Jee Main 2014

Upload
gouravsom 
Category
Documents

view
28 
download
2
Embed Size (px)
description
Transcript of Jee Main 2014

JEEMAINS 2014 Question Paper & Solutions
1
This booklet contains 28 printed pages. Test Booklet Code
PAPER 1 : PHYSIC, MATHEMATICS & CHEMISTY
Do not open this Test Booklet until you are asked to do so.
Read carefully the Instructions on the Back Cover of this Test Booklet.
Important Instructions : 1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball point Pen. Use of pencil
is strictly prohibited. 2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the
Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three parts in the questions paper A, B, C consisting of Physics, mathematics and Chemistry having
30 questions in each part of equation weightage. Each question is allotted 4 (four) marks for correct response. 6. Candidates will be awarded marks as stated above in instruction No. 5 correct response of each question 1/4
(one fourth) marks will be deducted for indicating incorrect response of each questions. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.
7. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.
8. Use Blue/ Black Ball Point Pen only for writing particulars/marking responses on Side 1 and Side 2 of the Answer Sheet. Use of Pencil is strictly prohibited.
9. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination hall/room.
10. Rough work is to be done on the space provided for this purpose in the Test Booklet Only. This space is given at the bottom of each page and in 3 page (Pages 2123) at the end of the booklet.
11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.
12. The CODE for this Booklet is H. Make sure that the CODE printed on Side 2 of the Answer Sheet is the same as that on this booklet. 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.
13. Do not fold or make any stray marks on the Answer Sheet.
H

JEEMAINS 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 Youngs modulus is 2 1011 N m2 and coefficient of thermal
expansion is 1.1 105 K1) (1) 72.2 10 Pa (2) 62.2 10 Pa (3)
82.2 10 Pa (4) 92.2 10 Pa Key. 3 Sol. P T = 11 52 10 1.1 10 100 = 82.2 10 Pa 2. A conductor lies along the zaxis at 1.5 z < 1.5 m and carries a fixed
current of 10.0 A in za direction (see figure). For a field 4 0.2x
yB 3.0 10 e a T,
find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 103s. Assume parallel motion along the xaxis.
(1) 14.58 W (2) 29.7 W (3) 1.57 W (4) 2.97 W
Key. 4
Sol. 1P iBl dxt
2 4 0.2x3 01 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.

JEEMAINS 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
TI e 1
1000V
TI 1 e
1000V
T1000I e . VT
1000 5 1 0.01300
1000V
Te 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 Boyles Law to the air column in the glass tube, P1V1 = P2V2 76 8 A = {76 (54 x)} . x.A x = 16 cm

JEEMAINS 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) Xrays (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) 6 107 C/m2 (4) 3 107 C/m2
Key. 3
Sol. Charge density oK A .EdQ d
A A
= Ko E 76 10 C/m2

JEEMAINS 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) GM 1 2 2R (2) 1 GM 1 2 22 R
(3) GMR
(4) GM2 2R
Key. (2) Sol. The centripetal force will be provided by the resultant force i.e.
R
2 2
2
GM 1 1 MvR 4 R2
v = 1 GM 1 2 22 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. 40 100 80 1000I 15 5 5 1 P VI220 220 220 220
= 11.36 A Minimum capacity of the main fuse = 12 A

JEEMAINS 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
OA
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 t2
= 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, 2A 3a A a2 1
A A
2 2
2
2 A a AA 3aA 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

JEEMAINS 2014 Question Paper & Solutions
7
12. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 103 Am1. 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 = 33 10 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 PN 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
2E o
1U E4
and avg
2
Bo
1 BU4
Hence avg avgE B
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 ee
(3) e1 e
(4) 1
RA CB
L

JEEMAINS 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) 5g6
(2) g
(3) 2g3
(4) g2
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
T mg
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

JEEMAINS 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 211 gt2
(1)
And let t2 be the time taken to reach the highest point
u = gt2 t2 = u/g
O
h
+y

JEEMAINS 2014 Question Paper & Solutions
10
Since t1 = nt2
From (1), H = unt2 2 21 gn t2
H = 2
22
u 1 uun gng 2 g
2 2 2 22gH 2nu n u nu n 2 (1)
19. A thin convex lens made from crown glass 32
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 11f 2 R R 2 R R
1 1 2 1 2
1 3 3 1 1 1 1 11f 2 4 R R 8 R R
2 1 2 1 2
1 3 3 1 1 1 1 11f 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 crosssection 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)

JEEMAINS 2014 Question Paper & Solutions
11
Sol. 0.92A 100 T 0.12A T 0 0.26A T 046 12 13
92 100 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 4.8caldt 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 12504
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 12
dd
is:
d 1
d2
(1) 1 tan
1 tan
(2) 1 sin1 cos
(3) 1 sin1 sin
(4) 1 cos1 cos
Key: (1) Sol. P0 + d1g(R R sin ) = P0 + d2g (R sin + R cos ) + d1g (R R cos)

JEEMAINS 2014 Question Paper & Solutions
12
From above
12
d sin cos 1 tand cos sin 1 tan
23. A green light is incident from the water to the airwater 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
c1sin
(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. 2hc 313.6 Z4
21Z

JEEMAINS 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. 5E 13.6 1 eV 1.88 eV36
25pr , p rqB 4.8 10qB
2pK.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 3xy .
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) 1 m3
(2) 1 m2
(3) 1 m6
(4) 2 m3
Key: (3) Sol. mgsin mg cos tan At x = 1; y = 1/6 27. When a rubberband 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 rubberband by L is:
(1) 2 3aL bL
2 3 (2)
2 31 aL bL2 2 3
(3) 2 3aL bL (4) 2 31 aL bL2
Key: (1)

JEEMAINS 2014 Question Paper & Solutions
14
Sol. 2 3L
0
aL bLW F dx2 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

JEEMAINS 2014 Question Paper & Solutions
15
30. Assume that an electric field 2E 30x i
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 0V V 80J
MATHEMATICS
31. The image of the line x 1 y 3 z 43 1 5
in the plane 2x y z 3 0 is the line
(1) x 3 y 5 z 23 1 5
(2) x 3 y 5 z 23 1 5
(3) x 3 y 5 z 23 1 5
(4) x 3 y 5 z 23 1 5
Key. (1)
Sol. Any point on the given x 1 y 3 z 43 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 42 1 1
6t 2 t 3 5t 4 32 2
6
x 3t 3, y t 5,z 2 5t which is the line
x 3 y 5 z 23 1 5
So (1) is correct. 32. If the coefficients of x3 and x4 in the expansion of 1821 ax bx 1 2x in
powers of x are both zero, then (a, b) is equal to
(1) 25116,3
(2) 25114,3
(3) 27214,3
(4) 27216,3
Key. (4)

JEEMAINS 2014 Question Paper & Solutions
16
Sol. 1821 ax bx 1 2x 2 3 42 18 18 18 181 2 3 41 ax bx 1 C 2x C . 2x C 2x C 2x .... Coefficient of x3 = 0 18 18 183 2 1C .8 a. C .4 b C 2 0 51a 3b 544 0 . (i) Coefficient of x4 = 0 18 18 184 3 2C .16 a C 8 b C 4 0 32a 3b 240 0 . (ii) Solving (i) & (ii), we get
A = 16, 272b3
So Answer (4)
33. If a R and the equation 2 23 x x 2 x x a 0 (where [x] denotes the greatest integer x) has no integral solution, then all possible values of a lie in the interval (1) 1,0 0,1 (2) 1,2 (3) 2, 1 (4) , 2 2,
Key. (1) Sol. x x x , since x I , so 0 x 1 Let x t
So 2 2
2 2 1 1 3a 1 3a3t 2t a 0, t as t 03 3
Now 0 < t < 1 20 a 1
a 1,0 0,1 Option (1) is correct. 34. If
2a b b c c a a b c then is equal to
(1) 2 (2) 3 (3) 0 (4) 1 Key. (4) Sol. a b b c c a
a b . b c c a a b . b c a c b c c a

JEEMAINS 2014 Question Paper & Solutions
17
a b . a b c c
2
a b c
1 Option (4) is correct. 35. The variance of first 50 even natural numbers is
(1) 8334
(2) 833 (3) 437 (4) 4374
Key. (2) Sol. The first 50 no. are 2,4,6,8,.....100 Mean
2 4 6 ....100X 5150
Variance
222 i1 x Xn
2 22 2 21 2 4 6 .... 100 5150 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
CdB
h
xO
d
oBOA 45 oCOA 30
o h 20tan tan 45x x

JEEMAINS 2014 Question Paper & Solutions
18
201 x 20x
Now o htan tan 30x d
1 2020 d3
20 d 20 3 d 20 3 1
So velocity
20 3 1 md 20 3 1 m / st 1sec
37. The integral 20
x x1 4sin 4sin dx2 2
(1) 4 (2) 2 4 4 33 (3) 4 3 4 (4) 4 3 4
3
Key. (4)
Sol. 20
x xI 1 4sin 4sin dx2 2
2
0
xI 2sin 1 dx2
0
xI 2sin 1 dx2
/3
0 /3
x xI 1 2sin dx 2sin 1 dx2 2
/3
/30
x xx 4cos 4cos x2 2
3 34 0 4 0 43 2 2 3
4 3 43
38. The statement ~ p ~ q is (1) equivalent to p q (2) equivalent to ~ p q
(3) a tautology (4) a fallacy

JEEMAINS 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 TT F F T T F T FF T T F T F T FF F T T F T F T
Clearly ~ p ~ q equivalent to p q 39. If A is an 3 3 nonsingular matrix such that AA' A'A and 1B A A' then
BB' (1) I B (2) I (3) 1B (4) 1B
Key. (2) Sol. Now '1 1BB' A A ' A A ' '1 1A A ' A. A 'AB B'A ' 11A A 'A A '
11A A A'. A ' I .I I
40. The integral 1xx11 x e dx
x
(1) 1xxx 1 e c
(2)
1xxx e c
(3)
1xxx 1 e c
(4)
1xxx e c
Key. (2)
Sol. 1xx1I 1 x e .dx
x
1 1x xx x1I e dx x e dx
x
1 1x xx x
2
1I e dx x. 1 e dxx
1 1x xx x
2
1I e dx x. 1 e dxx
1xx
2
11. 1 e dx dxx
1 1 1x x xx x xI e dx x.e e dx
2
1 1x t 1 dx dtx x
1xxI x.e c

JEEMAINS 2014 Question Paper & Solutions
20
41. If z is a complex number such that z 2 , then the minimum value of 1z2
:
(1) is equal to 52
(2) lies in the interval (1, 2)
(3) is strictly greater than 52
(4) is strictly greater than 32
but less
than 52
Key. (2) Sol.
O 2 +2 1/2 A B
1z
2 = Distance between z and 1
2
OA = minimum value of 1z2
= 32
.
42. If g is the inverse of a function f and 'f (x) = 51
1 x, then 'g (x) is equal to:
(1) 1 + x5 (2) 5x4 (3) 51
1 {g(x)} (4) 51 {g(x)}
KEY. (4) Sol. f (g(x)) = x 'f (g(x)) . 'g (x) = 1
'f (g(x)) = 1'g (x)
'g (x) = 1 + g5(x).

JEEMAINS 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 ff f ff f f
= 2 2(1 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 + 1 +1 + 1 + 1 +
1 + 1 + 1 +
=
2
2 2
1 1 111
= 2 2 2 Hence, K = 1.
44. Let fk (x) = 1k
(sink x + cosk x) where x R and k 1. Then f4(x) f6(x)
equals:
(1) 16
(2) 13
(3) 14
(4) 112
Key. (4)
Sol. fk(x) = k k1 sin x cos xk
f4(x) f6(x) = 4 4 6 61 1sin x cos x sin x cos x4 6
= 2 2 2 21 11 2sin x cos x 1 3sin x cos x4 6
= 112
.
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 is:
(1) 619
(2) 2 179
(3) 349
(4) 2 139
Key. (4)

JEEMAINS 2014 Question Paper & Solutions
22
Sol. Let , are the rots of px2 + qx + r = 0
1 1 4
q 4r
q = 4r
p, q, r are in A. P. p = 9r Hence the given equation becomes 9x2 + 4x 1 =0
16 4 52 2 1381 9 81 9

46. Let A and B be two events such that 1P(A B)6
, 1P(A B)4
and 1P(A)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(A B) P(A) P(B) P(A B)
5 3 1x6 4 4 where P(B) = x
1x3
P(A B) 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

JEEMAINS 2014 Question Paper & Solutions
23
h(0) = h(1) h(x) verify all conditions of Rolles 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 et/2 (3) 600 500 et/2 (4) 400
300 et/2 Key. (1) Sol. t /2I.F e So solution is t /2 t /2 t /2P t e 200.e dt 400e k P 0 100 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) 32
(2) 32
(3) 12
(4) 14
Key. (4)
Sol.
BA
O A = (0, y) B = (1, 1) Since AB 1 y
y = 14
Hence, 22 3

JEEMAINS 2014 Question Paper & Solutions
24
50. The area of the region described by A = {(x, y) : x2 + y2 1 and y2 1 x} is:
(1) 42 3 (2) 4
2 3 (3) 2
2 3 (4) 2
2 3
Key. (1)
Sol.
Required area =
1 2
02 ( 1 x 1 x )dx
1
21 3/2
0
x 1 x 1 22 sin x (1 x)2 2 3
2 424 3 2 3
51. Let a, b, c and d be nonzero numbers. If the point of intersection of the
lines 4 2 0ax ay c and 5 2 0bx by d lies in the fourth quadrant and is equidistant from the two axes then:
(1) 2 3 0bc ad (2) 2 3 0bc ad (3) 3 2 0bc ad (4) 3 2 0bc ad Key: (3) Sol. Point (k, k) satisfies both lines 4 2 0ak ak c . . . . . (1) 5 2 0bk bk d . . . . . (2) c = 2ak and d = 3bk
2 3c 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) 4 7 11 0x y (2) 2 9 7 0x y (3) 4 7 3 0x y (4) 2 9 11 0x y Key: (2)

JEEMAINS 2014 Question Paper & Solutions
25
Sol.
1 213 922
psm
21 19
y x
9 2 7 0y x
53. 2
20
sin( cos )limx
xx
is equal to :
(1) 2 (2) 1 (3) (4)
Key : (4) Sol.
2
20
sin coslimx
xx
=
2 2
220
sin sin sinlimsinx
x xxx
= 54. If 4 3 1 :nX n n N and 9 1 :Y n 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. 4 3 1n n = 1 3 3 1n n = 2 32 33 3 ........
n nC C = 2 39 3 .......
n nC C
Set X contains numbers which are multiples of 9 but not all multiples of 9 XY = Y

JEEMAINS 2014 Question Paper & Solutions
26
55. The locus of the foot of perpendicular drawn from the centre of the ellipse 2 23 6x y on any tangent to it is :
(1) 22 2 2 26 2x y x y (2) 22 2 2 26 2x y x y (3) 22 2 2 26 2x y x y (4) 22 2 2 26 2x y x y Key : (3) Sol. 26 2K mh m . . . . . (1)
1K mh
. . . . . . (2)
Eliminating m from (1) and (2) Locus is 22 2 2 26 2x y x y 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 arr (r > 1)
r 2 3 but as r > 1 r 2 3
57. If 9 1 8 2 7 9 910 2 11 10 3 11 10 .... 10 11 10k , then k is equal to :
(1) 12110
(2) 441100
(3) 100 (4) 110
Key : (3) Sol. S = 109 + 2(11) (10)8 + . . . . . . + 10(11)9
11S10
= (11.108) + . . . . . + . . . + (11)10

JEEMAINS 2014 Question Paper & Solutions
27
109
10
1110 110S (11)
1110 110
S = 1011 = (100) 109 K = 100
58. The angle between the lines whose direction cosines satisfy the equations
0l m n and 2 2 2l m n 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
1 1cos22 2
= 3
59. The slope of the line touching both the parabolas 2 4y x and 2 32x y is :
(1) 12
(2) 32
(3) 18
(4) 23
Key : (1) Sol. Eq. of tangent at t & t respectively are 2yt x t and 2' 8 'xt y t
Comparing 2
2
1 2' 1 8 '
t t tt t
Hence slope = 12

JEEMAINS 2014 Question Paper & Solutions
28
60. If 1x and 2x are extreme points of 2( ) log  f x x x x then :
(1) 16,2
(2) 16,2
(3) 12,2
(4) 12,2
Key: (3) Sol. Differentiating
2' 2 1 0 2 0f x x x xx
Put 2x and 1 gives 2 1 0 and 8 2 0
Solving 12,2
CHEMISTRY
61. Which one of the following properties is not shown by NO? (1) It combines with oxygen to form nitrogen dioxide (2) Its 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) PbZ 1RT
(2) PbZ 1RT
(3) RTZ 1PB
(4) aZ 1VRT
Key (4) Sol. As pressure is low so volume must be large. i.e., V b V So Van der Waals equation reduces into
aPV RTV
aZ 1RTV

JEEMAINS 2014 Question Paper & Solutions
29
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 2Ca
64. Resistance of 0.2 M solution of an electrolyte is 50 . The specific
conductance of the solution is 1.4 S m1. 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 mol1 is :
(1) 35 10 (2) 25 10 (3) 45 10 (4) 35 10 Key. 3
Sol. 1 1.R / A
1 11.4.50 / A
1 1.280 / A
Solving we get 10.25Sm
C
4 2 10.25 5 10 Sm mol0.5 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
3r r a2
(2) Cs Cl
r r 3a
(3) Cs Cl
r r 3a (4) Cs Cl3ar r2

JEEMAINS 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 2 Cs 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 2H O 2H 2e 2H O
(b) 2 2 2H O 2e O 2H
(c) 2 2H O 2e 2OH (d) 2 2 2 2H O 2OH 2e O 2H O
(1) (a), (c) (2) (b), (d) (3) (a), (b) (4) (c), (d) Key. (2)
Sol. 1 0
22 2H O O 2e
68. In SN2 reactions, the correct order of reactivity for the following
compounds: 3 3 2 3 32 3CH Cl,CH CH Cl, CH CHCl and CH CCl (1) 3 2 3 3 32 3CH CH Cl CH Cl CH CHCl CH CCl (2) 3 3 2 3 32 3CH CHCl CH CH Cl CH Cl CH CCl (3) 3 3 3 2 32 3CH Cl CH CHCl CH CH Cl CH CCl (4) 3 3 2 3 32 3CH Cl CH CH Cl CH CHCl CH CCl Key. (4) Sol. In NS 2 reactivity order is
o o o3CH X 1 R X 2 R X 3 R X

JEEMAINS 2014 Question Paper & Solutions
31
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 blue1& Energy
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 M10
sulphuric acid. The unreacted acid required 20 mL of M10
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 3Eq.H SO Eq.NH Eq.NaOH
360 1 20 12 Eq.NH
1000 10 1000 10
3100Eq.NH 0.01
10000
Mole 3NH 0.01
Weight percent 20.01 14of N 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

JEEMAINS 2014 Question Paper & Solutions
32
72. For the reaction 2(g ) 2(g) 3(g)1SO O SO2
, if KP = KC(RT)x where the symbols
have usual meaning, then the value of x is (assuming ideality)
(1) 12
(2) 1 (3) 1 (4) 12
Key. (4) Sol. gnP CK K (RT)
x or g1n2
73. In the reaction, 54 PClLiAlH Alc. KOH3CH COOH A B C , the product C is (1) Ethylene (2) Acetyl chloride (3) Acetaldehyde (4) Acetylene Key. (1)
Sol.
5PClLAH3 3 2 3 2
(B)(A)CH COOH CH CH OH CH CH Cl
2 2(C)
CH CH
alc. KOH
74. Sodium phenoxide when heated with CO2 under pressure at 125C yields a
product which on acetylation produces C.
+ 1252 5 Atm
2
HCO B CAc O
The product C would be
(1) (2)
(3) (4)
Key. (3)
ONa
OH
COOCH3
O COCH3
COOHO COCH3
COOHOH
COCH3
COCH3

JEEMAINS 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 OHRCH 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 3I
ions Key. (4) Sol. 3 3CsI Cs I
77. The equation which is balanced and represents the correct product(s) is (1) excess NaOH2 4 22 6 2[Mg(H O) ] (EDTA) [Mg(EDTA) 6H O
(2) 4 2 4 2 4CuSO 4KCN K [Cu(CN) ] K SO (3) 2 2Li O 2KCl 2LiCl K O (4) 23 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
1255 atm
COONa
OH
H+
5 atm
COOH
OH
COOH
OCOCH3(AC) O2
Acetylation
Aspirin
Cl
Cl
CN
CN
OH
OH
SH
SH

JEEMAINS 2014 Question Paper & Solutions
34
Key. (2) Sol.
79. For the nonstoichiometre reaction 2A B C D , 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 LS) 0.1 M 0.1 M 0.2 M
0.1 M 0.2 M 0.1 M
1.2 103 1.2 103 2.4 103
The rate law for formation of C is
(A) 2dc k[A][B]dt
(B) dc k[A]dt
(C) dc k[A][B]dt
(D) 2dc k[A] [B]dt
Key. (2) Sol. Rate reaction,
n md(c) A Bdt
=
m 3
3
0.1 1.2 100.2 1.2 10
m = 0
n 3
3
0.1 1.2 100.2 2.4 10
n1 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) 2Cl , heat heat , air Zn3 2Fe FeCl FeCl Fe (B) 2O , heat CO, 600 C CO, 700 C3 4Fe Fe O FeO Fe
(C) 2 4 2 4 2dil. H SO H SO , O heat4 2 4 3Fe FeSO Fe (SO ) Fe (D) 2 2 4O , heat dil. H SO heat4Fe FeO FeSO Fe
N

JEEMAINS 2014 Question Paper & Solutions
35
Key. (2)
Sol. 3 4 22 3 2
Fe O 4CO 3Fe 4CO500 800 K
Fe O CO 2FeO CO
2FeO CO Fe CO 900 1500 K 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, + 12
(2) 5, 0, 1, + 12
(3) 5, 0, 0, + 12
(4) 5, 1, 0, + 12
Key . (3) Sol. Configuration : Ar 4s2 3d10 4p6 5s1
Its quantum numbers n, l, m and s will be 5, 0, 0, + 12
respectively.
84. The major organic compound formed by the reaction of 1, 1, 1 trichloroethane with silver powder is (1) 2butyne (2) 2 butane (3) acetylene (4) ethane
Key . (1) Sol.
CH3
Cl
Cl
Cl + 6Ag + ClCl
Cl
CH3  6AgClCH3 C C CH3
2butyne

JEEMAINS 2014 Question Paper & Solutions
36
85. Given below are the halfcell 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 22 2 2
O O O
N N N
n w / M x / 32 28 7 :32n 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 terephthalic 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

JEEMAINS 2014 Question Paper & Solutions
37
Key. (1)
Sol. 7 5 3 1
4 3 2HClO 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 2ClO 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 mol1 at 250C. Assuming ideality the Enthalpy of combustion, CH, for the reaction will be: (R = 8.314 kJ mol1) (1) 1460.50kJ mol1 (2) 1350.50 kJ mol1 (3) 1366.95 kJ mol1 (4) 1361.95 kJ mol1
Key . (3) Sol. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
H = E + nRT qp = qv + nRT
= 1 8.31 2981364.471000
= 1364.47 2.47638 =  1366.95 kJ mol1
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 PCCO
R C H , other oxidizing agents may convest RCHO to RCOOH.

JEEMAINS 2014 Question Paper & Solutions
38
IITJEE MAINS2014 Answer Key (CODEH)
PHYSICS MATHEMATICS CHEMISTRY
Q.N. Code H Q.N. CodeH Q.N. CodeH 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