IIT Section

Subject Topic Grand Test – Paper I Date

C + M + P Grand Test – 07 IIT – GT – 07

10th May 2014 I20140510

1

Key Answers:

1. b 2. b 3. c 4. a 5. b 6. c 7. b 8. b 9. abd 10. acd

11. ac 12. bc 13. cd 14. b 15. a 16. c 17. c 18. c 19. 5 20. 5

21. 3 22. 3 23. 9 24. 2 25. 1 26. 3 27. 3 28. 8 29. b 30. a

31. c 32. b 33. a 34. c 35. c 36. b 37. ab 38. ab 39. abc 40. bc

41. ab 42. a 43. d 44. b 45. b 46. c 47. 2 48. 1 49. 9 50. 6

51. 1 52. 3 53. 3 54. 5 55. 2 56. 2 57. a 58. d 59. d 60. c

61. b 62. c 63. b 64. c 65. acd 66. ac 67. abc 68. ac 69. abd 70. d

71. a 72. b 73. c 74. d 75. 5 76. 5 77. 2 78. 5 79. 3 80. 4

81. 3 82. 0 83. 4 84. 8

Solutions:

Chemistry

1. 2NH contains 2 lone pairs which causes maximum repulsion. Hence, minimum angle, whereas

3NH contains one lone pair whereas 4NH has no lone pair although all three are 3sp hybridised.

2. A liquid boils when its vapour pressure becomes equal to its atmosphere pressure. Given for

water at 298K VP=atm. 23P mm so it will boil at 298 K. A decrease in pressure results in

decrease in temperature at which a liquid boils.

3.

4. 2X Y

1 2

1

2

1

1

2

1

1

1

p

p

K

p

or Z P Q

1

IIT Section

Subject Topic Grand Test – Paper I Date

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10th May 2014 I20140510

2

Total moles at (equilibrium) 1

2

2 2

2

1 1

1

1

p

p p

K

p

1

2

2

1

2

2

4p

p

K p

K p

1

2

41

9

p

p or 1

2

1

36

p

p

5. Let R

6. 33

3 ;Cr OH Cr OH

3 4

OH Cr OH Cr OH

7. Let the volume of fcc unit cell V

Density of 4

,.

A

A

A

MA

N V

Density of 8

, B

B

A

MB

N V

30

0.32 2 50

A A

B B

M

M

3.33 /0.3

A

B g cc

Total density 1 3.33A B 4.33 g/cc

8. X uses 3 of its valence electrons for bonding and it is angular shaped suggesting nitrogen

CH3

H

D

BD2

BD3, THF

CH3

H

D

B

Repeat twice

R

R

18

2

H2O2, OH

CH3

H

D

OH 3

IIT Section

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10th May 2014 I20140510

3

Y is likely to be Group V element since the diagram shows that there are five valence electrons

forming bonds.

However Y cannot be N

Z could be hydrogen or halogen.

9. 2

*2 2 22 2 2 2x yKK s s p p p

Four electrons are present in 2 molecular orbital, that’s why double bond contains both

bonds.

Molecule is diamagnetic.

10. HVZ reaction is an electrophilic substitution reaction.

11. Hofmann bromamide degradation is intramolecular rearrangement.

12. 2 22 2Br Cl Br Cl

1.07 1.36 0.29 0cellE V

2

4 2 22 10 16 2 5 8MnO Cl H Mn Cl H O

1.52 1.36 0.16 0cellE V

3 2 2

4 2 2 76 10 11 6 5 2MnO Cr H O Mn Cr O H

1.52 1.33 0.19 0cellE V

1.33 1.52 0.19 0cellE V

13. HCl cannot displace 2 .H O

14. 2

1

2Na g Cl g NaCl s H Heat of formation

fH =-98.2 kcal/mol

15. For lattice energy we require the heat absorbed in the reaction:

NaCl s Na g Cl g

This reaction is the sum of the first five reactions in the table and so the lattice energy is the sum

of the corresponding heats of reaction,

25.9 29.0 119.9 86.8 98.2 186.2 kcal/mol

16. We require the heat absorbed in the following reaction:

Na g Cl g Na aq Cl aq …(i)

The reaction can be obtained with following reaction

NaCls Na aq Cl aq …(ii) (in question)

0.9H kcal/mol

IIT Section

Subject Topic Grand Test – Paper I Date

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10th May 2014 I20140510

4

NaCl s Na g Cl g …(iii)

186.2H kcal/mol

Eq (ii)-Eq(iii) = Eq (i)

So, the energy = 0.9-186.2=-185.3 kcal

17.

18. Phenol is not converted into salt by 3NaHCO

19.

20. 1 2 3 4 9 25

3 3

Pka Pka PkaPH

21. 1/2 1

1,

nt

a

For 3rd order 1/2 2

1t

a

22. The final product A is benzene.

23. Number of equivalents of oxygen 151.2

275.6

Number of equivalent of aluminium =27

Mass of aluminium 27 9 g [Eq. wt. of Al=9 g]

Number of moles 27 9

927

24. Resistance of 0.1 mol 1L520HAc

Resistance of 0.1 mol 1L HAc+0.1 mol 1L122NaCl

Conductance due to 0.1 mol 1LNaCl , 11 1

0.00627122 520

G

Conductivity of 0.1 mol 1LNaCl solution, mC

1 2 1 1126 0.1

1000

cm mol mol L

1 10.0126 cm

COOK

OH

230oC

OH

COOK

IIT Section

Subject Topic Grand Test – Paper I Date

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10th May 2014 I20140510

5

Cell constant: 10.01262.01

0.00627K cm

G

Conductivity of 0.1 mol 1LHAc solution,

11 2.01.

520

cmK

R

Molar conductivity 1 1 1

1

2.01/ 5200.03865

0.1

Kcm Lmol

C mol l

1 2 138.65 cm mol

According to Kohlrausch’s law

m m m mHAc NaAc NaCl HCL

1 2 1420 91 126 cm mol

14 2 1385 cm mol

38.65

0.1385

m

m

1 10.1 0.1 0.01H C mol L mol L

2pH

25. 1 g 2H or 11200 cc of

2H is obtained from 1 F or 96500C.

So, 112 mL of 2H is obtained from 965C

965

1965

Qi A

t

26. 2 3Fe Fe e

Amount of 2 31 10Fe mol = Number of moles of electrons lost

Amount of 4

4 2.5 10MnO mol

Amount of electrons gained = amount of electrons lost 31 10 mol

Electrons gained per mol of 4MnO

3

4

1 104 .

2.5 10mol

Hence, the oxidation state of Mn decreases by 4 units.

27. For 6 6 ,C H 1000 5.12

1.28100N

wT

M

And in 2

1000 1.86, 1.40

100

wH O T

Mex

1.40 5.12

31.28 1.86

NMi

Mex

As solute is 100% ionised. So total number of furnished ions=3.

IIT Section

Subject Topic Grand Test – Paper I Date

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10th May 2014 I20140510

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28.

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Mathematics

29. Let 0,1 , 0,0 , 1,0 , 1,1A B C D

Family of circles passing through ,A B is 2 2 0x y y x

1 2 3

30. Area 1 2

0 1

1 sin 3 sinx x dx x x dx

23 3cos1 1 3 cos2 cos1 2 cos2 2 2cos 1 1

2 2

21 2cos 1

/2 /4

0 0

sinsin2 2

sin cossin cos

2 2

x

xI dx dx

x x x x

31. Let sin sin cos cos sinx A x x B x x

1A B and 1 1

0, ,2 2

A B A B

32. Equation of circum-circle is 2 2

3 1 17

2 2 2x y

3 17 1 17cos , sin

2 2 2 2C

Circum centre of ABC is 3 1

,2 2

Centroid can be obtained.

In a triangle centroid, circum centre and ortho centre are collinear

33. Total number of matrices that can be formed is 93 .

Let 3 3ijA a

where 1,0,1ija

If ' 'A is symmetric then ,ij ija a i j

If ' 'A is skew symmetric then ,ij ija a i j

34. 33 2 2 3

1 1 2 1 2 2 1 23 3 2 11 2 11z z z iz z iz i z iz i

Similarly, 3

1 2 2 11z iz i

1/3 1/32 2

1 2 1 2 1 2 2 11 2 11 5z z z iz z iz i i

/4

0

1 12 2 log sin cos log 2

2 4 2 4I x x

IIT Section

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10th May 2014 I20140510

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35. 12

2

dy

dx y

dy

dx at 3,6 1P 45

Component velocityparallel to x-axis is cosv

36. Area of a hexagon PQRSTU ABC area of APQ area of

BTU area of CRS

1 1 1sin sin sin

2 3 3 2 3 3 2 3 3

b c a c a bA B C

37. 6 6

0 0

sin sin cos cos sin

x x

f x x x t x t dt t x t x t dt 6 6

0 0

sin cos cos sin

x x

x t t dt x t t dt

38. Solving the equation we get ˆ ˆˆ ˆ ˆ ˆ 2 ,r i j k i j k R

39. Conceptual

40. 1 , 0 2

3, 2 4

x xf x

x x

2 , 1 0

2 , 0 3

x xg x

x x

1 , 0 2

3, 2 4

g x gfog x

g x g x

1 , 1 0

1 , 0 2

x x

x x

!3

nn

n

put 300300 100 300!n …. (1)

150 150 150 3003 100 3 100 300 100

300 300300 100 …. (2)

from (1) & (2) 300 300300 100 300!

41. Equation of tangent at ,x y is Y y p X x

Where .dy

pdx

Then y

OA xp

and OB y px

IIT Section

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10th May 2014 I20140510

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1 1y

OA OB y px xp

42. . 4 2OAOB y px p

43. 2

11

pAB y px

p

44. 45. & 46.

Let ,0,0t be a point on the x axis through which a straight line L is drawn perpendicular to

the x axis and intersecting both the lines1 2,L L . D.R’s of L may be taken as 0,1,

L and 1L are coplanar

5 3 15

1 5 2 0

0 1

t

L and 1L are coplanar

1 6

2 5 3 0

0 1

t

Solving we get 3

, ,24

t

or 13 137

,25 5

47. ab d and .cd b Also, 0a c d a b c .

Thus 0b d and 1.a c

2b d

2a b c d

48.

5

2 2

0 0

1 1 1 51

4 4I f x dx f x dx

49.

7

3

4 6 3 4

1 2 5n n

n n n np p

n nC

50. Choose ' 'A as origin of vectors and let ,b c be the position vectors of B and C respectively.

Then equation of AC is r t c and that of BE is 7

8

b cr b s

.

These two lines intersect at ,F where 1

7AF c

7AC AF or 6CF AF .

So, 6CF

AF

IIT Section

Subject Topic Grand Test – Paper I Date

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10th May 2014 I20140510

10

51.

3/221 1

1 2 1

0 0

11

3

n n

n

xI x x x dx x dx

1

3/22 2

0

11 1

3

nn x x dx

23 1n n nI n I I or 2

11

2

n

n

I n

I n

as n

52. Let 5 /3.xy f x x e Then 0 1.y Also,

1'

'g y

f x and so,

1' 1 3

' 0g

f

53. 1 2

21, ,

2 2

n nC C are in A.P. 8n

8

1

1 16 3. . .

42r r r

rT C x

So 4 16 3 0 8r r only if 0,4,8r

2sin

cos

2 tan

n

n

n

.

54. The given circles will cut orthogonally if

10 22 2 7 1

2 2 2 2

p p p

or

2 5 6 0p p or if 2p or 3.

1 2 2 3 5p p

55.

2 2 2

2 2 2

2 2 1 3 3 1 12 1 3 1 1

2 1 3 1 1 2 2 1 3 3 1 1n

n nna

n n n

Since 2 21 1 1 1,k k k k we get

2 11 2.

1 3n

n na

n n

2lim

3n

na

56. As 2 ,A O 2.kA O k

Thus, 50

50A I I A

50

50A I A I

1,a 0,b 0,c 1d

IIT Section

Subject Topic Grand Test – Paper I Date

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10th May 2014 I20140510

11

Physics

57. For particle p , motion between AC will be an accelerated one while between CB a retarded

one. But in any case horizontal component of its velocity will be greater than or equal to V . On

the other hand, in case of particle Q . It is always equal to V . Horizontal displacements for both

the particles are equal. Therefore, P Qt t .

58. Velocity of ball to just reach the top of the tube should be given by;

200 2v gh

Here (2 )oh R h

and in critical case velocity will be zero at topmost point.

Thus 2 (2 )v g R h

59. Let x be the displacement of bead. Displacement of particle with respect to bead is (1 cos ),L

i.e., displacement of particle with respect to ground will be (1 cos )L x . Now net force in

horizontal direction on the system is zero. Therefore, the centre of mass will not move in

horizontal direction.

or 2 [ (1 cos ) ]mx m L x

or 3 (1 cos )mx mL

or (1 cos )3

Lx

60. 2T KV

2dT KVDV

2

dTdV

KV

Further, nRT

PV

Now 2

nRT dTW PdV

V KV

0

0

4

2

T

T

nRdT (as 2KV T )

03

2

nRT

61. When magnification is +2.

u x

2v x

20f

IIT Section

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10th May 2014 I20140510

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Using 1 1 1

v u f we have

1 1 1

2 20x x

or 10x cm

To have a magnification of -2.

u y

2v y

and 20f

1 1 1

2 20y y

or 30y cm

20y x cm

62. Net resistance 2 ( )

(2 )

R

R

dr l

rl A

1 (2)

2n

l

63. 2

/ 2net

e Bvl Bvli

R r r

Here, 2

( )(1) 2r

(2)(1)(2) 4i A

or 16mF ilB N

64. At resonance 3200

540

i A

1

LC

L CL

X X LC

6

5250

80 10

1 2200

0.8250

i i A 1 2i i and 1 3i i

65. Net external force on the system is zero. Therefore, linear momentum will remain conserved.

Initially both the particles have equal momentum in opposite directions. Hence, initial

momentum is zero i.e., final linear momentum is also zero or velocity of centre of mass 0cv .

Let be the angular velocity about centre of mass. Net external torque on system is also zero.

Hence, angular momentum about centre of mass will also be conserved i.e.,

IIT Section

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10th May 2014 I20140510

13

i fL L

or (2 4 )mva mva I

2

2 2(6 )2 (2 )( ) (2 )

12

am m a m a

or 2 2 26 {24 2 4 }mva ma ma ma 2(30 )ma or 5

v

a

Finally 2

2 21 1(30 )

2 2 5

vE I ma

a

23

5mv

66. At the level of boundary between 2 and 3 , pressures will be equal from both sides. Hence,

1 3 22

hgh g gh

3 2 12( )

From this expression itself it is clean that 2 1

67. Potential of innermost shell is zero

31 2 02 3

qq q

r r r

or 1 2 36 3 2 0q q q

similarly, potential on outermost shell is also zero.

31 2 03 3 3

qq q

r r r

or 1 3 2q q q

Solving Eqns. (1) and (2), we get

321

1

, 34

qqq

q

and 3

2

3

4

q

q

Options (a), (b) and (c) are correct.

68. The intensity of light is 20( ) cos

2I I

Where 2 2

( ) ( sin )x d

(i) For 30o

IIT Section

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10th May 2014 I20140510

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8

6

3 10300

10

c

v

m and 150d m

2 1(150)

300 2 2

2 4

2 00( ) cos

4 2

II I

(option a)

(ii) For 90o

2

(150)(1)300

or 2 2

and ( ) 0I

(iii) For 0 , 0 or 02

o

0( )I I (option c)

69. Work done by magnetic force is always zero. Therefore, from work energy theorem:

Work done by electrostatic force from P to Q

= Kinetic energy at Q – Kinetic energy at P

2 2 21 3( )(2 ) (4 )

2 2qE a m v v mv

23

4

mvE

qa

Rate of work done by magnetic field (power) is always zero.

Rate of work done by electric field at P

Fv qEv ( 0)

2 33 3

4 4

mv mvq v

qa a

And rate of work done by electric field at Q = 0 as ( 90 )o

Here, angle between andeF v .

70. sin cosa g g

23 410 0.5 10 2 /

5 5m s

Velocity of block at B or at C,

IIT Section

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20

2 2 2 / sin3

hv as h

For not to leave contact anywhere

2v gR

or 20

203

hR

or 3h R or 1.5h m

71. Since, 2sin cos 2 /g g m s

or sin cosg g

Block will never stop on rough inclined surface. So, it will cross point C infinite number of times.

72. Electrostatics force on A is zero, while on B is

(1)(10) 10B BF q E N

(along negative x – direction )

210 /BB

B

Fa m s

m

Just before collision,

2 2 10 1.8 6B Bv a s

Now, from conservation of linear momentum, velocity of combined mass will be,

( )A B B Bm m v m v or 3v m/s

73. At equilibrium position, net force on the system is zero. Let 0x be the compression in the spring

in equilibrium, then

0Kx electrostatic force

or 018 10x

or 05

9x m

or equilibrium position will be at 5

9x

74. Angular frequency of SHM will be :

18

32A B

K

m m

At 5

9x m, speed is 3 m/s. Therefore, from

2 2v A x

IIT Section

Subject Topic Grand Test – Paper I Date

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10th May 2014 I20140510

16

We have 2 23 3 A x

or 2 2 1A x

2 2 25 1061 1

81 81A x

106

9A

75. Rate of change of speed

dv

dt tangential acceleration

tangential force

mass

sin30

sin30o

omgg

m

sin30

sin30o

omgg

m

2 2110 / 5 /

2m s m s

76. Since, sin30 cos30o omg mg

The block has a tendency to slip downwards. Let F be the minimum force applied on it, so that

it does not slip. Then

cos30oN F mg

sin30 ( cos30 )o omg N F mg

or sin30

cos30o

omgF mg

(2)(10)(1/ 2) 3

(2)(10)0.5 2

or 20 17.32F

or 2.68F N

77. 1

Tg

i.e.., 2 1

1 2

T g

T g

Where 1g = acceleration due to gravity on earth’s surface = g

2g = acceleration due to gravity at a height h R from earth’s surface / 4g

T

sin 30omg

mg cos30omg

IIT Section

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2

Using ( )

1

gg h

h

R

2

1

2/ 4

T g

T g

78. Maximum acceleration in SHM is

2maxa A

this will be provided to the block by friction.

Hence, maxa g or 2a g

or 2 2

1(10)

20.05

(10)

gA

m = 5 cm

79. Range will become twice if velocity of efflux becomes twice. Now as,

2v gh

Therefore, h should become 4 times or 40 m.

Thus, an extra pressure equivalent to 30 m of water should be applied.

1 atm = 0.76 13.6 m of water

=10.336 m of water

30 m of water = 3.0 atm.

80. Energy E (amplitude)2 (frequency)2

Amplitude (A) is same in both the cases, but frequency 2 in the second case is two times the

frequency ( ) in the first case. Therefore, 2 14E E

81. Deviation by a sphere is 2 (i – r)

Here, deviation 60 2( )o i r

or 30oi r

30 60 30 30o o o or i sin sin 60

3sin sin30

o

o

i

r

82. Since, the capacitor plates are directly connected to the battery, it will take no time in charging.

83. 01

LC

2

1

L

C

X LLC

X

IIT Section

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18

Here, 02

2LC

4L

C

X

X

84. 2v

ar

2( )

(1/ )

za

z (for 1n )

or 3a z 3

1

2

28

1

a

a