m . Its final KE will

be :

(1) 250 J (2) 475 J

(3) 450 J (4) 275 J

Sol.(2) W Fdx= −∫30

200.1W xdx= −∫

302

20

0.12

xW

= −

900 4000.1 25

2W

− = − = −

From work energy theorem f iW K K= −

⇒ 21

25 10(10)2

fK− = −

⇒ 475fK =

125. A conducting square frame of side ' 'a and a

long straight wire carrying current I are located

in the same plane as shown in the figure. Theframe moves to the right with a constant velocity

' 'V . The emf induced in the frame will be

proportional to :

V

I

(1) 1

(2 )(2 )x a x a− +(2) 2

1

x

(3) 2

1

(2 )x a−(4) 2

1

(2 )x a+

Sol.(1) emf Induced in side (1)

1 1BVlε =

emf Induced in side (2)

2 2B Vlε =

emf in the frame 1 2BVl B Vl= −

1 2[ ]Vl B Bε = −

⇒ 1 2B Bε ∝ − Since 1

rβ ∝

So

1 1

2 2

a ax x

ε

∝ − − +

⇒1 1

(2 ) (2 )x a x aε

∝ − − +

126. Three identical spherical shells, each of mass

m and radius r are placed as shown in figure.

Consider an axis 'XX which is touching to two

shells and passing through diameter of thirdshell.

11.



Moment of inertia of the system

'X

X

consisting of these three

spherical shells about 'XX axis

is:

(1) 24 mr (2) 211

5mr

(3) 23mr (4) 216

5mr

Sol.(1)

'X

1

32

X

' 1 2 3xxI I I I= + +

2 2 2 2 22 2 2

3 3 3mr mr mr mr mr

+ + + +

⇒ 2 2 2

' 2 2 4xxI mr mr mr= + =

127. Which logic gate is represented by the followingcombination of logic gates ?

A

2Y

Y

1Y

B

(1) NOR (2) OR

(3) NAND (4) AND

Sol.(4)1

y A= , 2

y B= , 1 2y y y A B= + = +

(using De-morgan’s theorem)

y A B= ⋅

Hence this logic gate represents AND gate.

128. A block A of mass 1m rests on a horizontalal

table. A light string connected to it passes overa frictionless pulley at the edge of table and from

its other end another block B of mass 2m is

suspended. The coefficient of kinetic friction

between the block and the table is kµ . When

the block A is sliding on the table, the tension

in the string is :

(1) 1 2

1 2

(1 )

( )

km m g

m m

µ−

+(2)

2 1

1 2

( )

( )

km m g

m m

µ+

+

(3) 2 1

1 2

( )

( )

km m g

m m

µ−

+(4)

1 2

1 2

(1 )

( )

km m g

m m

µ+

+

Sol.(4)

1m

2m

2m g

1km g

k A

B

T

Tµ

µ

For the motion of both blocks

2 2m g T m a− =

1 1kT m g m aµ− =

⇒ 2 1

1 2

( )km m ga

m m

µ−=

+

For the block of mass 2' 'm

2 12 2

1 2

km m

m g T m gm m

µ −− =

+

2 12 2

1 2

km m

T m g m gm m

µ −= −

+

12.



1 1

2

1 2

km mm g

m m

µ +=

+

⇒ 1 2

1 2

(1 )k

m m gT

m m

µ+=

+

129. A certain metallic surface is illuminated with

monochromatic light of wavelength, λ . The

stopping potential for photo-electric current for

this light is 03V . If the same surface is illuminated

with light of wavelength 2λ , the stopping

potential is 0V . The threshold wavelength for this

surface for photo-electric effect is :

(1) 6

λ(2) 6λ

(3) 4λ (4) 4

λ

Sol.(3)s

eV E φ= − ⇒

0

s

hc hcV

e eλ λ= −

here 0

0

3hc hc

Ve eλ λ

= − ...(1)

and 0

02

hc hcV

e eλ λ= − ...(2)

equation (i) 3− × equation (2)

⇒0

02

hc hc

e eλ λ= − +

⇒ 0 4λ λ=

130. When two displacements represented by

1 sin ( )y a tω= and 2 cos ( )y b tω= are

superimposed the motion is :

(1) simple harmonic with amplitude ( )

2

a b+

(2) not a simple harmonic

(3) simple harmonic with amplitude a

b

(4) simple harmonic with amplitude 2 2a b+

Sol.(4) 1 siny a tω=

& 2 cos sin2

y b t b tπ

ω ω

= = +

since the frequencies for both SHM are same,

resultant motion will be SHM . Now

Amplitude 2 2

1 2 1 22 cosA A A A A φ= + +

here 1A a= , 2A b= & 2

πφ =

so 2 2A a b= +

131. A potentiometer wire has length 4 m and

resistance 8Ω . The resistance that must be

connected in series with the wire and an

accumulator of e.m.f. 2V , so as to get a

potential gradient 1mV per cm on the wire is :

(1) 48Ω (2) 32Ω

(3) 40Ω (4) 44Ω

Sol.(2) Potential gradient

3 1110 / 10 /

mVV cm V m

cm

− −= = =

Let the resistance to be connected is R then

2

8I

R=

+Potential drop across the potentiometer wire

8 2 16

8 8R R

×= =

+ +

Potential gradient 16 1

/8 4

V mR

= ×

+

40.1

8 R= =

+⇒ 32R = Ω

13.



132. Two spherical bodies of mass M and 5M and

radii R and 2R are released in free space withinitial separation between their centres equal to

12R . If they attract each other due togravitational force only, then the distancecovered by the smaller body before collision is :

(1) 1.5 R (2) 2.5 R

(3) 4.5 R (4) 7.5 R

Sol.(4)12R

Initial distance between their centers 12R=

2RR

At time of collision the distance between their

centrers 3R=

So total distance travelled by both

12 3 9R R R= − =

Since the bodies move under mutual forces,center of mass will remain stationary so

1 1 2 2m x m x=

5 (9 )mx m R x= −

45 5x R x= −

6 45x R=

45

6x R=

7.5x R=

133. A resistance ' 'R draws power ' 'P when

connected to an AC source. If an inductance

is now placed in series with the resistance, such

that the impedance of the circuit becomes ' 'Z ,

the power drawn will be :

(1) P (2)

2R

PZ

(3) R

PZ

(4) R

PZ

Sol.(2)

2V

PR

= ⇒ 2V PR=

' cosV

P VZ

φ

= ⋅ ⋅

2

'V R

PZ Z

= ⋅

2

( )'

PR RP

Z=

2

'R

P PZ

=

134. Across a metallic conductor of non-uniform crosssection a constant potential difference is applied.The quantity which remains constant along theconductor is :

(1) electric field (2) current density

(3) current (4) drift velocitySol.(3) Fact

135. A parallel plate air capacitor of capacitance C

is connected to a cell of emf V and then

disconnected from it. A dielectric slab of dielectric

constant K , which can just fill the air gap of the

capacitor, is now inserted in it. Which of thefollowing is incorrect ?

(1) The charge on the capacitor is not conserved

(2) The potential difference between the plates

decreases K times.

(3) The energy stored in the capacitor decreases

K times.

(4) The change in energy stored is

21 11

2CV

K

−

Sol.(1) Once the capacitor is charged, its charge will be

constant Q CV=

question paper with solution of aipmt- 2015in a double slit experiment, the two slits are 1mm apart...

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