IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

1

Key Answers:

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

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

21. d 22. c 23. a 24. c 25. a 26. 2 27. 1 28. 1 29. 1 30. 1

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

41. a 42. c 43. c 44. c 45. 3 46. 2 47. 4 48. 4 49. 9 50. c

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

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

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

56. A –r, B – qr, C – ps, D – qr; 57. A – pq, B – pqs, C – pqr, D – pqr;

Solutions:

Chemistry

1. (a) Mixture of 100 ml of /10M HCl and 100 ml of 10

MNaOH is an exact neutralisation.

Hence 7pH

(b) After neutralisation , left 1010

MHCl ml

Total volume=100 , Dilution 10timesml H

= 210 2or pH

(c) After neutralisation, left 10 times10

MNaOH

Total volume 100 ml, 7PH

(d) After neutralisation, left 505

MHCl ml

total volume 100 ml, dilution 2 times

1110 or 1

10H pH

2. Ans: (b )

3. Selective reduction of one nitro group of a dinitro compound can often be achieved by the use of

hydrogen sulphide in aqueous or alcoholic ammonia. The reduction is favour at ortho position

with respect to OH group and para position with respect to 3CH group. (more electrons

deficient site is more readily reducible)

4. Taking a ratio of 0

/dP NO dt both experiments gives

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

2

0.2 2

10.1

/

/

xdP NO dt P NO

P NOdP NO dt

Taking logarithms, solving for x , and substituting the data gives

1 1log 200 / 1372.10 2

log 0.479 / 0.400

Pa s Pa s

bar bar

5. 40

100100

PP V

or 250V cc Total volume of gas mixture

volume of bulb 250 100 150B cc

6. Same magnetic moment=same number of unpaired electrons 2n n

Where n number of unpaired electrons

2 73 , 3Co d unpaired electrons

2 43 , 4Cr d unpaired electrons

2 53 , 5Mn d unpaired electrons

2 63 , 4Fe d unpaired electrons

7.

8. Balanced chemical equation is 2 2 2 2 22 5 2 2 5 6ClO H O OH Cl O H O

2 mol of 2 2 25 ofClO mol H O

9. Heat of neutralisation for strong acid with strong base 13.7 kcal /mol

3( ) 12.5 ( 13.7) 1.2 /CH COOHH kcal mol

3 413.7 . . 10.5CH COOH NH OHI E I E

4. 13.7 1.2 10.5 2NH OHI E

10. ' "X is prepared by the action of 2HNO on organic fertilizer

i.e., 2 2 2 2 2 22 2 3H NCONH HNO N CO H O

CH3CH2 C CH

CH3

CH3

O

(A) -ve Tollens Test

CH3CH2CH CHCH3

OH CH3

(B)

CH3 CH2 CH C CH3

CH3

5 4 3 2 1

(C)

CH3CH2CHO + CH3COCH3

+ve Tollens test

-ve Iodoform

-ve Tollens test

+ve Iodoform

-H2O

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

3

Bond order calculation of 2N

Bond order10 4

32

11. Since pyrolusite 2MnO is taken we know that A will be potassium manganate, a green coloured

compound. The purple coloured compound B will be potassium permanganate. Potassium

permanganate is a good oxidising reagent both, in alkaline and acidic medium, getting

decolourised with KOH

2 2 4 22MnO KOH O K MnO H O

(pyrolusite) A (dark green)

with dilute 2 4H SO

2 4

2 4 2 4 23 2 2 4dil H SO

K MnO H O KMnO MnO KOH

with KI (alkaline) ( )B

4 2 3 22 2 2KMnO KI H O KIO MnO KOH

C

Oxidation state of Mn in compound C is 4

So, EC 2 2 6 2 6 31 2 2 3 3 3s s p s p d

So number of unpaired electrons is 3

12. 3

23 2 3 3 2 32 2.

O

Zn H OCH CH CH C CH CH CH CHO CH CO

13. 3

22 2 3 3 3 2 3 2 3 3.

( ) ( ) ( )O

Zn H OCH C CH CH CH CH CH CH O CH CH COCH CH CH

14. 3

23 3 32

O

H OCH C C CH CH COOH

15. Heat liberated 17.7 0.5 / 0.01 8.85 /kJ mol kJ mol

heat liberated for 1 mol of 48.85

16 8850.16

CH kJ

or 1combustion 885E kJ mol

16. 885

55.316

kJ

17. 4( ) 2( ) 2( ) 2 ( )2 2g g g lCH O CO H O

gH E n RT

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

4

3885 ( 2 8.31 10 300)H

1889.986 kJ mol

18. Butter of tin is 4 25SnCl H O

Dry Ice is solid 2CO

Sugar Lead is 3 2Pb CH COO

White lead is 3 22PbCO Pb OH

19. Conceptual

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

5

Mathematics

20. 22f x x x ax b

20,2 0x x ax b has distinct real roots

0

2 28 0 8a b a b

min of 1b

min of 3a

min 4a b

21. Number of subsets containing 21101 C

Number of subsets containing 21102 C

Sum of elements 21 21 2110 10 101 2....... 22C C C

21 2110 101 2 ..... 3 253C C

22. 1, 1f x g x

1,2,3......g x

111, 1 1 tan 1

2f x g x f x g x x

1 1 11 1 11 tan 1 tan 0 0 tan 1

2 2 2x x x

10 tan 2 0, tan 2x x

tan 2 tan 2x and , 0,22

x x

No of real sol of x in 10 ,2 15

23. 2

1 2 1 2z i i

2 1 2i 2

1 21

ii

2 1 2 1 0i i

Locus of is from bisector of 1, 1 , 0,0 i.e., 1 0x y

24. ,f x y g x h y

2 24 , 6g x x x h y y y

1 2 4 0 2 2g x x x 1 2 6 2 3h y y y

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

6

in 0,1g x in 0,1h y

min of 1g x g min of h 0y h

min , 1 0 3 0 3f x y g h

25. 6 6

1 1

1 1

sin cos 6 62

i ii i

x y

1 1sin ,cos , 1......62

i ix y i

1, 1i ix y

6

2

26

log 11

x

x

eI x x dx C

e

2

2log 1 log is odd

1

x

x

ex x

e

26. 4 4sin cos 2 4sin cos 0x y x y

2 2

2 2 2 2sin 1 cos 1 2sin 2cos 4sin cos 0x y x y x y

2 2

2 2 2 2sin 1 cos 1 2 sin cos 2sin cos 0x y x y x y

2 2 22 2sin 1 cos 1 2 sin cos 0x y x y

It is true if 2 2sin 1,cos 1 & sin cos sin cos 1x y x y x y

sin cos 2x y

27. cotx y is

22 2

2 cos sincot cos sin

sin

y yy y y

y

2 2 1 511 cot 1 1

50 50y x

501

51

28. Let 0,0p

1 386 2 386SS ae

1 2S p Sp a

25 13 2a

2 12a

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

7

386

12e

121

386

29.

2

2

1 cos

1

sin

2 2

t b b

a at

I x f x x dx f x dx f a b c d

2 2

2 2

1 cos 1 cos

2 1

sin sin

2 2 2 2

t t

t t

f x x dx xf x x dx I I

11 2

2

2 2 1I

I II

30. 500 5002000 42 2 17 1

499500 500 5001 49917 17 ....... 7 1 17 1C C m

Remainder = 1

31. 1 3

1, 1,0 , 0, ,2 2

A B

A lies on 3 4

32. Projection of AB on x axis is

1 3

1 0 1 0 1 0 0 12 2

33. C is foot of on O AB

Equation of AB is 1 1

2 1 3

x y z

C can be taken as 2 1, 1, 3C

2 2 1 1 1 3 3 0OC AB 3 4 11 9

, , , ,14 7 14 14

C p q r

7 14 14 2p q r

34. Let ,x a y b be the asymptotes perpendicular tangents are intersecting at (2,2) 2, 2x y

Equation of hyperbola is 2 2 0x y k

0,0 reason it 4k

Equation of hyperbola is 2 2 4 0x y

Equation of conjugate hyperbola is 2 2 4x y

35. Equation of tangent at (4,4) is

8x y

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

8

2 6 2,6x y A

2 6 6,2y x A

4 2AB

36. Equation of chord is 1 11 4 0S S x y (1)

is tangent to conjugate hyperbola 2, 2 satisfies both (1) and conjugate hyperbola 2, 2 is

point of contact.

37. (a) Number of matrices 4! 24N

a bA

c d

Possible non negative value be 0,2,4,8 , ,A p q s

(b) de value 8 are 4

de value 8 are 4

de value 4 are 4

de value 4 are 4

de value 2 are 4

de value 2 are 4

sum of value of all dets = 0 B s

(c) 3

12

nads ads adjA A A

the absolute value of is least ,A C p r

(d) Algebraically least of 8A

1 1 1616 16 2

8A

A

D r

38. 2 2 2t x x

and 3 2 2and 2 2 1f t x at bt c t x x

0f x has roots less than or equal to 0

1f x has two negative roots 23 2 0x ax b has two negative roots

20, 0, 0 3a b a b

Possible values of , ,a b c be 3, 2, 0a b c

1 23 6 12 0f x k x x k has equal roots

0

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

9

36 4 3 2 0 3 2 0k k

1k

(a) - r (b) - q (c) - p (d) - s

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

10

Physics

39. 2

AR

X V t i

BX Ri Rj

2BAX R R i Rj

AV Vi

BV Vi

0BAV i

40. The temperature of land rises rapidly as compared to sea because of specific heat of land is 5

times less than that of sea water. Thus the air above the land becomes hot & light so rises up so

pressure drops over land. To compensate the drop of pressure, the cooler air from sea starts

blowing towards lands, setting up sea breeze. During night land as well sea radiate heat energy.

The temperature of land falls more rapidly as compared to sea water, as sea water consists of

higher specific heat capacity. The air above sea water being warm and light rises up & to take its

place the cold air from land starts blowing towards sea and set up breeze.

41. 1 sin90 .sin 90y

1 .cosy

2 21 .

dxy

dy dx

2

1y m y

42. After two and half time periods, it is at a distance 2R0 on the negative z-axis. Y-coordinate will be

zero And the x-coordinate =2.5p0 .i.e. it is at a distance 7.5P0 from the mirror, hence its image will

be at 2(7.5P0)+2.5P0=17.5P0.

43.

2 2 2 2

0

1.

4x

q RdE

R x R x

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

11

3/22 2

0

1. 2

4

qRd x dx

R x

2 2 2R x t

2 . 2 .x dx t dt

30

0

2 .

4

x R

x

qR t dtd

t

0

1 12 .2

4 2curved

qR

R R

0

11

2curved

q

0 0

11

2flat

q q

02

flat

q

44. Conservation of angular momentum

1 2 2c cI w m r h v I w mrv

2 2

2

2

r h vw

r

Conservation of energy

2 2

2

1

2cI mr w mgh

45. From the diagram the forces along the line

perpendicular to the inclined plane are

balanced.

3cos37 sin37

4N mg ma mg

-25ms

6a

46. Let the piston be displaced by x

2

0

02

qPA P A

A

2 2

20

2 0

0

1

2

APA P A

A L x

a𝑚𝑎

37

𝑁 𝑇

cos37mg

sin 37ma

cos37ma

mgsin 37mg

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

12

2 2 2

0 0 0

2 2 20 0 0

2

2

nP P P P

L LL x

, 2n

47. Let be the heat generated per unit volume per sec.

3 244

3

dTr k r

dr

3

dT r

dr k

2

6

wrT C

k

At 0, 20r R T C

2

020 406

C

RT C

k

48. Solving for each reflection we get distance from the pole of the lens.

424 .

2 1 2 4 1.25 1.25 1

Rcm

n

49. Wavelength of the incident sound is

10192

2i

uu

u

f f

Frequency of the incident sound is 10 18

1910

2

i r

u uf f f f

u

where when rf is the frequency of the reflected sound.

Wavelength of the reflected sound is

10 11 11 1919

18 18r

r

u u u u

f f f

19 18 9

2 11 19 11

i

r

u f

f u

50.

1 2 3 23

e

E E EE

6

13

e

e

Ei

R r

51. Current through 2V cell 6 13 6

3 3 39

iA

IIT Section

Subject Topic Grand Test – Paper II Date

C + M + P Grand Test – 09 IIT – GT – 09

14th May 2014 I20140508

13

52. 1 2 3 2

3 3e

E E EE

2 3 2 3 2

4 1 3 13 3 13

e

e

Ei A

R r

53. 2 2 2 2

1 2 3, , ,/ 4 /16

v v v va a a a

r r r r

2

1 2 3, ,4 16

r rt t t

v v v

54. 2

2c c

va v a r

r

1 2 11 2 3, ,

4 2 16 2 4c c c

r v r v vv a r v a v a

1 21 2 3

1

22, , ...........

2 24c c

r r t tt t t

v a a r

55. At a 11 2 1

2

vJ m v v m along Z axis

At b 1 1 12 3 2

4 2 4

v v mvJ m v v m along Z axis

13 4 3 along

8

mvJ m v v Zaxis

So, total impulses = 1 2 3 ........J J J

56.

Draw the ray diagrams.

57. (A) X is resistor

2 2

1dv B v

m Fm Bdt R

/

2 2

t T mRv ve T

B

Energy is dissipated as heat is the resistor at the cost of kinetic energy. Total energy is

conserved

(B) X is an inductor 2 2

0 sinvdv B x B

m x tdt L mL

Rod will oscillate simple harmonically and KE is converted into magnetic energy

(C) X is capacitor

Q CV CB V 2 2 0

0 2 2ln .

Fdv dvF B C a

dt dt m B C

constant

v at (D) 0F mg 2 2

constantmg

am B C