Physics - RG Academy Ajmer

JEE Main Test 2015
[1]
Physics
1.
Two stones are thrown up simultaneously from the
edge of a cliff 240 m high with initial speed of 10 m/s
and 40 m/s respectively. Which of the following graph
best represents the time variation of relative position
of the second stone with respect to the first?
(Assume stones do not rebound after hitting the
ground and neglect air resistance, take g = 10 m/s2)
Sol. [2] g 

4 2 L
T2
LM
N
g
T
Max.   2
g
T

Max.
OP
Q

 Max. .
LM FG 1 IJ  0.1OP  100%
N H 90 K 20 Q
 2
(1)
= 3%
3.
(2)
Given in the figure are two blocks A and B of weight
20 N and 100 N, respectively. These are being pressed
against a wall by a force F as shown. If the coefficient
of friction between the blocks is 0.1 and between block
B and the wall is 0.15, the frictional force applied by the
wall on block B is
(1) 100 N
(2) 80 N
(3) 120 N
(4) 150 N
(3)
Sol. [3] For block A's equilibrium
f1  m A g  20 N
and for B's equilibrium
f 2  f 1  mB g  120 N
4.
(4)
Sol. [3] Initially, when both are in flight, relative velocity will
remains constant and hence curve would be straight
line and after one of the stone comes to ground, it will
become parabolic and hence option (3) is correct.
2.
Two period of oscillation of a simple pendulum is
T  2
L
. Measured value of L is 20.0 cm known to
g
1 mm accuracy and time for 100 oscillations of the
pendulum is found to be 90 s using a wrist watch of 1 s
resolution. The accuracy in the determination of g is
(1) 2%
(2) 3%
(3) 1%
(4) 5%
A particle of mass m moving in the x direction with
speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v. If the collision is perfectly
inelastic, the percentage loss in the energy during the
collision is close to
(1) 44%
(2) 50%
(3) 56%
(4) 62%

Sol. [3] Let v f be the final velocity, then

m 2v i  2m v j  3mv f
b g



vf
bg
2
 dvi  vj i
3
Klos s
Ki
LMR 1
MNST 2 mb2vg

2
b g UVW b gFGH
1
1
mb2v g  b2mgv
2
2

1
1
2 2
2m v 2  3m
v
2
2
3
2
= 0.56 = 56%.
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2
I
JK
2
OP
PQ
JEE Main Test 2015
[2]
5.
Distance of the centre of mass of a solid uniform cone
from its vertex is z0. If the radius of its base is R and its
height is h then z0 is equal to
h2
(1)
4R

3h
(2)
4
8.
3h 2
(4)
8R
5h
(3)
8
Sol. [2]
6.
(3)
MR
2
(2)
32 2
4 MR 2
(4)
9 3
MR

3a  2 R or a 
M cube 
GM
.
R
A pendulum made of a uniform wire of cross sectional
area A has time period T. When an additional mass M
is added to its bob, the time period changed to TM. If
the Young's modulus of the material of the wire is Y
then
LF T I O A
(1) MGH T JK  1P Mg
MN
PQ
LM1  F T I OP A
(3)
MN GH T JK PQ Mg
2
M
2
16 2
2
4 MR 2
M
3 3
Sol. [3] Here body diagonal of such cube is equal to diameter
of sphere
2
M
2
M
3
F I
GH JK
3
2
for second part,

2
a
4 MR

6
9 3
I req  M cube
7.
From a solid sphere of mass M and radius R, a spherical
9.
R
is removed, as shown in figure.
2
Taking gravitational potential V = 0 at r   , the
potential at the centre of the cavity thus formed is
(G = gravitational constant)
portion of radius
LM
MN
GM
2R
(2)
GM
R
(3)
2GM
3R
(4)
2GM
R
Sol. [2] V  Vcomplete  Vcavity / percent
LM FG M IJ OP
O
I
R

P  M 3G H 8 K P
4 JK PQ M 2 F R I P
MN GH 2 JK PQ
L
MgL
AY
g
OP
PQ
Consider a spherical shell of radius R at temperature T.
The black body radiation inside it can be considered as
an ideal gas of photons with internal energy per unit
FG IJ
H K
1 U
U
 T 4 and pressure p 
. If the
3 V
V
shell now undergoes an adiabatic expansion the relation
between T and R is
volume u 
(3) T 
(1)
Tm  2
L  L
 2
g
1
T 2
A
 m2  1
Y
Mg
T
(1) T  e  R
Sol. [3] Here
or
T
 T4
V

T
1
R
(2) T  e 3R
(4) T 
1
R3
p  T4
or T 
1.
R
2
2
L
g
T  2
2R

3
LF T I O Mg
(2) MGH T JK  1P A
MN
PQ
LM F T I OP A
(4) M1  GH T JK P Mg
N
Q
Sol. [1] For first part,
M
8R
2M


4 3
3 3
3
R
3
LM GM F 3R
MN 2 R GH
M
)
8
1
is equal to
Y
(g = gravitational acceleartion)
From a solid sphere of mass M and radius R a cube of
maximum possible volume is cut. Moment of inertia of
cube about an axis passing through its center and
perpendicular to one of its faces is
(1)

(as M cavity 
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JEE Main Test 2015
[3]
10.
A solid body of constant heat capacity 1 J / 0C is being
heated by keeping it in contact with reservoirs in two
ways
(i) Sequentially keeping in contact with 2 reservoirs
such that each reservoir supplies same amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs
such that each reservoir supplies same amount of heat
In both the cases body is brought from initial
temperature 1000C to final temperature 2000C. Entropy
change of the body in the two cases respectively is
(1) n2, 4n2
12.
For a simple pendulum, a graph is plotted between its
kinetic energy (KE) and potential energy (PE) against
its displacement d. Which one of the following
represents these correctly?
(graphs are schematic and not drawn to scale)
(1)
(2)
(3)
(4)
(4) n2, n2
(3) n2, 2n2
(4) 2n2, 8n2
Sol. [2] Here data should be 100 K and 200 K.
In both cases
SC
z dTT  ln2 .
Tf
Sol. [2] Potential energy is maximum at extremes while kinetic
is maximum at mean position.
Ti
11.
Consider an ideal gas confined in an isolated closed
chamber. As the gas undergoes an adiabatic expansion,
the average time of collision between molecules
13.
increases as V q , where V is the volume of the gas. The
value of q is
FG  C IJ
H CK
A train is moving on a straight track with speed
20 m s–1 . It is blowing its whistle at the frequency of
1000 Hz. The percentage change in the frequency heard
by a person standing near the track as the train passes
him is (speed of sound = 320 m s–1) close to
(1) 6%
(2) 12%
(3) 18%
(4) 24%
p
Sol. [2] Here change in frequency,
V
(1)
3  5
6
(2)
3  5
6
(3)
 1
2
(4)
 1
2
Sol. [3] Average time of collision, t 

T
1
V

nvrms Nvrms
f  2 f 0
FG v IJ  2  1000  FG 20 IJ = 125 Hz
HvK
H 320 K
s

Percentage change is 12.5%.
14.
A long cylindrical shell carries positive surface charge
 in the upper half and negative surface charge 
in the lower half. The electric field lines around the
cylinder will look like figure given in
(figures are schematic and not drawn to scale)
V
T
Now for adiabatic process
TV  1  const.
T
(1)
(2)
1
V
 1

t  V  1

q
 1 .
2
(3)
(4)
Sol. [1] Treating system as analogue to dipole, at equatorial
point, field must point downwards which is satisfied
by (1).
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JEE Main Test 2015
[4]
15.
A uniformly charged solid sphere of radius R has
potential V0 (measured with respect to  ) on its surface.
For this sphere the equipotential surfaces with
3V0 5V0 3V0
,
,
potentials
2 4 4
17.
2.5  104 ms1 . If the electron density in the wire is
V0
and
have radius
4
8  1028 m3 , the resistivity of the material is close to
R1 , R2 , R3 and R4 respectively. Then
b
(1) R1  0 and R2  R4  R3
b
g
g b
(2) R1  0 and R2  R1  R4  R3
b
(3) R1  0 and R2  R4  R3
g
g

Sol. [3, 4] The values will be,
R
2
; R3 
.  107  m
(2) 16
(3) 16
.  106  m
.  105  m
(4) 16
i  neAvd 

R4  R3  R2
so both option satisfy.
16.
In the given circuit, charge Q2 on the 2F capacitor

V VA

R 
V
5

28
19
nevd  8  10  16
.  10  2.5  104  01
.
 16
.  105  m .
4
R; R4  4 R
3
and
.  108  m
(1) 16
Sol. [4] Current,
(4) 2R < R4
R1  0; R2 
When 5 V potential difference is applied across a wire
of length 0.1 m, the drift speed of electrons is
R4  2 R .
18.
changes as C is varied from 1F to 3F . Q2 as a
function of 'C' is given properly by
(Figure are drawn schematically and are not to scale)
In the circuit shown, the current in the 1  resistor is
(1) 1.3 A from P to Q
(3) 0.13 A, from Q to P
(2) 0 A
(4) 0.13 A, from P to Q
Sol. [3]
(1)
(2)
Applying Kirchhoff's loop law on both loops,
b
g
b
g
92 x y  x 3 x y  0
i.e.,
(3)
b
i.e.,
q

q
LM
N
OP
Q
2
3C
2E

E 
3
3 3 C
1
C
g
b
g
9  2 x  y  6  3y  3 x  y  0
(4)
Sol. [2] Here charge on 2 F capacitor is given by,
6 x  5 y  9 . . .(i)
for second,
15  5x  8 y . . .(ii)
on solving we get,
x
3
 013
. A.
23
Hence current is from Q  P.
A as c A and it will tend to a constant value q = 3E
when c   .
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JEE Main Test 2015
[5]
19.
Two coaxial solenoids of different radii carry current I

in the same direction. Let F1 be the magnetic force on

the inner solenoid due to the outer one and F2 be the
21.
magnetic force on the outer solenoid due to the inner
one. Then


(1) F1  F2  0


(2) F1 is radially inwards and F2 is radially outwards


(3) F1 is radially inwards and F2  0


(4) F1 is radially outwards and F2  0
A rectangular loop of sides 10 cm and 5 cm carrying a
current I and 12 A is placed in different orientations as
shown in the figures below
(a)
(b)
Sol. [1] Here net force on both is zero. But for inner solenoid,
every small section on it will experience force radially
outwards, so option (4) might also be considered
though Fnet is surely zero.
(c)
20.
(d)
Two long current carrying thin wires, both with current
I, are held by insulating threads of length L and are in
equilibrium as shown in figure, with threads making an
angle  with the vertical. If wires have mass  per
unit length then the value of I is
(g = gravitational acceleration)
gL
(1) sin   cos
0
gL
(3) 2  tan 
0
Sol. [2] For equilibrium,
gL
(2) 2 sin  cos
0
(4)
gL
tan 
0
If there is a uniform magnetic field of 0.3 T in the positive
z direction, in which orientations the loop would be in
(i) stable equilibrium and (ii) unstable equilibrium?
(1) (a) and (b), respectively
(2) (a) and (c), respectively
(3) (b) and (d), respectively
(4) (b) and (c), respectively


Sol. [3] For stable equilibrium, angle between m and B is 0°
and for unstable it is 180°.
22.
b
are connected in series to a battery of 15 V EMF in a
circuit shown below. The key K1 has been kept closed
for a long time. Then at t = 0, K1 is opened and key K2 is
closed simultaneously. At t = 1 ms, the current in the
circuit will be: ( e5 ~ 150 )
T cos  g
T sin  
0I 2
2  2 L sin 
4Lg sin 2 
 0 cos

I2 
or
I  2 sin 
gL .
 0 cos
g
. k
An inductor (L = 0.03 H) and a resistor R  015
(1) 100 mA
(3) 6.7 mA
(2) 67 mA
(4) 0.67 mA
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JEE Main Test 2015
[6]
Sol. [4] Before t  0 , current flowing

 100 mA
R
After that
Sol. [1] For ray to emerge from AC,
r2   c
i0 
b
g
2
 mA  0.67 mA .
3

FG 1 IJ
H K
sin    sinb A   g
or
  sin 1  sin A  sin 1
A red LED emits light at 0.1 watt uniformly around it.
The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(1) 1.73 V/m
(2) 2.45 V/m
(3) 5.48 V/m
(4) 7.75 V/m
c
LM F
MN GH
Sol. [2] Intensity at distance 1 m,
I
P
01
.

4r 2 4 1 2
bg
25.
Half of the intensity is due to electric field,
e
j

1
1
01
.
2
0 E rms
c 
2
2 4 1 2

Erms  3  173
. V/ m

E0  2 Erms  2.45 V / m .
24.
Monochromatic light is incident on an glass prism of
angle A. If the refractive index of the material of the
prism is  , a ray, incident at an angle  , on the face
AB would get transmitted through the face AC of the
prism provided
bg
LM F
MN GH
LM sinF A  sin
MN GH
LM sinF A  sin
MN GH
LM sinF A  sin
MN GH
(1)   sin 1  sin A  sin 1
(3)   sin 1
(4)   cos1
1
1
FG 1 IJ I OP
H  K JK PQ
FG 1 IJ I OP
H  K JK PQ
FG 1 IJ I OP
H  K JK PQ
FG 1 IJ I OP
H  K JK PQ
FG 1 IJ I OP .
H  K JK PQ
On a hot summer night, the refractive index of air is
smallest near the ground and increases with height form
the ground. When a light beam is directed horizontally,
the Huygen's principle leads us to conclude that as it
travels, the light beam
(1) becomes narrower
(2) goes horizontally without any deflection
(3) bends downwards
(4) bends upwards
Sol. [4] For horizontally directed beam, wavefront would be a
vertical plane and as we go down, speed of the points
keeps on increasing. Thus, after a short time it takes
the shape as shown. Hence, we can say that light bends
upwards as it travels.
26.
1
(2)   sin
r1  A   c .
1
Also  c  sin
i  i0e  tR / L  100 mA e 5
23.
or
Assuming human pupil to have a radius of 0.25 cm and
a comfortable viewing distance of 25 cm, the minimum
separation between two objects that human eye can
resolve at 500 nm wavelength is
(1) 1 m
(2) 30 m
(3) 100 m
(4) 300 m
Sol. [2] Let separation be x, then
x
122
. 

25 cm
D

x  30 m .
1
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JEE Main Test 2015
[7]
27.
As an electron makes a transition from an excited state
to the ground state of a hydrogen like atom / ion
(1) its kinetic energy increases but potential energy
and total energy decrease
(2) kinetic energy, potential energy and total energy
decrease
(3) kinetic energy decreases, potential energy increases
but total energy remains same
(4) kinetic energy and total energy decreases but
potential energy increases
Match List - I (Fundamental Experiment) with List- II(its
conclusion) and select the correct option from the
choices given below the list
List I
List II
(A) Franck-Hertz
(i) Particle nature of light
Experiment
(B) Photo-electric
(ii) Discrete energy levels of
experiment
atom
(C) Davison -Germer
(iii) Wave nature of electron
Experiment
(iv) Structure of atom
(1) (A) - (i), (B)- (iv), (C)- (iii)
(2) (A) - (ii), (B)- (iv), (C)- (iii)
(3) (A) - (ii), (B)- (i), (C)- (iii)
(4) (A) - (iv), (B)- (iii), (C)- (ii)
Sol. [3]
29.
e
j
2
charge Qmax on the capacitor with time (t) for the two
b
g
different values L1 and L2 L1  L2 of L then which of
the following represents this graph correctly? (plots
are schematic and not drawn to scale)
Sol. [1]
28.
If a student plots graphs of the square of maximum
A signal of 5 kHZ frequency is amplitude modulated
on a carrier wave of frequency 2 MHz. The frequencies
of the resultant signal is/are
(1) 2 MHz only
(2) 2005 kHz, and 1995 kHz
(3) 2005 kHz, 2000 kHz and 1995 kHz
(4) 2000 kHz and 1995 kHz
(1)
(2)
(3)
Sol. [3] Frequencies present are,
f c , f c  f m and f c  f m .
30.
An LCR circuit is equivalent to a damped pendulum. In
an LCR circuit the capacitor is charged to Q0 and then
connected to the L and R as shown below
(4)
Sol. [1] Here decrease will be exponential and greater the L is
(means more inertia) less steeper will be the drop.
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