iit model paper answer 8

8/20/2019 Iit Model Paper Answer 8

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Max Marks: 43

KEY SHEET

Physics :

1) 1 2) 3 3) 3 4) 3 5) 4 6) 1 7) 3 8) 3 9) 2 10) 3

11) 3 12) 3 13) 4 14) 4 15) 4 16) 3 17) 2 18) 3 19) 2 20) 2

21) 1 22) 1 23) 2 24) 2 25) 3 26) 3 27) 3 28) 2 29) 3 30) 3

Mathematics :

31)2 32)2 33)1 34)3 35)1 36)1 37)2 38)3 39)3 40)1

41)1 42)3 43)2 44)4 45)2 46)3 47)1 48)4 49)4 50)1

51)2 52)2 53)1 54)4 55)3 56)2 57)1 58)2 59)1 60)1

Chemistry:

61) 1 62)2 63)3 64)2 65)2 66)2 67)3 68)1 69)2 70)3

71) 4 72)3 73)4 74)1 75)1 76)3 77)2 78)4 79)3 80)3

81) 3 82)4 83)3 84)3 85)1 86)4 87)2 88)3 89)4 90)1


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SOLUTIONS

Physics

1. 1 1

2 2

A N

A N

3

2

110 500

2 A

2. Let Am and Bm be the mass of blocks A and B respectively.

As the force F increases from 0 to s Am gm , the frictional force f on block A is such that f=F. When

s AF m gm= the frictional force f attains maximum value s f mgm= .

As F is further increased to ( )s A Bm m gm + , the block A does not move. In this duration frictionalforce on block A remains constant at

s Am gm .

As F further increased, system will start moving and kinetic friction ( )k Am gm will start acting on

( )s k A m m> . Hence C is correct choice.

3. The free body diagram of cylinder is as shown.

Since net acceleration of cylinder is horizontal,

0cos30 AB N mg= or2

3 AB N mg= …… (1)

And 0sin30 BC AB N N ma- = or0

sin30 BC AB N ma N = + …… (2)

Hence AB N remains constant and BC N increases with increase in a.

4. A D A C A BW W W

Hence 3 2 1θ θ θ

5. 1 x x= and 3 x x= are not equilibrium positions because 0du

dx¹ at these points.

2 x x= is unstable, as U is maximum at this point.

6.1.5 1 0.5

OP OP R

5OP R

7. Conceptual,2 sin2

2

mRw f æ öç ÷ç ÷è ø

8. v ev= -


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9. f = 4 ma …….. (1)

( ) )2 23mg f r mr mr a- = +

mg-f=4 ma …….. (2)

From (1) and (2)

88 8

g gma mg a

r aÞ = Þ = Þ =

10. The direction of L

is perpendicular to the line joining the bob to point C. Since this line keeps

changing its orientation in space, direction of L

keeps changing however as w is constant,

magnitude of L

remains constant.

Aliter : The torque about point is perpendicular to the angular momentum vector about point C.Hence only change the direction of L, and not its magnitude.

11. The maximum static frictional force is

cos 2 tan cos 2 sin f mg mg mgm q q q q= = =

Applying Newton’s second law to upper block at lower extreme position2sin f mg m Aq w- = Þ

2 sin f m A mgw q= +

Or 2 sin A gw q= or3 sinmg

Ak

q=

12. 2 4 x ay=

Differentiating w.r.t. y, we get

2

dy x

dx a=

( )2 , , 1dy

At a adx

= Þ hence 045q =

The component of weight along tangential direction is mg sinq .

Hence tangential acceleration is g sin2

gq =

13. At t = 2 seconds, the position of both pulses are separately given by fig. (a) and fig. (b) : the

superposition of both pulses is given by fig. (c)


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2cm

-1 0

1

2 Fig. (a)

Fig.(b)

Fig,(c)

0

0

1

1

2

2-1

14. For interference at B, over all path difference =2

l

2 2(2 ) B I k a a ka= - =

For interference at A overall path difference = 02 2(2 ) 9 A I k a a ka

15. Heat absorbed by gas in three process in given by ACB ACBQ U W = D +

ADBQ U = D

AEB AEBQ U W = D +

The change in internal energy in all the three cases in same. And ACBW is ve+ , AEBW is – ve.

Hence ACB ADB AEBQ Q Q> >

16.3

2 E KT =

3(273 )

2

E k t = +

17. By argument of symmetry, it will be half of the potential produced by the full sphere

Charge on hemisphere Q= , so charge on sphere 2Q=

( )21.

2

K Q KQ

R RÞ =

9 9

2

9 10 5 10300

15 10

KQV V

R

-

-

́ ́ ´= = =

´

18. qi

T = Now 2 2 6 3

3

1T t n i T n

nµ µ Þ µ Þ µ

19. In figure all resistance are connected in parallel.


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So2 / 2

2 / 2eq

R R R

R R

´=

+and current in all resistance flow from positive terminal of battery (means A end)

to negative terminal of battery (means B end).

20. 0 / t use V V e t -=

21. (i) 0

/ 4

V V t

T

= 04V

V t

T

=

/42

2 0 0 0

/4

0

4

3

T

rms T

t dt V V

V V T

dt

ì üï ïï ï

Þ = < > = =í ýï ïï ïî þ

ò

ò

22. 0 sine e t w=

0where e NBAw=

23. The charged particle moves in a circle of radius2

a

2

/ 2

mvqvB

a\ = or

2mv B

qa=

24. OEH is an equipotential surface, the uniform E.F. must be perperpendicular to it pointing from

higher to lower potential as shown

O A B

C

D

E F

G H

y

x

E

v

v

1v

Hence,ˆ ˆ

2

i j E

æ ö-ç ÷= ç ÷è ø

( ) ( )0 22 / 2

E BV V

E V m EB

- - -

= = =

( )( )

ˆ ˆˆ ˆ. 2 /

2

i j E E E i j V m

-= = = -

25. When the rod rotates, there will be an induced current in the rod. The given situation can be treated

as if a rod ‘A’ of length ‘3l’ rotating in the clockwise direction, while an other say rod ‘B’ of length

‘2l’ rotating in the anti clockwise direction with same angular speed ' 'w .

As we know that 21

2e B lw=

For ‘A’ : A : For ‘B’ : B

( )21 3

2 A

e B lw= & ( ) ( )21 2

2 B

e B lw= -

Resultant induced emf will be :


6/12

( )21 9 4

2 A B

e e e B lw= + = - 25

2e B lw=

26. By snell’s law :

2

1

sin

sin

i

r

m

m=

For 1 4,i r q q= = and 1 1m =

11

2

sin

sin

qm

q= .

27. BE CC B BV V I R= -

5.5 5 0.5 BE V V = - =

28.

In the first case K.E. of H-atom increases due to recoil whereas in the second case K.E. decreases

due to recoil but 1 1 2 2 E KE E KE + = + .

2 1 E E \ >

29. Here path difference will be :( )2 1 X t m mD = -

( )2 12

t p

d m ml

Þ = -

Hence (C)

30. At a distance x consider small element of width dx.

dx

x

Magnetic moment of the small element is :2

.2

qdx

dm x

w

p p

æ öç ÷ç ÷è ø=

/22

/2 2

q M x dx

w

-

= ò

2 2

24 12

q q f M

w p= =


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Mathematics

31.

Domain of

32.

33..

= 0

34..

35..

36..

37.. 26 600

1sec tan

3 I x dx x

π π

38..

Let

2

2 2 21 18,

18 18i i d i xd x d diσ σ

=

2

1 9 5 1 94518 18 2 4 4

3 / 2 xσ

39..

(1,3),(1,5)(2,3)(2,5),(3,5),(4,5) R

1 (3,1)(5,1)(3,2),(5,2)(5,3),(5,4) R \

1 (3,3),(3,5),(5,3),(5,5)then ROR

40..


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41.

First 2place can be filled by (41 or 42 or 46 or 64 or 62) 5ways

Remaining by ways

Total ways = 5 1680= 8400.

42.. Let be one of the lines represented by given

Let be rootsIf

43..

Radius =

44.. is harmonic mean between b and C

=

45. .

P lies on ellipse

46.. Let Ellipse be

Director circle is ---------(1)

Now (0,0) lies on (1)

Locus of c 2 2 2 2, is x y a bα β

47..


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10/12

2 +x- k=0⇨ x=

For least + ve non _ integral ution

X= = ,p is odd integer

56.. <

= x< π ≃ 5.5

, maximum value of x is 5

57.. r2

cot cot cot

= . =

58.. p(s)=

=p(F)== P(F S or F F F S or F F F F F F S or……)

= P(F).P(s)+P .P(S) + P .P(S)+.................= .

59.. 2

a b

.= . 2 . .a a a b b b

=2(1+cos )

=2(2cos2

)

cos =1

2a b

60.. ( b

+ c

) = ( a

. c

) b

– ( a

. b

) c

⇨1

.2

a b

b

-

1.

2a b c

=0

Since , ,a b c

are linearly independent

a

. b

= - Cos =. 1

2

a b

a b

=


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Chemistry61. 3 2 3 2 2 3 1 0.732void metal metal void metal metal void void metala r r r r r r r r r

62.2

1 2

2

2 1

Z

Z

l

l =

64.

2 4 22

g g N O NO

1 0

1- 0.8= 0.2 1.6

1.8Totaln =

2 4

0.21.5

1.8 N

P O = ´ ,2

1.61.5

1.8 NO

P = ´

1.6 1.61.5 1.5

1.8 1.80.2

1.51.8

pk

́ ́ ´=

´

67. 0.0591 3 0.0591 4 0.0591 E = - ´ + ´ =

68. 2C C C C

H O H OH

H OH C

Conductivity of water2 H O H OH

k k K

. .1000 1000

C C k x y

1000 18

1000 18 1000

C k k x y x y

x y

,

2 6

2

10.w

k k C C

x yα α

71. Colloid with high zeta potential (negative or positive) are electrically more stable.

73. It is diazotization reaction and initially formed carbonation is

2CH

and forms

2CH OH

, or expands to new carbo cation

which forms

OH


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76. 2 2 2 36 5 2 10 I H O Cl HIO HCl

81. When 0P ,all real gases come near to ideal behaviour. At least S.T.P i.e., 1 bar,

273.15KVm = 22.711 dm3

82. Both 2 R NH and 2Ph NH gives positive iso-cynide test with 3 / CHCl NaOH . So given

reagents not useful to distinguish

83. 1

2

2.303 logvap

b

H PS R

T P

84. 42 3 2 3 3[ ( ) ] LiAlH ozonolysis

A CH C CH HCHO CH CO CH

3 3 3CH OH CH CH CH

OH

85.12 1

0.74 1.8660 0.5

i= ´ ´ ´0.74 5 0.5

11.86

i ́ ´

Þ = =

86. 10.693

40K = , 0

1.386

2 20k =

´

1

0

1

2

k

k =

88. Flame temperature is directly proportional to the amount of heat released per mole of

products

90. The ubility of salts of weak acids like phosphates, acetates increases with decreasing

the pH. This is because at lower pH, concentration of the anion decreases due to its

protonation.

iit model paper answer 8

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