JEE(Main)-2015

Part A – CHEMISTRY

1. Which of the following is the energy of a possible excited state of hydrogen?

(A) 6.8eV (B) 3.4eV

(C) 6.8eV (D) 13.6eV

Ans. B

Sol.

2

2n

2n

2

ZE 13.6 eV

n

For hydrogen Z 1

1E 13.6 eV

n

13.6E 3.4eV

4

2. In the following sequence of reactions:

4 2 2

4

KMnO SOCl H /Pd

BaSO,Toluene A B C

The product C is:

(A) 6 5 3

C H CH (B) 6 5 2

C H CH OH

(C) 6 5

C H CHO (D) 6 5

C H COOH

Ans. C

Sol.

3. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?

(A) (A)

(B)

(C)

(D)

Ans. A

Sol.

4. The ionic radii ( in3 2

A ) of N , O and F

are respectively:

(A) 1.36, 1.71 and 1.40 (B) 1.71, 1.40 and 1.36

(C) 1.71, 1.36 and 1.40 (D) 1.36, 1.40 and 1.71

Ans. B

Sol. Ionic radii

3 2

N O F

5. The color of 4

KMnO is due to :

(A) d d transition (B) chL M arge transfer transition

(C) transition

(D) chM L arge transfer transition

Ans. B

Sol. chL M arge transfer transition

6. Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.

Reason : The reaction between nitrogen and oxygen requires high temperature.

(A) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion

(B) The assertion is incorrect, but the reason is correct

(C) Both the assertion and reason are incorrect

(D) Both assertion and reason are correct, and the reason is the correct explanation for the assertion

Ans. D

7. Which of the following compounds is not an antacid?

(A) Cimetidine (B) Phenelzine

(C) Ranitidine (D) Aluminium hydroxide

Ans. B

Sol. Phenelzine is anti-depressant

8. In the context of the Hall-Heroult process for the extraction of Al, which of the following statements is false ?

(A) 2 3

Al O is mixed with CaF2

which lowers the melting point of the mixture and brings conductivity

(B) 3+

Al is reduced at the cathode to form Al

(C) 63

Na AlF serves as the electrolyte

(D) CO and CO2

are produced in this process

Ans. C

Sol. 2 3 2

* Al O + 3C 4Al + 3CO

3 6 2

* Na AlF or CaF is mixed with purified 2 3

Al O to lower the melting point and brings conductivity.

* Oxygen librated at anode reacts with carbon anode to liberate CO and CO2

.

9. Match the catalysts to the correct processes:

Catalyst Process

(A) TiCl3 (i) Wacker process

(B) PdCl2 (ii) Ziegler – Natta polymerization

(C) CuCl2 (iii) Contact process

(D) V2O5 (iv) Deacon’s process

(A) (A) (ii), (B) (i), (C) (iv), (D) (iii) (B) (A) (ii), (B) (iii), (C) (iv), (D) (i)

(C) (A) (iii), (B) (i), (C) (ii), (D) (iv) (D) (A) (iii), (B) (ii), (C) (iv), (D) (i)

Ans. A

Sol. 4 2 5 3

TiCl + (C H ) Al Ziegler Natta catalyst, used for coordination polymerization.

Wacker process

H O + PdCl2 2

2 2 3CH = CH CH CHO + 2HCl

2

CuCl used as catalyst in Deacon’s process of production of 2

Cl .

2 5

V O used as catalyst in Contact Process of manufacturing of 42

H SO .

10. In the reaction:

the product E is

(A) (A)

(B)

(C)

(D)

Ans. B

Sol.

11. Which polymer is used in the manufacture of paints and lacquers? (A) Glyptal (B) Polypropene

(C) Poly vinyl chloride (D) Bakelite

Ans. A

Sol. Glyptal Manufacture of paints and lacquers

Polypropene Manufacture of ropes, toys

Poly vinyl chloride Manufacture of raincoat, handbags

Bakelite Making combs and electrical switch

12. The number of geometric isomers that can exist for square planar

+

23[ Pt (Cl) (py) (NH ) (NH OH)] is py = pyridine :

(A) 3 (B) 4

(C) 6 (D) 2

Ans. A

Sol.

13. Higher order (>3) reactions are rare due to:

(A) increase in entropy and activation energy as more molecules are involved

(B) shifting of equilibrium towards reactants due to elastic collisions

(C) loss of active species on collision

(D) low probability of simultaneous collision of all the reacting species

Ans. D

Sol. Higher order (>3) reactions are less probable due to low probability of simultaneous collision of all the reacting species.

14. Which among the following is the most reactive?

(A) 2

Br (B) I2

(C) ICl (D) 2

Cl

Ans. C Sol. Inter halogen compounds are more reactive than corresponding halogen molecules due to polarity of

bond.

15. Two Faraday of electricity is passed through a solution of 4

CuSO . The mass of copper deposited

at the cathode is: ( at. mass of Cu = 63.5 amu ) (A) 63.5 g (B) 2 g

(C) 127 g (D) 0 g

Ans. A

Sol. 2

Cu 2e Cu

2F of electricity will give 1 mole of Cu.

16. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it

was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed

(per gram of charcoal) is:

(A) 36 mg (B) 42 mg

(C) 54 mg (D) 18 mg

Ans. D

Sol. Amount adsorbed

3

3

= (0.060 - 0.042) 50 10 60

= 0.018 50 60 10

= .018 3

= .054 gm = 54 mg

Amount adsorbed per gram of activated charcoal 54

18mg3

17. The synthesis of alkyl fluorides is best accomplished by:

(A) Sandmeyer’s reaction (B) Finkelstein reaction

(C) Swarts reaction (D) Free radical fluorination

Ans. C

Sol. Δ

2 2 2 22R - Cl + Hg F 2R F + Hg Cl

18. The molecular formula of a commercial resin used for exchanging ions in water softening is

8 7 3

C H SO Na Mol. Wt. 206 . What would be the maximum uptake of Ca2

ions by the resin when

expressed in mole per gram resin?

(A) 1

206 (B)

2

309

(C) 1

412 (D)

1

103

Ans. C

Sol.

2

8 7 3 8 7 3 22C H SO Na Ca C H SO Ca

2 mole 1 mole

412 gm 1 mole

Maximum uptake of Ca2

ions by the resin = 1/412 (mole per gm resin)

19. Which of the vitamins given below is water soluble?

(A) Vitamin D (B) Vitamin E

(C) Vitamin K (D) Vitamin C

Ans. D

Sol. Vitamin C is water soluble

20. The intermolecular interaction that is dependent on the inverse cube of distance between the molecule is:

(A) ion-dipole interaction (B) London force

(C) hydrogen bond (D) ion-ion interaction

Ans. A

Sol. For ion-dipole interaction

dE

Fdr

(where is dipole moment of dipole and r is distance between ion and dipole)

2

3

d 1

dr r

r

21. The following reaction is performed at 298 K .

2 2

2NO g O g 2NO g

The standard free energy of formation of NO g is 86.6 kJ / mol at 298K . What is the standard free

energy of formation of 12

p2NO g at 298K ? K 1.6 10

(A) 1286600 + R 298 ln 1.6 × 10 (B)

12ln 1.6 × 10

86600R 298

(C) 120.5 2 × 86,600 R 298 ln 1.6×10

(D) 12

R 298 ln 1.6 × 10 86600

Ans. C

Sol.

n

n

Re x

(Re x )

f f f

f f f

f

° ° ° °

2 2

° ° ° °

2 2

°p

p

°p

°

2

° 12

2

ΔG = 2ΔG NO - 2ΔG NO - ΔG O

2ΔG NO = ΔG + 2ΔG NO + ΔG O

ΔG = ΔG + RTln k

At equilibrium,

ΔG = 0, Q = k

ΔG = RTln k

ΔG O = 0

ΔG NO = 0.5 2 × 86,600 R 298 ln 1.6 × 10

22. Which of the following compounds is not colored yellow?

(A) 3 2 6

K [Co(NO ) ] (B) 4 3 3 10 4

(NH ) [As(Mo O ) ]

(C) 4

BaCrO (D) 2 6

Zn [Fe(CN) ]

Ans. D

Sol. 2 6

Zn [Fe(CN) ] is bluish white and rest are yellow colored compounds.

23. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The

percentage of bromine in the compound is: (at. Mass Ag = 108; Br = 80)

(A) 36 (B) 48

(C) 60 (D) 24

Ans. D

Sol. Let the organic compound is R – Br

3

R Br AgNO AgBr

So number of moles of AgBr Number of moles of R – Br Number of moles of Br

33

3

141 10number of moles of Br 0.75 10

188

Mass of bromine 0.75 80 10 g 60 mg

60Percentage of bromine 100 24%

250

24. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 A

. The radius of

sodium atom is approximately:

(A) 3.22 A

(B) 5.72 A

(C) 0.93 A

(D) 1.86 A

Ans. D

Sol. For body center unit cell.

4r a 3

3 4.29So radius of Na 1.857 1.86 A

4

25. Which of the following compounds will exhibit geometrical isomerism?

(A) 3-Phenyl-1-butene (B) 2-Phenyl-1-butene

(C) 1,1-Diphenyl-1-propane (D) 1-Phenyl-2-butene

Ans. D

Sol. Only 1 phenyl -2-butene can show geometrical isomerism.

26. The vapour pressure of acetone at 20 C is 185 torr. When 1.2 g of a non-volatile substance was

dissolved in 100 g of acetone at 20 C , its vapour pressure was 183 torr. The molar mass 1g mol

of

the substance is:

(A) 64 (B) 128

(C) 488 (D) 32

Ans. A

Sol. Vapour pressure of pure acetone 0p = 185 torr

Vapour pressure of solution (p) = 183 torr

Then from Raoult’s law,

solute

solvent

0 np p

p n

185 183 1.2 /MW

183 100 / 58

2 1.2 58

183 100 MW

1.2 58 183MW 63.68 64

200

27. From the following statements regarding 22

H O , choose the incorrect statement:

(A) It decomposes on exposure to light

(B) It has to be stored in plastic or wax lined glass bottles in dark

(C) It has to be kept away from dust

(D) It can act only as an oxidizing agent

Ans. D

Sol. 22

H O can act both as oxidizing agent as well as reducing agent and 22

H O decomposes on exposures to

light and dust; so as to kept in plastic or wax lined glass bottles in dark .

28. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its

lattice enthalpy?

(A) 4

BeSO (B) 4

BaSO

(C) 4

SrSO (D) 4

CaSO

Ans. A

Sol. Because of smallest size of 2

Be

, its hydration energy is maximum and is greater than the lattice energy

of 4

BeSO .

29. The standard Gibbs energy change at 300 K for the reaction 2A B C is 2494.2 J. At a given

time, the composition of the reaction mixture is 1 1

A , B 2 and C2 2

. The reaction proceeds in

the : R = 8.314 J / K / mol, e = 2.718

(A) reverse direction because Q > Kc

(B) forward direction because Q < Kc

(C) reverse direction because Q < Kc

(D) forward direction because Q > Kc

Ans. A

Sol.

2.718

°

1

c 2 2

cc

ΔG = RT ln K

2494.2 = 8.314 × 100 ln K

ln K = 1

log K = 1

1K = 2.718 = = 0.3679

2.718

12 ×B C 2Q = = = 41A

2

Q > k Reverse direction .

30. Which one has the highest boiling point?

(A) Ne (B) Kr

(C) Xe (D) He

Ans. C

Sol. The boiling point of Noble gases in increasing order:

He < Ne < Ar < Kr < Xe < Rn

Boiling point (K) : 4.2< 27.1< 87.2< 119.7< 165< 211

Part B – MATHEMATICS

31. The sum of coefficients of integral powers of x in the binomial expansion of 50

1 2 x is :

(A) 5013

2 (B) 501

3 12

(C) 5012 1

2 (D) 501

3 12

Ans. D

Sol.

50 50 r 1/2

50

5050 50 502

2

r 1

r /2

r

r

rr

r

25r

rr 0

t C . 1 . 2x

C .2 .x 1 r even integer.

1 1Sum of coefficient C .2 1 2 1 2 3 1

2 2

32. Let f(x) be a polynomial of degree four having extreme values a t x = 1 and x = 2. If

0x 2

f xlim 1 3

x

,

then f(2) is equal to :

(A) 4 (B) 0

(C) 4 (D) 8

Ans. B

Sol.

x 0

2

2

x f xlim 3

x

, since limit exits hence 2 4 3 2x f x ax bx 3x

4 3 2

3 2

43 2

f x ax bx 2x

f ' x 4ax 3bx 4x

also f ' x 0 at x 1, 2

1a , b 2

2

xf x 2x 2x

2

f x 8 16 8 0

33. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is

deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the

resultant data, is :

(A) 16.0 (B) 15.8

(C) 14.0 (D) 16.8

Ans. C

Sol.

i

New sum y 16 16 16 3 4 5 252

Number of observation 18

New mean

252y 14

18

34. The sum of first 9 terms of the series 3 3 3 3 3 3

1 1 2 1 2 3 .......1 1 3 1 3 5

is :

(A) 96 (B) 142

(C) 192 (D) 71

Ans. A

Sol.

9

r 1

23 22

r 2

2

9

102

t 1

r r r 1 1t r 1

2r 1 44r

1S r 1 , let t r 1

4

1t 1 96

4

35. Let O be the vertex and Q be any point on the parabola, 2

x 8y . If the point P divides the line segment

OQ internally in the ratio 1 : 3, then the locus of P is :

(A) 2

y x (B) 2

y 2x

(C) 2

x 2y (D) 2

x y

Ans. C

Sol.

2 2

2

4th t

4

2t tk

4 2

x 2y

36. Let and be the roots of equation 2 n n

nx 6x 2 0. If a , for n 1 , then the value of

10 8

9

a 2a

2a

is equal to :

(A) 6 (B) 3

(C) 3 (D) 6

Ans. B

Sol.

810 9

810

9

2 2

10 9 8

10 9 8

x 6x 2 6 2

6 2 .... 1

and 6 2 .... 2

Subtract 2 from 1

a 6a 2a

a 2a3

2a

37. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes

contains exactly 3 balls is :

(A)

102

553

(B)

121

2203

(C)

111

223

(D)

1155 2

3 3

Ans. Correct Option is not available

Sol. Required probability

1212 9

33 2 3

12

3 3 9

1C C 2 C C C

3

38. A complex number z is said to be unimodular if z 1 . Suppose 1 2

z and z are complex numbers such

that 1 2

1 2

z 2z

2 z z

is unimodular and

2z is not unimodular. Then the point

1z lies on a :

(A) straight line parallel to y-axis. (B) circle of radius 2.

(C) circle of radius 2 . (D) straight line parallel to x-axis.

Ans. B

Sol.

1 2

2

21 1

22

2

2

1

2

1 1

2 2

1 2 1 1 2

2 2

1 2 1 1 2

2 2 2 2

1 2 1 2

2 2 2

1 2 2

1 2

z 2z1

2 z z

z 2z z 2z

2 z z 2 z z

z 2z z 2z z 4 z

4 2z z 2z z z z

z 4 z 4 z z 0

z 1 z 4 1 z 0

z 2 as z 1

39. The integral

3/4

2 4

dx

x x 1 equals :

(A) 1/4

4x 1 c (B)

1/44

x 1 c

(C)

1/44

4

x 1c

x

(D)

1/44

4

x 1c

x

Ans. C

Sol.

3/45

4

4

5

3/4

1/41/4

4

dx

1x 1

x

11 t

x

4dx dt

x

1 1dt

4 t

1 14t c 1 c

4 x

40. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with

vertices (0, 0), (0, 41) and (41, 0), is :

(A) 861 (B) 820

(C) 780 (D) 901

Ans. C

Sol. x y 41, x 0, y 0 is bounded region.

Number of positive integral solutions of the equation x y k 41 will be number of integral co-ordinates

in the bounded region.

4041 1

3 1 2C C 780

41. The distance of the point (1, 0, 2) from the point of intersection of the line x 2 y 1 z 2

3 4 12

and the

plane x y z 16 , is :

(A) 8 (B) 3 21

(C) 13 (D) 2 14

Ans. C

Sol. Let the point of intersection be 2 3 , 4 1, 12 2

2 2

2 3 4 1 12 2 16

11 11

1

point of intersection is 5, 3, 14

distance 5 1 9 12

16 9 144 13

42. The equation of the plane containing the line 2x 5y z 3; x y 4z 5 , and parallel to the plane,

x 3y 6z 1 , is :

(A) x 3y 6z 7 (B) x 3y 6z 7

(C) 2x 6y 12z 13 (D) 2x 6y 12z 13

Ans. B

Sol. Let equation of plane is 2x 5y z 3 x y 4z 5 0

As plane is parallel to x 3y 6z 1 0

2 5 1 4

1 3 6

6 3 5

11 2

11

2

Also, 6 30 3 12

6 33

11

2

so the equation of required plane is

4x 10y 2z 6 11 x y 4z 5 0

7x 21y 42z 49 0

x 3y 6z 7 0

43. The area (in sq. units) of the region described by 2x, y : y 2x and y 4x 1 is :

(A) 5

64 (B)

15

64

(C) 9

32 (D)

7

32

Ans. C

Sol. The required region

1

2

11

2 2

1 1

1 2

2 3

y 1 ydy

4 2

1 y 1 yy

4 2 2 3

1 1 1 1 1 1 11 1

4 2 8 2 2 3 8

1 15 3 9

4 8 16 32

44. If m is the A.M. of two distinct real numbers 1 2 3and n , n 1 and G , G and G are three

geometric means between 44 4

21 3and n, then G 2G G equals :

(A) 2

4 m n (B) 2

4 mn

(C) 2 2 2

4 m n (D) 2

4 mn

Ans. A

Sol. Given

1 2 3

1

2

2

3

3

4

1/4

4 444 4 4 2 3

1 2 3

4 4 4 8

24 4 4

24

2

2

l + n = 2m ... i

l , G , G , G , n are in G.P.

G = lr let r be the common ratio

G = lr

G = lr

n = lr

nr =

l

G + 2G + G = lr + 2 lr + lr

= 1 × r 1 + 2r + r

= 1 × r r + 1

n n +1= 1 ×

1 1

= ln × 4m

= 4lnm

45. Locus of the image of the point (2, 3) in the line 2x 3y 4 k x 2y 3 0, k R , is a :

(A) straight line parallel to y-axis. (B) circle of radius 2 .

(C) circle of radius 3 . (D) straight line parallel to x-axis.

Ans. B Sol. Let M is mid point of BB' and AM is bisector of

BB' (where A is the point of intersection given lines)

2 2

2 2

x 2 x 1 y 2 y 3 0

h 2 h 2 k 3 k 32 1 2 3 0

2 2 2 2

h 2 h k 1 k 3 0

x 2x y 4y 3 0

x 1 y 2 2

46. The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the

ellipse 2 2

x y1

9 5 , is:

(A) 18 (B) 27

2

(C) 27 (D) 27

4

Ans. C Sol.

a 3, b 5

5 2e 1

9 3

foci 2, 0

2x 5ytangent at P 1

9 3.5

2x y1

9 3

2x 3y 9

Area of quadrilateral

4 area of triangle QCR

1 93 4 27

2 2

47. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is :

(A) 192 (B) 120

(C) 72 (D) 216

Ans. A

Sol. Four digit numbers will start from 6, 7, 8

3 4 3 2 72

Five digit numbers 5! 120

Total number of integers = 192 48. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the

set A B , each having at least three elements is :

(A) 256 (B) 275

(C) 510 (D) 219

Ans. D

Sol.

n A 4, n B 2

n A B 8

number of subsets having atleast 3 elements

8 8 8

212 1 C C 219

49. Let 1 1 1

2

2x 1tan y tan x tan , where x

31 x

. Then a value of y is :

(A)

3

2

3x x

1 3x

(B)

3

2

3x x

1 3x

(C)

3

2

3x x

1 3x

(D)

3

2

3x x

1 3x

Ans. D

Sol.

1 1 1

2

21

2

31

2 2

31 1

2

3

2

2xtan y tan x tan

1 x

2xx

1 xtan2x

1 x1 x

x x 2xtan

1 x 2x

3x xtan y tan

1 3x

3x xy

1 3x

50. The integral

4

2

2

2 2

logxdx

logx log 36 12x x is equal to :

(A) 4 (B) 1

(C) 6 (D) 2

Ans. B Sol. Apply the property

b b

a a

f x dx f a b x dx

And then add

4

2

2I = 1dx

2I = 2

I = 1

51. The negation of s r s is equivalent to :

(A) s r s (B) s r s

(C) s r (D) s r

Ans. C

Sol.

s r s

s r s

s r s s r s s

s r F

s r

52. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the

foot of the tower, are 30°, 45° and 60° respectively, then the ratio, AB : BC, is :

(A) 3 : 2 (B) 1: 3

(C) 2 : 3 (D) 3 :1

Ans. D Sol. AB 3x x

xBC x

3

AB 3x x 3

xBC 1x

3

53.

x 0

1 cos2x 3 cos xlim

x tan4x

is equal to :

(A) 3 (B) 2

(C) 1

2 (D) 4

Ans. B

Sol.

x 0

22sin x 3 + cosx×

limtan 4x

x × × 4x4x

2 42

4

54. Let a, b and c be three non-zero vectors such that no two of them are collinear and

1

a b c b c a3

. If is the angle between vectors b and c , then a value of sin is :

(A) 2

3

(B)

2

3

(C) 2 3

3

(D)

2 2

3

Ans. D

Sol.

1

a c b b c a b c a3

1b c b c

3

1b c cos b c

3

1cos

3

2 2sin

3

55. If

1 2 2

A 2 1 2

a 2 b

is a matrix satisfying the equation T

A A = 9I , where I is 3 3 identity matrix, then

ordered pair a, b is equal to :

(A) 2, 1 (B) 2, 1

(C) 2, 1 (D) 2, 1

Ans. C

Sol.

T

T

3 3ij

13

23

1 2 2 1 2 a

A 2 1 2 , A 2 1 2

a 2 b 2 2 b

A A b

b 0 0 a 4 2b

b 0 0 2a 2 2b

3a 6 0 a 2, b 1

56. If the function , k x 1 , 0 x 3

g xmx 2 , 3 x 5

is differentiable, then the value of k + m is :

(A) 16

5 (B)

10

3

(C) 4 (D) 2

Ans. D

Sol.

for f x to be continuous

2k 3m 2

2k 3m 2 .... i

for f x to be differentiable

km

4

k 4m

from i , 8m 3m 2

5m 2

2m

5

2 8k 4

5 5

2 8 10k m 2

5 5 5

57. The set of all values of for which the system of linear equations :

1 3 12

1 2 3 2

1 2 3

2x 2x x x

2x 3x 2x x

x 2x x

has a non-trivial solution, (A) is a singleton. (B) contains two elements.

(C) contains more than two elements. (D) is an empty set.

Ans. B

Sol.

1 32

1 2 3

1 2 3

2

2

2 x 2x x 0

2x 3 x 2x 0

x 2x x 0

2 2 1

2 3 2 0

1 2

2 3 4 2 2 2 1 4 3 0

2 3 4 4 1 1 0

2 4 1 5 1 0

1 4 2 5 0 1, 1, 3

58. The normal to the curve, 2 2x 2xy 3y 0, at 1, 1 :

(A) meets the curve again in the second quadrant.

(B) meets the curve again in the third quadrant.

(C) meets the curve again in the fourth quadrant.

(D) does not meet the curve again.

Ans. C

Sol.

2 2x 3xy xy 3y 0

x x 3y y x 3y 0

x 3y x y 0

Equation of normal is y 1 1 x 1

x y 2

It intersects x 3y 0 at 3, 1

And hence meets the curve again in the 4th

quadrant.

59. The number of common tangents to the circles 2 2 2 2

x y 4x 6y 12 0 and x y 6x 18y 26 0 ,

is : (A) 2 (B) 3

(C) 4 (D) 1

Ans. B

Sol.

1 1

2 2

1 2

1 2 1 2

C 2, 3 ; r 4 9 12 5

and C 3, 9 ; r 9 81 26 8

C C 25 144 13

C C r r touching externally.

3 common tangents.

60. Let y(x) be the solution of the differential equation dy

xlogx y 2xlogx, x 1dx

. Then y(e) is equal to:

(A) 0 (B) 2

(C) 2e (D) e

Ans. B

Sol.

1dx

x ln x ln ln x

dy y 2x ln x+ =

dx x ln x x ln x

I.F. = e = e = ln x

y lnx = 2 ln x dx

y ln x = 2 x ln x x + c

For x = 1, c = 2

y ln x = 2 x lnx x +1

put x = e y e = 2

Part C – PHYSICS

61. As an electron makes a transition from an excited state to the ground state of a hydrogen – like atom/ion: (A) kinetic energy, potential energy and total energy decrease

(B) kinetic energy decreases, potential energy increases but total energy remains same

(C) kinetic energy and total energy decrease but potential energy increases

(D) its kinetic energy increases but potential energy and total energy decrease

Ans. D

Sol. As electron goes to ground state, total energy decreases.

TE KE

PE 2TE

So, kinetic energy increases but potential energy and total energy decreases.

62. The period of oscillation of a simple pendulum is L

T 2g

. Measured value of L is 20.0 cm known to

1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s

resolution. The accuracy in the determination of g is:

(A) 3% (B) 1%

(C) 5% (D) 2%

Ans. A

Sol.

2

2

4 Lg

T

g L T2

g L T

L 0.1 T 0.01,

L 20 T 0.9

g L T100 100 2 100 3%

g L T

63. A long cylindrical shell carries positive surface charge in the upper half and negative surface charge in

the lower half. The electric field lines around the cylinder will look like figure given in: (figures are schematic and not drawn to scale)

(A) (A)

(B)

(C)

(D)

Ans. D Sol. It originates from + Ve charge and terminates at –Ve charge. It can not form close loop. 64. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies

of the resultant signal is/are: (A) 2005 kHz, and 1995 kHz (B) 2005 kHz, 2000 kHz and 1995 kHz

(C) 2000 kHz and 1995 kHz (D) 2 MHz only

Ans. B

Sol. R C

R C

m

m

f f f 2000 kHz 5 kHz 2005 kHz

f f f 2000 kHz 5 kHz 1995 kHz

So, frequency content of resultant wave will have frequencies 1995 kHz, 2000 kHz and 2005 kHz

65. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered

as an ideal gas of photons with internal energy per unit volume 4U

u TV

and pressure 1 U

p3 V

. If the

shell now undergoes an adiabatic expansion the relation between T and R is:

(A) 3R

T e

(B) 1

TR

(C) 3

1T

R (D)

RT e

Ans. B

Sol.

1/3

4 1/3

dQ dU dW

dU pdV

dU 1 Up

dV 3 V

dU 1 dV

U 3 V

1nU nV nC

3

U V C

VT V C

1T

R

'

66. An inductor (L = 0.03 H) and a resistor (R = 0.15 k ) are connected

in series to a battery of 15V EMF in a circuit shown. The key 1

K has

been kept closed for a long time. Then at 1

t = 0, K is opened and

key 2

K is closed simultaneously. At t = 1 ms, the current in the

circuit will be: 5e 150

(A) 67 mA (B) 6.7 mA

(C) 0.67 mA (D) 100 mA

Ans. C

Sol. When 1

K is closed and 2

K is open,

0

EI =

R

When 1

K is open and 2

K is closed, current as a function of time ‘t’ in L.R. circuit.

0I I

tR

L

5

e

1 1e 0.67mA

10 1500

67. A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional

mass M is added to its bob, the time period changes to M

T . If the Young’s modulus of the material

of the wire is Y then 1

Y is equal to: ( g = gravitational acceleration )

(A) M

2T Mg

1T A

(B) M

2T A

1T Mg

(C) M

2

T A1

T Mg

(D) M

2T A

1T Mg

Ans. D

Sol. Time period, T 2g

When additional mass M is added to its bob

M

M

M

M

2

2

T 2g

Mg

AY

Mg

AYT 2g

T Mg1

T AY

T1 A1

Y Mg T

68. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a

distance of 1 m from the diode is: (A) 2.45 V/m (B) 5.48 V/m

(C) 7.75 V/m (D) 1.73 V/m

Ans. A

Sol. Intensity, 2

00 0

1I = ε E C, where E

2 is amplitude of the electric field of the light.

2

00

1= ε E C

2

E0

0

2

2

P

4 r

2P2.45 V/m

4 r C

69. Two coaxial solenoids of different radii carry current I in the same direction. Let 1

F be the magnetic

force on the inner solenoid due to the outer one and 2

F be the magnetic force on the outer solenoid due

to the inner one. Then:

(A) 1

F is radially inwards and 2

F is radially outwards

(B) 1

F is radially inwards and 2

F 0

(C) 1

F is radially outwards and 2

F 0

(D) 1 2F F 0

Ans. D

Sol. Both solenoid are in equilibrium so, net

Force on both solenoids due to other is

zero.

So , 1 2F F 0

70. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion

the average time of collision between molecules increases as q

V , where V is the volume of the gas. The value of q is :

p

v

C

C

(A) 3 5

6

(B)

1

2

(C) 1

2

(D)

3 5

6

Ans. B

Sol. Average time between collision rms

Mean freePath

V

1

2

1

2 2

2

2

21

2

1

2

1 CV Mt ; t where C constan t

Td N / V d N 3R

3RT

M

VT

t

For adiabatic

TV k

VV k

t

Vk

t

t V

1so, q

2

71. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to

0Q and then connected to the L and R as shown.

If a student plots graphs of the square of maximum charge 2

MaxQ on the

capacitor with time (t) for two different values 21 1 2L and L L L of L

then which of the following represents this graph correctly? (Plots are

schematic and not drawn to scale)

(A) (A)

(B)

(C)

(D)

Ans. D

Sol. 2 2

1 221

1 1As L L , therefore L i L i ,

2 2

Rate of energy dissipated through R from 1

L will be slower as compared to2

L .

72. From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure. Taking gravitational potential V 0 at r , the potential at the centre of the cavity thus formed is (G

= gravitational constant)

(A) GM

R (B)

2GM

3R

(C) 2GM

R (D)

GM

2R

Ans. A

Sol.

required M/8M

222

3 3

V V V

GM R GM / 83R 3 R / 2

42R 2 R / 2

11GM 3GM GM

8R 8R R

73. A train is moving on a s traight track with speed 1

20 ms

. It is blowing its whistle at the frequency of

1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train

passes him is ( speed of sound = 3201

ms

) close to :

(A) 12 % (B) 18 %

(C) 24 % (D) 6 %

Ans. A

Sol.

1

2

1 2

1

320 320f train approaches 1000 1000 Hz.

320 20 300

320 320f train recedes 1000 1000 Hz.

320 20 340

f f 300 40f 100% 1 100% 100%

f 340 340

11.7% 12%

74. Given in the figure are two blocks A and B of weight 20 N and 100 N

respectively. These are being pressed against a wall by a force F as shown.

If the coefficient of friction between the blocks is 0.1 and between block B

and the wall is 0.15, the frictional force applied by the wall on block B is

(A) 80 N (B) 120 N

(C) 150 N (D) 100 N

Ans. B Sol. Normal force on block A due to B and between B and wall will be F. Friction on A due to B = 20 N Friction on B due to wall = 100 + 20 = 120 N

75. Distance of the centre of mass of a solid uniform cone from its vertex is 0

z . If the radius of its base is R

and its height is h then 0

z is equal to :

(A) 3h

4 (B)

5h

8

(C)

23h

8R (D)

2h

4R

Ans. A

Sol. 0

h 3hz h

4 4

76. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations

as shown in the figures below:

(A) (A)

(B)

(C)

(D)

If there is a uniform magnetic field of 0.3 T in the positive z direction , in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium ?

(A) (a) and (c), respectively (B) (b) and (d), respectively

(C) (b) and (c), respectively (D) (a) and (b), respectively Ans. B

Sol. Since B is uniform, only torque acts on a current carrying loop. τ = IA ×B

A A k for b and A Ak for d .

0 for both these cases.

The energy of the loop in the B field is : U = I A B , which is minimum for (b).

77. In the circuit shown, the current in the 1 resistor is :

(A) 0 A (B) 0.13 A, from Q to P

(C) 0.13 A, from P to Q (D) 1.3 A, from P to Q

Ans. B Sol. Taking the potential at Q to be 0 and at P to

be V, we apply Kirchoff’s current law at Q :

V 6 V V 90

3 1 5

3V 0.13 volt

23

The current will flow from Q to P.

78. A uniformly charged solid sphere of radius R has potential 0

V (measured with respect to ) on its

surface. For this sphere the equipotential surfaces with potentials 0 0 0 0

3V 5V 3V V, , and

2 4 4 4 have

1 2 3 4radius R , R , R and R respectively. Then

(A) 1 2 1 4 3R 0 and R R R R (B) 1 2 4 3

R 0 and R R R

(C) 4

2R R (D) 1 2 4 3R 0 and R R R

Ans. B, C

Sol. The potential at the centre R

000

3

2

4

3

Q 4 r dr 3 kQ 3 1k V ; k

r 2 R 2 4R

0

0

1

2 2

3 33

4 44

R 0

kQPotential at surface, V

R

5V RPotential at R R

4 2

kQ 3 kQ 4RPotential at R , R

R 4 R 3

kQ kQSimilarly at R , R 4R

R 4R

Both options (2) and (3) are correct.

79. In the given circuit, charge 2

Q on the 2 F capacitor changes as C is

varied from 2

1 F to 3 F. Q as a function of ‘C’ is given properly by :

( figures are drawn schematically and are not to scale )

(A) (A)

(B)

(C)

(D)

Ans. A Sol. Let the potential at P be V,

2

Then, C E V 1 V 2 V we take C in F

CEOr, V

3 C

2CEQ

3 C

80. A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

(A) 50% (B) 56%

(C) 62% (D) 44%

Ans. B

Sol.

initial

final

2 2 2

2 2 2

1 1E m 2v 2m v 3mv

2 2

1 4 4 4E 3m v v mv

2 9 9 3

43

53Fractional loss 56%3 9

81. Monochromatic light is incident on a glass prism of angle A. If the refractive

index of the material of the prism is , a ray, incident at an angle , on the

face AB would get transmitted through the face AC of the prism provided :

(A) 1 1 1

sin sin A sin

(B) 1 1 1

cos sin A sin

(C) 1 1 1

cos sin A sin

(D) 1 1 1

sin sin A sin

Ans. D Sol. At face AB,

C

1

1

1

1

1 1

sin sin r

At face AC r '

1A r sin

1r A sin

1sinr sin A sin

sin 1sin A sin

1sin sin A sin

82. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of

inertia of cube about an axis passing through its center and perpendicular to one of its faces is :

(A)

2MR

16 2 (B)

24MR

9 3

(C)

24MR

3 3 (D)

2MR

32 2

Ans. B Sol. For maximum possible volume of cube

2R 3 a , a is side of the cube.

Moment of inertia about the required axis I

3

23

34

a Ma , where

6R

I

5 5 2

3 3

3M 1 2R 3M 1 32R 4MR

6 63 9 3 9 34 R 4 R

83. Match List – I (Fundamental Experiment) with List – II (its conclusion) and select the correct option from

the choices given below the list:

List – I List – II

(A) Franck-Hertz Experiment (i) Particle nature of light

(B) Photo-electric Experiment (ii) Discrete energy levels of atom

(C) Davison – Germer Experiment (iii) Wave nature of electron

(iv) Structure of atom

(A) (A) – (ii) (B) – (iv) (C) – (iii) (B) (A) – (ii) (B) – (i) (C) – (iii)

(C) (A) – (iv) (B) – (iii) (C) – (ii) (D) (A) – (i) (B) – (iv) (C) – (iii)

Ans. B 84. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is

4 12.5 10 ms

. If the electron density in the wire is

28 38 10 ms

, the resistivity of the material is

close to:

(A) 7

1.6 10 m

(B) 6

1.6 10 m

(C) 5

1.6 10 m

(D) 8

1.6 10 m

Ans. C

Sol.

d

d

1928d

4

5

5

J ne v

A Vnev

A

V 5

nev 0.1 8 10 1.6 10 2.5 10

1.56 10

1.6 10 m

85. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly?

( Graphs are schematic and not drawn to scale )

(A) (A)

(B)

(C)

(D)

Ans. A Sol. At mean position, K.E. is maximum where as P.E. is minimum 86. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s

and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?

( Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 102

m / s )

( The figures are schematic and not drawn to scale )

(A) (A)

(B)

(C)

(D)

Ans. B

Sol.

2

1

2

2

2 1

1y 10t gt

2

1y 40t gt

2

y y 30 t straight line

but stone with 10 m/s speed will fall first and the other stone is still in air. Therefore path will become parabolic till other stone reaches ground .

87. A solid body of constant heat capacity 1 J / C is being heated by keeping it in contact with reservoirs in

two ways: (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of

heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of

heat.

In both the cases body is brought from initial temperature 001 C to final temperature 200 C . Entropy

change of the body in the two cases respectively is:

(A) ln2, ln2 (B) ln2, 2ln2

(C) 2ln2, 8ln2 (D) ln2, 4ln2

Ans. Correct option is not available

Sol.

423 473

423373

385.5 398 410.5 423 435.5 448 460.5 473

373 423398385.5 410.5 435.5 448 460.5

dT dTCase i dS C n 473 / 373

T T

dT dT dT dT dT dT dT dTCase ii dS C n 473 / 373

T T T T T T T T

l

l

Note: If given temperatures are in Kelvin then answer will be option (1) .

88. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the

minimum separation between two objects that human eye can resolve at 500 nm wavelength is :

(A) 30 m (B) 100 m

(C) 300 m (D) 1 m

Ans. A

Sol.

2

1.22D

Minimum separation 25 10 30 m

89. Two long current carrying thin wires, both with current I, are held by

insulating threads of length L and are in equilibrium as shown in the figure,

with threads making an angle ' ' with the vertical. If wires have mass

per unit length then the value of I is : ( g = gravitational acceleration )

(A) 0

gL2sin

cos

(B)

0

gL2 tan

(C) 0

gLtan

(D)

0

gLsin

cos

Ans. A

Sol. I

I

0

0

2

4 L sinF

tang g

Lg2sin

cos

90. On a hot summer night, the refractive index of air is smallest near the ground and increases with height

from the ground. When a light beam is directed horizontally, the Huygens’ principle leads us to conclude that as it travels, the light beam:

(A) goes horizontally without any deflection (B) bends downwards

(C) bends upwards (D) becomes narrower

Ans. C Sol. According to Huygens’ principle, each point on wavefront behaves as a point source of light .