Jee main 2014 qpaper key solutions

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2014 JEE-MAIN Q. PAPER WITH SOLUTIONS

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2

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 question paper A, B, C consisting of Mathematics, Physics and Chemistry having

30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct

response.

6. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question

1/4 (one fourth) marks will be deducted for indicating incorrect response of each question. 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 each 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 pages (Pages 21 – 23) 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 G. 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 mark on the Answer Sheet.



3

MATHEMATICS

1. 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 : x 1 y 3 z 4 is parallel to 2x y z 3 03 1 5

Image h 1 k 3 l 4 22 1 1

h, k, l 3, 5, 2

Required equation x 3 y 5 z 23 1 5

2. If the coefficients of 3 4x and x 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



4

Key : 4

Sol : Equating coefficients of 3 4x and x is zero 32 17 17a b 03

1615 – 32 a + 3b = 0

Solving (a, b) = 27216,3

3. 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 : 2 23f 2f a 0 where x x

21 1 3af 1

3

1 a 1; a 0

a 1, 0 0, 1

4. If 2

a b b c c a a b c then is equal to:

1) 2

2) 3

3) 0

4) 1



5

Key : 4

Sol : 2

a b b c c a a b c

5. The variance of first 50 even natural numbers is :

1) 8334

2) 833

3) 437

4) 4374

Key : 2

Sol : Variance of first ‘n’ even natural numbers is equal to 2n 13

250 1 8333

6. 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 045 . It flies off horizontally straight away from the

point ‘O’. After one second, the elevation of the bird from ‘O’ is reduced to 030 .

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



6

Key : 4

Sol :

P Q

A B C

450

300

0A 45 AB PB 20 mts and CQ 20

0 20Tan30 BC 20 3 1 PQ20 BC

dis tan cespeed 20 3 1 mts

time

7. The integral 2

0

x x1 4sin 4sin dx2 2

equals:

1) 4

2) 2 4 4 33

3) 4 3 4

4) 4 3 43

Key : 4

Sol : 3

0 03

x x x1 2sin dx 1 2sin dx 2sin 1 dx2 2 2

3

03

x xx 4cos 4cos x2 2

4 3 43



7

8. The statement ~ p ~ q is :

1) equivalent to p q

2) equivalent to ~ p q

3) a tautology

4) a fallacy

Key : 1

Sol : The statement ~ p ~ q is

P q ~q p ~ q ~ p ~ q p q

T T F F T T

T F T T F F

F T F T F F

F F T F T T

~ p ~ q is equivalent to p q

9. If A is an 3 3 non-singular matrix such that 1AA' A 'A and B A A', then BB' equals :

1) I + B

2) I

3) 1B

4) 1B

Key : 2

Sol : 11 1 1 1BB' A A A A

11 1 1A A A A I. I I



8

10. The integral 1xx11 x e dx

x is equal to :

1) 1xxx 1 e c

2) 1xxxe c

3) 1xxx 1 e c

4) 1xxxe c

Key : 2

Sol : 1xx11 x e dx

x

1xx

2

11 x 1 e dxx

1 1x xx x

2

1e dx x 1 e dxx

By parts = 1xxxe c

11. 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 53

4) is strictly greater than 32

but less than 52



9

Key: 2

Sol : z r cis then | z | r 2

2 2

2 21 1z r cos r sin2 2

2 1r r cos4

17 9 252cos which lies in ,4 4 4

1 3 5z ,2 2 2

Minimum value 32

lies in (1, 2)

12. If ‘g’ is the inverse of a function ‘f’ and 5

1f ' x , then g ' x1 x

is equal to:

1) 51 x

2) 45x

3) 5

11 g x

4) 51 g x

Key : 4

Sol : If ‘g’ is the inverse of a function ‘f’, then x f g x f ' g x g ' x 1

51g ' x 1 g xf ' g x



10

13. If

n n

3 1 f 1 1 f 2, 0, and f n and 1 f 1 1 f 2 1 f 3

1 f 2 1 f 3 1 f 4

=

2 2 2K 1 1 , then K is equal to :

1)

2) 1

3) 1

4) -1

Key : 3

Sol : Given determinant = 2

2 2 2

1 1 1 1 1 11 . 11 1

K 1

14. Let k kk

1f x sin x cos xk

where x R and k 1 . Then 4 6f x f x equals:

1) 16

2) 13

3) 14

4) 112

Key : 4

Sol : 4 6f x f x = 2 2 2 21 1[1 2sin x cos x] 1 3sin x cos x4 6



11

15. Let and be the roots of equation 2px 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

Sol : qp

r.p

2q p q

By given condition 1 1 4

q 4r

2 4

16. Let A and B be two events such that

1 1 1P A B , P A B and P A , where A6 4 4

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



12

Key : 3

Sol : 1 5P A B 16 6

1 5P A P B

4 6

3 1 5P B

4 4 6

5 1 5 3 1P B

6 2 6 3

3 1 1P A .P B P A B

4 3 4

A, B are independent but not equally likely P A P B

17. 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 :

f ' c f 1 f 0

for c 0, 1g ' c g 1 g 0

18. Let the population of rabbits surviving at a time ‘t’ be governed by the

differentiable equation dp t 1 p t 200

dt 2 . If p(0) = 100, then p(t) equals :

1) t /2400 300 e

2) t /2300 200 e

3) t /2600 500 e

4) t /2400 300 e



13

Key : 1

Sol : dp 1 p t 200dt 2

This is linear d.e in ‘p’1 1t t2 2p.e 400e c

By given condition p(0) = 100, c = -300

1 t2p 400 300e

19. Let ‘C’ be the circle with centre at (1, 1) and radius = 1. 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 :

1) 32

2) 32

3) 12

4) 14

Key : 4

Sol :

1 2c 1, 1 c 0, y 1 2r 1, r y 1 2 1 2c c r r

2 21 y 1 1 y 1= 4y



14

20. The area of the region described by 2 2 2A x, y :x y 1 and y 1 x is

1) 42 3

2) 42 3

3) 22 3

4) 22 3

Key : 1 Sol.

Required area = 1

0

42 1 x dx2 2 3

21. Let a, b, c and d be non-zero numbers. If the point of intersection of the line

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 of intersection 2ad 2bc 5bc 4ad,2ab 2ab

Equidistance from the both axis in

fourth quadrant 3bc 2ad 0



15

22. 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 7y 11 0 2) 2x 9y 7 0 3) 4x 7y 3 0 4) 2x 9y 11 0 Key : 2

Sol : Mid point 13 ,12

Slope PS 29

23. 2

2x 0

sin cos xlim

x

is equal to :

1) 2

2) 1

3)

4)

Key : 4

Sol : 2

2x 0

sin sin xlim

x

2 2

2 2x 0

sin sin x sin xlim 1 1sin x x

24. If nX 4 3n 1: n N and Y 9 n 1 :n N , where N is the set of natural

numbers, then X Y is equal to:

1) N

2) Y – X

3) X

4) Y



16

Key : 4

Sol : X and Y are multiplies of 9 and X Y

X Y Y

25. The locus of the foot of perpendicular drawn from the centre of the ellipse 2 2x 3y 6 on any tangent to it is :

1) 22 2 2 2x y 6x 2y

2) 22 2 2 2x y 6x 2y

3) 22 2 2 2x y 6x 2y

4) 22 2 2 2x y 6x 2y

Key : 3

Sol. Given that ellipse 2 2x y 1

6 2

C

P

A

B

Slope CP = 1

1

yx

Slope of AB = 1

1

xy

= m

Equation of the tangent is 2y mx 6m 2

By above equations we get 22 2 2 2x y 6x 2y



17

26. Three positive numbers from 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 : 2a,ar,ar G.P .

2a ar2ar2

24ar a ar

2r 4r 1 0

4 16 4r

2

r 2 3

r 2 3

27. If 9 1 8 2 7 9 910 2 11 10 3 11 10 ..... 10 11 k 10 , then ‘k’ is equal to

1) 12110

2) 441100

3) 100

4) 110



18

Key : 3

Sol : 9 1 8 2 7 9S 10 2 11 10 3 11 10 .....10 11 ……… (1)

2 3 6 108 711S 11.10 2 11 10 3 11 10 .....9 11 1110

……… (2)

(2) ... (1)

29 8 7 9 101 S 10 11.10 11 10 .... 11 1110

9S 100 10

K 100 28. The angle between the lines whose direction cosines satisfy the equations

2 2 2m n 0 and m nl l is :

1) 3

2) 4

3) 6

4) 2

Key : 1

Sol : 2 2 2l m n 0

l m n

2 2 2m n m n 0 2mn = 0 m = 0 (or) n = 0 If m = 0 l = -(0 + n) l = -n l : m : n = -n : 0 : n = -1 : 0 : 1 similarly if n = 0 then

l : m : n = -1, : 1 : 0 1 0 0cos2 2

3



19

29. The slope of the line touching both the parabolas 2 2y 4x and x 32y is :

1) 12

2) 32

3) 18

4) 23

Key : 1

Sol : Common tangent of 2/32 2 1/3 1/3y 4ax and x 4by is a x b y ab 0

Here 4a = 4 and 4b = -32

Slope of the common tangent = 1/2

30. If x = -1 and x = 2 are extreme points of 2f x log | x | x x then:

1) 16,2

2) 16,2

3) 12,2

4) 12,2

Key : 3

Sol : f ' 1 0 and f ' 2 0

Solving above equations we get 12,2



20

PHYSICS 31. When a rubber-band is stretched by a distance x , it exerts a restoring force of

magnitude 2F ax bx where a and b are constants. The work done in stretching

the unstretched rubber-band by L is:

1) 2 31

2 2 3aL bL

2) 2 3aL bL

3) 2 312

aL bL

4) 2 3

2 3aL bL

Key: 4

Sol: 2 3

2

0 2 3

L aL bLw Fdx ax bx dx

32. The coercivity of a small magnet where the ferromagnet gets demagnetized is3 13 10 Am . 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) 6A

2) 30 mA

3) 60 mA

4) 3 A

Key: 4

Sol: 3100 I=3 100.1

nI H

3I A



21

33. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 Wand 1heater 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) 14A

2) 8A

3) 10A

4) 12A

Key: 4

Sol: 1 40 15 100 5 80 5 1000220

P IV I

2500 11.4 12A220

A

34. 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 46cm. What will be length of the air column above mercury in the tube now?

(Atmospheric pressure = 76 cm of Hg)

1) 6 cm

2) 16 cm

3) 22 cm

4) 38 cm

Key: 2

Sol: 76 8 22x x

By trail and error method 16x



22

35. A bob of mass m attached to an inextensible string of length l is suspended 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 both in direction and magnitude

2) angular momentum is conversed

3) angular momentum changes in magnitude but not in direction

4) angular momentum changes in direction but not in magnitude

Key: 4

Sol: Angular Momentum Changes in directions

36. The current voltage relation of diode is given by 1000 / 1 ,V TI e mA where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring 0.01V while measuring the current of 5mA at 300 K, what will be the error in the value of current in mA?

1) 0.05 Ma

2) 0.2 mA

3) 0.02 mA

4) 0.5 mA

Key: 2

Sol: I+1 = 1000 /V Te

1000Vln I 1T

1000 VI+1 T

I

31000I= 6 10 0.013

= 0.2 mA



23

37. 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) 22gH n u

2) 2 22gH n u

3) 2 22gH n u

4) 22 2gH nu n

Key: 4

Sol:

H

u

Let 1t is the time taken to reach max height 1utg

Let 2t is the time taken to reach ground from tower equation

22 2

12

H ut gt 2

22 4 8

2u u gH

tg

2

2

2u u gHg g

Given that

2 1t nt

2

2

2u u gH nug g g

,

2 2

2 2

2 1u gH ung g

222 1 1gH u n

2 22 2gH u n n

22 2gH u n n



24

38. A thin convex lens made from crown glass 32

has focal length f . When it is

measured in two different liquids having refractive indices 43

and 53

, it has the

focal lengths 1f and 2f respectively. The correct relation between the focal lengths is:

1) 1f and 2f both become negative

2) 1 2f f f

3) 1f f and 2f becomes negative

4) 2f f and 1f becomes negative

Key: 3

Sol: From 1 21f R

1 1f R 3 2/

f R

For medium 4 3/

3

1 2 12 14 3 4f / R R

1 4f R

for medium 2

1 3 2 25 3 15 3

/M /f / R

15R

2 5f R



25

39. 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 43 10 /V m , the charge density of the positive plate will be close to:

1) 4 26 10 /C m

2) 7 26 10 /C m

3) 7 23 10 /C m

4) 4 23 10 /C m

Key: 2

Sol: 4

0

3 10K

12 42.2 8.8 10 3 10

76 10

40. 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:

L

R

B

A C

1) 1 ee

2) 1

ee

3) 1

4) 1

Key: 4

Sol: 0R LV V always for 0t



26

41. 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 30makes the two beams appear equally bright. If the initial intensities of two beams are AI and BI respectively, then A

B

II

equals:

1) 13

2) 3

3) 32

4) 1

Key: 1 Sol: 1 1 2 230 60A B A BI I I cos I cos

2

2

cos 60 1cos 30 3

A

B

II

42. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities 1d and 2d are filled in the tube. Each liquid subtends 90 angle at centre.

Radius joining their interface makes an angle with vertical. Ratio 1

2

dd

is:

1d

2d

1) 1 sin1 cos

2) 1 sin1 sin

3) 1 cos1 cos

4) 1 tan1 tan



27

Key: 4

Sol: A BP P

1 1 2 2h d g h d g

1 2cos sin cos sinR d R d

1

2

cos sin 1 tancos sin 1 tan

dd

A B

R2h

2d

1d

1h

R

43. 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 11 22 10 N m and coefficient of thermal expansion is 5 11.1 10 K )

1) 62.2 10 Pa

2) 82.2 10 Pa

3) 92.2 10 Pa

4) 72.2 10 Pa

Key: 2

Sol: Pressure 11 5 8α 2 10 1.1 10 100 2.2 10y



28

44. A block of mass m is placed on a surface with a vertical cross section given by 3

6xy . 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) 12

m

2) 16

m

3) 23

m

4) 13

m

Key: 2

Sol: At the instant of just sliding

H=?

tan

0.5 dy

dx

2

12x x

1' ' 16

H y at x m



29

45. Three rods of Copper, Brass and Steel are welded together to form a Y – shaped structure. Area of cross – section of each rod 24cm . End of copper rod is maintained at 100 C where as ends of brass and steel are kept at 0 C . 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 conductivites 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) 6.0 cal/s

2) 1.2 cal/s

3) 2.4 cal/s

4) 4.8 cal/s

Key: 4

Sol: Let T is the junction temperature then

cu Brass steelH H H

100cu B S

cu B S

K A T K A T K ATl l l

, 0 96 0 26 0 12100

46 13 12. . .T T T

200 5T

40T C

Heat flow per sec through copper

100KA Tl

0 96 4 60

46.

4 8. cal/sec

0 C0 C

100 C

Brass

Copp

er

Steell=13 cmk=0.26l=12 cm

k=0.12

l=46

k=0.

96

T



30

46. 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?

m

R

m

1) g

2) 23g

3) 2g

4) 56g

Key: 3

Sol: mg T ma

2 2. aTR mR mR T maR

2ma mg or / 2a g



31

47. 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) Ultra violet rays (iv) Absorbed by the ozone layer

of the atmosphere

(a) (b) (c) (d)

1) (i) (ii) (iii) (iv)

2) (iv) (iii) (ii) (i)

3) (i) (ii) (iv) (iii)

4) (iii) (ii) (i) (iv)

Key: 1

Sol: Conceptual

48. 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 43 10 T . If the radius of largest circular path followed by these

electrons is 10.0 mm, the work function of the metal is close to:

1) 1.6 eV

2) 1.8 eV

3) 1.1 eV

4) 0.8 eV



32

Key: 3

Sol: In magnetic field , 2 KE mr

Bq

28 19 42 2 2

31 19

9 10 1.6 10 10 12 2 9 10 1.6 10

B q rKEm

0.8ev

1 113.6 1.894 9

E eV

0 1.89 0.8 1.1W E KE eV

49. During the propagation of electromagnetic waves in a medium:

1) Both electric and magnetic energy densities are zero

2) Electric energy density is double of the magnetic energy density

3) Electric energy density is half of the magnetic energy density

4) Electric energy density is equal to the magnetic energy density

Key: 4

Sol: Conceptual 50. A green light is incident from the water the air – water interface at the critical

angle . Select the correct statement.

1) The entire spectrum of visible light will come out of the water at various angles

to the normal

2) The entire spectrum of visible light will come out of the water at an angle of

90 to the normal

3) The spectrum of visible light whose frequency is less than that of green light

will come out to the air medium

4) The spectrum of visible light whose frequency is more than that of green light

will come out to the air medium



33

Key: 3 Sol: With increase in frequency, critical angle c decreases i.e., present angle of

incidence in water will be more than respective critical angles of high frequencies

radiation. Hence high frequency radiations will get TIR and low frequencies

radiations will come out in to air

51. 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) 1 1 2 22

GMR

2) GMR

3) 2 2 GMR

4) 1 2 2GMR

Key: 1

Sol:

V

R

R 2

2

22 2

mv 1 1 122 42

GmR R R

2 1v 2 2 14

GmR

1v 2 2 12

GmR



34

52. 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) time period of oscillations is 6

2) amplitude of motion is 3a

3) time period of oscillations is 8

4) amplitude of motion is 4a

Key: 1

Sol:

a a a

0 26T

53. A conductor lies along the z – axis at 1.5 1.5z m and carries a fixed current of

10.0 A in ˆza direction (see figure). For a field 4 0.2 ˆ3.0 10 xyB e a T

, find the power

required to move the conductor at constant speed to 2.0x m , 0y m in 35 10 s .

Assume parallel motion along the x axis.

x

yB

I

2.0

1.5

1.5

1) 29.7 W

2) 1.57 W

3) 2.97 W

4) 14.85 W



35

Key: 3

Sol: 4 0.2Instant 3

23 10 35 10

xP B V e

1 0.212 105

xP e

2.97range

pdxP W

dx

54. The forward biased diode connection is:

1) 2V 2V

2)

2V2V

3) 3V3V

4) 2V 4V

Key: 2

Sol: Conceptual

55. Hydrogen 1

1 H , Deuterium 21 H , singly ionized Helium 4

2 He and doubly

ionized lithium 63 Li

all have one electron around the nucleus. Consider an

electron transition from 2n to 1n . If the wavelengths of emitted radiation are

1 2 3, , and 4 respectively then approximately which one of the following is

correct?

1) 1 2 3 42 3 4

2) 1 2 3 44 2 2

3) 1 2 3 42 2

4) 1 2 3 44 9



36

Key: 4

Sol: 22

1 1 114

Rzz

1 2 as 1z for both 11H and 2

1H

231

1 323 1

4 4zz

2

1 41 42

4 1

9 9zz

1 2 3 44 9

56. On heating water, bubbles being formed at the bottom of the vessel detach 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 detach is: (density of water is w )

R

2r

1) 2 3 wgRT

2) 2

3wgRT

3) 2

6wgRT

4) 2 wgRT



37

Key: 3(This is an ambigious Question, so expected key is 3) Sol: sinTdl buoyancy force on bubble

Rr r

2 rT rR

343 wR g

4

2 23

wR grT

2 23

wgr RT

57. 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) 4

2) 12

3) 8

4) 6

Key: 4

Sol: From 2

340 1004 4 85 10Vn HZ

l

Possible frequencies in closed pipe is odd harmonics only So

100 300 500 700 900HZ , HZ , HZ , HZ , HZ and 1100 1300 1500HZ , HZ , HZ..... etc below

1250HZ , 6 natural oscillations are possible



38

58. Assume that an electric field 2ˆ30E x i

exists in space. Then the potential

difference A OV V , where OV is the potential at the origin and AV the potential at

2x m is:

1) 80 J

2) 120 J

3) – 120 J

4)– 80 J

Key: 4

Sol: 0

22

00 0

30 80AV A

AV

dv Edx V V x dx J

59. A student measured the length of a rod and wrote it as 3.50 cm. Which

instrument did he use to measure it?

1) A screw guage having 50 divisions in the circular scale and pitch as 1mm

2) A meter scale

3) A vernier calipers where the 10 divisions in vernier scale matches with 9

division in main scale has 10 divisions in 1cm

4) A screw guage having 100 divisions in the circular scale and pitch as 1mm



39

Key: 3

Sol: Least count of vernier

1

10mm

60. 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:

P

V

A 400K C

B 800 K

600 K

1) The change in internal energy in the process BC is 500R

2) The change in internal energy in whole cyclic process is 250 R

3) The change in internal energy in the process CA is 700 R

4) The change in internal energy in the process AB is 350R

Key: 1

Sol: 1) du for isochoric process VnC dT 31 400 6002R R

2) for Process BC 1 2 200500

1 1 7 / 5nR T T R

du Rr

3) for whole process 0du



40

CHEMSITRY 61. Which one is classified as a condensation polymer?

1) Acrylonitrile

2) Dacron

3) Neoprene

4) Teflon

Key: 2

Sol: Dacron is the condensation polymer

nNeopreneAdd polymers

Teflon

62. Which one of the following properties is not shown by NO?

1) It’s bond order is 2.5

2) It is diamagnetic in gaseous state

3) It is a neutral oxide

4) It combines with oxygen to form nitrogen dioxide

Key: 2

Sol: NO is paramagnetic



41

63. Sodium phenoxide when heated with 2CO under pressure at 0125 C yields a product

which on acetylation produces C.

2CO

01255Atm

2

HAC OB C

ONa

The major product C would be:

1) 3OCOCH

COOH

2) 3OCOCH

COOH

3) OH

3COCH

3COCH

4) OH

3COOCH

Key: 2



42

Sol:

ONa0125 C

5atm2CO

OHCOOH

HAC O2

COOH

3OCOCH

(B) (C)Aspirin

64. Given below are the half-cell reactions:

2 02 ; 1.18Mn e Mn E V

3 2 02 ; 1.51Mn e Mn E V

The 0E for 2 33 2Mn Mn Mn will be:

1) - 0.33 V; the reaction will occur

2) – 2.69 V; the reaction will not occur

3) – 2.69 V; the reaction will occur

4) – 0.33 V; the reaction will not occur

Key: 2

Sol: 0 1.18 1.51E

= –2.69 V



43

65. For complete combustion of ethanol. 2 5 2 2 23 2 3C H OH l O g CO g H O l , the

amount of heat produced as measured in bomb calorimeter, is 1364.47 1kJmol at

025 C . Assuming ideally the Enthalpy of combustion, cH , for the reaction will be:

18.314R kJmol

1) 11350.50kJmol

2) 11366.95kJmol

3) 11361.95kJmol

4) 11460.50kJmol

Key: 2

Sol: H U nRT

31364 1 8.314 298 10

66. 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 of10M

sulphuric acid. The unreacted acid required 20 mL of10M sodium hydroxide for

complete neutralization. The percentage of nitrogen in the compound is:

1) 5%

2) 6%

3) 10%

4) 3%



44

Key: 3

Sol: Meq of 2 4

1H SO 60 2 1210

Meq of base used for back titration 120 210

= Meq of acid left

Meq of acid reacted with liberated

Ammonia = 12-2 = 10 meq

= meq of 3NH

Millimoles of 3NH liberated = 10 [ n-factor = 1]

Moles of 33NH 10 10

Moles “N” 310 10

Wt of “N” 310 10 14

% of “N” in the organic compound 310 10 14 100 10%

1.4

67. The major organic compound formed by the reaction of 1, 1, 1 – trichloroethane

with silver powder is:

1) 2 – Butene

2) Acetylene

3) Ethene

4) 2 – Butyne

Key: 4



45

Sol:

32CH C Cl

Cl

Cl

3 36Ag CH C C CH

2-Butyne+ 6 AgCl

68. 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) 3 : 16

2) 1 : 4

3) 7 : 32

4) 1 : 8

Key: 3

Sol: Ratio of the molecules = 1 4:32 28

= 7: 32

69. The metal that cannot be obtained by electrolysis of an aqueous solution of its salt

is:

1) Cr

2) Ag

3) Ca

4) Cu

Key: 3

Sol: Ca cannot be obtained by the electrolysis of aqueous solution of its salt.



46

70. 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

(where the constant B is positive)

1) C B C 2) C B C 3) C B C 4) C B C Key: 4

Sol: C B C

71. The correct set of four quantum numbers for the valence electrons of rubidium

atom (Z=37) is:

1) 15,0,1,2

2) 15,0,0,2

3) 15,1,0,2

4) 15,1,1,2

Key: 2 Sol: 15s

72. Consider separate solutions of 0.500 M 2 5 ,C H OH aq 3 4 20.100 ,M Mg PO aq

0.250M KBr aq and 3 40.125M Na PO aq at 025 C . Which statement is true about these

solutions, assuming all salts to be strong electrolytes?

1) 2 50.500 M C H OH aq has the highest osmotic pressure

2) They all have the same osmotic pressure

3) 3 4 20.100 M Mg PO aq has the highest osmotic pressure

4) 3 40.125M Na PO aq has the highest osmotic pressure



47

Key: 2

Sol: iCST 73. The most suitable reagent for the conversion of 2R CH OH R CHO is:

1) PCC (Pyridinium Chlorochromate)

2) 4KMnO 3) 2 2 7K Cr O 4) 3CrO Key: 1

Sol: Most suitable reagent for the conversion of 01 alcohol to aldehyde is “PCC”

74. CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which

of the following expressions is correct?

1) 3Cs Cl

r r a

2) 3Cs Cl

r r a

3) 32Cs Cl

ar r

4) 32Cs Cl

r r a

Key: 4

Sol: c a3a 2(r r )

c a3ar r2



48

75. In which of the following reactions 2 2H O acts as reducing agent?

(a) 2 2 22 2 2H O H e H O (b) 2 2 22 2H O e O H (c) 2 2 2 2H O e OH (d) 2 2 2 22 2 2H O OH e O H O 1) (b), (d)

2) (a), (b)

3) (c), (d)

4) (a), (c)

Key: 1

Sol: While acting as reducing agent it gives up electrons and release oxygen gas.

76. For which of the following molecule significant 0 ?

a)

Cl

Cl b)

CN

CN c)

OH

OH d)

SH

SH

1) (c) and (d)

2) Only (a)

3) (a) and (b)

4) Only (c)

Key: 1



49

Sol:

Cl

Cl

0

CN

CN

0 ;

0 0 ;

O H

OH

S H

SH

But dipole moment of

"SH"

SH is less than dipolemoment of

OH

OH , but is significant

77. On heating an aliphatic primary amine with chloroform and ethanolic potassium

hydroxide, the organic compound formed is:

1) an alkyl isocyanide

2) an alkanol

3) an alkanediol

4) an alkyl cyanide

Key: 1

Sol: 2 3R NH CHCl KOH R NC alkylisocyanide



50

78. In 2NS reactions, the correct order of reactivity for the following compounds:

3 3 2 3 2, ,CH Cl CH CH Cl CH CHCl and 3 3

CH CCl is

1) 3 3 2 3 32 3CH CHCl CH CH Cl CH Cl CH CCl

2) 3 3 3 2 32 3CH Cl CH CHCl CH CH Cl CH CCl

3) 3 3 2 3 32 3CH Cl CH CH Cl CH CHCl CH CCl

4) 3 2 3 3 32 3CH CH Cl CH Cl CH CHCl CH CCl

Key: 3

Sol: 3 3 2 3 32 3CH Cl CH CH Cl CH CHCl CH CCl

79. The octahedral complex of a metal ion 3M with four monodentate ligands

1 2 3, ,L L L and 4L absorb wavelengths in the region of red, green, yellow and blue,

respectively. The increasing order of ligand strength of the four ligands is:

1) 1 2 4 3L L L L

2) 4 3 2 1L L L L

3) 1 3 2 4L L L L

4) 3 2 4 1L L L L

Key: 3

Sol: Order of wavelength of colours

Red >Yellow > Green > Blue



51

80. The equation which is balanced and represents the correct product(s) is:

1) 4 2 2 444CuSO KCN K Cu CN K SO 2) 2 22 2Li O KCl LiCl K O 3) 2

3 455 5CoCl NH H Co NH Cl

4) 2 24

2 266excess NaOHMg H O EDTA Mg EDTA H O

Key: 3

Sol: 23 45

CoCl NH 5H Co 5NH Cl

81. In the reaction, 54 .3 ,PClLiAlH Alc KOHCH COOH A B C the product C is:

1) Acetyl chloride

2) Acetaldehyde

3) Acetylene

4) Ethylene

Key: 4

Sol: 54

3 3 2 3 2 2 2PClLiAlH Alc .KOHCH C OH CH CH OH CH CH Cl CH CH

O

82. The correct statement for the molecule, 3CsI , is:

1) it contains ,Cs I and lattice 2I molecule

2) it is a covalent molecule

3) it contains Cs and 3I ions

4) it contains 3Cs and I ions



52

Key: 3 Sol: 3CsI contains Cs and 3I ions

83. For the reaction 2 2 3

12g g g

SO O SO , if xp cK K RT where the symbols have

usual meaning then the value of x is: (assuming ideality) 1) 1

2) 1

3) 12

4) 12

Key: 3

Sol: nP CK K (RT)

3 1n 12 2

84. For the non-stoichiometre reaction 2 ,A 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 L S

0.1M 0.1 M 0.2 M

0.1 M 0.2 M 0.1 M

31.2 10 31.2 10 32.4 10

The rate law for the formation of ‘C’ is?

1) dc k Adt

2) dc k A Bdt

3) 2dc k A Bdt

4) 2dc k A Bdt



53

Key: 1 Sol: m mr K(A) (B) 85. Resistance of 0.2 M solution of an electrolyte is 50 . The specific conductance of

the solution is 11.4 S m . 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 2 1S m mol is:

1) 25 10

2) 45 10

3) 35 10

4) 35 10

Key: 2

Sol: lRa

1 l50

1.4 a 70

al

280 (70) 4

K = 0.25 3

410 0.25 5 100.5

86. Among the following oxoacids, the correct decreasing order of acid strength is:

1) 2 4 3HClO HClO HClO HOCl

2) 2 3 4HOCl HClO HClO HClO

3) 4 2 3HClO HOCl HClO HClO

4) 4 3 2HClO HClO HClO HOCl

Key: 4

Sol: With increase in unprotonated oxygen atoms acidic strength increases



54

87. Which one of the following bases is not present in DNA?

1) Thymine

2) Quinoline

3) Adenine

4) Cytosine

Key: 2

Sol: Qunoline is the base which is not present in DNA

88. Considering the basic strength of amines in aqueous solution, which one has the

smallest bpK value?

1) 6 5 2C H NH 2) 3 2CH NH 3) 3 2CH NH 4) 3 3CH N Key: 2

Sol: 3 2CH NH Stronger base in aqueous medium than 3Me N . So its pKb value is

smallest

89. If Z is a compressibility factor, vander Waals equation at low pressure can be

written as:

1) 1 PbZRT

2) 1 RTZPb

3) 1 aZVRT

4) 1 PbZRT



55

Key: 3

Sol: 2

aP V RTV

aPV RTV

aZ 1RTV

90. Which series of reactions correctly represents chemical relations related to iron

and its compound?

1) 0 02 , ,600 ,700

3 4O heat CO C CO CFe Fe O FeO Fe

2) 2 4 2 4 2dil ,4 2 4 3

H SO H SO O heatFe FeSO Fe SO Fe

3) 2 2 4, dil 4

O heat H SO heatFe FeO FeSO Fe

4) 2 , ,3 2

Cl heat heat air ZnFe FeCl FeCl Fe

Key: 1

Sol: O heat CO CO2600 7003 4Fe Fe O FeO Fe

Jee main 2014 qpaper key solutions

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