-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 1 ]IITians TAPASYA...IITians creating IITians
SOLUTION OF JEE MAIN. 2015
PART - B [PHYSICS]31. In the circuit shown, the current in the 1
resistor is :
(1) 0.13 A, from P to Q(2) 1.3 A, from P to Q(3) 0A(4) 0.13 A,
from Q to P
Ans. (4) For calculation of current, circuit can be drawn as
:
Apply Nodal methodx 0 x 6 x 9 0
1 3 5
6
6
3 x 5 9
9
0
13x
23
Current in 1 =x 0 3 0.13
1 23
Current is 0.13 A from Q to P.32. 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 :
(1)23h
8R(2)
2h4R
(3)3h4 (4)
5h8
Ans. (3) 223Mdm r dxR H
x
dx
Here Ycm =H
0
1 dmxM
IITians TAPASYA...IITians creating IITians
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 2 ]IITians TAPASYA...IITians creating IITians2H
20
1 3M Rx dx xM R H H
H3
30
3 x dxH 3H4
33. Match List - I (Fundamental Experiment) with List - II (its
conclusion) and select the correct optionfrom the choices given
below the list :
List - I List - II(1) Franck-Hertz Experiment (i) Particle
nature of light(B) Photo-electric experiment (ii) Discrete energy
levels of atom(C) Davision - Germer Experiment (iii) Wave nature of
electron
(iv) Structure of atom(1) (A) -(iv), (B) -(iii), (C) -(ii) (2)
(A) -(i), (B) -(iv), (C) -(iii)(3) (A) -(ii), (B) -(iv), (C) -(iii)
(4) (A) -(ii), (B) -(i), (C) -(iii)
Ans. (4)(1) Franck-Hertz experiment :- In 1914, Franck-Hertz
experiment demonstrated the existence of
excited states in mercury atoms.(2) Photo-electric effect is
explained by particle nature of light.(3) Davisson-Germer
experiment confirmed the De-broglie hypothesis which says that
particles of matter such as electron has wave nature also.
34. The period of oscillation of a simple pendulum is T =L2g
. Measured value of L is 20.0 cm
known to 1 mm accuracy and time for 100 oscillations of the
pendulum is found to be 90 s usinga wrist watch of 1s resolution.
The accuracy in the determination of g is :(1) 5% (2) 2% (3) 3% (4)
1%
Ans. (3)g L T2
g L T
L 0.1cm, L 20cm
t 1 90T 0.01s, T 0.9sn 100 100
g 0.1 0.012 0.005 0.222 0.027g 20 0.9
g% Accuracy 100% 2.7% 3%g
35. A red Led emits light at 0.1 watt uniformly around it. The
amplitude of the electric field of the lightat a distance of 1 m
from the diode is :(1) 7.75 V/m (2) 1.73 V/m (3) 2.45 V/m (4) 5.48
V/m
Ans. (3) Intensity I 20 01 E C2
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 3 ]IITians TAPASYA...IITians creating IITians2
0 02P 1 E C
4 r 2
(where P = power of LED)
920 2 8 2
0
2P 2 0.1 9 10E 64 cr 3 10 (1)
0E 6 2.45 v /m
36. In the given circuit, charge Q2 on the 2 F capacitor changes
asC is varied from 1F to 3F. Q2 as a function of C is givenproperly
by : (figure are drawn shcematically and are not to scale)
(1) (2)
(3) (4)
Ans. (3) Charge in series are same.Let potential on c is V1
& potential on 3F is V2 .CV1 = 3V2 ...(1)
C
E
3FV1 + V2 = E ...(2)From (1) and (2)
2CEV
(C 3)
Now, Q2 = 2V2
22CEQ
(C 3)
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 4 ]IITians TAPASYA...IITians creating IITians37. Two long
current carrying thin wires, both with current I, are held by
insulating threads of length L
and and are in equiblibrium as shown in the figure, with threads
making an angle with thevertical. If wire have mass per unit length
then the value of I is :(g = gravitational acceleration)
(1)0
gL tan (2)
0
gLsincos
(3)0
gL2sinu cos
(3)0
gL2 tanu
Ans. (3) Making free body diagram of unit length of a wire
T
g
20I
2 (2Lsin )
Vertical equilibrium : Tcos g ....(1)
Horizontal equilibrium :2
0ITsin4 Lsin
....(2)
Using equation (1) and (2), we get
0
LgI 2sincos
38. A particle of mass m moving in the x direction with speed 2
is hit by another particle of mass 2mmoving in the y direction with
speed . If the collision is perfectly inelastic, the percentage
lossin the energy during the collision is close to :(1) 62% (2) 44%
(3) 50% (4) 56%
Ans. (4) Initial Momentum of System iP = 2 2(2mv) (2mv) 2 2mv
Final Momentum of System = PfConservation of Linear Momentum: iP =
Pf
1 12 22 2 mV 3m V V V
3
Initial kinetic energy Ki =2 2 21 1m(2V) 2mV 3mV
2 2
Final kinetic energy Kf =2 21 8 4(3m) V mV
2 9 3
2 2 2i f
4 5K K K 3mV mV mV3 3
2
2i
K 5 / 3 mV 5% loss 100 56%K 3mV 9
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 5 ]IITians TAPASYA...IITians creating IITians39. 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 theblocks 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) 150 N(2) 100 N(3) 80 N(4) 120 N
Ans. (4) FBD of A and B
FF F F
f1
f120 N
A B
100 N
f2
For A, f1 = 20 NFor B, f1 + 100 = f2 = 120 N
40. Consider an ideal gas confined in an isolated closed
chamber. As the gas undergoes an adia-batic expansion, the average
time of collision between molecules increases as Vq, where V is
the volume of the gas. The value of q is :p
v
CC
(1)1
2
(2)3 5
6
(3)3 5
6
(4)1
2
Ans. (4) Average time of collision
2
kT Tt t & TP2 P
T TtPTP
V T Vt tT T
For adiabatic process 1TV cons tan t 1t V V
1 112 2t V t V
Given qt V , Hence, q =1
2
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 6 ]IITians TAPASYA...IITians creating IITians41. A rectangular
loop of sides 10cm and 5cm carrying a current I of 12 A is placed
in different
orientations a shown in the figures below :
(A) (B)
(C) (D)
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) (b) and (c),
respectively (2) (a) and (b), respectively(3) (a) and (c),
respectively (4) (b) and (d), respectively
Ans. (4) Magnetic moment M
& Magnetic field B
(i) If M
|| B
, then loop will be in stable equilibrium.(ii) If M
antiparallel to B
, then loop will be in un stable equilibrium.
Hence, (b) shows stable equilibrium(d) shows unstable
equilibrium
42. 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 volume u =UV T4 and
pressure p =1 U3 V
. If the shell now undergoes an adiabatic expansion the relation
between T
and R is :
(1) 31T
R (2) RT e (3) 3RT e (4)
1TR
Ans. (4) 4 4U 1 Uu T & P P T ..........(i)V 3 V
from Ideal gas equation :TPV nRT P ..............(ii)V
from equation (i) & (ii)
31/ 3
1 1T TV V
Volume 3 1/ 3V R V R
Hence,1TR
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 7 ]IITians TAPASYA...IITians creating IITians43. As an
electron makes a transition form an excited state to the ground
state of a hydrogen- like
atom/ion :(1) kinetic energy and total energy decrease but
potential energy increases(2) its kinetic energy increases but
potential energy and total energy decreases.(3) kinetic energy,
potential energy and total energy decreases.(4) kinetic energy
decreases, potential energy increases but total energy remains
same.
Ans. (2) VelocityZVn , As electron goes to lower orbit, the
velocity of electron increases.
hence, kinetic energy increases Total energy decreases because
electron emits photon. Total energy (T.E) = Potential energy (P.E)
+ kinetic energy (K.E)
If K.E & T.E P.E 44. On a hot summer night, the refractive
index of air is smallest near the ground and increases with
height from the ground. When a light beam is directed
horizontally, the Huygens principle leadsu sto conclude that as it
travels, the light beam :(1) bends upwards (2) becomes narrower(3)
goes horizontally without any deflection (4) bends downwards.
Ans. (1)If a wavefront is moving in horizontal direction then
upward rays will move slower then lower rays
hence if rays are going in forward direction then wavefront will
have a tendency to bend backward and the light beam will bend
upwards.45. From a solid sphere of mass M and radius R, a spherical
portion of radius R/2 is removed, as
shown in the figure. Taking gravitational potential V = 0 at r ,
the potential at the centre of thecavity thus formed is :(G =
gravitational constant)
(1)2GMR
(2)GM2R
(3)GMR
(4)2GM3R
Ans. (3) Potential inside a solid sphere is given by
2 22V G (3R r )3
In the cavity we can assume that both and density material is
filledV = V solid sphere + Vcavity
2 222 R 2 RV G 3R G 3 0
3 2 3 2
+ &
24V R G3
By Putting3
M4 R3
, We get GMVR
46. Monochromatic light is incident on a glass prism of angle A.
If the refractive index of the material
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 8 ]IITians TAPASYA...IITians creating IITiansof the prism is ,
a ray, incident at an angle , on the face AB would get transmitted
through theface AC of the prism provided.
(1) 1 1 1cos sin A sin
(2) 1 1 1sin sin A sin
(3) 1 1 1sin sin A sin
(4) 1 1 1cos sin A sin
Ans. (2) If 2 Cr (critical angle)1
21r sin ...........(i)
Then ray will always transmit through face AC
At face AB : 11 11sin sinr r sin sin
r1 r2
A
B C
We know for prism r 1 + r2 = A
r2 = A r1 = A 1 1sin sin
................ (ii)
from equation (i) and (ii)
1 11 1A sin sin sin
1 11 1A sin sin sin
1 1 1sin A sin sin
1 1 1sin sin A sin
47. Two stones are thrown up simultaneously from the edge of a
cliff 240 m high with intial speed of10 m/s and 40 m/s
respectively. Which of the following graph best represents the time
variationof relative position of the second stone with respect to
the first?
(1) (2)
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 9 ]IITians TAPASYA...IITians creating IITians
(3) (4)
Ans. (4)Initially both the particles will be in air and moving
under gravity. Their relative acceleration will bezero. So relative
separation Vs time will be straight line.When one particle reaches
the ground then relative acceleration will be g.Now relative
separation vs time is parabolic with concave downwards as relative
speed shouldincrease with time.
48. 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)
Ans. (3) At Extreme positions K.E = 0At Mean position K.E is
maximumAt Extreme positions P.E is maximumAt Mean position P.E is
minimumSince, T.E. in S.H.M remains conserved.d = 0 Mean positiond
= A Extreme positions (A Amplitude)
49. A train is moving on a straight track with speed 20 ms1. It
is blowing its whistle at the frequencyof 1000 Hz. The percentage
change in the frequency heard by a person standing near the trackas
th e train passes is (speed of sound = 320 ms1) close(1) 24% (2) 6%
(3) 12% (4) 18%
Ans. (3) For approaching case :
app oVf f
V 20
For moving away case :
away oVf f
V 20
oapp away o 2
40f V1 1f f f f VV 20 V 20 V 400
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 10 ]IITians TAPASYA...IITians creating IITians% change =
2of 40 x320 x100 40 x320 x100x100% 12.54% 12%
f 102000320 400
50. An LCR circuit is equivalent to a damped pendulum. In an LCR
circuit the capacitor is charged toQ0 and then connected to the L
and R as shown below :
If a student plots graphs of the square of maximum charge 2MaxQ
on the capacitor with time (t)for two different values L1 and L2
(L1 > L2) of L then which of the following represents this
graphcorrectly ? (plots are schematic and not drawn to scale)
(1) (2)
(3) (4)
Sol. (2) Damping is more, if inductance is less
51. A solid body of constant heat capacity 1 J/oC is being
heated by keeping it in contact with reservoirsin 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 100oC to final temperature 200oC.Entropy change of the
body in two cases respectively is :
(1) l l2 n2,8 n2 (2) l ln2,4 n2 (3) l ln2, n2 (4) l ln2, 2
n2
Ans. (3)dq CdTdsT T
, temp should be given in kelvin
For both the case : 200100
s 1 lnT | 200100s 1 lnT | ln 252. An inductor (L = 0.03H) and
resistor (R = 0.15 k ) are connected in series to a battery of
15V
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 11 ]IITians TAPASYA...IITians creating IITiansEMF in a circuit
shown below. The key K1 has been kept closed for a long time. Then
at t = 0, K1is opened and key K2 is closed simultaneously. At t =
1ms, the current in the circuit will be :( 5e 150 )
(1) 0.67 mA(2) 100 mA(3) 67 mA(4) 6.7 mA
Ans. (1) Current at t =0
o 315I 0.1 A
0.15 x 10
Now current will decrease with time as per equation3 30.15x10
x10
Rt /L 50.03o
0.1I I e 0.1 e 0.1 e 0.67 mA150
53. A uniformly charged solid sphere of radius R has potential
V0(measured with respect to ) o n
its surface. For this sphere the equipotential surfaces with
potentials 0 0 0 03V 5V 3V V, , and2 4 4 4
have radius R1, R2, R3 and R4 respectively. Then(A) 2R < R4
(B) R1 = 0 and R2 > (R4 R3)
(C) R1 0 and (R2 R1) > (R4 R3) (D) R1 = 0 and R2 < (R4
R3)
Sol. (1, 4)
2o
o 2VKQ 3 rV , V
R 2 R for r < R and 0V RV , r R
r
0centre 1
3VV R 04
02
5V RV R4 2
03
3V 4RV R4 3
for 0 4VV R 4R4
Hence, R4 > 2R and 4 3 2R R R
54. 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 :
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 12 ]IITians TAPASYA...IITians creating IITians
(1) (2) (3) (4)
Sol. (2)(i) Electric field line curves should be smooth(ii)
Electric field lines originates from positive charge &
terminates at negative charge(iii) Tangent at any point on curve
shows direction of resultant electric field.(iv) Preference of
termination should be at negative charge but in option (4) no such
termination
is shown.55. 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 car
resolve at 500 nm wavelength is :(1) 300 m (2) 1 m (3) 30 m (4) 100
m
Sol. (3)
object 1
object 225 cm
For just resolve the images
0.61a
where a radius of pupilFrom figure
d 0.61 d 3025cm a
56. A signal of 5 kHz frequency is amplitude modulated on a
carrier wave of frequency 2 MHz. Thefrequencies of the resultant
signal is/are :(1) 2000 kHz and 1995 kHz (2) 2 MHz only(3) 2005 kHz
and 1995 kHz (4) 2005 kHz, 2000 kHz and 1995 kHz
Sol. (4) Amplitude modulated wave carries three frequencies = WC
W, WC, WC + WWC - frequency of carrier waveW - frequency of
signalWC = 2 MHz or 2000 KHzW = 5 KHzHence, three frequencies in
resultant signal are 2005 KHz, 2000 KHz, 1995 KHz
57. Two coaxial solenoids of different radii carry current I in
the same direction. Let 1F
be the magneticforce on the inner solenoid due to the outer one
and 2F
be the magnetic force on the outer
solenoid due to the inner one. Then :
(1) 1F
is radially outwards and 2F
= 0
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 13 ]IITians TAPASYA...IITians creating IITians(2) 1F
= 2F
= 0
(3) 1F
is radially inwards and 2F
is radially outwards
(4) 1F
is radially inwards and 2F
= 0
Sol. (2) Inner solenoid is kept in uniform magnetic field of
outer solenoid so force on all currectloops is zeroInner solenoid
produce zero magnetic field outside. So force on outer solenoid is
also zero.
58. A pendulum made of a uniform wire of cross sectional area A
has time period T. When an additionalmass M is added to its bob,
the time period changes to TM. If the Youngs modulus of the
material
of the wire is Y then1Y is equal to :
(g = gravitational acceleration)
(1)2
M
T A1T Mg
(2)2
MT A1T Mg
(3)
2MT Mg1
T A
(4)2
MT A1T Mg
Sol. (2)
M
lT 2 ...(1)g
Mg lY ....(2)A l
l lT 2 ....(3)g
59. From a solid sphere of mass M and radius R a cube of maximum
possible volume is cut. Momentof inertia of cube about an axis
passing through its center and perpendicular to one of its facesis
:
(1)24MR
3 3 (2)2MR
32 2 (3)2MR
16 2 (4)24MR
9 3
Sol. (4) From symmetry it is obvious that for maximum volume
3a 2R
Mass of cube = 33
2cube
M 2Ma4 3R3
M aI6
60. When 5V potential difference is applied across a wire of
length 0.1 m, the drift speed of electrons
-
Head Office : Pushpanjali Place, Boring Road Crossing, Opp.
Alankar Place, Patna-1Branch Office : Rukanpura Flyover, Near Sri
Sai Yamaha, Bailey Road, Patna- 14
Ph.: 0612-3225103 / 7677 133 770 / 7857 966 777visit us on
-www.iitianstapasya.com-
[ 14 ]IITians TAPASYA...IITians creating IITiansis 2.5 x 104
ms1. If the electron density in the wire is 8 x 1028 m3, the
resistivity of the material isclose to :
(1) 1.6 x 105 m (2) 1.6 x 108 m (3) 1.6 x 107 m (4) 1.6 x 106 m
.
Sol. (1) l
l
d
5
d
II neAV , V IRA
V 1.56 10neV