IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

1

Key Answers:

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

11. 3 12. b 13. b 14. b 15. c 16. a 17. b 18. 19. 20. b

21. c 22. c 23. d 24. d 25. c 26. 3 27. 4 28. 7 29. 9 30. 1

31. d 32. b 33. a 34. c 35. a 36. c 37. 38. 39. b 40. a

41. a 42. a 43. d 44. c 45. 7 46. 2 47. 2 48. 4 49. 2 50. c

51. a 52. d 53. c 54. d 55. c 56. 57.

18. A – r, B – p, C – s, D – q; 19. A – q, B – r, C – p, D – s;

37. A – p,q, B – r, C – r,s, D – p,q; 38. A – q, B –r, C – s, D-r;

56. A – p,r, B – p,q, C – s, D – p; 57. A – p,q,s, B – p,q,r, C – p,q,s, D p,q,r;

Solutions:

Chemistry

1. No of lone pair of 0Xe and number of bond pair 5

Hybridisation of 3Xe sp d

Hence shape 3 2XeO F should be trigonal bipyramidal and not octahedral

2. For the desired reaction

1 ,

1

GAgCl e Ag Cl E

F

the needed ,G can be obtained by adding the values of ,G for the recton

' 1 ;Ag e Ag G nFE

; ln ,spAgCl Ag Cl G RT K giving

,G = 4 1 101 9.648 10 (298 )ln 1.56 10mol J mol K

77.10 55.95 21.15kJ kJ kJ

The potential is

21.150.2192

1 96.485E V

3.

4. Alkenes with even number of cumulative double bonds are optically active if both sides are

dissymmetric

5. None of the acids evolve 2H gas with alkali metals

CH3

C2H5

O

OH

OH

OCH3

C2H5 O

-H2O

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

2

6. Given 0.1 , 18.0HCl HClM M V mL

3 30.125 , ?NaHCO NaHCOM M V

on applying , 3 3HCl HCl NaHCO NaHCOM V M V

30.1 18 0.125 NaHCOV

314.4NaHCOM mL

Thus, 14.4mL of 31.25M NaHCO solution is needed to neutralise 18.0 mL of the 0.100

M HCl solution

7. The destruction reaction for 3O is given as 3 22 3O O

The net rate of production of 3O = 15 1 17.2 10 rate of this reactionmol h

Now

23

2 ;d O

k a xdt

but initially

0x

So

23 2 15 82 27.2 10 2 10

d Ok a k

dt

151 1

2 16

7.2 10

4 10k mol h

1 1 1 12

1818

60 60k mol h mol s

3 1 15 10 .mol s So value of 5x

8. We know that , d d

p RT or M RTM P

given, 3 7752.64 / ;

760d g dm P atm

3 1 10.0821 , 310 273 583R dm atmK mol T K

2.64 0.0821 583 760123.9 /

775M g mol

Atomic mass of 31 /P g mol

Number of P atoms in a molecule

123.94

31M

9. There are 2 bonds in C which is and B is 2 3Na SO

10. Total number of unpaired electrons 6

11. C is an amide which on heating with 2 5P O gives propane nitrole and so C is propanamide

2 5

3 2 2 3 2PO

CH CH CONH CH CH CN

NaO S

S

O

ONa

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

3

C is formed by the action of 3NH on acid B , so acid B is propanoic acid

3 2 3 3 2 4 3 2 2CH CH COOH NH CH CH COONH CH CH CONH

B C

(iii) Acid B is formed from hydrolysis of A as well as A fumes in moist air, so A is

acid halide. Thus A is

3 2 3 2CH CH COCl HOH CH CH COOH HCl

Also, (A) 3 2 (propanoylchloride)CH CH COCl

(B) 3 2 (Propanoicacid)CH CH COOH

(C) 3 2 2(propanamide)CH CH CONH

(i) 3 2CH CH COCl (ii) 3 2CH COCH Cl (iii) 2 2ClCH CH CHO

12. Na metal is soluble in liquid ammonia. The reaction takes place in a homogeneous solution

phase

13. More highly substituted alkenes are adsorbed on metal catalyst effectively so are not readily

reduce

14. The least stable alkenes has highest heat of hydrogenation

15. Yellow curdy precipitate with Ag confirms the presence of I ion in the compound The

precipitate AgI is insoluble in 4NH OH

16. Compound A gives brown precipitate with 4CuSO which becomes white when 2 2 3Na S O is

added. Thus, compound A should be KI

17. 4 2 2 42CuSO KI CuI K SO

A

2 2 2 22CuI Cu I I

brown

2 2 3 2 2 42 2Na S O I Na S O NaI

white

(sodium tetrahtionate)

18. Magnetic moment, 2n n BM

(where n number of unpaired electrons)

For 2 , 0, 0 0 2 0.00Fe n BM

For 3 , 1, 1 1 2 1.73Ti n BM

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

4

For 2 , 2, 2 2 2 2.83Ni n BM

For 3 , 5, 5 5 2 5.91Fe n BM

19. , order of reactionndCKC n

dt

or, 1/ time /n

mol lit k mol lit

or, 1 1 1time

n nK mol lit

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

5

Mathematics

20. Favourable number of ways 12 47 41 1 11C C C

Required probability 51 51

52 51 221

21. For the vector a and b to be inclined at an obtuse angle, we must have

. 0 0,a b for all x

2

2 2log 12 6 log 0c x c x for all 0,x

2 6 12 0cy cy for all y R

Where 2logy x

20 and 36 48 0c c c

(using 2

0 if 0 and 0ax bx c x R a D

4

0 3 4 0 ,03

c and c c c

22. We have 2 2 0 0a b c a b c a b c

Line 0ax by c passes through either of the points 1, 1 and ( 1,1)

23. Let 2 and 3y t z u then the equation becomes , where 1, 2, 3........... 2, 4, 6x y u n x t

and 3, 6, 9u

Hence, the number of solution of this equation

= 2 3coefficient of ......nx in x x x 2 4 6 3 6 9..... ....x x x x x x

= 6 2 2 4 3 6coefficient of 1 ... 1 .... 1 .......nx in x x x x x x x

1 116 2 3coefficient of in 1 1 1nx x x x

24. The given quadratic equation is satisfied by , andx a x b x c Hence the quadratic

equation has three roots, which is only possible if it is an identify, hence it has infinite roots

25. Let z i

z i

2Arg Argz

3arg

4z

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

6

26. 3 2. 5 7A A A A A I

25 7A A

5 5 7 7A I A

18 35A I

Now 4 3. 18 35A A A A A I

218 35A A

18 5 7 35A I A

55 126A I

5 4. 55 126A A A A A I

255 126A A

55 5 7 126A I A

149 385a and b

2 3 2 149 3 385a b

298 1155 1453

27. The period of

sin12

x

is 2

24

as 24 24

sin sin12 12

x x

12sin 2 sin

12

x x

similarly the period of

tan3

x

is 3 and the period of cos4

x

is 2

8

4

Hence the period of the given function

24,8,3 24LCM of

201 201 24 4824

28.

2 3 97

1

.... 1 1 1 ....1lim

1x

x x x x

x

1 2 3 .... 97

97

1 97 97 49 47532

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

7

29. If ‘r’ be the radius and ' 'h be the height of the cylinder

Volume 2 3 3 2 32 2 4

3 3 3V r h r r r h r

2 242 .3

3V r h h r v tr r

2 22 4r h rh r r r

2

2 4

4

3

r r h h r r rV

Vr rh r

2

1.5 0.05 2 4 0.01 4 1.5 0.01

41.5 4 1.5

3

0.215

9

0.215 21.5

100 100 2.389%9 9

V

V

2389abcd

30. Let 1 2 3 4, 5, , ,E E E E E and 6E be the events of occurrence of 1, 2, 3, 4, 5 and 6 on the dice

respectively, and let E be the even of getting a sum of numbers equal to 9

1 2 31 1 2 1

, ; ;6 6 6

k k kP E P E P E

4 5 61 1 2 1

; ;6 6 6

k k kP E P E P E

and 1 2

( )9 9

P E

Then, E= 3,6 6,3 4,5 5,4

Hence, 3 6 6 3 4 5 5 4( )P E P E E P E E P E E P E E

6 6 3 4 5 5 43P E P E P E P E P E P E P E P E

= 62 3 2 4 5P E P E P E P E

1, 2 3 4Since , , are independentE E E E

1 1 1 1 22 2

6 6 66

k k k k

212 2

18k k

since, 1 2

9 9P E

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

8

21 1 22 2 ...............(1)

9 18 9k k

22 2 2 4k k

22 2 2k k

22 2 2k k

22 0k k

2 1 0k k

10

2K

Hence, integral value of k is zero and for 0k from equation (1)

Set of integral value of 0k

Number of integral solutions of k is 1

31. Here 1 0a therefore we make the substitution 2 2 2x x t x Squaring both sides of this

equality and reducing the similar terms, we get

2 22

2

2 2 22 2 2 ;

2 1 2 1

t t tx tx t x dx dt

t t

2 22 2 4 4

1 2 2 12 1 2 1

t t tx x t

t t

substituting into the integral, we get

22 2

22

2 1 2 2 2 2

4 4 2 1 1 2

t t t t t dtI dt

t t t t t

Now let us expand the obtained proper rational fraction into per tail fractions

2

3 2

2 2

1 21 2 2

t t A B D

t tt t t

32. 2 1

dxI

x x x

Since here 1 0,c we can apply the second Euler substitution

2 1 1x x tx 2 2

2

2 12 1 1 ;

1

tt x t x x

t

substituting into I, we can get an integral of a rational fraction

2

22

2 2 2,

1 11

dx t tdt

t t tx x x

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

9

Now,

2

2

2 2 2

1 9 1 1 11

t t A B D E

t t t t t tt

33. In this case 0 0a and c therefore, nether the first, nor the second Euler substitution is

applicable, But the quadratic 27 10x x has the real roots 2, 5 . Therefore we

use the third Euler substitution

27 10 2 5 2x x x x x t

25 2 ;x x t

2

2

5 2

1

tx

t

34. Ans (c)

35. Ans (a)

36. Let the equation of the circle be 2 2 2 2 0x y gx fy c ………..(1)

The line 1lx my will touch the circle (1) if the length of perpendicular from the centre

,g f of the circle on the line is equal to its radius,

i.e.,

2 2

2 2

1gl mfg f c

l m

2 2 2 21gl mf l m g f c

2 2 2 2 2 2 2 1 0 ( )c f l c g m gl fm gflm ii

But the given condition of tangency is

2 24 5 6 1 0l m l ……………….(iii)

Comparing equations (ii) and (iii), we get 2 24 5, 2 6, 2 ,2 0c f mc g g f c gf

solving we get 0, 3, 4f g c

substituting these values in equation (i), the equation of the circle is 2 2 6 4 0x y x .

any point on the line 1 0x y is ,1 ,t t t R

Chord of the contact generated by this point for the circle is 1 3 4 0tx y t t x or

3 3 4 0t x y x y which are concurrent at point of intersection of the 3 0x y and

3 4 0x y for all values of ‘t’. Hence, lines are concurrent at 1 5

,2 2

Also point 2, 3 lies outside the circle from which two tangents can be drawn

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

10

37. (A) 2 7z i z i k is ellipse if 7 2 5k i i or k

(B) We have, for zC

2 2 2i z i z z i z

2 2z z i

Thus, minimum value of 2z z i is 2 and it is attained for any z lying on the segment joining

2 .z i

(C) 3 4z z i k is hyperbola, if 3 4 0 5k i k

(D) 3 450

kz i az az b

3 45 2 3 4

az az bkz i

i

This is hyperbola if 1 55

kk

38. 1

.12

P A B P A P B

1 1 1 4 3 1 6 1

3 4 12 12 12 12P A B

1

23/ /1 3

2

P A A B P A P A B

/ ' ' 0

' '

PP B A B

P A B

''/

P A s P B P A BP A B

P B P B

1

1 2121 11 3 3

4

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

11

Physics

39. The centre of mass will follow the same path.

2

Rmx m mR

or 3 3 100

1502 2

Rx

m

40. 4 4 4

1 2 31 2 3

Pr Pr Pr, ,

8 8 8V V V

l l l

and

4Pr

8V

l

Now 1 2 3V V V V

Substituting the values, we get

1 2 3

1 2 1 2 2 3

l l ll

l l l l l l

1 2 3 6

1 2 1 3 2 3 11

m

41. 1 2( )

1

nR T T

2/3 2/31 2(5.6) (0.7)T T

2 14T T

11

3 9

2 2

3

nR TnRT

But 1

4n

1

9

8RT

42. 2

2 21

1 113.6(1)

2 3

hC

2

2 22

1 113.6(2)

2 4

hC

Dividing

2

1

1 5 / 36

4 3 /16

2 15

121527

oA

xR/2

R

m m m

COM

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

12

43. 212

2iu v

2 2(2 )

2 10 5t

V Vu

Loss of energy 2 2

2 4

5 5

V VV

% loss 2

2

4 / 5100

V

V = 80%

44. Ratio of magnetic moment and angular momentum is given by

2

M q

L m

Which is a function of q and m only. This can be derived as follows :

2( )( )M iA qf r

2

2( ) ( )2 2

q rq r

and 2( )L I mr

2

2

2

2

rq

M q

L mmr

45. Distance between two cars leaving from station X is,

1

60 106

d

km

Man meets the first car after time,

160 1

60 60 2t h

He will meet the next car after time,

210 1

60 60 12t h

In the remaining half an hour, number of cars he will meet again is, 1/ 2

61/12

n

Total number of cars would be meet on route will be 7.

46. From work energy theorem

netKE W

or f iK K pdt

or 2

2 2

0

1 3

2 2mv t dt

( 2m kg)

or

23

2

02

tv

or 2v m/s

2

s

1

2 F

8 F

V

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

13

47. 24 cos2

oI I

24 cos2

o oI I

1

cos2 2

or 2 3

or 2 2

3x

or 1 1

.3

d

D

yd

xD

7

4

6 10

3 103

yd

D

32 10 m = 2 mm

48. / ,E

v m s t sm v

7 922 10 2 10

For 200 neutrons generation: T t 200

s 74 10

49. d

dq idtR

Area under i - t graph

d (Area under i - t graph) ( R )

1

(4)(0.1)(10) 22

Wb

50. For rod 2

( 3 )1:

(2 )(16 ) / 3

N l

m l

or 32

3 3

mlN

N

A

3l

N

mg

30oB

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

14

51. (2 ) (2 )(2 ) 4cN F m a m l ml

32 12 3

43 3

F N ml ml

52. 22 2

2

( cos30 ) (2 sin 30 )

( )(2 ) / 3

m gl N l

m l

2

2

32( 3 / 2)

3 3 83 3

8 3(4 / 3)

mlmgl

g

lml

53. (4)(2)(1) 8V Bvl V

,R C and L all are in parallel. Therefore, ,R CV V and LV all are 8V .

Ldi

V Ldt

or 8 (4)di

dt 2

di

dt A/s

i.e.., rate of exchange of current through L is constant. After 2s current will be 4A.

54. ( ) ( )applied mF rightwards F leftwards

IlB 1 2 3( )I I I lB

3I we have already calculated at 2 which is 4 A.

2 (8) 0Cdq d d

I CVdt dt dt

and 18

42

RVI A

R

(4 4)(1)(4) 32appliedF N

55. . (32)(2) 64 /P F v J s

21

1( )

2

d dip Li Li

dt dt

3(4)(4)(2) 32 / ( )J s i I

2 22 1(4) (2) 32 / ( )P i R J s i I

N

a

2l

2ca lC

F

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

15

56. (A) Since 0E and 0,B so path will be straight line if velocity is parallel to B . Or path will be

circular if V B . Or path will be helical (with uniform pitch) if V is at some other angle to B .

Hence. (A) – p, r

(B) Since 0E and 0,B so path will be straight line if V B or parabola otherwise.

Hence (B) – p, q

(C) 0E , 0,B E B

Helical path with non-uniform pitch Hence, (C) – s

(D) Straight line path if V B E

Hence. (D) – (p)

57.

F f ma

,a

Fx fR Ia aR

Solve to get 2

( )F MxR If

I MR

B

E

90o

v

B

E

V

F

a

xa

R

kf

,M l

IIT Section

Subject Topic Grand Test – Paper II Date

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

10th May 2014 I20140510

16

0,f if 1

xMR

0,f if 1

xMR

So friction will act in forward direction.

0F if /x I MR , so friction will act in backward direction.

Also solved to get

2

( )FR x Ra

I MR

A is positive for all cases, hence body will accelerate in forward direction in all cases.

It means will be in clockwise direction. Hence bodies will rotate in clockwise direction.