JEE-Adv Grand Test Solutions (P 1)

IIT Section

Subject Topic Grand Test – Paper I Date

C + M + P Grand Test – 08 IIT – GT – 08

12th May 2014 I20140508

1

Key Answers:

1. a 2. d 3. c 4. b 5. a 6. b 7. c 8. c 9. ab 10. ac

11. ac 12. a 13. d 14. a 15. b 16. b 17. b 18. a 19. 3 20. 8

21. 3 22. 8 23. 4 24. 9 25. 8 26. 6 27. 3 28. 5 29. c 30. a

31. b 32. a 33. b 34. a 35. a 36. d 37. abc 38. ad 39. abc 40. bc

41. ad 42. a 43. c 44. b 45. b 46. c 47. 2 48. 6 49. 7 50. 1

51. 4 52. 2 53. 2 54. 1 55. 4 56. 3 57. b 58. a 59. b 60. d

61. c 62. a 63. b 64. a 65. acd 66. a 67. ad 68. acd 69. ac 70. b

71. a 72. c 73. a 74. d 75. 6 76. 2 77. 7 78. 2 79. 2 80. 5

81. 6 82. 3 83. 6 84. 1

Solutions:

Chemistry

1.

2. Tertiary carbocation is formed first, which undergoes D 1,6- migration to give, more resonance

stablised carbocation, and it undergoes deprotonation in the last step.

3. 2 2 3 2 26 4 2 3NaOH S Na S O Na S H O

4. 2

X Na Ag CN Y = Ag

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5. log.

aE

3

1 15

303 3332 303 2 10

31log5 303 333 2.303 2 10

10.82 333 303

aE k cal mol

6.

2

wh

b

HKK

K C

214

4 3

10

4 10 3.3 10h

HK

6.5pH

7. thE H of graph (c) > thE H of graph (d)

8. .

. l

30 6 273

22 4373

9. In the first reaction, tertiary carbocation intermediate in formed and the product in a tertiary

bromide. In the second reaction 1NS hydrolysis takes place to give same carbocation

intermediate and the product is a monohydric alcohol.

10. Neighboring group participation of -bond pair of electrons of benzene ring.

11. 3

Bi OH ppt is formed, but 2

3 4Zn NH

soluble complex is formed.

Ferrous hydroxide green ppt, Ferric hydroxide brown ppt are formed.

12. O.N of Nickel in the complex is zero. Two electrons of 4s orbital are shifted to 3d - orbital so

that 3sp hybridistion occurs.

13. Hint No.of disintegrations per sec

Total no. of atoms in one gram of Ra

0.693

1600 365 24 60 60

or No. of disintegrations per sec 230.693 6.023 10

1600 365 24 60 60 226

14. Double inversion is taking place to give retention in the configuration, because OH group

present in C atom

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15.

16.

17.

nnC H COOH C H COOH

n

6 5 6 5

1

ni

11 1

1

n

n

1

11

18. In (a) .m 0 03

bT m

19. 2 3

0 0 0

/ /cell

Fe Fe Ag AgE E E

0.771 0.799 0.028volt

At equilibrium, 0cellE

3

0

2

0.05910 log

1cell

FeE

Fe Ag

H5C2 S

CH2

CH CH3

Cl

H5C2 S CH CH3

CH2 OH-

CH2 CH SC2H5

CH3

OH

(H5C2)2 N CH C2H5

CH2 OH-

CH2 CH N(C2H5)2

C2H5

OH

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0 10.0591logcellE

Ag

0.34Ag

0

log0.0591

3.0

nEK

K

20. Let the molecular formula be 22Ba OH xH O

Mol. Mass of 22. 137.4 2 16 2 1 18Ba OH xH O x

171.4 18x

Equilibrium mass of 22

171.4 18.

2

xBa OH xH O

.

. x

30 789 2 20 10

171 4 18 4

x 8

Thus, 8g moles of water molecules are present in one g mole of the base

21. .

.x

x

2 5 3600 22 2

2 97 396500 177

22. pH of salt after hydrolysis may be calculated as,

1

log2

w apH pK pK C

4114 log10 log0.01

2

114 4 2

2

8

23. 3 2

x x

RT RT

m m

3 400 2 60

40

4 3030

4

R R

m

m

m

24. log P

400

10 9373

xP x 910 10 9

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25. SO Cl

PSO Cl

p pK

p

2 2

2 2

SOp

2

10 15

2

Total pressure SO Cl SO Clp p p 2 2 2 2

5 2 1 8

26. Hint: 6 5 2 6 5 2 26 2

Nitrobenzene Aniline

C H NO H C H NH H O

So, 6n

27. Hint: 1000

;Af

A B

wT K A solute B Solvent

m w

1200 10006 1.86

62 Bw

6000Bw g

weight of iceI total weight of 2H O - weight of 2H O at 6 C

9000 6000 3000g

3 kg ice

28. Hint: Original solution 1500

283..............( )760

wV S i

m

After dilution 2105.3

298......................( )760

wV S ii

m

or 2

1

105.3 298

500 283

V

V

or 2

1

298 5005

283 105.3

V

V

i.e., the solution is diluted to the extent that 2V is five times the original volume 1V

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Mathematics

29. Let ,z x iy x y

,z x y

,iz y x

,z iz x y y x

21 2 1 2 2 2

1 3 1 3

1 1 33

2 2 2

x x y yx y z

x x y y

30. Solving two equations

2 21 2 3 1 1 0x x x x

Only one root 0

2 4 1 1 0

2

2 0 2

31. Total number of triangles 3

10C

number of triangles with one side common 1

10 6C

number of triangles with two sides common 10

3 1required number of traingles 10 10 6 10C C

32. Continuous at 0

10 lim sin 0 0n

xx n d

x

not differentiable at 10 0x f does not exist

1

0 0 0

1sin

0 1lim lim lim sin

0

n

n

x x x

xf x f x

xx x x

does not exist 1 0 1 0, 1n n n

33.

2 2 4, 1, 3x y P

TN TA AN

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1

3m

1

1

yy m

m

21 1

3 32 3

1

2PTN TN AP

2 3 11 1

1

2

yy y m

m

21 1

2

ym

m

34.

1P be image of P in

1 , ,P A Q are collinear 1 1

13

5AP QPm m a

35. 2 1 ________ 1x t t by eliminating t

2 1 ________ 2y t t

2

x yx t x t td

2

21 1 2 2

4 2

x y x yx x y x y

. . 2L L R

36. cos2 sin 2 7x a x a

21 2sin sin 2 7x a x a

2 4 2 2 3sin

2 2

a a ax

4sin or 2

2

ax

1 sin 1x

41 1

2

a

2 6a

axisx

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37. 1 1 1 1 2 1 1

b a c b b a c

2

2 1 1 2 1. ________ 1G E a

c b b bc b

3 2 1.G E

b a b

2

3 2___________ 2 c

abb

1 1 1 1 1 1 1 1 1 1.

2 2G E

a c c a c a c a

2 2

1 3 2 1________ 3

4b

acc a

38. 2 1

3 3 5 , 0 1n

Let 2 1

3 3 5 , 0 1n

Consider

22 112 3 3 5 .........

nn C

10 K

also 1 1 and an integer is an interger

0

10K is even

2 12 2 12 26 15 3nn

not divisible by 2 12 n

39. 1 2 3

1

2P E P E P E

1 2 2 3 3 1

1

4P E E P E E P E E

1 2 3

1

4P E E E

40.

AC a b

A B

CD

a

b

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2 2 2AC a b a b

2 216 4 16 112

4 7AC

also BC a b

4 3BC

41. 0 1, 1 2x x

2 3 2 3x x x x

2 3

3 42 2x x I I

1 2I I

42.

50 2

100 !100

50 !C

7 2

100 100100 ! 16

7 7E

7 2

50 5050 ! 8

7 7E

Exponent of 50

7 in100 0C

43. 2 108 ! 104E

5 108 ! 25E

member of zero’s = exponent of 10 = min 104, 25 25

44. 18 8 4 2 1 1 1 120 ! 2 3 5 7 11 13 17 19

4 14 8 2 1 1 1 110 2 3 7 11 13 17 19

It ends with 4 as non-zero digit

Exponent of 12 = 48

The D.E. for 31 21 2 3

m xm x m xy c e c e c e is

3 2

1 2 3 1 2 2 3 3 1 1 2 33 2

0 7 6

0

b c d

d y d y dym m m m m m m m m m m m y

dxdx dx

1a

45. order=3, degree=1

46. 1 0 6 7a b d

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47. 2 2 2 22 21 2 1 2 1 2az bz bz az a b z z

2 2 2 215 13a b

225 169

394

1 1 1

2 4 8

= 394 197k , 2k

48. 2 1000

1000. 1 1 .......

1 1 1

x x xG E x

x x x

1000

10001

11

11

x

xx

x

x

100110011001

1001

11

1

x xx

x

1001 10011 x x

coefficient of

50

50 1001 !1001

50! 951 !Cx

1001, 50, 951V

R.V 1001 100 28543 3954 659 6k k

49. 2 5 7A A I

3 2 180 35A A A A I

5 3 2 18 35 5 7A A A A I A I

149 385A I aA bI

149, 385a b

2 3 1453, 207 7a b k k

50. 1nf x x

10 1001f

31 10 1nx

3n

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320 20 1 8000 1f

8000 1k k

51. L.H.L = . .2

f R H L

3

2 2

2 2

1 sin1 sinPut 2

3cos 2x y

b xxLt a Lt x y

x x

2

20

2

1 cos1 sin 1 sin sin 2

3 1 sin 1 sin yx

ybx x x

Lt a Ltx x y

1

3 4

3 2 2a b

1 1

2 8 24

ba a

b

5/35/3 81 324

8 8 8

b

a

52. cos sinx a t t t

sin cosy a t t t

cos sin cos

sin cos sin

a t t t tdy

dx a t t t t

sintan

cos

dyat tt

dtat tdx

dt

2

2 2

2

1sec sec

cos

d y dtt tdx at tdx

3sec t

at

2 2

2 2

/3

8 24120 24 120

3t

d y d ya

adx dxa

2880 1440 , 2k k

53. 1 23 12 9 0f x x x

2 4 3 0x x

1, 3 but 3 0, 2x

0 1f

1 1 6 9 1 5f

2 8 24 18 1f 3

5, 1

4 8 625 1 626 313 2k k

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54. Let 2 2 2x y a be the circle

2 2 2 0x y a

Put 2 2

2

1 1, 0mi mi ami mi

1 2 3 4 1m m m m

55. Let 3 4, 5 6, 2 1P r r r be a point in the first line. This lies on both planes of second line

3 3 4 2 5 6 2 1 5 0r r r

2r 2,4, 3P in 2 3 4 0x y z k 4 12 12 0k

4k

56. 1 1 1sin sin sin 2 12

x y x y z

P.V 216

2500 33

7500 72 7428 2476 , 3k k

4 2 2 1 0mi a mi

1m

2m

3m

4m

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Physics:

57.

2

2

1

2 4

2 2

lw

TPo h pg

R

2

2

2

3

2 4

2 2

lw

TPo h g p

R

Solving we get 2 2

2 1

4

4

l w R Th h

prg

58. When spring extended by x

20m L x kx

202

m Lx

K m

The new length L will be 0L L x

Tension in the spring 20T m L x

22

0 21

mm L

K m

202

m KLT

K m

59. Emf induced Blv

Q Cv

Q CBlv

60. I max 2

R

at 0t

IR

at t so charge on the capacitor is C , when current is 50% of maximum current.

61. Let the length of edge be ' 'x

Net magnitude of displacement from A to 3D x

Total time 1 2 3

x x x

v v v

Average velocity

1 2 3 1 2 3

3 3

1 1 1x x x

v v v v v v

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62. When frequency is less 1

wlwc

As frequency increases, capacitive reactance decreases but inductive reactance increases. So phase

difference decreases. Then L becomes greater than 1

c. Now as frequency increases, the

magnitude of phase difference again increases.

63. 2T V

2TV C

or 1PV C

1

2 1

1

nR T TW

450W R

64. Let at any time the speed of the block along the induce up wards be V . from 2N L

sin cosp mdv

mg mgv dt

The speed is maximum when 0dv

dt

maxsin cos

pV

mg mg

65. Charge of 1l is 1 1 1q l rd

Charge of 2l 2is 2 2 2q l r d

1 1

2 2

q r

q r

1 1

1 2 2

11 1

kq k r d k dE

rr r

2 2

2 2 2

22 2

kq k r d k dE

rr r

So net electric field can’t be zero

1E is more, so direction is towards 2l

Potential due to 1l is 1 1

1

1 1

kq k r dv k d

r r

Potential due to 2l is 2 2

2

2 2

kq k r dv

r r

2v k d 1 2v v

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66. EMF across 2

2

BwlPR

EMF across 2

2

BwlPQ

So emf across 0QR

67. At

1, maxz x Upper end open lower end closed.

0.99 , minz m x

left2 2

0.80.8

m

0.99 0.01 1m

5, 5 five half loops Second overtone

4 4

ll

68. Velocity is more in rarer medium

v f , where f is constant.

69. sinX A t

12

sin2

AA t

T

112

Tt

1 22

sin2

AA t t

T

1 28

Tt t

70. FBD of small section of ring subtending angle d at center of ring is as shown

,22 2

N N mdTd g

2

mgT

71. T

v

22

mggR

m

R

72. Since there is no friction between ring and cone the tension in the ring will remain same

73. 2 1

3

4v v

We know that

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1

2

i r i

vA A A

v …(1)

2 1

2 1

r i

v vA A

v v

and 2

2 1

2i i

vA A

v v

…(2)

From equation (1) and (2), 1: 7r

i

A

A

74. 6 : 7i

i

A

A

75. Let x be the temperature of block in steady state10 5 3

0 6x x x

x CR R R

76. 1 0

s

v vn n

v v

77. 1 2 3 0Q Q Q

1 2 02

Q QIR

c c

3 2 2 0

2

Q QIR

c c

1 2 3

5 7, and

4 2 4

IRCQ IRC Q Q IRC

78. When S is closed 1t RC

When S is opened 1

2

2

3 2

2 3

tRCt

t

79. Y bright n D

d

Y Dark 1

2

Dn

d

80. 2

2 2

1 2

1 113.6E z

n n

2

2 2

1 147.2 13.6 5

2 3z z

81. 5 12 12 3 8C BV V

6C BV V V

82. 17h ev

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Now for metal maxh KE

17 2ev ev

15ev

3K

83. In steady state let T be the temperature of all the three blocks

10 2 5 2 5 0mc T mc T mc T

84. Considering reflection at the curved surface

1 2

3; 1

2

20 ; 10u cm v xcm

20R cm

from 2 1 2 1

v u R

1 3 0.5

10 40 20x

1 1 3

10 40 40x

10 10x

1x

JEE-Adv Grand Test Solutions (P 1)

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