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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
Our Top class IITian/Doctor faculty team promises to give you an authentic solutions whichwill be fastest in the whole country.
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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.
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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.
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(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.
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(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.
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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.
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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.
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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.
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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.
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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.
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(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.
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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.
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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.
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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=
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