JEE-MAINS – 2014 Question Paper & Solutions
1
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JEE-MAINS – 2014 Question Paper & Solutions
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PHYSICS
1. The pressure that has to be applied to the ends of a steel wire of length 10
cm to keep its length constant when its temperature is raised by 100oC is :
(For steel Young’s modulus is 2 × 1011 N m–2 and coefficient of thermal
expansion is 1.1 × 10–5 K–1)
(1) 2.2107Pa (2) 2.2106Pa (3) 2.2108Pa (4) 2.2109Pa
Key. 3
Sol. P T
= 21011 1.1105 100
= 2.2108Pa
2. A conductor lies along the z-axis at –1.5 z < 1.5 m and carries a fixed
current of 10.0 A in z aˆ direction (see figure). For a field 4 0.2x
y B 3.010 e aˆ T,
find the power required to move the conductor at constant speed to x =
2.0 m, y = 0 m in 5 × 10–3s. Assume parallel motion along the x-axis.
(1) 14.58 W (2) 29.7 W (3) 1.57 W (4) 2.97 W
Key. 4
Sol. P 1 iBl dx
t
2 4 0.2x
3 0
1 10 3.0 10 e 3 dx
5 10
= 2.97 W
3. A bob of mass m attached to an inextensible string of length l is suspened
from a vertical support. The bob rotates in a horizontal circle with an
angular speed rad/s about the vertical. About the point of suspension:
(1) angular momentum changes in direction but not in magnitude.
(2) angular momentum changes both in direction and magnitude.
(3) angular momentum is conserved.
(4) angular momentum changes in magnitude but not in direction.
JEE-MAINS – 2014 Question Paper & Solutions
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Key. 3
Sol. The torque of all the forces about the point of suspension is zero therefore
angular momentum is conserved.
4. The current voltage relation of diode is given by I = (e1000V/T – 1) mA,
where the applied voltage V is in volts and the temperature T is in degree
Kelvin. If a student makes an error measuring 0.01 V while measuring
the current of 5mA at 300 K, what will be the error in the value of current
in mA?
(1) 0.5 mA (2) 0.05 A (3) 0.2 mA (4) 0.02 mA
Key. 3
Sol.
1000V
I e T 1
1000V
I 1 e T
1000V
T I 1000 e . V
T
1000 5 10.01
300
1000V
e T 5 1
= 0.2 mA
5. An open glass tube is immersed in mercury in such a way that a length of 8
cm extends above the mercury level. The open end of the tube is then
closed and sealed and the tube is raised vertically up by additional 46 cm.
what will be length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
(1) 38 cm (2) 6 cm (3) 16 cm (4) 22 cm
Key. 3
Sol.
8 cm 54 cm
x
Applying Boyle’s Law to the air column in the glass tube,
P1V1 = P2V2
76 × 8 × A = {76 – (54 – x)} . x.A
x = 16 cm
JEE-MAINS – 2014 Question Paper & Solutions
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6. Match List – I (Electromagnetic wave type) with List – II (Its
association/application) and select the correct option from the choices
given below the lists:
List – I List- II
(a) Infrared waves (i) To treat muscular strain
(b) Radio waves (ii) For broadcasting
(c) X-rays (iii) To detect fracture of bones
(d) Ultraviolet rays (iv) Absorbed by the ozone
layer of the
atmosphere
(a) (b) (c) (d)
(1) (iii) (ii) (i) (iv)
(2) (i) (ii) (iii) (iv)
(3) (iv) (iii) (ii) (i)
(4) (i) (ii) (iv) (iii)
Key. 2
Sol. Conceptual
7. A parallel plate capacitor is made of two circular plates separated by a
distance of 5 mm and with a dielectric of dielectric constant 2.2 between
them. When the electric field in the dielectric is 3 × 104 V/m, the charge
density of the positive plate will be close to :
(1) 3 × 104 C/m2 (2) 6 × 104 C/m2 (3) 6 × 10–7 C/m2 (4) 3 × 10–7 C/m2
Key. 3
Sol. Charge density
o K A Q .Ed d
A A
= Ko E
6107 C/m2
JEE-MAINS – 2014 Question Paper & Solutions
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8. A student measured the length of a rod and worte it as 3.50 cm. Which
instrument did he use to measure it ?
(1) A screw gauge having 100 divisions in the circular scale and pitch as 1
mm.
(2) A screw gauge having 50 divisions in the circular scale and pitch as 1
mm.
(3) A meter scale.
(4) A vernier caliper where the 10 divisions in vernier scale matches with
9 division in main scale and main scale has 10 divisions in 1 cm.
Key. 4
Sol. The least count of the observation is 0.01 cm which corresponds to option
(4)
9. Four particles, each of mass M and equidistant from each other, move
along a circle of radius R under the action of their mutual gravitational
attraction. The speed of each particle is
(1) GM1 2 2
R
(2) 1 GM1 2 2
2 R
(3) GM
R
(4) 2 2 GM
R
Key. (2)
Sol. The centripetal force will be provided by the resultant force i.e.
R
2 2
2
GM 1 1 Mv
R 2 4 R
v = 1 GM1 2 2
2 R
10. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of
80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. the
minimum capacity of the main fuse of the building will be :
(1) 12 A (2) 14 A (3) 8 A (4) 10 A
Key. 1
Sol. I 15 40 5 100 5 80 1 1000 P VI
220 220 220 220
= 11.36 A
Minimum capacity of the main fuse = 12 A
JEE-MAINS – 2014 Question Paper & Solutions
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11. A particle moves with simple harmonic motion in a straight line. In first
s, after starting from rest it travels a distance a, and in next s it travels
2a, in same direction, then
(1) amplitude of motion is 4a (2) time period of oscillations is 6
(3) amplitude of motion is 3a (4) time period of oscillations is 8
Key. (2)
Sol.
A
O
A
t = 0
Since it is given that the particle starts from rest
So, the particle starts from an extreme position.
Let the particle start from positive extreme at t = 0.
x = A sin t
2
= A cos t
After t = distance travelled by particle is a
x = A – a
A – a = A cos (1)
After t = 2, distance travelled by particle is a + 2a = 3a x = A – 3a
A – 3a = A cos (2) = A cos (2) (2)
Using cos 2 = 2cos2 - 1 and putting cos and cos 2 from equation
(1) and (2),
We get,
2 A 3a 2 A a 1
A A
2 2
2
A 3a 2 A a A
A A
A(A – 3a) = 2(A – a)2 – A2
A2 – 3aA = 2A2 + 2a2 – 4aA – A2 3aA = 2a2 – 4aA
2a2 = aA a = A/2 (3)
Putting a = A/2 in equation (1) we get A – (A/2) = A cos
(A/2) = A cos cos = (1/2)
= /3 = (/3)
T = 2/ = 2
/ 3
= 6
JEE-MAINS – 2014 Question Paper & Solutions
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12. The coercivity of a small magnet where the ferromagnet gets
demagnetized is 3 103 Am-1. The current required to be passed in a
solenoid of length 10 cm and number of turns 100, so that the magnet gets
demagnetized when inside the solenoid is
(1) 3 A (2) 6 A (3) 30 mA (4) 60 mA
Key. (1)
Sol. H = nI = (N/L)I
I = HL/N =
3 103 0.1 3A
100
13. The forward biased diode connection is
(1) 2 V 4 V (2) 2 V +2 V
(3) +2 V 2 V (4) 3 V 3 V
Key. (3)
Sol. For forward biasing of P-N junction diode, P side potential should be
higher than N side.
14. During the propagation of electromagnetic waves in a medium
(1) Electric energy density equal to the magnetic energy density
(2) Both electric and magnetic energy densities are zero
(3) Electric energy density is double of the magnetic energy density
(4) Electric energy density is half of the magnetic energy density
Key. (1)
Sol. In electromagnetic waves,
avg
2
E o
U 1 E
4
and
avg
2
B
o
U 1 B
4
Hence
Eavg Bavg U U
15. In the circuit shown here, the point ‘C’ is kept connected
to point ‘A’ till the current flowing through the circuit
becomes constant. Afterward, suddenly, point ‘C’ is
disconnected from point ‘A’ and connected to point ‘B’
at time t = 0. Ratio of the voltage across resistance and
the inductor at t = L/R will be equal to
(1) 1 (2) 1 e
e
(3) e
1 e
(4) 1
A C R
B
L
JEE-MAINS – 2014 Question Paper & Solutions
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Key. (1)
Sol. Using KVL, during decay of current in RL circuit at any time, we get
VR + VL = 0
VL/VR = 1
16. A mass ‘m’ is supported by a
massless string wound around a
uniform hollow cylinder of mass
m and radius R. If the string does
not slip on the cylinder, with
what acceleration will the mass
fall on release?
(1) 5g
6
(2) g
(3) 2g
3
(4) g
2
m
R
m
Key. (4)
Sol. F. B. D of block is
T
mg
a
mg – T = ma (1)
F. B. D of pulley is
mg T
T
from = I, we get
TR = mR2 (2)
Due to no slipping a = R (3)
Solving (1), (2) and (3), we get a= g/2
JEE-MAINS – 2014 Question Paper & Solutions
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17. One mole of diatomic ideal gas undergoes a cyclic
process ABC as shown in figure. The process BC is
adiabatic. The temperatures, at A, B and C are 400 K, 800
K and 600 K respectively. Choose the correct statement
(1) The change in internal energy in the process AB is
350 R
(2) The change in internal energy in the process BC is
500R
(3) The change in internal energy in whole cyclic process
is 250 R
(4) The change in internal energy in the process CA is
700R
A
B
C
P
V
800 K
400 K 600 K
Key. (2)
Sol. In process AB,
U = nCvT = 1. (5R/2) 400 = 1000R
In process BC,
U = nCvT = 1 (5R/2)(200) = 500R
In process CA,
U = nCvT = 1 (5R/2) (200) = 500R
Hence option (2) is correct.
18. From a tower of height H, a particle is thrown vertically upwards with a
speed u. The time taken by the particle, to hit the ground, is n times that
taken by it to reach the highest point of its path. The relation between H, u
and n is
(1) 2gH = nu2(n – 2) (2) gH = (n – 2)u2
(3) 2gH = n2u2 (4) gH = (n – 2)2u2
Key. (1)
Sol. Let height of tower be H, and time taken to reach
ground be t1
H = ut1 2
1
1 gt
2
(1)
And let t2 be the time taken to reach the highest
point
u = gt2 t2 = u/g
O
h
+y
JEE-MAINS – 2014 Question Paper & Solutions
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Since t1 = nt2
From (1), H = unt2 2
2
1 gn t
2
H =
2
2
2
un u 1 gn u
g 2 g
2gH 2nu2 n2u2 nu2 n 2
(1)
19. A thin convex lens made from crown glass 3
2
has focal length f. When
it is measured in two different liquids having refractive indices (4/3) and
(5/3), it has the focal lengths f1 and f2 respectively. The correct relation
between the focal lengths is
(1) f2 > f and f1 becomes negative (2) f1 and f2 both become negative
(3) f1 = f2 < f (4) f1 > f and f2 becomes negative
Key. (4)
Sol.
1 2 1 2
1 3 1 1 1 1 1 1
f 2 R R 2 R R
1 1 2 1 2
1 3 3 1 1 1 1 1 1
f 2 4 R R 8 R R
2 1 2 1 2
1 3 3 1 1 1 1 1 1
f 2 5 R R 10 R R
f : f1 : f2 :: 2 : 8 : 10
20. Three rods of Copper, Brass and Steel are welded together to form a Y –
shaped structure. Area of cross-section of each rod = 4 cm2. End of copper
rod is maintained at 100oC where as ends of brass and steel are kept at
0oC. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms
respectively. The rods are thermally insulated from surroundings except
at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26
and 0.12 CGS units respectively. Rate of heat flow through copper rod is
(1) 4.8 cal/s (2) 6.0 cal/s (3) 1.2 cal/s (4) 2.4 cal/s
Key. (1)
JEE-MAINS – 2014 Question Paper & Solutions
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Sol. 0.92A100 T 0.12AT 0 0.26AT 0
46 12 13
92100 T 12T 26T
46 12 13
2(100 – T) = T + 2T
200 = 5T
T = 40
Now in copper rod
dQ 0.92 4 60 24 4.8cal
dt 46 5
/s
100oC
0oC
0oC
Copper Brass
Steel
21. A pipe of length 85 cm is closed from one end. Find the number of possible
natural oscillations of air column in the pipe whose frequencies lie below
1250 Hz. The velocity of sound in air is 340 m/s.
(1) 6 (2) 4 (3) 12 (4) 8
Key: (1)
Sol. Given l =85 cm
nv 1250
4
n 12.5
Where n = 1, 3, 5, 7, 9, 11
22. There is a circular tube in a vertical plane. Two liquids which do not mix
and of densities d1 and d2 are filled in the tube. Each liquid subtends 900
angle at centre. Radius joining their interface makes an angle with
vertical. Ratio 1
2
d
d
is:
d 1
d2
(1) 1 tan
1 tan
(2) 1 sin
1 cos
(3) 1 sin
1 sin
(4) 1 cos
1 cos
Key: (1)
Sol. P0 + d1g(R – R sin ) = P0 + d2g (R sin + R cos ) + d1g (R – R cos)
JEE-MAINS – 2014 Question Paper & Solutions
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From above
1
2
d sin cos 1 tan
d cos sin 1 tan
23. A green light is incident from the water to the air-water interface at the
critical angle(). Select the correct statement.
(1) The spectrum of visible light whose frequency is more than that of
green light will come out to the air medium.
(2) The entire spectrum of visible light will come out of the water at
various angles to the normal
(3) The entire spectrum of visible light will come out of the water at an
angle of 900 to the normal
(4) The spectrum of visible light whose frequency is less than that of
green light will come out to the air medium.
Key: (4)
Sol. 1
…………….(i) By Cauchy relation
c
sin 1
…………(ii)
From both eq. c
24. Hydrogen (1H1), Deuterium (1H2), singly ionized Helium (2He4)+ and
doubly ionized lithium (3Li6)++ all have one electron around the nucleus.
Consider an electron transition from n = 2 to n = 1. If the wave lengths of
emitted radiation are 1, 2, 3 and 4 respectively then approximately
which one of the following is correct?
(1) 1 = 2 = 43 = 94 (2) 1 = 22 = 33 = 44
(3) 41 = 22 = 23 = 4 (4) 1 = 22 = 23 = 4
Key: (1)
Sol. hc 13.6 Z2 3
4
2
1
Z
JEE-MAINS – 2014 Question Paper & Solutions
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25. The radiation corresponding to 3 2 transition of hydrogen atom falls on
a metal surface to produce photoelectrons. These electrons are made to
enter a magnetic field of 3 × 104. If the radius of the largest circular path
followed by these electrons is 10.0 mm, the work function of the metal is
close to :
(1) 0.8 eV (2) 1.6 eV (3) 1.8 eV (4) 1.1 eV
Key: (4)
Sol. E 13.6 1 5 eV 1.88 eV
36
r p , p rqB 4.8 10 25
qB
p2 K.E 0.79 eV
2m
E K.E 1.09
1.1eV
26. A block of mass m is placed on a surface with a vertical cross section given
by
x3 y .
6
If the coefficient of friction is 0.5, the maximum height above the
ground at which the block can be placed without slipping is:
(1) 1m
3
(2) 1m
2
(3) 1m
6
(4) 2m
3
Key: (3)
Sol. mgsin mg cos
tan
At x = 1; y = 1/6
27. When a rubber-band is stretched by a distance x, it exerts a restoring
force of magnitude
F = ax + bx2 where a and b are constants. The work done in stretching the
unstretched rubber-band by L is:
(1)
aL2 bL3
2 3
(2)
1 aL2 bL3
2 2 3
(3) aL2 bL3 (4) 1 aL2 bL3
2
Key: (1)
JEE-MAINS – 2014 Question Paper & Solutions
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Sol.
L 2 3
0
W F dx aL bL
2 3
28. On heating water, bubbles being formed at the bottom of the vessel
detatch and rise. Take the bubbles to be spheres of radius R and making a
circular contact of radius r with the bottom of the vessel. If r << R, and the
surface tension of water is T, value of r just before bubbles detatch is:
(1) 2 wg
R
T
(2) 2 w 3 g
R
T
(3) 2 wg
R
3T
(4) 2 wg
R
6T
Key: (No Correct Option)
Sol.
3
w
Tsin 2 r 4 R g
3
sin r
R
2 w 2 g
r R
3T
q
q
T T
O
c
F = buoyant force B
29. Two beams, A and B, of plane polarized light with mutually perpendicular
planes of polarization are seen through a Polaroid. From the position
when the beam A has maximum intensity (and beam B has zero intensity),
a rotation of Polaroid through 300 makes the two beams appear equally
bright. If the initial intensities of the two beams are IA and IB respectively,
then A
B
I
I
equals:
(1) 1 (2) 1
3
(3) 3 (4) 3
2
Key: (2)
Sol. IA cos2 = IBcos2(90 ) (I = I0 cos2 ) by malus Eq
A 2
B
I tan 1
I 3
Where = 300
JEE-MAINS – 2014 Question Paper & Solutions
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30. Assume that an electric field E 30x2iˆ
exists in space. Then the potential
difference
(VA - VO), where VO is the potential at the origin and VA the potential at x =
2 m is:
(1) – 80 J (2) 80 J (3) 120 J (4) – 120J
Key: (1)
Sol. A
O
V 2
V 0
dV E dx
2
2
0
30x dx
A 0 V V 80J
MATHEMATICS
31. The image of the line x 1 y 3 z 4
3 1 5
in the plane 2x y z 3 0 is the line
(1) x 3 y 5 z 2
3 1 5
(2) x 3 y 5 z 2
3 1 5
(3) x 3 y 5 z 2
3 1 5
(4) x 3 y 5 z 2
3 1 5
Key. (1)
Sol. Any point on the given x 1 y 3 z 4
3 1 5
will be 3t 1, t 3,5t 4 its image in
the plane 2x y z 3 0 is given by
x 3t 1 y t 3 z 5t 4
2 1 1
6t 2 t 3 5t 4 3
2 2
6
x 3t 3, y t 5, z 2 5t which is the line
x 3 y 5 z 2
3 1 5
So (1) is correct.
32. If the coefficients of x3 and x4 in the expansion of 1 ax bx2 1 2x18 in
powers of x are both zero, then (a, b) is equal to
(1) 16, 251
3
(2) 14, 251
3
(3) 14, 272
3
(4) 16, 272
3
Key. (4)
JEE-MAINS – 2014 Question Paper & Solutions
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Sol. 1 ax bx2 1 2x18
2 18 18 2 18 3 18 4
1 2 3 4 1 ax bx 1 C 2x C . 2x C 2x C 2x ....
Coefficient of x3 = 0
18 18 18
3 2 1 C .8 a. C .4 b C 2 0
51a 3b 544 0 …. (i)
Coefficient of x4 = 0
18 18 18
4 3 2 C .16 a C 8 b C 4 0
32a 3b 240 0 …. (ii)
Solving (i) & (ii), we get
A = 16, b 272
3
So Answer (4)
33. If aR and the equation 3 x x 2 2 x x a2 0 (where [x] denotes the
greatest integer x) has no integral solution, then all possible values of a
lie in the interval
(1) 1,00,1 (2) 1,2 (3) 2,1 (4) ,22,
Key. (1)
Sol. x x x , since xI , so 0 x 1
Let x t
So
2 2
3t2 2t a2 0, t 1 1 3a 1 3a as t 0
3 3
Now 0 < t < 1 0 a2 1
a1,00,1
Option (1) is correct.
34. If 2 a b bc c a a b c
then is equal to
(1) 2 (2) 3 (3) 0 (4) 1
Key. (4)
Sol. a b bc ca
a b.b cc a
a b.b c a c b c c a
JEE-MAINS – 2014 Question Paper & Solutions
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a b.a b cc
2 a b c
1
Option (4) is correct.
35. The variance of first 50 even natural numbers is
(1) 833
4
(2) 833 (3) 437 (4) 437
4
Key. (2)
Sol. The first 50 no. are 2,4,6,8,.....100
Mean
X 2 4 6 ....100 51
50
Variance
2 2 2
i
1 x X
n
1 22 42 62 ....1002 512
50
3434 2601 833
36. A bird is sitting on the top of a vertical pole 20 m high and its elevation
from a point O on the ground is 45o. It flies off horizontally straight away
from the point O. After one second, the elevation of the bird from O is
reduced to 30o. Then the speed (in m/s) of the bird is
(1) 40 2 1 (2) 40 3 2
(3) 20 2 (4) 20 3 1
Key. (4)
Sol. A A
h
B d C
h
x
O
d
BOA 45o
COA 30o
tan tan 45o h 20
x x
JEE-MAINS – 2014 Question Paper & Solutions
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1 20 x 20
x
Now tan tan 30o h
x d
1 20
3 20 d
20 d 20 3
d 20 3 1
So velocity d 20 3 1 m 20 3 1 m / s
t 1sec
37. The integral 2
0
1 4sin x 4sin x dx
2 2
(1) 4 (2) 2 4 4 3
3
(3) 4 3 4 (4) 4 3 4
3
Key. (4)
Sol. 2
0
I 1 4sin x 4sin xdx
2 2
2
0
I 2sin x 1 dx
2
0
I 2sin x 1 dx
2
/3
0 /3
I 1 2sin x dx 2sin x 1 dx
2 2
/3
0 /3
x 4cos x 4cos x x
2 2
4 3 0 4 0 4 3
3 2 2 3
4 3 4
3
38. The statement ~ p~ q is
(1) equivalent to pq (2) equivalent to ~ pq
(3) a tautology (4) a fallacy
JEE-MAINS – 2014 Question Paper & Solutions
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Key. (1)
Sol.
p q ~ p ~ q p ~ q ~ p ~ q ~ p q p q
T T F F F T F T
T F F T T F T F
F T T F T F T F
F F T T F T F T
Clearly ~ p~ q equivalent to pq
39. If A is an 33 non-singular matrix such that AA' A'A and B A1A' then
BB'
(1) I B (2) I (3) B1 (4) B1
Key. (2)
Sol. Now BB' A1A' A1A' '
A1A'A.A1 ' ' AB B'A'
A 1A'AA' 1
A 1AA'.A' 1
I.I I
40. The integral
x 1
x 1 x 1 e dx
x
(1)
x 1
x 1 e x c (2)
x 1
x e x c (3)
x 1
x 1 e x c (4)
x 1
x e x c
Key. (2)
Sol.
x 1
x I 1 x 1 e .dx
x
x 1 x 1
x x I e dx x 1 e dx
x
x 1 x 1
x x
2
I e dx x. 1 1 e dx
x
x 1 x 1
x x
2
I e dx x. 1 1 e dx
x
x 1
x
2
1. 1 1 e dx dx
x
x 1 x 1 x 1
I e x dx x.e x e xdx 2
x 1 t 1 1 dx dt
x x
x 1
I x.e x c
JEE-MAINS – 2014 Question Paper & Solutions
20
41. If z is a complex number such that z 2 , then the minimum value of z 1
2
:
(1) is equal to 5
2
(2) lies in the interval (1, 2)
(3) is strictly greater than 5
2
(4) is strictly greater than 3
2
but less
than 5
2
Key. (2)
Sol.
2 O +2
A 1/2 B
z 1
2
= Distance between z and –1
2
OA = minimum value of z 1
2
= 3
2
.
42. If g is the inverse of a function f and f '(x) = 5
1
1 x
, then g ' (x) is equal to:
(1) 1 + x5 (2) 5x4 (3) 5
1
1{g(x)}
(4) 1{g(x)}5
KEY. (4)
Sol. f (g(x)) = x
f ' (g(x)) . g ' (x)= 1
f ' (g(x)) = 1
g '(x)
g ' (x)= 1 + g5(x).
JEE-MAINS – 2014 Question Paper & Solutions
21
43. If , , and f (n) = n n and
3 1 (1) 1 (2)
1 (1) 1 (2) 1 (3)
1 (2) 1 (3) 1 (4)
f f
f f f
f f f
= 2 (1 2 ( 2 , then K is equal to:
(1) (2) 1
(3) 1 (4) –1
Key. (3)
Sol.
2 2
2 2 3 3
2 2 3 3 4 4
3 1 α + β 1 α + β
1 α + β 1 α + β 1 α + β
1 α + β 1 α + β 1 α + β
=
2
2 2
1 1 1
1
1
= 2 2 2
Hence, K = 1.
44. Let fk (x) = 1
k
(sink x + cosk x) where x R and k 1. Then f4(x) – f6(x)
equals:
(1) 1
6
(2) 1
3
(3) 1
4
(4) 1
12
Key. (4)
Sol. fk(x) = 1 sink x cosk x
k
f4(x) – f6(x) = 1 sin4 x cos4 x 1sin6 x cos6 x
4 6
= 1 1 2sin2 x cos2 x 11 3sin2 x cos2 x
4 6
= 1
12
.
45. Let and be the roots of equation px2 + qx + r = 0, p 0. If p, q, r are in A.
P.
and 1
+ 1
=4, then the value of α – β is:
(1) 61
9
(2) 2 17
9
(3) 34
9
(4) 2 13
9
Key. (4)
JEE-MAINS – 2014 Question Paper & Solutions
22
Sol. Let , are the rots of px2 + qx + r = 0
1 1 4
q 4
r
q = 4r
p, q, r are in A. P.
p = 9r
Hence the given equation becomes
9x2 + 4x – 1 =0
| 16 4 52 2 13
81 9 81 9
|
46. Let A and B be two events such that P(A B) 1
6
, P(A B) 1
4
and P(A) 1
4
,
where A stands for the complement of the event A. Then the events A and
B are:
(1) mutually exclusive and independent.
(2) equally likely but not independent.
(3) Independent but not equally likely.
(4) Independent and equally likely.
Key. (3)
Sol. P(AB) P(A) P(B) P(AB)
5 3 x 1
6 4 4
where P(B) = x
x 1
3
P(AB) P(A).P(B)
Hence A and B are independent but not equally likely.
47. If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0)
= 0 and f(1) = 6, then for some c]0,1[ :
(1) 2f’(c) = g’(c) (2) 2f’(c) = 3g’(c) (3) f’(c) = g’(c) (4) f’(c) = 2g’(c)
Key. (4)
Sol. Let h(x) = f(x) – 2g(x)
h(x) is continuous and differentiable in [0, 1].
h(0) = f(0) – 2g(0) = 2 – 0 = 2
h(1) = f(1) – 2g(1) = 6 – 4 = 2
JEE-MAINS – 2014 Question Paper & Solutions
23
h(0) = h(1)
h(x) verify all conditions of Rolle’s theorem.
Hence, at least one c(0, 1) such that
h '(c) 0
f '(c) 2g'(c)
48. Let the population of rabbits surviving at a time t be governed by the
differential equation dP(t) 1 p(t)
dt 2
– 200.
If p(0) = 100, then p(t) equals:
(1) 400 – 300 et/2 (2) 300 – 200 e–t/2 (3) 600 – 500 et/2 (4) 400 –
300 et/2
Key. (1)
Sol. I.F et/2
So solution is Ptet/2 200.et/2dt 400et /2 k
P0 100 k = –300
P(t) = 400 – 300 et/2
49. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle
centre at (0, y), passing through origin and touching the circle C
externally, then the radius of T is equal to:
(1) 3
2
(2) 3
2
(3) 1
2
(4) 1
4
Key. (4)
Sol.
B
A
O
A = (0, y) B = (1, 1)
Since AB 1 y
y = 1
4
Hence, 2
2 3
JEE-MAINS – 2014 Question Paper & Solutions
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50. The area of the region described by A = {(x, y) : x2 + y2 1 and y2 1 – x}
is:
(1) 4
2 3
(2) – 4
2 3
(3) 2 –
2 3
(4) 2
2 3
Key. (1)
Sol.
Required area = 1 2
0
2 ( 1 x 1 x )dx
1
2
1 3/2
0
2 x 1 x 1 sin x 2 (1 x)
2 2 3
2 2 4
4 3 2 3
51. Let a, b, c and d be non-zero numbers. If the point of intersection of the
lines 4ax 2ay c 0 and 5bx 2by d 0 lies in the fourth quadrant and is
equidistant from the two axes then:
(1) 2bc 3ad 0 (2) 2bc 3ad 0 (3) 3bc 2ad 0 (4) 3bc 2ad 0
Key: (3)
Sol.
Point (k, -k) satisfies both lines
4ak 2ak c 0 . . . . . (1)
5bk 2bk d 0 . . . . . (2)
c = -2ak and d = -3bk
2 3
c d
a b
2ad - 3bc = 0
3bc – 2ad = 0
52. Let PS be the median of the triangle with vertices P(2, 2), Q(6, - 1) and R(7,
3). The equation of the line passing through (1, -1) and parallel to PS is:
(1) 4x 7 y 11 0 (2) 2x 9y 7 0 (3) 4x 7 y 3 0 (4) 2x 9 y 11 0
Key: (2)
JEE-MAINS – 2014 Question Paper & Solutions
25
Sol.
1 2
2 13 9
2
ps m
1 2 1
9
y x
9y 2x 7 0
53.
2
0 2
lim sin( cos )
x
x
x
is equal to :
(1)
2
(2) 1 (3) (4)
Key : (4)
Sol.
2
0 2
sin cos
lim
x
x
x
=
2 2
0 2 2
sin sin sin lim
x sin
x x
x x
=
54. If X 4n 3n 1 : n N and Y 9 n 1 : n N, where N is the set of natural
numbers, then XY is equal to :
(1) N (2) Y – X (3) X (4) Y
Key : (4)
Sol.
4n 3n 1
= 1 3 3 1 n n
= 2 3
2 3 nC 3 nC 3 ........
= 2 3 9 nC 3 nC .......
Set X contains numbers which are multiples of 9 but not all multiples of 9
XY = Y
JEE-MAINS – 2014 Question Paper & Solutions
26
55. The locus of the foot of perpendicular drawn from the centre of the ellipse
x2 3y2 6 on any tangent to it is :
(1) x2 y2 2 6x2 2 y2 (2) x2 y2 2 6x2 2 y2
(3) x2 y2 2 6x2 2y2 (4) x2 y2 2 6x2 2 y2
Key : (3)
Sol.
K mh 6m2 2 . . . . . (1)
K m 1
h
. . . . . . (2)
Eliminating m from (1) and (2)
Locus is
x2 y2 2 6x2 2y2
56. Three positive numbers form an increasing G.P. If the middle term in this
G.P. is doubled, the new numbers are in A.P. Then the common ratio of the
G.P is :
(1) 2 3 (2) 3 2 (3) 2 3 (4) 2 3
Key : (4)
Sol.
4a = a ar
r
(r > 1)
r 2 3
but as r > 1
r 2 3
57. If 9 1 8 2 7 9 9 10 2 11 10 3 11 10 .... 10 11 k 10 , then k is equal to :
(1) 121
10
(2) 441
100
(3) 100 (4) 110
Key : (3)
Sol. S = 109 + 2(11) (10)8 + . . . . . . + 10(11)9
S 11
10
= (11.108) + . . . . . + . . . + (11)10
JEE-MAINS – 2014 Question Paper & Solutions
27
10
9
10
10 11 1
S 10 (11)
10 11 1
10
S = 1011
= (100) 109
K = 100
58. The angle between the lines whose direction cosines satisfy the equations
l m n 0 and l2 m2 n2 is :
(1)
3
(2)
4
(3)
6
(4)
2
Key : (1)
Sol. Solving two equation
l = -(m + n) then (m + n)2 = m2 + n2
mn = 0
If m = 0 , If n = 0
l = -n l = -m
hence dr are 1, 0, -1 and 1, -1, 0
cos 1 1
2 2 2
=
3
59. The slope of the line touching both the parabolas y2 4x and x2 32y is :
(1) 1
2
(2) 3
2
(3) 1
8
(4) 2
3
Key : (1)
Sol. Eq. of tangent at t & t respectively are yt x t2 and xt ' y 8t '2
Comparing
2
2
1 2
' 1 8 '
t t t
t t
Hence slope = 1
2
JEE-MAINS – 2014 Question Paper & Solutions
28
60. If x 1 and x 2 are extreme points of f (x) log | x | x2 x then :
(1) 6, 1
2
(2) 6, 1
2
(3) 2, 1
2
(4) 2, 1
2
Key: (3)
Sol. Differentiating
f 'x 2 x 1 0 2 x2 x 0
x
Put x 2 and -1 gives
2 1 0 and 8 2 0
Solving 2, 1
2
CHEMISTRY
61. Which one of the following properties is not shown by NO?
(1) It combines with oxygen to form nitrogen dioxide
(2) It’s bond order is 2.5
(3) It is diamagnetic in gaseous state
(4) it is a neutral oxide
Key. (3)
Sol. Nitric Oxide, NO is paramagnetic.
62. If Z is a compressibility factor, van der Waals equation at low pressure can
be written as:
(1) Z 1 Pb
RT
(2) Z 1 Pb
RT
(3) Z 1 RT
PB
(4) Z 1 a
VRT
Key (4)
Sol. As pressure is low so volume must be large.
i.e., V b V
So Van der Waal’s equation reduces into
PV a RT
V
Z 1 a
RTV
JEE-MAINS – 2014 Question Paper & Solutions
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63. The metal that cannot be obtained by electrolysis of an aqueous solution
of its salts is :
(1) Cu (2) Cr
(3) Ag (4) Ca
Key. (4)
Sol. In case of aqueous solution of calcium salt hydrogen gas will
preferentially evolve due to selective electrolysis. H has higher reduction
potential than that of Ca2
64. Resistance of 0.2 M solution of an electrolyte is 50 . The specific
conductance of the solution is 1.4 S m–1. The resistance of 0.5 M solution of
the same electrolyte is 280 . The molar conductivity of 0.5 M solution of
the electrolyte in Sm2 mol–1 is :
(1) 5103 (2) 5102
(3) 5104 (4) 5103
Key. 3
Sol. 1 . 1
R / A
1 1.4. 1
50 / A
1 . 1
280 / A
Solving we get
0.25Sm1
C
0.25 5 10 4 Sm2 mol 1
0.5 1000
65. CsCl crystallizes in body centred cubic lattice. If ‘a’ is its edge length then
which of the following expression is correct?
(1) Cs Cl
r r 3 a
2 (2) Cs Cl r r 3a
(3) Cs Cl r r 3a (4) Cs Cl
r r 3a
2
JEE-MAINS – 2014 Question Paper & Solutions
30
Key. (1)
Sol. In CsCl lattice, centrally placed Cs remains in contact with all eight Cl
ions present at the vertices of the cube.
Body diagnonal = 3a
3a 2Cs Cl .
66. Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2
(aq), 0.25 M KBr(aq) and 0.125 M Na3PO4(aq) at 25oC. Which statement is
true about these solutions, assuming all salts to be strong electrolytes?
(1) 0.125 M Na3PO4(aq) has the highest osmotic pressure.
(2) 0.500 M C2H5OH(aq) has the highest osmotic pressure.
(3) The all have the same osmotic pressure.
(4) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure.
Key. (3)
Sol. Considering complete dissociation of electrolytes, the effective
concentration of solute particles is same (0.5 M) in all the solution.
67. In which of the following reaction H2O2 acts as a reducing agent?
(a) 2 2 2 H O 2H 2e 2H O (b) 2 2 2 H O 2e O 2H
(c) 2 2 H O 2e 2OH (d) 2 2 2 2 H O 2OH 2e O 2H O
(1) (a), (c) (2) (b), (d)
(3) (a), (b) (4) (c), (d)
Key. (2)
Sol. 1 0
2 2 2 H O O 2e
68. In SN2 reactions, the correct order of reactivity for the following
compounds:
3 3 2 3 2 3 3 CH Cl,CH CH Cl, CH CHCl and CH CCl
(1) 3 2 3 3 2 3 3 CH CH Cl CH Cl CH CHCl CH CCl
(2) 3 2 3 2 3 3 3 CH CHCl CH CH Cl CH Cl CH CCl
(3) 3 3 2 3 2 3 3 CH Cl CH CHCl CH CH Cl CH CCl
(4) 3 3 2 3 2 3 3 CH Cl CH CH Cl CH CHCl CH CCl
Key. (4)
Sol. In N S 2 reactivity order is o o o
3 CH X 1 R X 2 R X 3 R X
JEE-MAINS – 2014 Question Paper & Solutions
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69. The octahedral complex of a metal ion M3+ with four monodentate ligands
L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and
blue, respectively. The increasing order of ligand strength of the four
ligands is :
(1) L3 < L2 < L4 < L1 (2) L1 < L2 < L4 < L3
(3) L4 < L3 < L2 < L1 (4) L1 < L3 < L2 < L4
Key. (4)
Sol. As the order of wavelength
red yellow green blue
& Energy 1
70. For the estimation of nitrogen, 1.4 g of an organic compound was digested
by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of
M
10
sulphuric acid. The unreacted acid required 20 mL of M
10
sodium
hydroxide for complete neutralization. The percentage of nitrogen in the
compound is:
(1) 3 % (2) 5 %
(3) 6 % (4) 10 %
Key. (4)
Sol. 2 4 3 Eq.H SO Eq.NH Eq.NaOH
3
60 1 2 Eq.NH 20 1
1000 10 1000 10
3
Eq.NH 100 0.01
10000
Mole 3 NH 0.01
Weight percent 2
of N 0.01 14 100 10%
1.4
71. The equivalent conductance of NaCl at concentration C and at infinite
dilution are C and , respectively. The correct relationship between C
and is given as
(1) C (B) C (2) C (B) C
(3) C (B)C (4) C (B)C
Key. (1)
Sol. It is Debye Huckel onsager equation i.e.
C B C
JEE-MAINS – 2014 Question Paper & Solutions
32
72. For the reaction 2(g) 2(g) 3(g)
SO 1 O SO
2
, if KP = KC(RT)x where the symbols
have usual meaning, then the value of x is (assuming ideality)
(1) 1
2
(2) 1 (3) –1 (4) 1
2
Key. (4)
Sol. ng
P C K K (RT)
x or g
n 1
2
73. In the reaction,
LiAlH4 PCl5 Alc. KOH
3 CH COOHABC ,
the product C is
(1) Ethylene (2) Acetyl chloride (3) Acetaldehyde (4) Acetylene
Key. (1)
Sol.
LAH PCl5
3 3 2 3 2
(A) (B)
CH COOHCH CH OHCH CH Cl
2 2
(C)
CH CH
alc. KOH
74. Sodium phenoxide when heated with CO2 under pressure at 125°C yields a
product which on acetylation produces C.
+ 125
2 5 Atm
2
CO B H C
Ac O
The product C would be
(1) (2)
(3) (4)
Key. (3)
ONa
OH
COOCH3
O COCH3
COOH
O COCH3
COOH
OH
COCH3
COCH3
JEE-MAINS – 2014 Question Paper & Solutions
33
Sol.
75. On heating an aliphatic primary amine with chloroform and ethanolic
potassium hydroxide, the organic compound formed is
(1) an alkyl cyanide (2)an alkyl isocyanide
(3) an alkanol (4) an alkanediol
Key. (2)
Sol. Carbylamine reaction
3
2 5
(CHCl KOH)
2 2 C H OH RCH NH R N C
76. The correct statement for the molecule, CsI3, is
(1) it contains Cs3+ and I– ions
(2) it contains Cs+, I– and lattice I2 molecule
(3) it is a covalent molecule (4)it contains Cs+ and 3 I ions
Key. (4)
Sol. 3 3 CsI Cs I
77. The equation which is balanced and represents the correct product(s) is
(1) 2 4 excess NaOH 2
2 6 2 [Mg(H O) ] (EDTA) [Mg(EDTA) 6H O
(2) 4 2 4 2 4 CuSO 4KCNK [Cu(CN) ]K SO
(3) 2 2 Li O2KCl2LiCl K O
(4) 2
3 5 4 [CoCl(NH ) ] 5H Co 5NH Cl
Key. (3)
Sol. No proof available.
78. For which of the following molecule significant 0?
(A)
(B)
(C)
(D)
(1) only (C) (2) (C) and (D) (3) only (A) (4) (A) and (B)
ONa
+ CO2
125°
5 atm
COONa
OH
H+
5 atm
COOH
OH
COOH
OCOCH3
(AC) O 2
Acetylation
Aspirin
Cl
Cl
CN
CN
OH
OH
SH
SH
JEE-MAINS – 2014 Question Paper & Solutions
34
Key. (2)
Sol.
79. For the non-stoichiometre reaction 2ABCD, the following kinetic
data were obtained in three separate experiments, all at 298 K.
Initial
Concentration
(A)
Initial
Concentration
(B)
Initial rate of
formation of C
(mol L–S–)
0.1 M
0.1 M
0.2 M
0.1 M
0.2 M
0.1 M
1.2 × 10–3
1.2 × 10–3
2.4 × 10–3
The rate law for formation of C is
(A) dc k[A][B]2
dt
(B) dc k[A]
dt
(C) dc k[A][B]
dt
(D) dc k[A]2 [B]
dt
Key. (2)
Sol. Rate reaction,
d(c) A n B m
dt
=
m 3
3
0.1 1.2 10
0.2 1.2 10
m = 0
n 3
3
0.1 1.2 10
0.2 2.4 10
n 1 1
2 2
=
n = 1
rate equation dc k[A]
dt
80. Which series of reactions correctly represents chemical relations related
to iron and its compound?
(A) Cl2 , heat heat , air Zn
3 2 FeFeCl FeCl Fe
(B) O2 , heat CO, 600 C CO, 700 C
3 4 FeFe O FeOFe
(C) dil. H2SO4 H2SO4 , O2 heat
4 2 4 3 FeFeSO Fe (SO ) Fe
(D) O2 , heat dil. H2SO4 heat
4 FeFeOFeSO Fe
N
JEE-MAINS – 2014 Question Paper & Solutions
35
Key. (2)
Sol. 3 4 2
2 3 2
Fe O 4CO 3Fe 4CO
500 800 K
Fe O CO 2FeO CO
2 FeO COFe CO 900 1500K
81. Considering the basic strength of amines in aqueous solution, which one
has the smallest pKb values
(A) (CH3)3N (2) C6H5H2 (3) (CH3)2NH (4) CH3NH2
Key. (3)
Sol. The base which has maximum Kb will have smallest pKb
82. Which one of the following bases is not present in DNA?
(1) Cytosine (2) Thymine (3) Quinoline (4) Adenine
Key . (3)
Sol. In DNA, Nitrogenous base
(i) Cytosine (ii) Thymine
(iii) Adenine (iv) Guanine
83. The correct set of four quantum numbers for the valence electrons of
rubidium atom (Z = 37) is
(1) 5, 1, 1, + 1
2
(2) 5, 0, 1, + 1
2
(3) 5, 0, 0, + 1
2
(4) 5, 1, 0, + 1
2
Key . (3)
Sol.
Configuration : Ar 4s2 3d10 4p6 5s1
Its quantum numbers n, l, m and s will be 5, 0, 0, + 1
2
respectively.
84. The major organic compound formed by the reaction of 1, 1, 1 –
trichloroethane with silver powder is
(1) 2-butyne (2) 2- butane (3) acetylene (4) ethane
Key . (1)
Sol.
H3C
Cl
Cl
Cl + 6Ag + Cl
Cl
Cl
CH3 - 6AgCl
H3C C C CH3
2-butyne
JEE-MAINS – 2014 Question Paper & Solutions
36
85. Given below are the half-cell reactions :
Mn2+ + 2e- Mn; E0 = -1.18 V
2(Mn3+ + e- Mn2+); E0 = +1.51 V
The E0 for 3Mn2+ Mn + 2Mn3+ will be
(1) -0.33 V; the reaction will not occur
(2) -0.33V; the reaction will occur
(3) -2.69 V; the reaction will not occur
(4) -2.69V; the reaction will occur
Key . (3)
Sol. Mn2+ + 2e- Mn; G10 = 2 F (1.18) . . . . .(i)
2(Mn3+ + e- Mn2+); G20 = 2 F (+1.51) . . . . .(ii)
for
3Mn2+ Mn + 2Mn3+
G30 = G10 G20
2FE30 = 2F (1.18) + 2F(1.51)
E30 = 2.69 V
E30 = 2.69 V
Reaction will not occur
86. The ratio of masses of oxygen and nitrogen in a particular gaseous
mixture is 1 : 4. The ratio of number of their molecule is :
(1) 1 : 8 (2) 3 : 16 (3) 1 : 4 (4) 7 : 32
Key . (4)
Sol. 2 2 2
2 2 2
O O O
N N N
n w /M x / 32 28 7 :32
n w /M 4x / 28 4 32
87. Which one is classified as a condensation polymer ?
(1) Teflon (2) Acrylonitrile (3) Dacron (4) Neoprene
Key . (3)
Sol. Dacron is condensed polymer of tere-phthalic acid and glycol.
88. Among the following oxoacids, the correct decreasing order of acid
strength is :
(1) HClO4 > HClO3 > HClO2 > HOCl
(2) HClO2 > HClO4 > HClO3 > HOCl
(3) HOCl > HClO2 > HClO3 > HClO4
(4) HClO4 > HOCl > HClO2 > HClO3
JEE-MAINS – 2014 Question Paper & Solutions
37
Key. (1)
Sol.
7 5 3 1
4 3 2 HClO HClO HClO HOCl
Higher is the oxidation state, more will be the acidic strength of these
oxoacids.
Order of resonance stability of conjugate bases 4 3 2 ClO ClO ClO
89. For complete combustion of ethanol,
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
the amount of heat produced as measured in bomb calorimeter, is
1364.47 kJ mol-1 at 250C. Assuming ideality the Enthalpy of combustion,
CH, for the reaction will be:
(R = 8.314 kJ mol-1)
(1) -1460.50kJ mol-1 (2) -1350.50 kJ mol-1
(3) -1366.95 kJ mol-1 (4) -1361.95 kJ mol-1
Key . (3)
Sol. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
H = E + nRT
qp = qv + nRT
= 1364.47 1 8.31 298
1000
= -1364.47 – 2.47638
= - 1366.95 kJ mol-1
90. The most suitable reagent for the conversion of R – CH2 – OH R – CHO is :
(1) CrO3 (2) PCC (Pyridinium Chlorochromate)
(3) KMnO4 (4) K2Cr2O7
Key. (2)
Sol. R – CH2OH PCC
|| O
R C H , other oxidizing agents may convest RCHO to
RCOOH.
JEE-MAINS – 2014 Question Paper & Solutions
38
IITJEE MAINS-2014
Answer Key (CODE-H)
PHYSICS MATHEMATICS CHEMISTRY
Q.N. Code -H Q.N. Code-H Q.N. Code-H
1. 3 31. 1 61. 3
2. 4 32. 4 62. 4
3. 3 33. 1 63. 4
4. 3 34. 4 64. 3
5. 3 35. 2 65. 1
6. 2 36. 4 66. 3
7. 3 37. 4 67. 2
8. 4 38. 1 68. 4
9. 2 39. 2 69. 4
10. 1 40. 2 70. 4
11. 2 41. 2 71. 1
12. 1 42. 4 72. 4
13. 3 43. 3 73. 1
14. 1 44. 4 74. 3
15. 1 45. 4 75. 2
16. 4 46. 3 76. 4
17. 2 47. 4 77. 3(Not confirmed)
18. 1 48. 1 78. 2
19. 4 49. 4 79. 2
20. 1 50. 1 80. 2
21. 1 51. 3 81. 3
22. 1 52. 2 82. 3
23. 4 53. 4 83. 3
24. 1 54. 4 84. 1
25. 4 55. 3 85. 3
26. 3 56. 4 86. 4
27. 1 57. 3 87. 3
28. No answer 58. 1 88. 1
29. 2 59. 1 89. 3
30. 1 60. 3 90. 2
1
This booklet contains 28 printed pages. Test Booklet Code
PAPER -1 : PHYSIC, MATHEMATICS & CHEMISTY
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H
JEE-MAINS – 2014 Question Paper & Solutions
2
PHYSICS
1. The pressure that has to be applied to the ends of a steel wire of length 10
cm to keep its length constant when its temperature is raised by 100oC is :
(For steel Young’s modulus is 2 × 1011 N m–2 and coefficient of thermal
expansion is 1.1 × 10–5 K–1)
(1) 2.2107Pa (2) 2.2106Pa (3) 2.2108Pa (4) 2.2109Pa
Key. 3
Sol. P T
= 21011 1.1105 100
= 2.2108Pa
2. A conductor lies along the z-axis at –1.5 z < 1.5 m and carries a fixed
current of 10.0 A in z aˆ direction (see figure). For a field 4 0.2x
y B 3.010 e aˆ T,
find the power required to move the conductor at constant speed to x =
2.0 m, y = 0 m in 5 × 10–3s. Assume parallel motion along the x-axis.
(1) 14.58 W (2) 29.7 W (3) 1.57 W (4) 2.97 W
Key. 4
Sol. P 1 iBl dx
t
2 4 0.2x
3 0
1 10 3.0 10 e 3 dx
5 10
= 2.97 W
3. A bob of mass m attached to an inextensible string of length l is suspened
from a vertical support. The bob rotates in a horizontal circle with an
angular speed rad/s about the vertical. About the point of suspension:
(1) angular momentum changes in direction but not in magnitude.
(2) angular momentum changes both in direction and magnitude.
(3) angular momentum is conserved.
(4) angular momentum changes in magnitude but not in direction.
JEE-MAINS – 2014 Question Paper & Solutions
3
Key. 3
Sol. The torque of all the forces about the point of suspension is zero therefore
angular momentum is conserved.
4. The current voltage relation of diode is given by I = (e1000V/T – 1) mA,
where the applied voltage V is in volts and the temperature T is in degree
Kelvin. If a student makes an error measuring 0.01 V while measuring
the current of 5mA at 300 K, what will be the error in the value of current
in mA?
(1) 0.5 mA (2) 0.05 A (3) 0.2 mA (4) 0.02 mA
Key. 3
Sol.
1000V
I e T 1
1000V
I 1 e T
1000V
T I 1000 e . V
T
1000 5 10.01
300
1000V
e T 5 1
= 0.2 mA
5. An open glass tube is immersed in mercury in such a way that a length of 8
cm extends above the mercury level. The open end of the tube is then
closed and sealed and the tube is raised vertically up by additional 46 cm.
what will be length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
(1) 38 cm (2) 6 cm (3) 16 cm (4) 22 cm
Key. 3
Sol.
8 cm 54 cm
x
Applying Boyle’s Law to the air column in the glass tube,
P1V1 = P2V2
76 × 8 × A = {76 – (54 – x)} . x.A
x = 16 cm
JEE-MAINS – 2014 Question Paper & Solutions
4
6. Match List – I (Electromagnetic wave type) with List – II (Its
association/application) and select the correct option from the choices
given below the lists:
List – I List- II
(a) Infrared waves (i) To treat muscular strain
(b) Radio waves (ii) For broadcasting
(c) X-rays (iii) To detect fracture of bones
(d) Ultraviolet rays (iv) Absorbed by the ozone
layer of the
atmosphere
(a) (b) (c) (d)
(1) (iii) (ii) (i) (iv)
(2) (i) (ii) (iii) (iv)
(3) (iv) (iii) (ii) (i)
(4) (i) (ii) (iv) (iii)
Key. 2
Sol. Conceptual
7. A parallel plate capacitor is made of two circular plates separated by a
distance of 5 mm and with a dielectric of dielectric constant 2.2 between
them. When the electric field in the dielectric is 3 × 104 V/m, the charge
density of the positive plate will be close to :
(1) 3 × 104 C/m2 (2) 6 × 104 C/m2 (3) 6 × 10–7 C/m2 (4) 3 × 10–7 C/m2
Key. 3
Sol. Charge density
o K A Q .Ed d
A A
= Ko E
6107 C/m2
JEE-MAINS – 2014 Question Paper & Solutions
5
8. A student measured the length of a rod and worte it as 3.50 cm. Which
instrument did he use to measure it ?
(1) A screw gauge having 100 divisions in the circular scale and pitch as 1
mm.
(2) A screw gauge having 50 divisions in the circular scale and pitch as 1
mm.
(3) A meter scale.
(4) A vernier caliper where the 10 divisions in vernier scale matches with
9 division in main scale and main scale has 10 divisions in 1 cm.
Key. 4
Sol. The least count of the observation is 0.01 cm which corresponds to option
(4)
9. Four particles, each of mass M and equidistant from each other, move
along a circle of radius R under the action of their mutual gravitational
attraction. The speed of each particle is
(1) GM1 2 2
R
(2) 1 GM1 2 2
2 R
(3) GM
R
(4) 2 2 GM
R
Key. (2)
Sol. The centripetal force will be provided by the resultant force i.e.
R
2 2
2
GM 1 1 Mv
R 2 4 R
v = 1 GM1 2 2
2 R
10. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of
80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. the
minimum capacity of the main fuse of the building will be :
(1) 12 A (2) 14 A (3) 8 A (4) 10 A
Key. 1
Sol. I 15 40 5 100 5 80 1 1000 P VI
220 220 220 220
= 11.36 A
Minimum capacity of the main fuse = 12 A
JEE-MAINS – 2014 Question Paper & Solutions
6
11. A particle moves with simple harmonic motion in a straight line. In first
s, after starting from rest it travels a distance a, and in next s it travels
2a, in same direction, then
(1) amplitude of motion is 4a (2) time period of oscillations is 6
(3) amplitude of motion is 3a (4) time period of oscillations is 8
Key. (2)
Sol.
A
O
A
t = 0
Since it is given that the particle starts from rest
So, the particle starts from an extreme position.
Let the particle start from positive extreme at t = 0.
x = A sin t
2
= A cos t
After t = distance travelled by particle is a
x = A – a
A – a = A cos (1)
After t = 2, distance travelled by particle is a + 2a = 3a x = A – 3a
A – 3a = A cos (2) = A cos (2) (2)
Using cos 2 = 2cos2 - 1 and putting cos and cos 2 from equation
(1) and (2),
We get,
2 A 3a 2 A a 1
A A
2 2
2
A 3a 2 A a A
A A
A(A – 3a) = 2(A – a)2 – A2
A2 – 3aA = 2A2 + 2a2 – 4aA – A2 3aA = 2a2 – 4aA
2a2 = aA a = A/2 (3)
Putting a = A/2 in equation (1) we get A – (A/2) = A cos
(A/2) = A cos cos = (1/2)
= /3 = (/3)
T = 2/ = 2
/ 3
= 6
JEE-MAINS – 2014 Question Paper & Solutions
7
12. The coercivity of a small magnet where the ferromagnet gets
demagnetized is 3 103 Am-1. The current required to be passed in a
solenoid of length 10 cm and number of turns 100, so that the magnet gets
demagnetized when inside the solenoid is
(1) 3 A (2) 6 A (3) 30 mA (4) 60 mA
Key. (1)
Sol. H = nI = (N/L)I
I = HL/N =
3 103 0.1 3A
100
13. The forward biased diode connection is
(1) 2 V 4 V (2) 2 V +2 V
(3) +2 V 2 V (4) 3 V 3 V
Key. (3)
Sol. For forward biasing of P-N junction diode, P side potential should be
higher than N side.
14. During the propagation of electromagnetic waves in a medium
(1) Electric energy density equal to the magnetic energy density
(2) Both electric and magnetic energy densities are zero
(3) Electric energy density is double of the magnetic energy density
(4) Electric energy density is half of the magnetic energy density
Key. (1)
Sol. In electromagnetic waves,
avg
2
E o
U 1 E
4
and
avg
2
B
o
U 1 B
4
Hence
Eavg Bavg U U
15. In the circuit shown here, the point ‘C’ is kept connected
to point ‘A’ till the current flowing through the circuit
becomes constant. Afterward, suddenly, point ‘C’ is
disconnected from point ‘A’ and connected to point ‘B’
at time t = 0. Ratio of the voltage across resistance and
the inductor at t = L/R will be equal to
(1) 1 (2) 1 e
e
(3) e
1 e
(4) 1
A C R
B
L
JEE-MAINS – 2014 Question Paper & Solutions
8
Key. (1)
Sol. Using KVL, during decay of current in RL circuit at any time, we get
VR + VL = 0
VL/VR = 1
16. A mass ‘m’ is supported by a
massless string wound around a
uniform hollow cylinder of mass
m and radius R. If the string does
not slip on the cylinder, with
what acceleration will the mass
fall on release?
(1) 5g
6
(2) g
(3) 2g
3
(4) g
2
m
R
m
Key. (4)
Sol. F. B. D of block is
T
mg
a
mg – T = ma (1)
F. B. D of pulley is
mg T
T
from = I, we get
TR = mR2 (2)
Due to no slipping a = R (3)
Solving (1), (2) and (3), we get a= g/2
JEE-MAINS – 2014 Question Paper & Solutions
9
17. One mole of diatomic ideal gas undergoes a cyclic
process ABC as shown in figure. The process BC is
adiabatic. The temperatures, at A, B and C are 400 K, 800
K and 600 K respectively. Choose the correct statement
(1) The change in internal energy in the process AB is
350 R
(2) The change in internal energy in the process BC is
500R
(3) The change in internal energy in whole cyclic process
is 250 R
(4) The change in internal energy in the process CA is
700R
A
B
C
P
V
800 K
400 K 600 K
Key. (2)
Sol. In process AB,
U = nCvT = 1. (5R/2) 400 = 1000R
In process BC,
U = nCvT = 1 (5R/2)(200) = 500R
In process CA,
U = nCvT = 1 (5R/2) (200) = 500R
Hence option (2) is correct.
18. From a tower of height H, a particle is thrown vertically upwards with a
speed u. The time taken by the particle, to hit the ground, is n times that
taken by it to reach the highest point of its path. The relation between H, u
and n is
(1) 2gH = nu2(n – 2) (2) gH = (n – 2)u2
(3) 2gH = n2u2 (4) gH = (n – 2)2u2
Key. (1)
Sol. Let height of tower be H, and time taken to reach
ground be t1
H = ut1 2
1
1 gt
2
(1)
And let t2 be the time taken to reach the highest
point
u = gt2 t2 = u/g
O
h
+y
JEE-MAINS – 2014 Question Paper & Solutions
10
Since t1 = nt2
From (1), H = unt2 2
2
1 gn t
2
H =
2
2
2
un u 1 gn u
g 2 g
2gH 2nu2 n2u2 nu2 n 2
(1)
19. A thin convex lens made from crown glass 3
2
has focal length f. When
it is measured in two different liquids having refractive indices (4/3) and
(5/3), it has the focal lengths f1 and f2 respectively. The correct relation
between the focal lengths is
(1) f2 > f and f1 becomes negative (2) f1 and f2 both become negative
(3) f1 = f2 < f (4) f1 > f and f2 becomes negative
Key. (4)
Sol.
1 2 1 2
1 3 1 1 1 1 1 1
f 2 R R 2 R R
1 1 2 1 2
1 3 3 1 1 1 1 1 1
f 2 4 R R 8 R R
2 1 2 1 2
1 3 3 1 1 1 1 1 1
f 2 5 R R 10 R R
f : f1 : f2 :: 2 : 8 : 10
20. Three rods of Copper, Brass and Steel are welded together to form a Y –
shaped structure. Area of cross-section of each rod = 4 cm2. End of copper
rod is maintained at 100oC where as ends of brass and steel are kept at
0oC. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms
respectively. The rods are thermally insulated from surroundings except
at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26
and 0.12 CGS units respectively. Rate of heat flow through copper rod is
(1) 4.8 cal/s (2) 6.0 cal/s (3) 1.2 cal/s (4) 2.4 cal/s
Key. (1)
JEE-MAINS – 2014 Question Paper & Solutions
11
Sol. 0.92A100 T 0.12AT 0 0.26AT 0
46 12 13
92100 T 12T 26T
46 12 13
2(100 – T) = T + 2T
200 = 5T
T = 40
Now in copper rod
dQ 0.92 4 60 24 4.8cal
dt 46 5
/s
100oC
0oC
0oC
Copper Brass
Steel
21. A pipe of length 85 cm is closed from one end. Find the number of possible
natural oscillations of air column in the pipe whose frequencies lie below
1250 Hz. The velocity of sound in air is 340 m/s.
(1) 6 (2) 4 (3) 12 (4) 8
Key: (1)
Sol. Given l =85 cm
nv 1250
4
n 12.5
Where n = 1, 3, 5, 7, 9, 11
22. There is a circular tube in a vertical plane. Two liquids which do not mix
and of densities d1 and d2 are filled in the tube. Each liquid subtends 900
angle at centre. Radius joining their interface makes an angle with
vertical. Ratio 1
2
d
d
is:
d 1
d2
(1) 1 tan
1 tan
(2) 1 sin
1 cos
(3) 1 sin
1 sin
(4) 1 cos
1 cos
Key: (1)
Sol. P0 + d1g(R – R sin ) = P0 + d2g (R sin + R cos ) + d1g (R – R cos)
JEE-MAINS – 2014 Question Paper & Solutions
12
From above
1
2
d sin cos 1 tan
d cos sin 1 tan
23. A green light is incident from the water to the air-water interface at the
critical angle(). Select the correct statement.
(1) The spectrum of visible light whose frequency is more than that of
green light will come out to the air medium.
(2) The entire spectrum of visible light will come out of the water at
various angles to the normal
(3) The entire spectrum of visible light will come out of the water at an
angle of 900 to the normal
(4) The spectrum of visible light whose frequency is less than that of
green light will come out to the air medium.
Key: (4)
Sol. 1
…………….(i) By Cauchy relation
c
sin 1
…………(ii)
From both eq. c
24. Hydrogen (1H1), Deuterium (1H2), singly ionized Helium (2He4)+ and
doubly ionized lithium (3Li6)++ all have one electron around the nucleus.
Consider an electron transition from n = 2 to n = 1. If the wave lengths of
emitted radiation are 1, 2, 3 and 4 respectively then approximately
which one of the following is correct?
(1) 1 = 2 = 43 = 94 (2) 1 = 22 = 33 = 44
(3) 41 = 22 = 23 = 4 (4) 1 = 22 = 23 = 4
Key: (1)
Sol. hc 13.6 Z2 3
4
2
1
Z
JEE-MAINS – 2014 Question Paper & Solutions
13
25. The radiation corresponding to 3 2 transition of hydrogen atom falls on
a metal surface to produce photoelectrons. These electrons are made to
enter a magnetic field of 3 × 104. If the radius of the largest circular path
followed by these electrons is 10.0 mm, the work function of the metal is
close to :
(1) 0.8 eV (2) 1.6 eV (3) 1.8 eV (4) 1.1 eV
Key: (4)
Sol. E 13.6 1 5 eV 1.88 eV
36
r p , p rqB 4.8 10 25
qB
p2 K.E 0.79 eV
2m
E K.E 1.09
1.1eV
26. A block of mass m is placed on a surface with a vertical cross section given
by
x3 y .
6
If the coefficient of friction is 0.5, the maximum height above the
ground at which the block can be placed without slipping is:
(1) 1m
3
(2) 1m
2
(3) 1m
6
(4) 2m
3
Key: (3)
Sol. mgsin mg cos
tan
At x = 1; y = 1/6
27. When a rubber-band is stretched by a distance x, it exerts a restoring
force of magnitude
F = ax + bx2 where a and b are constants. The work done in stretching the
unstretched rubber-band by L is:
(1)
aL2 bL3
2 3
(2)
1 aL2 bL3
2 2 3
(3) aL2 bL3 (4) 1 aL2 bL3
2
Key: (1)
JEE-MAINS – 2014 Question Paper & Solutions
14
Sol.
L 2 3
0
W F dx aL bL
2 3
28. On heating water, bubbles being formed at the bottom of the vessel
detatch and rise. Take the bubbles to be spheres of radius R and making a
circular contact of radius r with the bottom of the vessel. If r << R, and the
surface tension of water is T, value of r just before bubbles detatch is:
(1) 2 wg
R
T
(2) 2 w 3 g
R
T
(3) 2 wg
R
3T
(4) 2 wg
R
6T
Key: (No Correct Option)
Sol.
3
w
Tsin 2 r 4 R g
3
sin r
R
2 w 2 g
r R
3T
q
q
T T
O
c
F = buoyant force B
29. Two beams, A and B, of plane polarized light with mutually perpendicular
planes of polarization are seen through a Polaroid. From the position
when the beam A has maximum intensity (and beam B has zero intensity),
a rotation of Polaroid through 300 makes the two beams appear equally
bright. If the initial intensities of the two beams are IA and IB respectively,
then A
B
I
I
equals:
(1) 1 (2) 1
3
(3) 3 (4) 3
2
Key: (2)
Sol. IA cos2 = IBcos2(90 ) (I = I0 cos2 ) by malus Eq
A 2
B
I tan 1
I 3
Where = 300
JEE-MAINS – 2014 Question Paper & Solutions
15
30. Assume that an electric field E 30x2iˆ
exists in space. Then the potential
difference
(VA - VO), where VO is the potential at the origin and VA the potential at x =
2 m is:
(1) – 80 J (2) 80 J (3) 120 J (4) – 120J
Key: (1)
Sol. A
O
V 2
V 0
dV E dx
2
2
0
30x dx
A 0 V V 80J
MATHEMATICS
31. The image of the line x 1 y 3 z 4
3 1 5
in the plane 2x y z 3 0 is the line
(1) x 3 y 5 z 2
3 1 5
(2) x 3 y 5 z 2
3 1 5
(3) x 3 y 5 z 2
3 1 5
(4) x 3 y 5 z 2
3 1 5
Key. (1)
Sol. Any point on the given x 1 y 3 z 4
3 1 5
will be 3t 1, t 3,5t 4 its image in
the plane 2x y z 3 0 is given by
x 3t 1 y t 3 z 5t 4
2 1 1
6t 2 t 3 5t 4 3
2 2
6
x 3t 3, y t 5, z 2 5t which is the line
x 3 y 5 z 2
3 1 5
So (1) is correct.
32. If the coefficients of x3 and x4 in the expansion of 1 ax bx2 1 2x18 in
powers of x are both zero, then (a, b) is equal to
(1) 16, 251
3
(2) 14, 251
3
(3) 14, 272
3
(4) 16, 272
3
Key. (4)
JEE-MAINS – 2014 Question Paper & Solutions
16
Sol. 1 ax bx2 1 2x18
2 18 18 2 18 3 18 4
1 2 3 4 1 ax bx 1 C 2x C . 2x C 2x C 2x ....
Coefficient of x3 = 0
18 18 18
3 2 1 C .8 a. C .4 b C 2 0
51a 3b 544 0 …. (i)
Coefficient of x4 = 0
18 18 18
4 3 2 C .16 a C 8 b C 4 0
32a 3b 240 0 …. (ii)
Solving (i) & (ii), we get
A = 16, b 272
3
So Answer (4)
33. If aR and the equation 3 x x 2 2 x x a2 0 (where [x] denotes the
greatest integer x) has no integral solution, then all possible values of a
lie in the interval
(1) 1,00,1 (2) 1,2 (3) 2,1 (4) ,22,
Key. (1)
Sol. x x x , since xI , so 0 x 1
Let x t
So
2 2
3t2 2t a2 0, t 1 1 3a 1 3a as t 0
3 3
Now 0 < t < 1 0 a2 1
a1,00,1
Option (1) is correct.
34. If 2 a b bc c a a b c
then is equal to
(1) 2 (2) 3 (3) 0 (4) 1
Key. (4)
Sol. a b bc ca
a b.b cc a
a b.b c a c b c c a
JEE-MAINS – 2014 Question Paper & Solutions
17
a b.a b cc
2 a b c
1
Option (4) is correct.
35. The variance of first 50 even natural numbers is
(1) 833
4
(2) 833 (3) 437 (4) 437
4
Key. (2)
Sol. The first 50 no. are 2,4,6,8,.....100
Mean
X 2 4 6 ....100 51
50
Variance
2 2 2
i
1 x X
n
1 22 42 62 ....1002 512
50
3434 2601 833
36. A bird is sitting on the top of a vertical pole 20 m high and its elevation
from a point O on the ground is 45o. It flies off horizontally straight away
from the point O. After one second, the elevation of the bird from O is
reduced to 30o. Then the speed (in m/s) of the bird is
(1) 40 2 1 (2) 40 3 2
(3) 20 2 (4) 20 3 1
Key. (4)
Sol. A A
h
B d C
h
x
O
d
BOA 45o
COA 30o
tan tan 45o h 20
x x
JEE-MAINS – 2014 Question Paper & Solutions
18
1 20 x 20
x
Now tan tan 30o h
x d
1 20
3 20 d
20 d 20 3
d 20 3 1
So velocity d 20 3 1 m 20 3 1 m / s
t 1sec
37. The integral 2
0
1 4sin x 4sin x dx
2 2
(1) 4 (2) 2 4 4 3
3
(3) 4 3 4 (4) 4 3 4
3
Key. (4)
Sol. 2
0
I 1 4sin x 4sin xdx
2 2
2
0
I 2sin x 1 dx
2
0
I 2sin x 1 dx
2
/3
0 /3
I 1 2sin x dx 2sin x 1 dx
2 2
/3
0 /3
x 4cos x 4cos x x
2 2
4 3 0 4 0 4 3
3 2 2 3
4 3 4
3
38. The statement ~ p~ q is
(1) equivalent to pq (2) equivalent to ~ pq
(3) a tautology (4) a fallacy
JEE-MAINS – 2014 Question Paper & Solutions
19
Key. (1)
Sol.
p q ~ p ~ q p ~ q ~ p ~ q ~ p q p q
T T F F F T F T
T F F T T F T F
F T T F T F T F
F F T T F T F T
Clearly ~ p~ q equivalent to pq
39. If A is an 33 non-singular matrix such that AA' A'A and B A1A' then
BB'
(1) I B (2) I (3) B1 (4) B1
Key. (2)
Sol. Now BB' A1A' A1A' '
A1A'A.A1 ' ' AB B'A'
A 1A'AA' 1
A 1AA'.A' 1
I.I I
40. The integral
x 1
x 1 x 1 e dx
x
(1)
x 1
x 1 e x c (2)
x 1
x e x c (3)
x 1
x 1 e x c (4)
x 1
x e x c
Key. (2)
Sol.
x 1
x I 1 x 1 e .dx
x
x 1 x 1
x x I e dx x 1 e dx
x
x 1 x 1
x x
2
I e dx x. 1 1 e dx
x
x 1 x 1
x x
2
I e dx x. 1 1 e dx
x
x 1
x
2
1. 1 1 e dx dx
x
x 1 x 1 x 1
I e x dx x.e x e xdx 2
x 1 t 1 1 dx dt
x x
x 1
I x.e x c
JEE-MAINS – 2014 Question Paper & Solutions
20
41. If z is a complex number such that z 2 , then the minimum value of z 1
2
:
(1) is equal to 5
2
(2) lies in the interval (1, 2)
(3) is strictly greater than 5
2
(4) is strictly greater than 3
2
but less
than 5
2
Key. (2)
Sol.
2 O +2
A 1/2 B
z 1
2
= Distance between z and –1
2
OA = minimum value of z 1
2
= 3
2
.
42. If g is the inverse of a function f and f '(x) = 5
1
1 x
, then g ' (x) is equal to:
(1) 1 + x5 (2) 5x4 (3) 5
1
1{g(x)}
(4) 1{g(x)}5
KEY. (4)
Sol. f (g(x)) = x
f ' (g(x)) . g ' (x)= 1
f ' (g(x)) = 1
g '(x)
g ' (x)= 1 + g5(x).
JEE-MAINS – 2014 Question Paper & Solutions
21
43. If , , and f (n) = n n and
3 1 (1) 1 (2)
1 (1) 1 (2) 1 (3)
1 (2) 1 (3) 1 (4)
f f
f f f
f f f
= 2 (1 2 ( 2 , then K is equal to:
(1) (2) 1
(3) 1 (4) –1
Key. (3)
Sol.
2 2
2 2 3 3
2 2 3 3 4 4
3 1 α + β 1 α + β
1 α + β 1 α + β 1 α + β
1 α + β 1 α + β 1 α + β
=
2
2 2
1 1 1
1
1
= 2 2 2
Hence, K = 1.
44. Let fk (x) = 1
k
(sink x + cosk x) where x R and k 1. Then f4(x) – f6(x)
equals:
(1) 1
6
(2) 1
3
(3) 1
4
(4) 1
12
Key. (4)
Sol. fk(x) = 1 sink x cosk x
k
f4(x) – f6(x) = 1 sin4 x cos4 x 1sin6 x cos6 x
4 6
= 1 1 2sin2 x cos2 x 11 3sin2 x cos2 x
4 6
= 1
12
.
45. Let and be the roots of equation px2 + qx + r = 0, p 0. If p, q, r are in A.
P.
and 1
+ 1
=4, then the value of α – β is:
(1) 61
9
(2) 2 17
9
(3) 34
9
(4) 2 13
9
Key. (4)
JEE-MAINS – 2014 Question Paper & Solutions
22
Sol. Let , are the rots of px2 + qx + r = 0
1 1 4
q 4
r
q = 4r
p, q, r are in A. P.
p = 9r
Hence the given equation becomes
9x2 + 4x – 1 =0
| 16 4 52 2 13
81 9 81 9
|
46. Let A and B be two events such that P(A B) 1
6
, P(A B) 1
4
and P(A) 1
4
,
where A stands for the complement of the event A. Then the events A and
B are:
(1) mutually exclusive and independent.
(2) equally likely but not independent.
(3) Independent but not equally likely.
(4) Independent and equally likely.
Key. (3)
Sol. P(AB) P(A) P(B) P(AB)
5 3 x 1
6 4 4
where P(B) = x
x 1
3
P(AB) P(A).P(B)
Hence A and B are independent but not equally likely.
47. If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0)
= 0 and f(1) = 6, then for some c]0,1[ :
(1) 2f’(c) = g’(c) (2) 2f’(c) = 3g’(c) (3) f’(c) = g’(c) (4) f’(c) = 2g’(c)
Key. (4)
Sol. Let h(x) = f(x) – 2g(x)
h(x) is continuous and differentiable in [0, 1].
h(0) = f(0) – 2g(0) = 2 – 0 = 2
h(1) = f(1) – 2g(1) = 6 – 4 = 2
JEE-MAINS – 2014 Question Paper & Solutions
23
h(0) = h(1)
h(x) verify all conditions of Rolle’s theorem.
Hence, at least one c(0, 1) such that
h '(c) 0
f '(c) 2g'(c)
48. Let the population of rabbits surviving at a time t be governed by the
differential equation dP(t) 1 p(t)
dt 2
– 200.
If p(0) = 100, then p(t) equals:
(1) 400 – 300 et/2 (2) 300 – 200 e–t/2 (3) 600 – 500 et/2 (4) 400 –
300 et/2
Key. (1)
Sol. I.F et/2
So solution is Ptet/2 200.et/2dt 400et /2 k
P0 100 k = –300
P(t) = 400 – 300 et/2
49. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle
centre at (0, y), passing through origin and touching the circle C
externally, then the radius of T is equal to:
(1) 3
2
(2) 3
2
(3) 1
2
(4) 1
4
Key. (4)
Sol.
B
A
O
A = (0, y) B = (1, 1)
Since AB 1 y
y = 1
4
Hence, 2
2 3
JEE-MAINS – 2014 Question Paper & Solutions
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50. The area of the region described by A = {(x, y) : x2 + y2 1 and y2 1 – x}
is:
(1) 4
2 3
(2) – 4
2 3
(3) 2 –
2 3
(4) 2
2 3
Key. (1)
Sol.
Required area = 1 2
0
2 ( 1 x 1 x )dx
1
2
1 3/2
0
2 x 1 x 1 sin x 2 (1 x)
2 2 3
2 2 4
4 3 2 3
51. Let a, b, c and d be non-zero numbers. If the point of intersection of the
lines 4ax 2ay c 0 and 5bx 2by d 0 lies in the fourth quadrant and is
equidistant from the two axes then:
(1) 2bc 3ad 0 (2) 2bc 3ad 0 (3) 3bc 2ad 0 (4) 3bc 2ad 0
Key: (3)
Sol.
Point (k, -k) satisfies both lines
4ak 2ak c 0 . . . . . (1)
5bk 2bk d 0 . . . . . (2)
c = -2ak and d = -3bk
2 3
c d
a b
2ad - 3bc = 0
3bc – 2ad = 0
52. Let PS be the median of the triangle with vertices P(2, 2), Q(6, - 1) and R(7,
3). The equation of the line passing through (1, -1) and parallel to PS is:
(1) 4x 7 y 11 0 (2) 2x 9y 7 0 (3) 4x 7 y 3 0 (4) 2x 9 y 11 0
Key: (2)
JEE-MAINS – 2014 Question Paper & Solutions
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Sol.
1 2
2 13 9
2
ps m
1 2 1
9
y x
9y 2x 7 0
53.
2
0 2
lim sin( cos )
x
x
x
is equal to :
(1)
2
(2) 1 (3) (4)
Key : (4)
Sol.
2
0 2
sin cos
lim
x
x
x
=
2 2
0 2 2
sin sin sin lim
x sin
x x
x x
=
54. If X 4n 3n 1 : n N and Y 9 n 1 : n N, where N is the set of natural
numbers, then XY is equal to :
(1) N (2) Y – X (3) X (4) Y
Key : (4)
Sol.
4n 3n 1
= 1 3 3 1 n n
= 2 3
2 3 nC 3 nC 3 ........
= 2 3 9 nC 3 nC .......
Set X contains numbers which are multiples of 9 but not all multiples of 9
XY = Y
JEE-MAINS – 2014 Question Paper & Solutions
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55. The locus of the foot of perpendicular drawn from the centre of the ellipse
x2 3y2 6 on any tangent to it is :
(1) x2 y2 2 6x2 2 y2 (2) x2 y2 2 6x2 2 y2
(3) x2 y2 2 6x2 2y2 (4) x2 y2 2 6x2 2 y2
Key : (3)
Sol.
K mh 6m2 2 . . . . . (1)
K m 1
h
. . . . . . (2)
Eliminating m from (1) and (2)
Locus is
x2 y2 2 6x2 2y2
56. Three positive numbers form an increasing G.P. If the middle term in this
G.P. is doubled, the new numbers are in A.P. Then the common ratio of the
G.P is :
(1) 2 3 (2) 3 2 (3) 2 3 (4) 2 3
Key : (4)
Sol.
4a = a ar
r
(r > 1)
r 2 3
but as r > 1
r 2 3
57. If 9 1 8 2 7 9 9 10 2 11 10 3 11 10 .... 10 11 k 10 , then k is equal to :
(1) 121
10
(2) 441
100
(3) 100 (4) 110
Key : (3)
Sol. S = 109 + 2(11) (10)8 + . . . . . . + 10(11)9
S 11
10
= (11.108) + . . . . . + . . . + (11)10
JEE-MAINS – 2014 Question Paper & Solutions
27
10
9
10
10 11 1
S 10 (11)
10 11 1
10
S = 1011
= (100) 109
K = 100
58. The angle between the lines whose direction cosines satisfy the equations
l m n 0 and l2 m2 n2 is :
(1)
3
(2)
4
(3)
6
(4)
2
Key : (1)
Sol. Solving two equation
l = -(m + n) then (m + n)2 = m2 + n2
mn = 0
If m = 0 , If n = 0
l = -n l = -m
hence dr are 1, 0, -1 and 1, -1, 0
cos 1 1
2 2 2
=
3
59. The slope of the line touching both the parabolas y2 4x and x2 32y is :
(1) 1
2
(2) 3
2
(3) 1
8
(4) 2
3
Key : (1)
Sol. Eq. of tangent at t & t respectively are yt x t2 and xt ' y 8t '2
Comparing
2
2
1 2
' 1 8 '
t t t
t t
Hence slope = 1
2
JEE-MAINS – 2014 Question Paper & Solutions
28
60. If x 1 and x 2 are extreme points of f (x) log | x | x2 x then :
(1) 6, 1
2
(2) 6, 1
2
(3) 2, 1
2
(4) 2, 1
2
Key: (3)
Sol. Differentiating
f 'x 2 x 1 0 2 x2 x 0
x
Put x 2 and -1 gives
2 1 0 and 8 2 0
Solving 2, 1
2
CHEMISTRY
61. Which one of the following properties is not shown by NO?
(1) It combines with oxygen to form nitrogen dioxide
(2) It’s bond order is 2.5
(3) It is diamagnetic in gaseous state
(4) it is a neutral oxide
Key. (3)
Sol. Nitric Oxide, NO is paramagnetic.
62. If Z is a compressibility factor, van der Waals equation at low pressure can
be written as:
(1) Z 1 Pb
RT
(2) Z 1 Pb
RT
(3) Z 1 RT
PB
(4) Z 1 a
VRT
Key (4)
Sol. As pressure is low so volume must be large.
i.e., V b V
So Van der Waal’s equation reduces into
PV a RT
V
Z 1 a
RTV
JEE-MAINS – 2014 Question Paper & Solutions
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63. The metal that cannot be obtained by electrolysis of an aqueous solution
of its salts is :
(1) Cu (2) Cr
(3) Ag (4) Ca
Key. (4)
Sol. In case of aqueous solution of calcium salt hydrogen gas will
preferentially evolve due to selective electrolysis. H has higher reduction
potential than that of Ca2
64. Resistance of 0.2 M solution of an electrolyte is 50 . The specific
conductance of the solution is 1.4 S m–1. The resistance of 0.5 M solution of
the same electrolyte is 280 . The molar conductivity of 0.5 M solution of
the electrolyte in Sm2 mol–1 is :
(1) 5103 (2) 5102
(3) 5104 (4) 5103
Key. 3
Sol. 1 . 1
R / A
1 1.4. 1
50 / A
1 . 1
280 / A
Solving we get
0.25Sm1
C
0.25 5 10 4 Sm2 mol 1
0.5 1000
65. CsCl crystallizes in body centred cubic lattice. If ‘a’ is its edge length then
which of the following expression is correct?
(1) Cs Cl
r r 3 a
2 (2) Cs Cl r r 3a
(3) Cs Cl r r 3a (4) Cs Cl
r r 3a
2
JEE-MAINS – 2014 Question Paper & Solutions
30
Key. (1)
Sol. In CsCl lattice, centrally placed Cs remains in contact with all eight Cl
ions present at the vertices of the cube.
Body diagnonal = 3a
3a 2Cs Cl .
66. Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2
(aq), 0.25 M KBr(aq) and 0.125 M Na3PO4(aq) at 25oC. Which statement is
true about these solutions, assuming all salts to be strong electrolytes?
(1) 0.125 M Na3PO4(aq) has the highest osmotic pressure.
(2) 0.500 M C2H5OH(aq) has the highest osmotic pressure.
(3) The all have the same osmotic pressure.
(4) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure.
Key. (3)
Sol. Considering complete dissociation of electrolytes, the effective
concentration of solute particles is same (0.5 M) in all the solution.
67. In which of the following reaction H2O2 acts as a reducing agent?
(a) 2 2 2 H O 2H 2e 2H O (b) 2 2 2 H O 2e O 2H
(c) 2 2 H O 2e 2OH (d) 2 2 2 2 H O 2OH 2e O 2H O
(1) (a), (c) (2) (b), (d)
(3) (a), (b) (4) (c), (d)
Key. (2)
Sol. 1 0
2 2 2 H O O 2e
68. In SN2 reactions, the correct order of reactivity for the following
compounds:
3 3 2 3 2 3 3 CH Cl,CH CH Cl, CH CHCl and CH CCl
(1) 3 2 3 3 2 3 3 CH CH Cl CH Cl CH CHCl CH CCl
(2) 3 2 3 2 3 3 3 CH CHCl CH CH Cl CH Cl CH CCl
(3) 3 3 2 3 2 3 3 CH Cl CH CHCl CH CH Cl CH CCl
(4) 3 3 2 3 2 3 3 CH Cl CH CH Cl CH CHCl CH CCl
Key. (4)
Sol. In N S 2 reactivity order is o o o
3 CH X 1 R X 2 R X 3 R X
JEE-MAINS – 2014 Question Paper & Solutions
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69. The octahedral complex of a metal ion M3+ with four monodentate ligands
L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and
blue, respectively. The increasing order of ligand strength of the four
ligands is :
(1) L3 < L2 < L4 < L1 (2) L1 < L2 < L4 < L3
(3) L4 < L3 < L2 < L1 (4) L1 < L3 < L2 < L4
Key. (4)
Sol. As the order of wavelength
red yellow green blue
& Energy 1
70. For the estimation of nitrogen, 1.4 g of an organic compound was digested
by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of
M
10
sulphuric acid. The unreacted acid required 20 mL of M
10
sodium
hydroxide for complete neutralization. The percentage of nitrogen in the
compound is:
(1) 3 % (2) 5 %
(3) 6 % (4) 10 %
Key. (4)
Sol. 2 4 3 Eq.H SO Eq.NH Eq.NaOH
3
60 1 2 Eq.NH 20 1
1000 10 1000 10
3
Eq.NH 100 0.01
10000
Mole 3 NH 0.01
Weight percent 2
of N 0.01 14 100 10%
1.4
71. The equivalent conductance of NaCl at concentration C and at infinite
dilution are C and , respectively. The correct relationship between C
and is given as
(1) C (B) C (2) C (B) C
(3) C (B)C (4) C (B)C
Key. (1)
Sol. It is Debye Huckel onsager equation i.e.
C B C
JEE-MAINS – 2014 Question Paper & Solutions
32
72. For the reaction 2(g) 2(g) 3(g)
SO 1 O SO
2
, if KP = KC(RT)x where the symbols
have usual meaning, then the value of x is (assuming ideality)
(1) 1
2
(2) 1 (3) –1 (4) 1
2
Key. (4)
Sol. ng
P C K K (RT)
x or g
n 1
2
73. In the reaction,
LiAlH4 PCl5 Alc. KOH
3 CH COOHABC ,
the product C is
(1) Ethylene (2) Acetyl chloride (3) Acetaldehyde (4) Acetylene
Key. (1)
Sol.
LAH PCl5
3 3 2 3 2
(A) (B)
CH COOHCH CH OHCH CH Cl
2 2
(C)
CH CH
alc. KOH
74. Sodium phenoxide when heated with CO2 under pressure at 125°C yields a
product which on acetylation produces C.
+ 125
2 5 Atm
2
CO B H C
Ac O
The product C would be
(1) (2)
(3) (4)
Key. (3)
ONa
OH
COOCH3
O COCH3
COOH
O COCH3
COOH
OH
COCH3
COCH3
JEE-MAINS – 2014 Question Paper & Solutions
33
Sol.
75. On heating an aliphatic primary amine with chloroform and ethanolic
potassium hydroxide, the organic compound formed is
(1) an alkyl cyanide (2)an alkyl isocyanide
(3) an alkanol (4) an alkanediol
Key. (2)
Sol. Carbylamine reaction
3
2 5
(CHCl KOH)
2 2 C H OH RCH NH R N C
76. The correct statement for the molecule, CsI3, is
(1) it contains Cs3+ and I– ions
(2) it contains Cs+, I– and lattice I2 molecule
(3) it is a covalent molecule (4)it contains Cs+ and 3 I ions
Key. (4)
Sol. 3 3 CsI Cs I
77. The equation which is balanced and represents the correct product(s) is
(1) 2 4 excess NaOH 2
2 6 2 [Mg(H O) ] (EDTA) [Mg(EDTA) 6H O
(2) 4 2 4 2 4 CuSO 4KCNK [Cu(CN) ]K SO
(3) 2 2 Li O2KCl2LiCl K O
(4) 2
3 5 4 [CoCl(NH ) ] 5H Co 5NH Cl
Key. (3)
Sol. No proof available.
78. For which of the following molecule significant 0?
(A)
(B)
(C)
(D)
(1) only (C) (2) (C) and (D) (3) only (A) (4) (A) and (B)
ONa
+ CO2
125°
5 atm
COONa
OH
H+
5 atm
COOH
OH
COOH
OCOCH3
(AC) O 2
Acetylation
Aspirin
Cl
Cl
CN
CN
OH
OH
SH
SH
JEE-MAINS – 2014 Question Paper & Solutions
34
Key. (2)
Sol.
79. For the non-stoichiometre reaction 2ABCD, the following kinetic
data were obtained in three separate experiments, all at 298 K.
Initial
Concentration
(A)
Initial
Concentration
(B)
Initial rate of
formation of C
(mol L–S–)
0.1 M
0.1 M
0.2 M
0.1 M
0.2 M
0.1 M
1.2 × 10–3
1.2 × 10–3
2.4 × 10–3
The rate law for formation of C is
(A) dc k[A][B]2
dt
(B) dc k[A]
dt
(C) dc k[A][B]
dt
(D) dc k[A]2 [B]
dt
Key. (2)
Sol. Rate reaction,
d(c) A n B m
dt
=
m 3
3
0.1 1.2 10
0.2 1.2 10
m = 0
n 3
3
0.1 1.2 10
0.2 2.4 10
n 1 1
2 2
=
n = 1
rate equation dc k[A]
dt
80. Which series of reactions correctly represents chemical relations related
to iron and its compound?
(A) Cl2 , heat heat , air Zn
3 2 FeFeCl FeCl Fe
(B) O2 , heat CO, 600 C CO, 700 C
3 4 FeFe O FeOFe
(C) dil. H2SO4 H2SO4 , O2 heat
4 2 4 3 FeFeSO Fe (SO ) Fe
(D) O2 , heat dil. H2SO4 heat
4 FeFeOFeSO Fe
N
JEE-MAINS – 2014 Question Paper & Solutions
35
Key. (2)
Sol. 3 4 2
2 3 2
Fe O 4CO 3Fe 4CO
500 800 K
Fe O CO 2FeO CO
2 FeO COFe CO 900 1500K
81. Considering the basic strength of amines in aqueous solution, which one
has the smallest pKb values
(A) (CH3)3N (2) C6H5H2 (3) (CH3)2NH (4) CH3NH2
Key. (3)
Sol. The base which has maximum Kb will have smallest pKb
82. Which one of the following bases is not present in DNA?
(1) Cytosine (2) Thymine (3) Quinoline (4) Adenine
Key . (3)
Sol. In DNA, Nitrogenous base
(i) Cytosine (ii) Thymine
(iii) Adenine (iv) Guanine
83. The correct set of four quantum numbers for the valence electrons of
rubidium atom (Z = 37) is
(1) 5, 1, 1, + 1
2
(2) 5, 0, 1, + 1
2
(3) 5, 0, 0, + 1
2
(4) 5, 1, 0, + 1
2
Key . (3)
Sol.
Configuration : Ar 4s2 3d10 4p6 5s1
Its quantum numbers n, l, m and s will be 5, 0, 0, + 1
2
respectively.
84. The major organic compound formed by the reaction of 1, 1, 1 –
trichloroethane with silver powder is
(1) 2-butyne (2) 2- butane (3) acetylene (4) ethane
Key . (1)
Sol.
H3C
Cl
Cl
Cl + 6Ag + Cl
Cl
Cl
CH3 - 6AgCl
H3C C C CH3
2-butyne
JEE-MAINS – 2014 Question Paper & Solutions
36
85. Given below are the half-cell reactions :
Mn2+ + 2e- Mn; E0 = -1.18 V
2(Mn3+ + e- Mn2+); E0 = +1.51 V
The E0 for 3Mn2+ Mn + 2Mn3+ will be
(1) -0.33 V; the reaction will not occur
(2) -0.33V; the reaction will occur
(3) -2.69 V; the reaction will not occur
(4) -2.69V; the reaction will occur
Key . (3)
Sol. Mn2+ + 2e- Mn; G10 = 2 F (1.18) . . . . .(i)
2(Mn3+ + e- Mn2+); G20 = 2 F (+1.51) . . . . .(ii)
for
3Mn2+ Mn + 2Mn3+
G30 = G10 G20
2FE30 = 2F (1.18) + 2F(1.51)
E30 = 2.69 V
E30 = 2.69 V
Reaction will not occur
86. The ratio of masses of oxygen and nitrogen in a particular gaseous
mixture is 1 : 4. The ratio of number of their molecule is :
(1) 1 : 8 (2) 3 : 16 (3) 1 : 4 (4) 7 : 32
Key . (4)
Sol. 2 2 2
2 2 2
O O O
N N N
n w /M x / 32 28 7 :32
n w /M 4x / 28 4 32
87. Which one is classified as a condensation polymer ?
(1) Teflon (2) Acrylonitrile (3) Dacron (4) Neoprene
Key . (3)
Sol. Dacron is condensed polymer of tere-phthalic acid and glycol.
88. Among the following oxoacids, the correct decreasing order of acid
strength is :
(1) HClO4 > HClO3 > HClO2 > HOCl
(2) HClO2 > HClO4 > HClO3 > HOCl
(3) HOCl > HClO2 > HClO3 > HClO4
(4) HClO4 > HOCl > HClO2 > HClO3
JEE-MAINS – 2014 Question Paper & Solutions
37
Key. (1)
Sol.
7 5 3 1
4 3 2 HClO HClO HClO HOCl
Higher is the oxidation state, more will be the acidic strength of these
oxoacids.
Order of resonance stability of conjugate bases 4 3 2 ClO ClO ClO
89. For complete combustion of ethanol,
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
the amount of heat produced as measured in bomb calorimeter, is
1364.47 kJ mol-1 at 250C. Assuming ideality the Enthalpy of combustion,
CH, for the reaction will be:
(R = 8.314 kJ mol-1)
(1) -1460.50kJ mol-1 (2) -1350.50 kJ mol-1
(3) -1366.95 kJ mol-1 (4) -1361.95 kJ mol-1
Key . (3)
Sol. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
H = E + nRT
qp = qv + nRT
= 1364.47 1 8.31 298
1000
= -1364.47 – 2.47638
= - 1366.95 kJ mol-1
90. The most suitable reagent for the conversion of R – CH2 – OH R – CHO is :
(1) CrO3 (2) PCC (Pyridinium Chlorochromate)
(3) KMnO4 (4) K2Cr2O7
Key. (2)
Sol. R – CH2OH PCC
|| O
R C H , other oxidizing agents may convest RCHO to
RCOOH.
JEE-MAINS – 2014 Question Paper & Solutions
38
IITJEE MAINS-2014
Answer Key (CODE-H)
PHYSICS MATHEMATICS CHEMISTRY
Q.N. Code -H Q.N. Code-H Q.N. Code-H
1. 3 31. 1 61. 3
2. 4 32. 4 62. 4
3. 3 33. 1 63. 4
4. 3 34. 4 64. 3
5. 3 35. 2 65. 1
6. 2 36. 4 66. 3
7. 3 37. 4 67. 2
8. 4 38. 1 68. 4
9. 2 39. 2 69. 4
10. 1 40. 2 70. 4
11. 2 41. 2 71. 1
12. 1 42. 4 72. 4
13. 3 43. 3 73. 1
14. 1 44. 4 74. 3
15. 1 45. 4 75. 2
16. 4 46. 3 76. 4
17. 2 47. 4 77. 3(Not confirmed)
18. 1 48. 1 78. 2
19. 4 49. 4 79. 2
20. 1 50. 1 80. 2
21. 1 51. 3 81. 3
22. 1 52. 2 82. 3
23. 4 53. 4 83. 3
24. 1 54. 4 84. 1
25. 4 55. 3 85. 3
26. 3 56. 4 86. 4
27. 1 57. 3 87. 3
28. No answer 58. 1 88. 1
29. 2 59. 1 89. 3
30. 1 60. 3 90. 2
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