IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

1

Key Answers:

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

11. 4 12. c 13. a 14. a 15. b 16. b 17. a 18. 19. 20. b

21. a 22. a 23. b 24. b 25. d 26. 5 27. 2 28. 3 29. 2 30. 4

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

41. c 42. d 43. d 44. a 45. 5 46. 8 47. 4 48. 3 49. 6 50. d

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

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

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

56. A –rs, B –p, C –rs, D -p ; 57. A –r, B –qr, C –ps, D -qr

Solutions:

Chemistry

1. Coagulation value is inversely proportional to their flocculating power

2. 2 22Ca Lac Ca Lac ower.

Concentration of …....after dissociation 2 0.3 0.9 0.54

1 114 5.6 7 log0.54

2 2apK

2.8 log0.54apK

3. Allylic C is more stable than 1 C . Hetero atom next to C makes it very stable. C2 is more

stable than C1 .

4. For aldose different arrangement of groups in C 1 anomer

Different arrangement of groups other than C 1 epimer

5. (A) 3 3 3 34 4,Cr NH ClBr Cl AgNO Cr NH Cl Br NO AgCl (white ppt)

(B) 3 2 3 3 2 34Cr NH Cl Br AgNO Cr NH Cl NO AgBr (pale yellow)

6.

7. Molarity of 2 4N H solution 0.16 1000

0.0132 500

M

2 4N H reacts with water in the following way.

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

2

Let the amount of 2 4N H reacted with water =

2 4 2N H H O 2 5N H OH

Moles at start 0.01 0 0

Moles at equilibrium 0.01

or 2

64 100.01

0.01 0.01

2 8

4

4 10

2 10

% of 2 4N H reacting with water 42 10 100

2%0.01

8. The relevant chemical equation is

22 3 3 2Ca Na CO CaCO Na

40g 106g

Thus 40g 22 3106 Ca g Na CO

Converting these into required units

240 2g Ca equivalents of 2 3Na CO Mol.wt 106

No.of Eq= =Eq.wt 53

40 1000mg of 2 2Ca equivalent

2 1000 milliequivalents of 2 3Na CO

20mg of 2 2 100020

40000Ca

milliequivalents of 2 3Na CO

= 1 milliequivalents of 2 3Na CO

Hence, no. of meq. Required to soften 1 litre of sample of water = 1

9. 2 2 33 2N g H g NH g

0 032 0 0

2 46 92

fH H NH

kJ

Determination of entropy change, we know that

0 0products ReactantsS S S

= 2 192 191 3 130 = 384 191 390

Now according to Gibbs Helimholtz equation.

0 0 0

3 92 298 0.197 33.294 33.294 10

G H T S

kJ j

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

3

Now we know that

0 ln

2.303 log

p

p

G RT K

RT K

333.294 10 2.030 8.3 298 log pK

333.294 10

log 62.303 8.3 298

pK

10. 2 2 2 2x y yC H O xO xCO yH O

Since oxygen taken is 2x litre and thus x litre 2O is left at STP after reaction. Also x litre of

2CO is formed by 1 mole of . .O C

2 2.24x

1.12x litre 2 2 21.12

molee 0.05mole 22.4

CO CO CO

Mole of 2H O formed 0.9

0.0518

y

: 0.05: 0.05 1x y x

1,y i.e., ratio of x and y

Empirical formula of 2. .O C CH O

And Empirical formula wt. of . . 30O C

Now Mol. Wt. of compound is derived by Raoul’s law

s

s

P P w M

P m w

0.104sP P mm and 17.5P mm

0.104 50 18

150.517.396 1000

mm

Molecular formula 2 5CH O 5 10 5C H O

150.5

530

n

11. Volume of the unit cell 3

85A 5 10 cm

24 3125 10 cm 22 31.25 10 cm

Density of FeO 34.0g cm

Mass of the unit cell 22 3 31.25 10 4cm g cm

225 10 g

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

4

Mass of all molecules of FeO = 23

72

6.22 10

221.195 10 g

No. of FeO molecules/units cell 22

22

5 10

1.195 10

g

g

4.19 4

12. Fe 2 in 1

8 of T.V. One Fe 3 in another th

1

8 of T.B. and one Fe 3 in th

1

4 of O.V.

13. Divalent cation 2Zn occupy 1

8th of tetrahydral voids

14. Charge 2O charges of

8

3X

ions

8

4 23

n

3n

4 3 1 (ionic vacancy)

15. R.D. step is the formation of C

16. This reaction is intramolecular rearrangement. Benzyl C is more stable than alkyl C .

17.

18. Conceptual

19. 2 2 45 5( )

:

Violet Colour

A S Na Fe CN NO Na S Na Fe CN NOS

3 3( )

: 3

Blood red colour

B R NaSCN FeCl Fe SCN NaCl

34 46 6 3

( )

: 3 4 12

Blue colur

C Q Na Fe CN Fe Fe Fe CN Na

33 4 4 4 4 4 32 3

( )

: 12HNO

Yellow ppt

D P Na PO NH MaO NH PO MoO

CH3 CH3

OH

CH3

N2

+

CH3

CH3 CH3

O

CH3CH3

H

CH3 CH3

O CH3

CH3

-N2

+

-H+

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

5

Mathematics

20. Put 1 28t A B C D E

Put 0 39t E

67A B C D E

21. 0 0y 2

64 5 4 16 7 5 0a a 15, 2a

n s length of the interval 20, 0

20n s

n A length of the interval 15, 2

13

13

20P A

22. 3 2 21 2 1 2 3 1 2f x x x x

22 1 5 7x x x

max at 1x & min at 7

5x

23. Equation of BC is b

y xa

________(1)

Equation of OA is a

y xb

________(2)

Put 2 2 2 iny x y r

2

2 2 2 2 2 2 2

21 , liesan

ax r a b a b x y r

b

2 2x b

x b

y a

, or ,A b a b a

24. 4 4 4 2 2 2 2 2 2 2 22 2 2 3a b c a b c a b c b c

2

2 2 2 2 23b c a b c

,a b

,a bO

A

B

C

1 7

5

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

6

2 2 2 3

2 2

b c a

bc

3cos

2A

1sin

2A

25. ,a b is obtuse . 0a b

214 8 0,x x x also ,6

b k

22 0x x

2 1 0x x

1

0, ______ 12

x

(1) & (2) have no common x

3

cos ,2

b k

2

3

253

x

x

2 22 3 53x x

2 22 3 53x x

2 159 ______ 2x

26. Let 0.4, 0.6, 11p q n

1p q it follows binomial distribution

require the porosities 5 6p x p x

5 6 6 511 11

5 60.4 0.6 0.4 0.5c c

5 511

5 0.4 0.6 0.6 0.4c

511

5 0.24c

0.37

5000 1850 370 ,  5k k

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

7

27.

4

5

2 21 2 1 2

2 1 1 2

127 ,   2k k

28. 3, 10dx

xdt

23

4A x

3 53

2 20 3 15 3 64 4

dA dxx

dt dt

5 3 225 75 ,   3k k

29.

3

1

xx x dx

2 3

1 2

1 2

1 2x dx x dx

2 32 3

1 2

1 2

2 3

x x

1 10

2 3

5

6

12965

12966

216 5     1080 540 ,  2k k

30. Point of intersection 12 12

,7 7

Q

Coefficient of AB with P as its mid-point is

1 1

2,x y

x y but Q liesan it

1 1

12 122

7 7x y 1 1 1 16 7x y x y

locus of P is

12

77

xy x

12

7k

343 49 12 588k 147  p 4p

1

5

2 16 1 0

1____ 13 4

1____ 24 3

x y

x y

2 1 5 0x x

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

8

31. 20

sin 2 2lim

33t

x x

x

32. 0 0

1lim

sincos sin 2cos

t t

t tLt

tt t tt t

t

33.

lim1

x

a

t a

xf x f t dt

af a

Passage - 1

2 ______ 1xy c

Point an (1) can be , , ,i i ii

c cP ct x y ct

t t

Consider of normal at P to (1) is

4 3 0ct t x ty c

it passes through ,

34. i ix c t cc

35. 1

ii

y c ct c

1 2 3 4

1 1 1 1c

t t t t

3

4

/s cc c

cs

c

36. xi cti 41 2 3 4c t t t t 4 c

cc

4c

4

1 2 3 4

1yi c

t t t t

4 1c 4c

37. (A) a b c d = a c d c a b c d

a c d b a b c a R

(B) a c b d a c d b a c b d a c d b a b c d P

(C) a d b c a d c b a d b c a b c d d b c a Q

4 3 0ct t t c

1t

2t

3t

4t

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

9

(D) a d c b a d b c a d c b S

38. (A)

2 21 1

x x

xx

dx e e dxI

e e

x xe t e dx td

2 1

t dtI

t

21

log 12

t c

21log 1

2

xe c R 21log 1

2

xx e c T d

(B) 2

2log 1

1

dtI t t c

t

2log 1x xe e c P d

2log 1 1 xx e c s

(C) /2

2,

1 1

x x

xx

x

e edx dx P S

eee

(D)

2 2

2 2

1 .

1 1

x

x x x

e dxI dx

e e e

Put tanxe

2secxe dx d

2

22

sec

tan sec

a dI

3cos

sinI d

21 sincos

sind

cot sin cosd d

2sin

log sin2

c

2

22 1

1log

2 1

x x

xx

e ec

ee

2

2

1 1 1log 1

2 2 1

x

xx e c Q

e

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

10

Physics

39. Net external torque on the system is zero. Therefore, angular momentum is conserved. Force

acting on the system is only conservative. Therefore, total mechanical energy of the system is also

conserved.

40. f v T

2AB CDf f

4AB CDT T …(1)

Further 0p

AB CDT x T l x

or 4x l x (as 4AB CDT T )

or / 5x l

41. From Y to X charge flows to

plates a and ,b

0, 27a b a bi fq q q q C

27 C charge flows from Y to

.X

Correct option is (c).

42. When 3S is closed, due to

attraction with opposite charge, no flow of charge takes place through3S . Therefore, potential

difference across capacitor

plates remains unchanged or

1 30V V and 2 20V V .

Alternate Solution

Charge on the capacitor is

1 30 2 60q pC and

2 20 3 60q pC

or 1 2q q q (say)

The situation is similar as the two capacitor in series are first charged with a battery of emf 50V

and then disconnected.

When 3S is closed, 1 30V V and 2 20V V

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

11

43. Net magnetic field due to both the wires will

be downward as shown in the figure.

Since, angle between v

and B

is 180

Therefore, magnetic force

0mF q v B

44. Momentum of striking electrons

hP

Kinetic energy of striking electrons

2 2

22 2

P hK

m m

This also, maximum energy of X -ray photons. Therefore, 2

2

0 2

hc h

m or

2

0

2m c

h

45. Normal reaction in vertical direction 1N mg

Normal reaction from side to the groove 2 sin37N ma

Therefore, acceleration of block with respect to discs 1 2cos37r

ma N Na

m

Substituting the values we get, 210 /ra m s

46. 2

2net

e Bvl Bvli

R r r

Here, 1 2r

2 1 2 4i A

16mF ilB N

47. Electron passes undeviated. Therefore,

e mF F

or eE eBv

or /E V d

Bv v

(V potential difference between

the plates)

or V

Bdv

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

12

Substituting the values, we have 3 6

12002

0.3 10 2 10B T

Further, direction of eF

should be opposite ofmF

. or e E e v B

E v B

Here, E

is in positive x direction.

There v B

should be in negative x direction or B

should be in negative z direction (or

perpendicular) to paper inwards, because velocity of electron is in positive y direction

48. Using Kirchhoff’s first law at junction a and ,b we have found the current in other wires of the

circuit on which currents were not shown.

Now, to calculate the energy stored in the capacitor we will have to first find the potential

difference abV across it.

3 5 3 1 3 2a bV V

a b abV V V 12 volt

21

2abU CV

2614 10 12 0.288

2J mJ

49. If we assume that mass of nucleus >> mass of mu-meson, then nucleus will be assumed to be at

rest, only mu-meson is revolving round it.

In thn orbit the necessary centripetal force to the mu-meson will be provided by the electrostatic

force between the nucleus and the mu-meson. Hence,

2

2

0

1

4

Ze emv

r r …(1)

Further, it is given that Bohr model is applicable to the system also.

Hence,

Angular momentum in thn orbit 2

nh

or 2

hmvr n

We have two unknowns v and r (in thn orbit). After solving these

two equations, we get

2 2

0

2

n hr

Z me

Substituting 3Z and 208 ,em m we get 2 2

0

2624 e

n hr

m e

IIT Section

Subject Topic Grand Test – Paper II Date

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

12th May 2014 I20140508

13

The radius of the first Bohr orbit for the hydrogen atom is: 2

0

2

e

h

m e

Equating this with the radius calculated in part (i), we get 2 624n or 25n

50. qE

am

or q

am

51. 2

2

y

y

uh

a . Here

y

qEa

m and 100sin30 .yu

52. 2

,y

x x

y

uR u T u R

a

will increase when ya will decrease.

Here .y

qEa

m

53. mv

rBe

…(i)

2

nhmvr

…(ii)

Solving these two equations, we get

2

hnr

Be

and

22

nhBev

m

4

nhBeK

m

54. Ans: (a)

55.

2 2

2 /

e eM iA r r

T r v

2 4

evr neh

m

Now, cos1804

nehBU MB

m

56. If fork loaded with wax the frequency may increase or decrease.

If the tension in the string increases then the frequency may increase or decrease.

If fork is filed or string tension decreases then the beat frequency always increases.

57. Draw the ray diagrams