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JEE-MAIN 2014
...

Read carefully the Instructions on the Back Cover of this Test Booklet
...
Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen
...

2
...
When you are directed to open the Test Booklet, take
out the Answer Sheet and fill in the particulars carefully
...
The test is of 3 hours duration
...
The Test Booklet consists of 90 questions
...

5
...
Each question is allotted 4 (four) marks for each
correct response
...
Candidates will be awarded marks as stated above in instruction No
...
1/4 (one fourth) marks will be deducted for indicating incorrect response of each question
...

7
...
Filling up more than one response in each question
will be treated as wrong response and marks for wrong response will be deducted accordingly as per
instruction 6 above
...
Use Blue/Black Ball Point Pen only for writting particulars/ marking responses on Side−1 and Side−2
of the Answer Sheet
...

9
...
, except the Admit Card inside the examination hall/room
...
Rough work is to be done on the space provided for this purpose in the Test Booklet only
...

11
...
However, the candidates are allowed to take away this Test Booklet with them
...
The CODE for this Booklet is H
...
In case of discrepancy, the candidate should immediately report the
matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet
...
Do not fold or make any stray marks on the Answer Sheet
...
1)

JEE-MAIN 2014 : Paper and Solution (2)

Read the following instructions carefully :
1
...

2
...

3
...

4
...

5
...
No deduction from the total score, however, will be made if no response
is indicated for an item in the Answer Sheet
...
Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy
in Test Booklet Code and Answer Sheet Code), another set will be provided
...
The candidates are not allowed to do any rough work or writing work on the Answer Sheet
...
This space is given at the bottom of each page and in 3 pages (Pages
21 – 23) at the end of the booklet
...
On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in
the Room/Hall
...

9
...

10
...

11
...
Cases where a candidate has not signed the
Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with
as an unfair means case
...

12
...
is prohibited
...
The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct
in the Examination Hall
...

14
...

15
...


(Pg
...
Questions and Solutions
...
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 100°C is :
(For steel Young’s modulus is 2 × 1011 N m−2 and coefficient of thermal expansion is 1
...
2 × 107 Pa
(2) 2
...
2 × 108 Pa
(4) 2
...
(3)

F
=
A
Δ

=

P

=
=

Y

Δ

αΔT
F
= YαΔT
A
2 ×1011 ×1
...
2 × 108 N/m 2

2
...
5 ≤ z < 1
...
0 A in − a z
direction (see figure)
...
0 × 10−4 e−0
...
0 m, y = 0 m in 5 × 10−3 s
...


(1) 14
...
7 W
(3) 1
...
97 W
2
...
V
...
2
10 × 3 × 3 × 10−6 ⎡
1 − e −0
...
2
1
...
16 0
...
⎟ ⎟
⎜ 1 − ⎜ 1 − (0
...
2 ⎝ ⎝
2
16
⎠⎠

=

0
...
88
2
On Exact evaluation P = 2
...
3)

JEE-MAIN 2014 : Paper and Solution (4)
3
...
The
bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical
...

(2) angular momentum changes both in direction and magnitude
...

(4) angular momentum changes in magnitude but not in direction
...
(1)

τnet = τT + τmg
'
= τmg ≠ 0

L ≠ constant

But as ω is constant
L = MVL = constant
4
...
If a student makes an error measuring
± 0
...
5 mA
(2) 0
...
2 mA
(4) 0
...
(3)
⎛

⎞
− 1⎟
⎝
⎠
1000 1000V
dI
=
e T
dV
T
⎛ 1000 1000V ⎞
e T ⎟ dV
dI = ⎜
T
⎝
⎠

I = ⎜e

1000V
T

I = 5 MA
1000V
= 6 MA
T
10
1
⎛ 1000 ⎞
dI = ⎜
⎟ (0
...
2 mA
5
...
The open end of the tube is then closed and sealed and the tube is raised vertically up by additional
46 cm
...
(1)

P0

(54 − h)

P′

h

(Pg
...
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

(1)
(2)
(3)
(4)
6
...
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
...
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
7
...
2
d = 5 mm

E = 3 × 104

v
m

σ
= 3 × 104
k ∈o

σ

= k∈o × 3 × 104
= 2
...
85 × 10−12 × 3 × 104
= 58
...
84 × 10−7 = 6 × 10−7 c/m2 (approx
...
5)

JEE-MAIN 2014 : Paper and Solution (6)
8
...
50 cm
...

(2) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm
...

(4) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale &
main scale has 10 divisions in 1 cm
...
(4)
Least count = 0
...
e
...
1 mm
for vernier calliper of
10 v
...
d
...
s
...

1 v
...
d
...
9 M
...
d
...
C
...
s
...
1) M
...
d
...
s
...
= 0
...
c = 0
...
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
...
(2)
F⎞
⎛
⎜ 2F + 2 ⎟
⎝
⎠
⎛ 2 2 + 1⎞
⎜
⎟
⎜
2 ⎟
⎝
⎠

=

mv 2
R

a

⎛ Gm2 ⎞
mv 2
⎜ 2 ⎟ = a
⎝ a ⎠
2

a

⎛ 2 2 + 1⎞ Gm
⎛ 2 2 + 1⎞ Gm
=⎜
⎜ 2 2 ⎟ 2R
⎟
a
⎝
⎠

V2 = ⎜
⎜ 2 2 ⎟
⎟
⎝
⎠

R

a

a

⎛ 2 2 + 1⎞ Gm
⎟
4 ⎟ R
⎝
⎠

V2 = ⎜
⎜

10
...

The voltage of the electric mains is 220 V
...
(1)
1
1
1
=
+
R eq R1 R 2
P=

V2 V2 V2
=
+
+
...
A particle moves with simple harmonic motion in a straight line
...
6)

(7) VIDYALANKAR : JEE-MAIN 2014 : Paper and Solution
11
...
The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 × 103 A m−1
...
(1)
B = μ0 H
μ0 n i = μ0 H

i=

H 3 × 103
=
= 3A
n
1000

13
...
(3)

+2V

−2V

p side should be at higher potential for forward bias
...
During the propagation of electromagnetic waves in a medium :
(1) Electric energy density is equal to the magnetic energy density
...

(3) Electric energy density is double of the magnetic energy density
...

14
...

15
...
Afterward, suddenly, point ‘C’ is disconnected from point ‘A’ and connected to
point ‘B’ at time t = 0
...
7)

e
1− e

(4) 1

JEE-MAIN 2014 : Paper and Solution (8)
15
...
A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and
radius R
...
(4)

T
T

mg − T =
T· R =
a=

ma
m R2 α
Rα
g
Solving a =
2

(3)

2g
3

(4)

g
2

…(1)
…(2)
…(3)
…(4)

mg
17
...
The process BC is
adiabatic
...
Choose the correct
statement :

(1)
(2)
(3)
(4)

The change in internal energy in the process AB is − 350 R
...

The change in internal energy in whole cyclic process is 250 R
...


(Pg
...
(2)
5R
(800 − 400) = 1000R
AB
ΔU = nCv ΔT = 1 ⋅
2
Option (1) is not correct

CA

5R
(400 − 600) = −500R
2
Option (4) is not correct

ΔU = nCv ΔT = 1 ⋅

ΔUcycle = 0
∴ ΔUAB + ΔUBC + ΔUCA
1000 R + ΔUBC − 500 R
ΔUBC = −500 R

= 0
= 0

18
...
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
...
(1)
time taken to reach highest point
u
u
+ve
t1 =
g
For time taken to reach the ground
g
H
1
−H = ut − gt 2
2
2
∴ gt − 2ut − 2H = 0

t=

2u ± 4u 2 + 8gH
2g

u + u 2 + 2gH
g
Given t = n t1
t=

(− ve sign not acceptable)

u + u 2 + 2gH nu
=
g
g
2
Solving 2g H = nu (n − 2)

⎡

3⎤

19
...
When it is measured in two
2⎦
⎣

different liquids having refractive indices

4
5
and , it has the focal lengths f1 and f2 respectively
...
(4)
1 ⎛ 3 ⎞⎛ 1
1 ⎞
= ⎜ − 1⎟ ⎜
−
⎟
f ⎝ 2 ⎠ ⎝ R1 R 2 ⎠
1 1⎛ 1
1 ⎞
= ⎜
−
…(1)
⎟
f 2 ⎝ R1 R 2 ⎠

(2) f1 and f2 both become negative
(4) f1 > f and f2 becomes negative

(Pg
...
Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure
...
End of copper rod is maintained at 100 °C where as ends of brass and steel
are kept at 0 °C
...
The rods
are thermally insulated from surroundings except at ends
...
92, 0
...
12 CGS units respectively
...
8 cal/s
(2) 6
...
2 cal/s
(4) 2
...
(1)
0°C
0°C
K3 = 0
...
20
13cm
12cm
100 − T
T−0
T−0
Brass
=
+
1

2

k1A

3

k2 A

k1(100 − T)

=

1

k3A

k2 T

+

2

A = 4 cm2

k3 T

T

3

0
...
92
0
...
92
100°C

0
...
8 cal/s
21
...
Find the number of possible natural oscillations of air
column in the pipe whose frequencies lie below 1250 Hz
...

(1) 6
(2) 4
(3) 12
(4) 8

(Pg
...
(1)

L

= n

λ λ
λ
= (2n + 1)
+
2 4
4
v
( 2n + 1)

⇒ 0
...
85× 4

⇒ f = (2n + 1) 100 H2

The possible frequencies are : 100 HZ, 300 HZ, … 1100 HZ
...
There is a circular tube in a vertical plane
...
Each liquid subtends 90°
angle at centre
...
Ratio

d1
is
d2

1 + tan α
1 − tan α
1 + sin α
(3)
1 − sin α

1 + sin α
1 − cos α
1 + cos α
(4)
1 − cos α
(2)

(1)

22
...

d1 (1 − sin α)
⇒

= d1 (1 − cos α) + d2 [cos α + sin α]

d1
sin α + cos α
=
d2
cos α − sin α

α

α
α

d2
0

23
...
Select the
correct statement
...

(2) The entire spectrum of visible light will come out of the water at various angles 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
...
(4)
μr < μg < μv ⇒ sin (θr) > sin (θg) > sin (θv)
⇒ θr > θg > θv [θr, θg, θv are critical angles of corresponding colors]
Thus all colors from red to green will emerge out of water
24
...

Consider an electron transition from n = 2 to
n = l
...
(1)
RZ2
1
=
λ
⎛ me ⎞
⎜ 1+ M ⎟
⎝
⎠
1
1
∝
λ
⎛ me ⎞
⎜ 1+
⎟
⎜
Mp ⎟
⎝
⎠

⎡ 1⎤
⎢1 − 4 ⎥
⎣
⎦
1
1
∝
λ2
⎛
me
⎜ 1+
⎜ 2M
p
⎝

⇒

⎞
⎟
⎟
⎠

Z2
3
1
=
R0 ⋅
λ
4
⎛ me ⎞
⎜ 1+ M ⎟
⎝
⎠

[M = mass of mucleus]

[Mp = mass of H − atom]

(Pg
...
The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce
photoelectrons
...
If the radius of the
largest circular path followed by these electrons is 10
...
8 eV
(2) 1
...
8 eV
(4) 1
...
(4)

r=

mν
=
eB

2mEp

[E = electron’s K
...
] [m = electron mass, e = electron charge]

eB

E = Ep − φ

Ep = photon energy = 1
...


(reB)2
2m
−4
−19
10 × 1
...
9 eV −
2 × 9
...
9eV − 0
...
1eV

26
...
If the coefficient
6

of friction is 0
...
(3)

N
Mg sinθ

fs
Mg cosθ

mg sinθ = μs mg cos θ

[when partied is just balanced]

⇒ tan θ = μs
tanθ =

⎛ x2 ⎞
dy
=⎜ ⎟
dx
⎝ 2 ⎠

x2
= 0
...
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
...
(1)
L

w = ∫ F
...
12)

(4)

1 2
(aL + bL3 )
2

(13) VIDYALANKAR : JEE-MAIN 2014 : Paper and Solution
28
...
Take the bubbles to be spheres of radius R and making a circular
contact of radius r with the bottom of the vessel
...
⎢ r =

⎣

2 ρg 2 ⎤
⋅R ⎥
3 T
⎦

The down ward force on the bubble due to surface tension = 2πr
...


⎡

r
θ

4
2πTr 2
πR 3ρg =
∴
3
R

θ

δT

2 ρg 2 ⎤
⋅R ⎥
3 T
⎦

⇒ ⎢r =

⎣

29
...
From the position when the beam A has maximum intensity (and beam B has zero
intensity), a rotation of polaroid through 30° makes the two beams appear equally bright
...
(2)
IA cos2 30° = IB cos2 60°

⇒

IA
IB

=

(3) 3
IA

IA
equals :
IB
(4)

3
2

I

1
3

IB

30
...
Then the potential difference VA − VO, where
i
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) −120 J
30
...
dx = − 80 J
0

(Pg
...
The image of the line

31
...
If the coefficients of x3 and x4 in the expansion of (1+ ax+ bx2) (1−2x)18 in powers of x are both
zero, then (a, b) is equal to:
251 ⎞
251 ⎞
272 ⎞
272 ⎞
⎛
⎛
⎛
⎛
(2) ⎜14,
(3) ⎜14,
(4) ⎜16,
(1) ⎜16,
⎟
⎟
⎟
⎟
3 ⎠
3 ⎠
3 ⎠
3 ⎠
⎝
⎝
⎝
⎝
32
...


33
...
14)

(15) VIDYALANKAR : JEE-MAIN 2014 : Paper and Solution
33
...
a ∈ (−1, 0) ∪ (0, 1)
2

⎡
⎤
34
...
(4)

⎡a × b b × c c × a ⎤
⎣
⎦

{(

)

Let u = b × c

}

= (a × b ) ⋅ b × c × ( c × a )

(

)

= a × b ⋅ {u × (c × a)}
= (a × b) ⋅

{( u ⋅ a ) c − ( u ⋅ c ) a }

{

}

= (a × b ) ⋅ ⎡b c a ⎤ c − ⎡b c c ⎤ a
⎣
⎦
⎣
⎦
= ⎡ b c a ⎤ ⎡a b c ⎤
⎣
⎦⎣
⎦
⇒

⎡
⎤
= ⎣a b c ⎦
=1

λ

2

35
...
(2)

Variance

=

∑x

2
i

n

()

− x

2

⇒

⎛ 22 + 4 2 + 62 +
...
+ 100 ⎞
σ =⎜
⎟−⎜
⎟
50
50
⎠
⎝
⎠ ⎝

⇒

σ2 = 3434 − 2601

2

= 833

(Pg
...
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 45°
...
After one second, the
elevation of the bird from O is reduced to 30°
...
(4)
Here,

⇒

)

(

)

(

Q

⇒ OB − OA = 20 ( 3 − 1)
Hence distance covered in one second by the bird is

P

B

AP = QB = 20 m
∠POA = 45°, ∠QOB = 30°
OA = 20; OB = 20 3

)

A

O

AB = 20 ( 3 − 1)
Thus, speed of bird = 20 ( 3 − 1) m/s
π

37
...
(4)
π

I=

∫

1 + 4 sin 2

0
π

I=

x
x
− 4 sin
2
2

x

∫ 1 − 2 sin 2 dx
0

π
3

I=

π

x⎞
π⎞
⎛
⎛
∫ ⎜1 − 2 sin 2 ⎟ dx + ∫ − ⎜1 − 2 sin 2 ⎟ dx
⎠
⎠
π ⎝
0⎝
3

⎡
⎣

x⎤

π
3

⎡
⎣

x⎤

π

I = ⎢ x + 4 cos ⎥ + ⎢ − x − 4 cos ⎥
2
2 π

⎦0

⎦

3

⎛ 3 ⎞ ⎛
⎛
π
π⎞
3⎞
+ 4⎜
⎜ 2 − 1⎟ − ⎜ π − 3 ⎟ − 4 ⎜ 0 − 2 ⎟
⎟ ⎝
⎜
⎟
3
⎠
⎝
⎠
⎝
⎠
π
π
2π
+2 3 = 4 3−4−
I= +2 3−4−
3
3
3
I=

38
...
(1)

(2) equivalent to ~p ↔ q
(4) a fallacy

Truth table for ~(p ↔ ~q) is
p
T
T
F
F

q
T
F
T
F

~q
F
T
F
T

(1)
p → ~q
F
T
T
T

Thus ~(p ↔ ~q) is equivalent to p ↔ q
...
16)

(2)
~q → p
T
T
T
F

(3)
(1) ∧ (2)
F
T
T
F

~(3)
T
F
F
T

(17) VIDYALANKAR : JEE-MAIN 2014 : Paper and Solution

39
...
(2)

B B T = ( A −1 A T ) ( A −1 A T )

(

T

= ( A −1 A T ) A ⋅ ( A −1 )

)
= ( A −1 A ) A T ⋅ ( A −1 ) T
T
= ( A −1 A ) = I

)

= A −1 ( A A T ) ( A −1 ) T

= A −1 ⋅ ( A T A ) ( A

−1 T

= A T ⋅ ( A −1 )

T

T

1

1 ⎞ x+
⎛
40
...
(2)
1

1 ⎞ x+
⎛
⎜1 + x − ⎟ e x dx
∫⎝
x⎠
Put x e

x+

1
x

xe

=t

x+

1
x

1

1 ⎫ x+
⎧
⎨1 − 2 ⎬ + e x dx = dt
⎩ x ⎭
1

⎧ ⎛
1 ⎞ ⎫ x+
⎨ x ⎜1 − 2 ⎟ + 1⎬ e x dx = dt
⎩ ⎝ x ⎠ ⎭
1

1 ⎞ x+
⎛
⎜1 + x − ⎟ e x dx = dt
x⎠
⎝
=

∫ dt = t + c = x e

x+

1
x

+C

41
...
(2)

z+

y

1
⎛ 1⎞
= z −⎜− ⎟
2
⎝ 2⎠

B (0, 2)

1
3
is
The minimum value of z +
2
2

(−2,0)
x′

Q
A′

1
− ,0
2

(Pg
...
If g is the inverse of a function f and f′ (x) =

(1) 1 + x5

, then g′ (x) is equal to :

1 + x5

(2) 5x4

(3)

1

(4) 1 + {g(x)}5

1 + {g (x)}

5

42
...
f′(x) = 1

1
f ′(x)

⇒ g′{f(x)} = 1 + x 5

g′ {f {g(x)}} = 1 + {g(x)} ⇒ g′(x) = 1 + {g(x)}5
5

3
n

1 + f (1) 1 + f (2)

n

43
...
(3)

1+1+1

1+ α + β

1 + α 2 + β2

1
= 1

1 + α 2 + β 2 1 + α 3 + β3

1+ α + β

1 + α 2 + β 2 1 + α 3 + β3 1 + α 4 + β 4

1 1 1 1
β × 1 α α2

1
α

1 α 2 β2

1 β

β2

1 1

1 1

1

1

α2 × 1 α α2

= 1 α

1 β

β2

1 1

1

= 1 α

α2

1 β

β2

1 β

β2

2

2
2
2
= (1 − α ) ( α − β ) ( β − 1)

∴

44
...
Then f4(x) − f6(x) equals :
k
1
1
1
(2)
(3)
(4)
3
4
12

44
...
18)

)

(19) VIDYALANKAR : JEE-MAIN 2014 : Paper and Solution

45
...
If p, q, r are in A
...
and

then the value of |α − β| is :
61
2 17
(2)
(1)
9
9

(3)

34
9

(4)

1 1
+ = 4,
α β

2 13
9

45
...
P
...
Then the events A and B are :
(1) mutually exclusive and independent
...

(3) independent but not equally likely
...


46
...
(3)

(

)

P( A ∪ B =

1
6

1 − P( A ∪ B) =

1
6

5
6
1
P (A ∩ B) =
4
1
P ( A) =
4
3
P (A) =
4
P(A ∪ B) =

… (1)
… (2)

… (3)

P(A ∪ B) = P(A) + P (B) − P (A ∩ B)

5 3
1
=
+ P(B) −
6 4
4

(Pg
...
A and B has different probability so if is not
equally likely
...
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)
47
...

F(0) = F(1)
So, according to Rolle’s theorem, there is at least are root between 0 and 1
...

f′(x) − 2g′(x) = 0
f′(c) − 2g′(c) = 0
f′ (c) = 2g′ (c)

48
...

dt
2
If p(0) = 100, then p(t) equals :
(2) 300 − 200 e−t/2
(1) 400 − 300 et/2
(3) 600 − 500 et/2
(4) 400 − 300 e−t/2
48
...

Using p(0) = 100, we get
k = −300
∴ the relation is
p(t) = 400 − 300 et/2

49
...
If T is the circle centred at (0, y), passing
through origin and touching the circle C externally, then the radius of T is equal to :
3
3
1
1
(3)
(4)
(1)
(2)
2
2
4
2
(Pg
...
(4)
Equation of C is
(x − 1)2 + (y − 1)2 = 1
…(1)
Also let (0, y) ≡ (0, k) then equation of T is
…(2)
x2 + (y − k)2 = k2
From the figure and equations (1) and (2) we get
(1 + k)2 = 12 + (1 − k)2

1
⇒ k=
4

y

(1, 1)

T

C

(0, k)

Hence radius of T is

O

1

...
The area of the region described by A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 − x} is
π 4
π 4
π 2
π 2
(2) −
(3) −
(4) +
(1) +
2 3
2 3
2 3
2 3
y

50
...
Let a, b, c and d be non−zero numbers
...
(3)

Solving the given lines we get the point of intersection P

⎛ bc − ad 4ad − 5bc ⎞
P⎜
,
⎟
2ab ⎠
⎝ ab
As point is equidistant from the axes
...
Let PS be the median of the triangle with vertices P(2, 2), Q(6, −1) and R (7, 3)
...
21)

JEE-MAIN 2014 : Paper and Solution (22)
52
...
lim

sin ( π cos 2 x)
x2

x→0

(1)

π
2

13
,1
2

R(7, 3)

is equal to :
(3) −π

(2) 1

53
...
If X = {4n − 3n − 1 : n ε N} and Y = {9(n − 1) : n ε N}, where N is the set of natural numbers,
then X ∪ Y is equal to :
(1) N
(2) Y − X
(3) X
(4) Y
54
...

y = 9(n − 1), n ∈ N
⇒ y = 0, 9, 18, …
⇒ x ∪ y = y
...
The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any
tangent to it is :
(2) (x2 − y2)2 = 6x2 − 2y2
(1) (x2 − y2)2 = 6x2 + 2y2
(4) (x2 + y2)2 = 6x2 − 2y2
(3) (x2 + y2)2 = 6x2 + 2y2
55
...
Three positive numbers form an increasing G
...
If the middle term in this G
...
is doubled, the
new numbers are in A
...
Then the common ratio of the G
...
is :
(1) 2 + 3
(2) 3 + 2
(3) 2 − 3
(4) 2 + 3
(Pg
...
(4)
Let the numbers be a, ar, ar2 is G
...

Given a, 2ar, ar2 are in A
...


a + ar 2
(a ≠ 0)
2
which gives r = 2 + 3 , as the G
...
is an increasing G
...


the

2ar =

57
...
(3)
2

⎛ 11 ⎞ ⎛ 11 ⎞
⎛ 11 ⎞
⎟ + 3 ⎜ ⎟ +
...
3 ⎜ ⎟ + 10 ⎜ ⎟
=
10
⎝ 10 ⎠
⎝ 10 ⎠
⎝ 10 ⎠

…(2)

(1) − (2) gives

−

k
10

⇒ −

k
10

⇒ −

k
10

2

⎛ 11
⎝ 10

9

10

⎞ ⎛ 11 ⎞
⎛ 11 ⎞
⎛ 11 ⎞
⎟ + ⎜ ⎟ +
...
⎛ 11 ⎞
=
⎜ ⎟
11
⎝ 10 ⎠
−1
10
= 1+ ⎜

10

10

⎛ 11 ⎞
⎛ 11 ⎞
⎟ − 10 − 10 ⎜ ⎟
⎝ 10 ⎠
⎝ 10 ⎠

= 10 ⋅ ⎜

⇒

k

= 100

58
...
(1)

⇒
⇒
⇒
⇒

+ m + n = 0;
= −(m + n)
2
m + n2 + 2mn = m2 + n2
mn = 0
m=0
or
n=0

Case I
m=0
+n=0
=k
⇒
m=0
n = −k
= 1/ 2
m=0
n = −1/ 2
⇒ cos θ =
⇒ θ

2

= m2 + n2

Case II
n=0
+m=0
=k
m = −k
n=0
= 1/ 2
m = −1/ 2
n=0

1
2

= π/3
...
23)

+ m + n = 0 and

JEE-MAIN 2014 : Paper and Solution (24)
2

59
...
( 1)

Now tangent with slope ‘m’ to
y2 = 4x is y = mx +

1
m

…(1)

Also tangent with slope ‘m’ to
x2 = − 32 y is y = mx + 8m2
…(2)
If (1) and (2) are equations of the same line then their coefficients should match
...
If x = −1 and x = 2 are extreme points of f(x) = α log |x| + βx2 + x then :
1
1
(1) α = −6, β =
(2) α = −6, β = −
2
2
1
1
(4) α = 2, β =
(3) α = 2, β = −
2
2
⇒ 8 m2 =

60
...
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
...
(3)
NO is paramagnetic in gaseous state and diamagnetic in solid state
...
If Z is a compressibility factor, van der Waals equation at low pressure can be written as :
Pb
Pb
RT
a
(2) Z = 1 +
(3) Z = 1 +
(4) Z = 1 −
(1) Z = 1 −
RT
RT
Pb
VRT
62
...
24)

(25) VIDYALANKAR : JEE-MAIN 2014 : Paper and Solution

63
...
(4)
Factual
64
...
2 M solution of an electrolyte is 50 Ω
...
4 S m−1
...
5 M solution of the same electrolyte is 280 Ω
...
5 M solution of the electrolyte in S m2 mol−1 is :
(2) 5 × 102
(3) 5 × 10−4
(4) 5 × 10−3
(1) 5 × 103
64
...
5

= 5 × 10−4 Sm2 mol−1

65
...
If ‘a’ is its edge length then which of the
following expressions is correct ?
3
3a
(1) r + + r − =
(4) r + + r − =
a (2) r + + r − = 3a (3) r + + r − = 3a
Cs
Cl
Cs
Cl
Cs
Cl
Cs
Cl
2
2
65
...
Consider separate solutions of 0
...
100 M Mg3(PO4)2(aq), 0
...
125 M Na3PO4(aq) at 25°C
...
125 M Na3PO4(aq) has the highest osmotic pressure
...
500 M C2H5OH(aq) has the highest osmotic pressure
...

(4) 0
...

66
...

67
...
(2)
H2O2 → O2 + 2H+ + 2e−
H2O2 + 2OH− → O2 + 2H2O + 2e−
68
...
(4)
Order of reactivity of alkyl halide towards SN-2 reaction is CH3X > 1° > 2° > 3°
69
...
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

(Pg
...
(4)
Factual
70
...
4 g of an organic compound was digested by Kjeldahl method
M
and the evolved ammonia was absorbed in 60 mL of
sulphuric acid
...
The percentage of nitrogen
required 20 mL of
10
in the compound is :
(1) 3 %
(2) 5 %
(3) 6 %
(4) 10 %
70
...
4 ⎜ 60 × × 2 − 20 × ⎟
10
10 ⎠
⎝
= 10%
% of Nitrogen =
1
...
The equivalent conductance of NaCl at concentration C and at infinite dilution are λC and λ∞,
respectively
...
(1)
Variation of λc for strong electrotype is given by λ c = λ ∞ − B C
1
72
...
(4)
Kp = K c ( RT )

Δn g

K p = K c ( RT )

−

1
2

⎛ 1⎞
Since Δn g = ⎜ − ⎟
⎝ 2⎠
LiAlH

PCl

Alc
...
In the reaction, CH3COOH ⎯⎯⎯⎯ A ⎯⎯⎯ B ⎯⎯⎯⎯→ C, the product C is :
(1) Ethylene
(2) Acetyl chloride (3) Acetaldehyde
(4) Acetylene

73
...
KOH
CH3 − C − OH ⎯⎯⎯⎯ CH3 CH2 OH ⎯⎯⎯ C2H5 C ⎯⎯⎯⎯→ CH2 = CH2
→
→
⎯

74
...


(Pg
...
(3)
OH

OH

ONa
+ CO2

1250
5 atm

COONa

H+

COOH

Ac2O
OCOCH3
COOH
(Aspirin)

75
...
(2)
CHCl3
R − NH 2 ⎯⎯⎯⎯ R − N C
→
alc
...

76
...

(2) it contains Cs+, I− and lattice I2 molecule
...

−
(4) it contains Cs+ and I3 ions
...
(4)
−
CsI3 → Cs + + I 3
77
...
(3)
Factual

(Pg
...
For which of the following molecule significant μ ≠ 0?

(a)

(b)

(c)

(d)

(1) Only (c)
(2) (c) and (d)
(3) Only (a)
(4) (a) and (b)
78
...

79
...

Initial concentration (A) Initial Concentration (B)
Initial rate of formation of C
(mol L−S−)
0
...
1 M
1
...
1 M
0
...
2 × 10−3
0
...
1 M
2
...
(2)
r = k [A]x [B]y
1
...
1)x [0
...
(I)
−3
x
y
1
...
1) [0
...
(II)
From (I) and (II)
y

⎛1⎞
1= ⎜ ⎟ ⇒ y = 0
⎝2⎠
Now from (I) and (III)

……
...
Which series of reactions correctly represents chemical relations related to iron and its
compound?
Cl2 ,heat
heat,air
Zn
(1) Fe ⎯⎯⎯⎯ FeCl3 ⎯⎯⎯⎯ FeCl2 ⎯⎯→ Fe
→
→
O , heat

CO,600°C

CO,700°C

2
→
→
→
(2) Fe ⎯⎯⎯⎯ Fe3O4 ⎯⎯⎯⎯⎯ FeO ⎯⎯⎯⎯⎯ Fe

dil H SO

H SO ,O

heat

2 4→
2 4 2→
→
(3) Fe ⎯⎯⎯⎯⎯ FeSO 4 ⎯⎯⎯⎯⎯ Fe 2 (SO 4 )3 ⎯⎯⎯ Fe

O ,heat

dil H SO

heat

2
2 4⎯
→
→
(4) Fe ⎯⎯⎯⎯ FeO ⎯⎯⎯⎯→ FeSO4 ⎯⎯⎯ Fe
80
...
28)

(29) VIDYALANKAR : JEE-MAIN 2014 : Paper and Solution

81
...
(3)
Less is the pKb value, more is the basic strength
...
medium is
(CH3)2NH > CH3NH2 > (CH3)3N > C6H5NH2
82
...
(3)

(4) Adenine

Quinoline is not present in DNA
...
The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is :
1
1
1
1
(2) 5, 0, 1 +
(3) 5, 0, 0, +
(4) 5,1, 0, +
(1) 5, 1, 1, +
2
2
2
2
83
...
C
...
The major organic compound formed by the reaction of 1, 1, 1 − trichloroethane with silver powder is :
(1) 2 − Butyne
(2) 2 − Butene
(3) Acetylene
(4) Ethene
84
...
Given below are the half − cell reaction :
Mn2+ + 2e− → Mn; E° = −1
...
51 V
...
33 V ; the reaction will not occur
(2) −0
...
69 V ; the reaction will not occur
(4) −2
...
(3)
0
0
0
Mn +2 + 2e− → Mn
E1 = −1
...
51V, ΔG 0 = −2FE 0
2
2
2
0
3Mn +2 → Mn + 2Mn +3 E3 = ?

0
0
, ΔG 3 = −2FE3

0
0
ΔG 3 = ΔG1 − ΔG 0
2
0
0
E3 = E1 − E 0 = −1
...
51 = −2
...
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4
...
(4)

n O2
n N2

=

WO2
M O2

×

M N2
WN2

1 28 7
= × =
4 32 32

(Pg
...
Which one is classified as a condensation polymer?
(1) Teflon
(2) Acrylonitrile
(3) Dacron

(4) Neoprene

87
...
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
88
...

89
...
47 kJ mol−1 at 25°C
...
314 kJ mol−1)
(1) −1460
...
50 kJ mol−1
−1
(3) −1366
...
95 kJ mol−1
89
...
47 kJ/mol
ΔH = ΔU + Δng
...
314 × 298
= −1364
...
95 kJ/mol
1000
90
...
(2)
PCC
R − CH 2 − OH ⎯⎯⎯ R − CHO
→

(Pg

---