1. education

JEE Main 2015
[1]
JEE Main 2015
Answer Key
Chemistry
31.
The molecular formula of a commercial resin used for
35.
exchanging ions in water softening is C8 H 7SO 3Na
2 NO (g)  O 2 (g)
(Mol. wt 206). Ca ions by the resin when expressed
in mole per gram resin?
1
(2)
206
2
(3)
306
1
(4)
412
e
.  1012
formation of NO 2 ( g) at 298 K ? K p  16
e
e
2
(3) 86600 
tice with a unit cell edge of 4.29Å . The radius of sodium atom is approximately.
(2) 3.22 Å
(3) 5.72 Å
(4) 0.93Å
Sol. [1] rB.C.C. 
33.
(3) 3.4 eV
(4)  6.8 eV
Sol. [3] E  13.6
34.
G   RT ln Kp
G   RT ln 16
.  1012
G  2  G NO2  2  G NO
E  3.4 eV
The intermolecular interaction that is dependent on the
inverse cube of distance between the molecules is:
(1) ion-ion interaction (2) ion-dipole interaction
(3) London fforce
(4) hydrogen bond
so, force is 
1
r3
b g e
G NO2  0.5 2  86,600  R 298 ln 16
.  1012
36.
z2
eV
n2
Sol. [2] Energy of ion-dipole interaction 
j
Sol. [4] 2 NO  O 2 
 2 NO 2
n  2 (First Excited state)

j
R (298)
e
Which of the following is the energy of a possible
excited state of a possible excited state of hydrogen?
(2) 6.8 eV
e
ln 16
.  1012
j
.  1012
(4) 0.5 2  86,600  R(298) ln 16
3a 173
.  4.29

 186
. Å
4
4
(1) 13.6 eV
j
.  1012
(2) 86600  R(298) ln 16
Sodium metal crystallizes in a body centered cubic lat-
(1) 186
. Å
j
.  1012  86600
(1) R(298) ln 16
Sol. [4] 2 mol of water softner require one mole of Ca
32.
2NO (g)
The standard free energy of formation of NO(g) is 86.6
kJ/mol at 298 K. What is the standard free energy of
2
1
(1)
103
The following reaction is performed at 298 K.
1
,
r2
j
The vapour pressure of acetone at 20 is 185 torr. When
1.2 g of a non-volatile substance was dissolved in
100 g of acetone at 20 C, its vaopour pressure was
183 torr. The molar mass (g mol–1) of the substance is
(1) 32
(2) 64
(3) 128
(4) 488
Sol. [2] O2
p o  p s w solute  M solvent

ps
M solute  w solvent
w solute  64 g mol 1
JEE Main 2015
[2]
37.
The standard gibbs energy change at 300 K for the
reaction 2A
B + C is 2494.2 J. At a given time,
the composition of the reaction mixture is
40.
1
1
, B  2 and C  . The reaction proceeds
2
2
in the : [R = 8.314 J/K/mol, e = 2.718]
A 
(1) forward direction because Q  K C
(2) reverse direction because Q  K C
Sol. [1] Initial moles of CH 3COOH 
(3) forward direction because Q  K C
Finial moles of CH 3COOH
(4) reverse direction because Q  K C
Kp  e
Kp  e 1 
QP 
2
e

0.9  60  103
 18 Mg
3
Adsorb per gram of charcoal
Mass 
1
1

 0.36
e 2.718
FG 1 IJ
H 2K
41.
Two Faraday of electricity is passed through a solution of CuSO 4 . The mass of copper deposited at the
cathode is: (at. mass of Cu = 63.5 amu)
(1) 0 g
(2) 63.5 g
(3) 2 g
(4) 127 g
Sol. [2] Cu 2 (aq )  2e  
 Cu(s)
2F  1 mol Cu = 63.5 g Cu
39.
Higher order (>3) reactions are rare due to :
(1) low probability of simultaneous collision of all the
reacting species
(2) increase in entropy and activation energy as more
molecules are involved
(3) shifting of equilibrium towards reactants due to elastic collisions
(4) loss of active species on collision
Sol. [1] According to collision theory
The ionic radii (in Å) of N 3 , O2 and F  are
respectively:
(1) 1.36, 1.40 and 1.71
(3) 1.71, 1.40 and 1.36
 Q p  K p backward
38.
0.042  50
 2.1  103
1000
 0.9  103
2494.2
8.314  300
1
2 4
2
0.06  50
 3  103
1000
CH 3COOH adsorbed  3.0  103  2.1  103
Sol. [2] G   RT ln Kp
 G  / RT
3 g of activated charcoal was added to 50 mL of acetic
acid solution (0.06N) in a flask. After qan hour it was
filered and the strenght of tyhe filtrate was found to be
0.042 N. The amount of acetic acid adsorbed (per gram
of charcoal) is:
(1) 18 mg
(2) 36 mg
(3) 42 gm
(4) 54 mg
(2) 1.36, 1.71 and 1.40
(4) 1.71, and 1.36 and 1.40
Sol. [3] N 3  O2  F
In isoelectronic series size decreases as
42.
z
increases .
e
In the context of the Hall - Heroult process for the extraction of Al, which of the following statements is
false?
(1) CO and CO2 are produced in this process
(2) Al 2 O 3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
(3) Al 3 is reduced at the cathode to form Al
(4) N a 3 A lF 6 serves as the electrolyte
Sol. [4] Refer to NCERT
JEE Main 2015
[3]
43.
Form the following statements regarding H 2 O 2 ,
choose the incorrect statement:
(1) It can act only as an oxidizing agent
(2) It decompose on exposure to light
(3) It has to be stored in plastic or was lined glass
bottles in dark
(4) It has to be kept away form dust
Sol. [1] It is disproportionating agent.
44.
Which one of the following alkaline earth metal
sulphates has its hydration enthalpy greater than its
lattice enthalpy ?
(1) CaSO4
(2) BeSO 4
(3) BaSO 4
(4) SrSO 4
47.
Which one has the highest boiling point ?
(1) He
(2) Ne
(3) Kr
(4) Xe

Sol. [4] Boiling point Molecular mass
48.
b
pyridine):
(1) 2
(3) 4
Sol. [2]
49.
enthalpy of Be
45.
and Mg
46.
(4)    * transition
(1) Cl 2
(2) Br2
50.
(3) I 2
(4) ICI
(i) Wacker process
(B) PdCl 2
(ii) Ziegler - Natta polymeron
(C) CuCl 2
(iii) Contact process
(D) V2 O5
(iv) Deacon's process
(1) A  (iii), B  (ii), C  (iv), D  (i)
(2) A  (ii), B  (i), C  (iv), D  (iii)
(3) A  (ii), B  (iii), C  (iv), D  (i)
(4) A  (iii), B  (i), C  (ii), D  (iv)
Sol. [2]
TiCl 3
- Ziegler - Natta polymeron
PdCl 2
- Wacker process
CuCl 2
- Deacon's process
V2 O5
- Contact process
(2) 3
(4) 6
(3) L  M charge transfer transtion
Sol. [3]
(A) TiCl 3
is (py =
The color of KMnO 4 is due to :
Which among the following is the most reactive ?
Match the catalysts to the correct processes:
Catalyst
Process

(2) d  d transition
.
Sol. [4] Interhalogen compounds are more reactive.
g
(1) M  L charge transfer transition
MgSO4 are water soluble due to high hydration
2
gb
square planar Pt ( Cl) ( py) NH 3 NH 2 OH
Sol. [2] Among alkaline earth metal sulphates BeSO 4 and
2
The number of geometric isomers that can exist for
Assertion: Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form
oxides of nitrogen.
Reason: The reaction between nitrogen and oxygen
requires high temperature.
(1) Both assertion and reason are correct, and the reason is the correct explanation for the assertion
(2) Both assertion and reason are correct, but the reason is the not correct explanation for the assertion
(3) The asswertion is incorrect, but the reason is correct
(4) Both the assertion and reason are incorrect
Sol. [1]
51.
In carius method of estimation of halogens, 250 mg of
an organic compound gave 141 mg of AgBr. The percentage of bromine in the comkpound is (at mass Ag =
108, Br = 80)
(1) 24
(2) 36
(3) 48
(4) 60
Sol. [1] % of Br 
Atomic mass of Br  Mass of AgBr
Molecular mass of AgBr  Mass of Compound
JEE Main 2015
[4]
52.
Which of the following compounds will exhibit geometrical isomerism?
(1) 1-Phenyl - 2 butene
(2) 3-Phenyl-1-butene
(3) 2-Phenyl-1-butene
(4) 1,1-Diphenyl-1-propane
56.
Sol. [1] H 3C  CH  CH  CH 2  Ph
53.
In the reaction
the product E is
Which compound would give 5-keto-2-methyl hexanal
upon ozonlysis?
(1)
(1)
(2)
(2)
(3)
(4)
(3)
Sol. [2]
CH3
o
o
Sol. [3] Sand mayer reaction
57.
CH3
54.
The synthesis of alkyl fluorides is best accomplished
by:
(1) Free radical fluorination
(2) Sandmeyer's reaction
(3) Finkelstein reaction
(4) Swarts reaction
Sol. [4] R  X  AgF 
 R  F  AgX
55.
(4)
SOCl
H / Pd
2
4  A  2  B  
C ,
Toluene  
BaSO
4
the product C is:
(1) C 6 H 5COOH
(2) C 6 H 5CH 3
(3) C6 H 5CH 2 OH
(4) C6 H 5CHO
Sol. [4] A = C 6 H 5COOH , B = CH 5COCl
C = C6 H 5CHO
58.
Which of the vitamins given below is water soluble?
(1) Vitamin C
(2) Vitamin D
(3) Vitamin E
(4) Vitamin K
Sol. [1] Vitamin B and C are water soluble.
59.
In the following sequence of reactions:
KMnO
Which polymer is used in the manufacture of paints
and lacquers ?
(1) Bakelite
(2) Glyptal
(3) Polypropene
(4) Poly vinyl chloride
Sol. [2]
Which of the following compounds is not an antacid ?
(1) Aluminium hydroxide
(2) Cimetidine
(3) Phenelzine
(4) Ranitidine
Sol. [3]
60.
Which of the following compounds is not colored yellow ?
b g
(2) K
(3) b NH g Asb Mo O g
(1) Zn 2 Fe CN
4 3
(4) BaCrO 4
Sol. [1]
6
3
10 4
3
b g
CO NO 2
6
[1]
JEE Main 2015
81-B/3, Lohagal Road, Ajmer 305 001 Tel: 0145-2628805, 2628178
Physics
1.
As an electron makes a transition from an excited state
to the ground state of a hydrgone-like atom/ion:
(1) kinetic energy, potential energy and total energy
decrease
(2) kinetic energy decreases, potential energy increases but total energy remains same
(3) kinetic energy and total energy decrease but potential energy increases
(4) its kinetc energy increases but potential energy
and total energy decrease
Sol. [4] d
2.
4.
A signal of 5 kHz freq2uency is amplitude modulated
on a carrier wave of frequency 2 MHz. The frequencies
of the resultant signal is/are:
(1) 2005 kHz, and 1995 kHz
(2) 2005 kHz, 200 kHz and 1995 kHz
(3) 2000 kHz and 1995 kHz
(4) 2 MHz only
Sol. [2]
5.
The period of oscillation of a simple pendulum is
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 gi s
(1) 3%
(2) 1%
(3) 5%
(4) 2%
l
4 2 l
or g  2
g
T
For error estimation,
FG IJ
H K
1 U
U
 T 4 and pressure p 
. If the
32 V
V
shell now undergoes an adiabatic expasnion the relation between T and R is
volume u 
L
T  2
. Measured value of L is 20.0 cm known to
g
Sol. [1] T  2
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
(1) T  e 3R
(3) T 
(2) T 
1
1
R
(4) T  e  R
R3
Sol. [2] For adiabatic expansion dQ  dU  dW  0
g l
T
 2
g
l
T
 dU  dW   PdV
 dU  
103
1

2
 0.005  0.022
2
90
20  10

 0.027  100  3 %
FG IJ
H K
1 U
dV
3 V
dU
1 dV
 
U
3 V
1
Integrating both sides, we get ln U   lnV  C
3
3.
A long cylindrical shell carries positive surface charge
or
U  CV 1/3
in the upper half and negative surface charge 
int he lower half. The electric field lines around the
cylinder will look like figure tgiven in:

U CV 1/ 3

 T 4 (Given)
V
V
(figure are schematic and not drawn to scale)
4
4 / 3

 T V
(1)
 T 4  R 4


(2)
6.
(3)
(4)
. . .(i)
FG 4 R IJ
H3 K
3
4 / 3
 T
b
g
1
R
b
g
. k
An inductor L  0.03 H and a resistor R  015
are connected in series to a bettery 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
Sol. [4]
is closed simultaneously. At t  1ms , the current in
e
j
5
the circuit will be : e  150
[2]
JEE Main 2015
81-B/3, Lohagal Road, Ajmer 305 001 Tel: 0145-2628805, 2628178
8.
(1) 67 mA
(3) 0.67 mA
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) 2.45 V / m
(2) 5.48 V / m
(3) 7.75 V / m
(4) 1.73 V / m
Sol. [1] Intensity of EMW is
(2) 6.7 mA
(4) 100 mA
1
P
v CE 02 
2
A
Sol. [3] In the saturated state before t  0,
i0 
15 V
 100 mA
0.15k
As k1 is opered and k 2 is closed, the inductor current
will decay in LR circuit according to the equation
i  i0 e


E 02 

E0 
Rt
L
i  i0 e

 i0 e 5 
7.
9.
0.15103 10 3
0.03
100
mA  0.67 mA
150
is added to its bob, the time period changes to TM . If
the Young's modulus of the material of the wire is Y
then
LF T I O Mg
(1) MGH T JK  1P A
MN
PQ
LM F T I OP A
(3) M1  GH T JK P Mg
N
Q
L FT I
(2) M1  GH T JK
MN
2
M
M
LMF T I
(4) GH T JK
MN
2
M
M
Sol. [4] T  
FG IJ
H K
TM2
'


T2
 
 ' 
Mg

A.Y
Mg

YA
FG T IJ
HTK
1
2
1
A
 Y  Mg
OP A
PQ Mg
OP A
PQ Mg
1
m
2
10.
Consider an ideal gas confined in an isolated closed
chamber. As the gas undergoes an adiabatic. As the
gas undergoes an adiabatic expansion, the average time
of collison between molecules increases as V q , where
F
GH
V is the volume of the gas. The value of q is :  
(1)
3  5
6
OP
PQ
(2)
 1
(4)
2
tavg 
1
0
2  01
.  9  109
Sol. [2] Vavg 
Mg
YA
LMF T I
MNGH T JK
2
Sol. [3]
(3)
'  
m
TM2

T2
2
2
e4d jc 
magnetic force ont he outer solenoid due to the inner
one. Then:


(1) F1 is radially inwards and F2 is radially outwards


(2) F1 is radially inwards and F2  0


(3) F1 is radially outwards and F2  0


(4) F1  F2  0
A pendulum made of a uniform wire of cross sectional
area A has time period T. When an additional mass M
1
is equal to
Y
(g = gravitational acceleration)
2P
 6  2.45 V / m
12  3  108
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

At t  1ms
2P
AC 0
 avg
vavg
 1
2
3  5
6
 avg
tavg

v
2Nd 2 T
For Adiabatic PV   const
PV  nRT
TV  1  const
tavg
1
V 2
 q
1 
2
I
C JK
Cp
v
[3]
11.
An LCR circuit is equivalent to a damped pendulum. In
an LCR circuit the capacitor is charged to Q0 and then
GM
connected to the L and R as shwon below:
V2  
FG M IJ LM3FG R IJ
H 8 K MN H 2 K
F RI
2G J
H 2K
So, V  V1  V2  
If a student plots graphs of the square of maximum
e
j
13.
2
charge Q Max on the capacitor with time (t) for 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)
(1)
(2)
(3)
(4)
Sol. [4]
From a solid sphere of mass M and radius R, a spherical porition of radius R / 2 is removed, as shown in the
figure. Taking gravitational potential V = 0 at r   ,
the potential at the centre of the cavity thus formed is:
(G = gravitational constant)
2
 r2
3
OP
PQ

11 GM
8 R
r 0
11 GM 3 GM
GM


8 R
8 R
R
A train is moving on a straight track with speed 20 ms –
1
. It is blowing its whistle at the frequency of 1000 hz.
The percentage change in the frequency heard by a
sperson standing near the track as the train passes him
is (speed of sound = 320 ms–1) close to:
(1) 12%
(2) 18%
(3) 24%
(4) 6%
Sol. [1]
14.
12.
JEE Main 2015
81-B/3, Lohagal Road, Ajmer 305 001 Tel: 0145-2628805, 2628178
f
2V
2  20
 100 
 100 
 100  12 %
f0
Vs
320
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 shwon. If the coefficient
of friction between teh 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) 80 N
(3) 150 N
(2) 120 N
(4) 100 N
Sol. [2]
(1)
GM
R
(2)
2GM
(3)
R
2GM
3R
F  N1 , FA  M A g  20 N  01
.  N1
GM
(4)
2R
 N1  200 N

Sol. [1]
O
P

O
P
–
P
15.
f B  100  f A  120 N
Distance of the centre of mass of a solid uniform cone
from its vertex is z0 . If the radius of tis base is R and its
height is h then z0 is equal to
V  V1  V2
V1  
e
GM 3R 2  r 2
2R
j

3
r ( R / 2 )
FG IJ
H K
11 GM
8 R
(1)
3h
4
(2)
5h
8
(3)
3h 2
8R
(4)
h2
4R
[4]
Sol. [1] Center of mass x0 
z
z
z
z
h
O
h
O
h

O
h
x dm
x2
I1
I2–I1
dm
(2)
(1)
2
LM x OP
N4Q

Lx O
dx M P
N3Q
x 3dx
I2
Sol. [2]
Loop -(1) 9  2 I 2  I1  3I 3  0
b
2
O
16.
z
z
FG R xIJ dx
Hh K
FR I
 G xJ dx
Hh K
x.  . 

JEE Main 2015
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4 h
o
3 h
o
g
Loop -(2) 6  3 I 2  I1  I1  0
By using (2) and (1) I1  Q13 A  Q to P
Fh I
GH 4 JK 3

Fh I  4h
GH 3 JK
4
3
A rectangular loop of sides 10 cm and 5 cm carrying a
current I of 12 A is placed in different orieintations as
shown in the figures below:
18.
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
potentials
V0
3V0 5V0 3V0
,
,
and
have radius R1 ,
4
2
4
4
R2 , R3 and R4 respectively. Then
b
b
(2) R1  0 and R2  R4  R3
(3) 2 R  R4
(a)
(b)
g b
(1) R1  0 and R2  R1  R4  R3
b
(4) R1  0 and R2  R4  R3
g
g
g
Sol. [2] ,[3] Potential due to uniformly charged solid sphere
V
(c)
(d)
V 
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 (c), respectively
(2) (b) and (c), respectively
(3) (b) and (d), respectively
(4) (a) and (b), respectively
Q
rR
4 0 r
e
Q 3R 2  r 2
4  0 2 R
Given V0 
3
jr  R
1 Q
(Potential at surface)
4  0 R
Sol. [3] Torque on the loop will be   M  B
Potential energy of the loop in the magnetic field
U  M B
The loop will be in equlibrium if Z  0 ; the equilibrium
will be stable when U is least.
17.
In the circuit shown, the current in the 1 resistro is
R4  R3  4 R 
R4  R3  R2
 R1  0
(1) 0 A
(3) 0.13 A, from P to Q
(2) 0.13 A, from Q to P
(4) 1.3 A, from P to Q
4
8
R
R R
 R2
3
3
2
[5]
19.
JEE Main 2015
81-B/3, Lohagal Road, Ajmer 305 001 Tel: 0145-2628805, 2628178
In the given circuit, charge Q2 on the 2 µF capacitor
changes as C is varied from 1 µF to 3 µF. Q2 as a
function of 'C' is given properly by : (figure are drawn
schematically and are not to scale)
Sol. [2]
By momentum conservation
 
pi  p j  m 2v i  m 2v j
b g
(1)
 
Let p  2mv then pi  p j  pi  pj . . .(i)
(2)
Kinetic energy before collision
Ki 
(3)
b g
p 2f
2(3m)

p2  p2 p2

6m
3m
So, the pericentage loss in K.E.,
Sol. [1]

Q2 

2
2
QNet   ECNet
2 1
3
FG
H
IJ
K
2
3C
2E
E

3
C3
1 3 / C
and as C
b g
A;QA
2
 It is a increasing curve.
Now,
dQ2
 2E  
dC
1
FG1  3 IJ
H CK
d 2 Q2
12 E

2
dC
C3
b
g
2
2

Ki  K f
Ki
3 p2 p2

 100  4 m 2 3m  100
3 p
4 m
3 1

4
3  100  500  5555

.  56%
3
9
4
Clearly the Q2 versus C graph will be a curve (Not a
straght line)
20.
p12
p2
p2
p2
3 p2
 2 


2m1 2m2 2m 2 2m 4 m
Kinetic energy after collision
(4)
Kj 
and
b g
3
6E

0
2
2
C
C3
b
21.
Monochromatic light is incident on a glass prism of
angle A. If the refractive index of the material of the
prims is µ, a ray, incident at an angle  , on the face AB
would get transmitted through the face AC of the prism
provided:
g
0
A particle of mass m movign 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) 50%
(2) 56%
(3) 62 %
(4) 44%
LM F
MN GH
LM sinF A  sin
MN GH
LM sinF A  sin
MN GH
FG 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
1
1 1
(1)   sin  sin A  sin 
1
(2)   cos
1
1
(3)   cos
1
[6]
(4)   sin
1
(A) Franck-Hertz
Experiment.
(B) Photo-electric
experiment
(C) Davison - Germer
LM sinF A  sin F 1 I I OP
GH  JK JK PQ
MN GH
1
Sol. [4]
Condition for incident light to transmit through face
AC, r2   C
22.
or
A  r1   C
or
r1  A   C
or
sin r1  sin A   C
or
sin   sin A   C
or
  sin 1  sin A  sin 1
b
[ r1  r2  A]
b
g
g
(3)
16 2
4 MR
(A) – (ii)
(A) – (ii)
(A) – (iv)
(A) – (i)
When 5V potential difference is applied across a wire
of length 0.1 m, the drift speed of electrons is
(2)
(4)
8  1028 m3 , the resistivity of the material is close to
FG 1 IJ I OP
H  K JK PQ
9 3
MR
2
(1) 16
.  107 m
(2) 16
.  10 6 m
(3) 16
.  10 5 m
(4) 16
.  10 8 m
Sol. [3] J  E

E
E
V .l


J n. evd n. evd
Putting values,   16
.  105 m
4 MR 2
2
3 3
24.
sin   sin r snells law
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
MR 2
(1)
(2)
(3)
(4)
Sol. [2]
(i) Particle nature of
light
(ii) Discrete energy
levels of atom
(iii) Wave nature of
electron
(iv) Structure of atom
(B) – (iv)
(C) – (iii)
(B) – (i)
(C) – (iii)
(B) – (iii)
(C) – (ii)
(B) – (iv)
(C) – (iii)
2.5  10 4 ms1 . If the electron density in the wire is
LM F
MN GH
(1)
JEE Main 2015
81-B/3, Lohagal Road, Ajmer 305 001 Tel: 0145-2628805, 2628178
25.
32 2
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 d.
Which one of the following represents these correctly?
(graphs are schematic and not drawn to scale)
Sol. [2]
R
23.
a
2R
(a is the length of side of cube)
3
I
M ' a2
6
M'

3M
a5
4R 3  6

3M
32 R 5
4 MR 2


4R 3  6 9  3 9 3
M
.a 3
4 3
R
3
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
(1)
(2)
(3)
(4)
Sol. [1]
KE 
26.
1
1
K A2  x 2  PE  Kx 2
2
2
e
j
Two stones are thrown up simultaneosuly fromt he
edge fo 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 second stone with respect to the first?
(Assume stones do nto rebound after hitting the
[7]
JEE Main 2015
81-B/3, Lohagal Road, Ajmer 305 001 Tel: 0145-2628805, 2628178
ground and neglect awir resistance, take g  10 m / s2 )
(The figures are schematic and not drawn to scale)
(1)
In both the cases body is brough from initial
temperatgure 100 oC to final temperature 200 oC.
entropy change of the body in the two cases respectively is
(1) ln2, ln2
(2) ln2, 2ln2
(3) 2ln2, 8ln2
(4) ln2, 4ln2
Sol.[1] Note:The temperature should have been 100 K and 200 K
CT  Q  T .S
T
S  C
T
200
S  ln T 100
S 1  ln 2
S 2  ln 2
z z
(2)
(3)
28.
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) 30 µm
(2) 100 µm
(3) 300 µm
(4) 1 µm
Sol. [1] Minimum angular separation that can be resolved
(4)
d 
Sol. [2]
122
. 
D
 Minimum dest between objects that can be resolved
d  rd 
1
y1  y0  u y0 t  gt 2
2
 240  10t  5t 2 . . . (i)

y2  240  40t  5t 2 . . . (ii)
j e
122
.  25  102  500  109
e0.5  10 j
Aslo, y1  0, y2  0
 122
.  25  106 m
For y1  0  0  240  10t  5t 2
 30 m
 t 2  2t  48  0
 t1  8 s
For y 2  0 , t 2  12 s (similarly)
When both the stores are in air t  8s
y2  y1  30t . . . (iii) (by (ii) - (i))
when first stone has reached the round but record is
still in air 8  t  12
y2  y1  y2  240  40t  5t 2 . . . (iv)
Putting (iii) and (iv) graphically
27.
e
122
. d
D
A solid body of constant heat capacity 1 J / oC 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.
29.
j
2
Two long current carrying thin wires, both with current
I, are held by insulating threasds of length L and are in
equilibrium as shown in the figure, with threads making and angle  with the vertical. if wires have mass
 per unit length then the vlaue of 1 is
(g = gravitational acceleration)
[8]
 g L
(1) 2 sin   cos
0
(3)
gL
tan 
0
81-B/3, Lohagal Road, Ajmer 305 001 Tel: 0145-2628805, 2628178
(2) 2
gL
tan 
0
gL
(4) sin   cos
0
Sol. [1]
Let   length of wiere
T sin   Fm
T cos  g
tan  
Fm
 g
  g tan  

30.
I  2 sin
 0 L2 
4  L sin 
gL
 0 cos
On a hot summer night, the refractive index of air is
amllest near the ground and increases with height from
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) goes horizontally without any deflection
(2) bends downwards
(3) bends upwards
(4) becomes narrower
Sol. [3]
JEE Main 2015
JEE Main 2015
[1]
Mathematics
61.
Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the
set A × B, each having at least hree elemnts is:
(1) 219
(2) 256
(3) 275
(4) 510
b
g
Sol. [3] x 2  6x  2  0
  is one of the roots
  2  6  2  0
  10  6 9  2 8  0
8
Sol. [1] Total no of subset of A  B  2  256
8
8
8
Given, a n   n   n
8
Required no of subsets  C 3  C 4  C5  ... C8
8
2 
8
8
Now
8
C 0  C1  C 2
e
e
8
 2  37  256  37  219
62.

A complex number z is said to be unimodular if z  1 .
6 9  6 9
z1  2 z2
2  z1z2 is unimodular and z2 is not unimodular. Then
the point z1 lies on a:
64.
(1) straight line parallel to x-axis.
(2) straight line parallel to y-axis.
(3) circle of radius 2.
(4) circle of radius
LM1
If A  M2
MNa
2
LM1
MM2
Na
LM
MM
N
2
 2  z1z 2
2
b
gb
g b
gb
g
 b z  2z gb z  2 z g  b2  z z gb2  z z g
 e1  z je z  4j  0
 z  2 c z  1h
1
1
1
2
1 2
1 2
2

2
 circle of radius 2
63.
OP
PP
Q
2
2
1 2
is a matrix satsfying the equation
2
b
b g
 z1  2 z 2 z1  2 z 2  2  z1 z 2 2  z1 z 2
2
6
3
2
b
Let  and  be the roots of equation x 2  6x  2  0
65.
OP LM1
PP MM2
b Q N2
OP LM
OP
P M
P
2
2 bPQ MN0 0 9PQ
 a  4  2 b O L9 0 0O
 2a  2  2 b PP  MM0 9 0PP
 a  4  b PQ MN0 0 9PQ
a  2 b  4  0U
V  a  2, b  1
2a  2  2 b  0W
2 2
1 2
2
2 x1  2 x2  x3  x1
a10  2a8
is equal tyo
2a9
2 x1  3x2  2 x3  x2
(2) – 6
(4) – 3
2
1
a
9 0 0
2  0 9 0
2
The sets of all values of  for which the system of
linear equations:
If an   n   n , for n  1, then the value of
(1) 6
(3) 3
g
(4) 2,  1
Sol. [4] AA T  9 I
squaring, z1  2z 2
2
j

(3) 2, 1
z1  2z 2  2  z1z 2
2
9
j
j
AAT  9 I , where I is 3 × 3 identity matrix, then the
ordered pair (a, b) is equal to :
(1) (2, –1)
(2) (–2, 1)
z1  2 z 2
Sol. [3] 2  z z  1
1 2
1
e
9
2  
Suppose z1 and z2 are complex numbers such that
j e
10  10  2  8  8
a 10  2a 8

2a 9
2  9  9
 x1  2 x2
 x3
has a non-trivial solution,
(1) is an empty set
(2) is a singleton.
(3) contains two elements
(4) contains more than two elements
JEE Main 2015
[2]
b g
2 x  b3   g x
On replacing x by 1, we get
Sol. [3] 2   x1  2x 2  x 3  0
1
e
350  1  2
 2x 3  0
2
 x1  2 x 2  x 3  0
System has non-trivial solution
2
2
b g
2
1

 3 
2
b  3gb  1g
66.
2
68.
2 0

  1, 1,  3
2
3
2

(1)
1 50
3 1
2
e
j
(2)
1 50
(3) 3  1
2
e
(4) 4 l 2 m2 n 2
n
. . . (i)
2
FG n IJ
H K
F nI
 G J
H K
F nI
 G J
H K
F nI
 G J
H K
e
j
50
50
 50 C 0  50
on adding we get,
e1  2 x j  e1  2 x j
2
e j e j
C e2 x j  ... C e2 x j
C e2 x j  C e2 x j 
C e2 x j  ... C e2 x j
50
50
j
 50 C 0  50 C1 2 x  50 C 2 2 x
50
e1  x j
69.
e
50
3
50
3
50
1
3
3
F
H
 213
50
50
50
2
2
50
50
50
e j
C 0  50 C 2 2 x
2

e j IK
... 50 C50 2 x
50
 G14  n 3
2/4
 G 2 4  n2 2
3/ 4
 G 34  n 3
G14  2G 2 4  G 34  n 3  2 n 2  2  n 3
b g
e j
(Common ratio)
1/ 4
 n 2 m
1 50
3
2
1/ 4
 n  2  2 n  n 2  n   n
is
1 50
2 1
(4)
2
j
Sol. [1] 1  2 x
50
j
G1 , G2 and G3 are three geometric
b g
The sum of coefficients of integral powers of x in the
e
j
(3) 4 lmn2
G3
no. of ways  5  4  3  2  1  120
Total no. of number  72  120  192
binomial expansion of 1  2 x
e
(2) 4 lm2 n
G2
4
1 50
3 1
2
(1) 4 l 2 mn
G1
(ii)
5
50
, G 1 , G 2 , G 3 , n are in GP
no. of ways  3  4  3  2  72
67.
bl, n  1g and
 R
3
bg j
 ... 50 C50 2
If m is the A.M. of two distinct real numbers l and n
Sol. [2] m 
Sol. [2] (i)
4
2
means between l and n, then G14  2G24  G34 equals.
The number of integers greater than 6,00 than can be
formed, using the digits 3, 5, 6, 7 and 8, without repetition, is
(1) 216
(2) 192
(3) 120
(4) 72
3
bg
C 0  50 C 2 2
 sum of coeff 
1
0 
50
2
2
 4 m 2 n
The sum of first 9 terms of the series
13
13  2 3  33

.... is
1
1 3  5
(1) 71
(2) 96
(3) 142
(4) 192
Sol. [2] Tn 
1
2
n 1
4
b g
 sum of first 9 terms 
1 2
2  32  4 2  ...102
4
1 2
1  2 2  ...102  12
4
e
j e j
1 L b10gb11gb21g O 1
 M
 1P 
b385  1g  96
4N
6
Q 45

JEE Main 2015
[3]
70.
b1  cos 2 xgb3  cos xg is equal to :
lim
bx  3gbx  1g  0
(2) 3
(4) 1/2
 x  3, 1
 Points of interesection are (3, –1) and (1, 1)
 Normal meets the curve again at (3, –1) i.e. in the
fourth quadrant.
b1  cos 2xgb3  cos xg
[3] lim
x tan 4 x
x 0
 lim
F 2 sin
GH x
2
x
2
x0
71.
x 2  4x  3  0
x tan 4 x
x0
(1) 4
(3) 2
Sol.
On eleminating y from (i) and (ii), we get

I b gb gFG IJ
JK
HK
4x
1
 4 2
2
4 tan 4 x
4
73.
LM
N
treme values at x  1 and x  2 . If lim 1
If the function.
g( x) 
|RSk x  1,
|T mx  2,
x0
Now
f x  ax 4  bx 3  2 x 2
x 0
bg
f ' b xg  4ax  3bx  4 x
f ' b1g  4a  3b  4  0
f ' b2g  32a  12 b  8  0
3
and
2
on solving, a 
b g
The normal to the curve, x 2  2 xy  3y 2  0 at 1, 1
(1) does not meet the curve again.
(2) meets the curve again in the second quadrant
(3) meets the curve again in the third quadrant
(4) meets the curve again in the fourth quadrant
2

OP
Q
dy
dy
 y  6y
0
dx
dx
dy
xy
 dx  3y  x
1
1, 1
 Equation of normal at (1, 1) is
b g
y  1  1 x  1
 x  y  2  0 . . . (ii)
1
; b  2
2
f x 
bg
1 4
x  2x 3  2x 2
2
bg
1
 16  2  8  2  4
2
f 2 
 8  16  8  0
2
Sol. [4] Curve is x  2xy  3y  0 . . . (i)
on differentiates w.r.t. x, we set
g
lim
ax 4  bx 3  cx 2  dx  e

 km2
F dy I
 GH dx JK
b
x2
2
x2
d and e must be zero and c = 2

2
8
,k
5
5
LM
N
(4) 4
x0
k
 m  k  4 m . . . (i)

4
(ii) g is also continuous
 2k  3m  2 . . . (ii)
solving (i) and (ii), we get
2x  2 x
(3) 0
bg
f b xg
lim
2
13
(3)
(4) 4
3
Sol. [1] (i) g is differentiable at x = 3
 LHD = RHD
72.
(2) 4
Sol. [3] Let f x  ax 4  bx 3  cx 2  dx  e
16
(2)
5
(1) 2
(1) 8
74.
z
The integral
F x  1I
(1) G
H x JK
1/ 4
e
1/ 4
4
4
4
j
(3)  x  1
dx
2
e
4
j
x x 1
3/ 2
equals:
e
j
1/ 4
c
(2) x 4  1
c
F x  1I
(4) G
H x JK
4
4
OP
Q
f ( x)
3
x2
then f (2) is equal to :
0 x 3
3 x 
is differentiable, then the value of k  m is
m
Let f ( x) be a polynomial of degree four having ex-
c
1/ 4
c
JEE Main 2015
[4]
z
Sol. [4]

z
1
e
e
x5 1  x
4
x
 t
75.
5
4
j
4 3/ 4
Required Area 
z
1
4
z
1
t
3/ 4
2
4
4
e
log x 2  log 36  12 x  x 2
Sol. [3]
z
2
I
z
z
2
4
I
2
j
dx
log x  log 36  12 x  x
log x
2
b g
logb6  xg
logb6  xg  log x
2
log x  log 6  x
2
2
j
One of the initial condition be x  1, y  0 .
Now, given diff equation is
dx
FG
H
dx
IF  e
2
dx (using property)
z
2I  1 dx
e

2
I  1
2
76.
The area (in sq. units) of the region described by
 0  2  c
 C2
 Solution is
2
b
log log x
g  log x
z
when x  1, y  0
2
(3)
1
dx
x log x
b gb g
y log x  2b x log x  xg  c

(1)
z
 solution is y log x  2 log xdx  c
4
o( x, y) : y
IJ
K
dy
1

y2
dx
x log x
2
2
(2) 0
(4) 2e
 y0
when x  1, 0  y  0
e
2
1/ 2
 y  2 x log x, b x  1g
bx log xg dy
dx
log x 2
4
3 1
Let y( x) be the solution of the differential equation
They y(e) is equal to :
(1) e
(3) 2
Sol. [3] (Seems incomplete)
(2) 4
(4) 6
I
2
2
b x log xg dydx  2 x log x, bx  1g
4
log x 2
4
or
77.
F x  1I  c
F 1I
 G 1  J  c  G
H xK
H x JK
(1) 2
(3) 1
or
dt
The integral
z
g
FG 1 y  1  1 y IJ dy  LM y  y  y OP

H 4 4 2 K MN 8 4 6 PQ
F 1 1 1I F 1 1 1 I 9
 G   J G   J 
H 8 4 6K H 32 8 48K 32
1
dx
t
4
OP
Q
1
1
y  1  y 2 dy
4
2
1/ 2
dx  dt  
1/ 4
z LMN b
1
dx
1/ 2
1
Let 1  x

j
x2 x4  1
3/ 4
b
t
FG IJ
H K
 2 x and y  4 x  1 is
7
32
(2)
15
64
(4)
g
y log x  2 x log x  x  2
x
 y log x  2 x log e  2
5
64
2 x log
9
32
y
FG x IJ  2
H eK
log x
bg
 y e 2
Sol. [4]
x
1 2
y
2
78.
1
A ,1
1 2
x  y 1
4
GH JK
b g
1
1
B ,
8
2
GH
The number of points, having both co-ordinates as
integers, that lie in the interior of the triangle with
b gb g
b g
vertices 0, 0 0, 41 and 41, 0 is :
p
JK
(1) 901
(3) 820
(2) 861
(4) 780
JEE Main 2015
[5]
LM F 12  3  2 I  4OP  0
MN GH 2    7 JK PQ
2 3
Sol. [4]
B
(0, 40)
(0, 39)
(0, 41)
  2   2  2  4  3  0
(1, 40)
(2, 39)
R39

b g
locus of Q ,  is
x 2  y 2  2x  4y  3  0
(0, 2)
R2
R1
(0, 1)
(39, 2)
(40, 1)
(41, 0)
D
A
O
This is circle of radius
80.
79.
b
g  780
39 39  1
2
Locus of the image of the point (2, 3) in the line
b2x  3y  4g  kbx  2y  3g  0 , k R , is a :
x 2  y 2  6x  18y  26  0 , is
(1) 1
(3) 3
2.
(4) circle of radius
3.
81.
Given line
(2+k)x–(3+2k)y+(4+3k)=0
Q(, )
(Image of P)
3
Slope of PQ is
2
Slope of given line is
(1)
27
4
(2) 18
(3)
27
2
(4) 27
FG 2  k IJ
H 3  2k K
x2 y2

1
9
5
given line
b2  kgFGH  2 2 IJK  b3  12kgFGH  2 3IJK  b4  3kg  0
using the value of k, we get
b  2gLMN122 3  27  2OPQ  b  gLMM2FGH 122 3  27 IJK  3OPP
N
Q
b2
5
 A 2,
a
3
JK GH JK
S (ae, 0)
Here e 
FG 5IJ is
H 3K
b g  yb5 / 3g  1
x2
9
FG   2 ,   3IJ will lie on the
H 2 2K
A ae,
Equation of tanget at A 2,
12  3  2
k
2    7
Mid point of PQ i.e. T
GH
Q
O
F 2  k IF   3I
 GH 3  2 k JK GH   2 JK  1

x2 y2

 1 , is
9
5
Sol. [4]
Both are perpendicular

The area (in sq. units) of the quadrilateral formed by
the tangents at the end points of the latera recta to the
ellipse
P(2, 3)
Sol. [3]
(2) 2
(4) 4
Sol. [3] C1C 2  r1  r2
 no of common tangents is 3
(1) straight line parallel to x-axis.
(2) straight line parallel to y-axis.
(3) circle of raidus
The number of common tangents to the circles
x 2  y 2  4x  6y  12  0 and
No of points, having both co-ordinates as integ ers,
that lie in the interior of the triangle AOB be as follows:
39  38  37  ...1 
2
5

x
y
 1
9/2 3

area of POQ 

Required Area  4
1 9
27
 3
2 2
4
FG 27 IJ  27
H 4K
P
2
3
JEE Main 2015
[6]
82.
Let O be the vertex and Q be any point on the parabola,
Sol. [3] Equation containing the line line is
x 2  8y. If the point P divides the line segment OQ
internally in the ratio 1 : 3, then the locus of P is
(1) x 2  y
(2) y 2  x
(3) y 2  2x
(4) x 2  2y
This is parallel to x  3y  6z  1 . . . (i)
2     5 1  4


1
3
6
11
2
 (i) becomes
x  3y  6z  7  0
 
2
x  8y
Sol. [4]

O
3
1P
e
Q 4 t, 2 t 2
j
85.
Let P(h, k) be the point whise locus is required. This
point divides OQ is the ratio 1 : 3 in ternally.
h

t  h and t 2  2k
On elemniatry t we get
e j
value of sin is
2
h  2k

locus is x 2  2y
83.
The distance of the point (1, 0, 2) from the point of
x  2 y 1 z  2


intersection of the line
and the
3
4
12
2 2
3
(2)
 2
3
(3)
2
3
(4)
2 3
3

(1) 2 14
(2) 8
(3) 3 21
(4) 13
FG IJ
H K


ec  bj a  ea  bjc  FGH 13 bcIJK a
RSec  b j  1 bcUV a  ea  b j c
2 W
T
FG bc cos  1 bcIJ a  ea  b j c
H
3 K
e j

x  2 y 1 z  2


k
3
4
12
Point on the line be
 no two of them are collinear
It is on the plane x  y  z  16
1
 Above exists if bc cos  bc  0
3

b3k  2, 4k  1, 12k  2g
3k  2  4k  1  12k  2  16
11k  11
 k 1
 Point of intersectionis (5, 3, 14)
 required distance

b5  1g  b3  0g  b14  2g
2
2
2
 16  9  144  169  13
84.
(1)
  

1
bc a
Sol. [1] a  b  c 
3
plane x  y  z  16, is


 
Let a, b and c be three non-zero vectors such that no
   1   
two of them are collinear and a  b  c  b c a .
3


If  is the angle between vectors b and c, then a
4t  0
2t 2  0
and k 
4
4

Sol. [4]
b2x  5y  z  3g  bx  y  4z  5g  0
b2  gx  b  5gy  b1  4g  b5  3g  0
The equation of the plane containing the line
2 x  5 y  z  3 ; x  y  4z  5 , and parallel to the
plane, x  3 y  6z  1, is
(1) 2 x  6 y  12 z  13
(2) x  3 y  6z  7
(3) x  3 y  6z  7
(4) 2 x  6 y  12 z  13
86.
 cos 
1
3
 sin 
2 2
3
If 12 identical balls are to be placed in 3 identical boxes,
then the probability that one of the boxes contains
excatly 3 balls is (Seems in complete)
FG IJ
HK
F 1I
(3) 220G J
H 3K
(1)
55 2
3 3
11
12
FG 2 IJ
H 3K
F 1I
(4) 22G J
H 3K
(2) 55
10
11
JEE Main 2015
[7]
Sol. [1] n  12
1
3
q = Probability of ball not going in that particular
1
1
1
Let tan y  tan x  tan
89.
p = Probability of a ball going in one particular box 
2
3
p (exactly 3 success)
box 
 12 C 3 p 3q 9  12 C 3
87.
(1)
FG 1IJ FG 2 IJ
H 3K H 3 K
3
9

FG IJ
HK
55 2

3 3
11
(3)
The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are
added to the data, then the mean of the resultant data,
is
(1) 16.8
(2) 16.0
(3) 15.8
(4) 14.0
S o l. [4] x 
x
n
16 
x
16
new  x 256  16  3  4  5

18
18
252

 14
18
If the angles of elevation of the top of a tower from
three collinear points A, B and C, on a line leading to
the foot of the tower, are 30o, 45o and 60o respectively,
then the ratio, AB : BC, is
(2)
3 :1
(3) 1 : 3
S o l. [1]
3: 2
(4) 2 : 3
D
60o
O
45o
C
30o
B
Let OB  x


OA  x 3 and OC 
x
3
AB OA  OB x 3  x


 3
x
BC OB  OC
x
3
A
2
. Then a value of y is
3x  x 3
(2)
1  3x 2
3x  x 3
(4)
1  3x 2
3x  x 3
1  3x 2
3x  x 3
1  3x 2
FG 2x IJ
H 1 x K
2
 tan 1 x  2 tan 1 x  3 tan 1 x
 tan 1
F 3x  x I
GH 1  3x JK
3
2
3x  x 3

y
90.
The negation of ~ s   r  s is equivalent to
1  3x 2
b
g
b
(1) s  ~ r
New mean 
(1)
3
1
1
1
Sol. [1] tan y  tan x  tan
 x  256
88.
1
where x 
FG 2x IJ ,
H 1 x K
g
(2) s  r  ~ s
b
g
(4) s  r
(3) s  r  ~ s
Sol. [4]
r
s
~r
T
T
F
F
T
F
T
F
F
F
T
T
b
g
~ r  s ~ s ~ s ~ s s
F
F
T
F
F
T
F
T
F
T
T
T
negative of
~ s  ~ r  s s ~ r
T
F
F
F
F
T
F
F
b
g
s r
T
F
F
F