iit model paper answer 5

8/20/2019 Iit Model Paper Answer 5

1/12

Max Marks:36

KEY SHEET

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

11) 1 12) 1 13) 1 14) 1 15) 1 16) 2 17) 3 18) 4 19) 2 20) 1

21) 2 22) 2 23) 3 24) 1 25) 3 26) 4 27) 2 28) 3 29) 3 30) 3

Chemistry :

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

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

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

Mathematics :

31) 3 32) 3 33) 1 34) 2 35) 1 36) 4 37) 2 38) 2 39) 2 40) 3

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

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

SOLUTIONS

PHYSICS:

1. Conceptual

2. c

71 10 LC

w = = ;710

2 f

p= ;

8

7

3 102 60

10l p p

´= ´ =

3. v0 =22 (8.5 0.8)g

r g h

-; v0 =

22 (2.5 0.8)gr

g h

-; = 17/77

4. B

kT is dimensionless

[B] = [kT]Also [I] = [AT2]

[A] = [IT – 2]

[AB2] = [IK2]5. Process AB and CD are isothermal

W = nRT f lni

V

V

æ ö÷ç ÷ç ÷ç ÷çè ø

Dotted lines are isochoric lines as they pass through origin

f i

V

V is same for both process

WAB = nRT1

f lni

V

V

æ ö÷ç ÷ç ÷ç ÷çè ø


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3/12

20. longest Emin

for Balmer longest is for n = 2 3

for Lyman longest is for n = 1 2 longest Balmer > longest Lyman

21. Conceptual

22. Force of buoyancy FB = A[x0 + (L – x)w]gWeight W = LA

sg For equilibrium F

B= W

LAsg = A[x

0+ (L – x)

w]g L

s= x

0+ (L – x)

w

s=

x

L(

0 –

w) +

w

x

L=

0

w s

w

r r

r r

-

-=

1000 800

1000 300

-

-=

2

7

23. Least count =1( )main scale

N ; 0.1 mn =

1( )

20

main scale

1 Main scale division = 2mm

24. Since area vector rotates in xy plane, therefore flux is due to by only.

At any instant (t) = r2 By

cos t|E

ind| = r2 B

y sin t

25. Field will be uniform at the outside and, E =2

(2 )

(4 )

k Q

a=

28

kQ

a

26. 4maximum 100 10 0.01 I A A

gV I G R

10 0.01 100 R 900 R

27.

The Boolean expression of the given circuit . X A B=

A B= + (Using De morgan theorem )=A+B (Using Boolean identity)

This is same as the Boolean expression of OR gate.

Alternative method

The truth table of the given circuit is as shown is the table

A B A B . A B . X A B=0 0 1 1 1 0

0 1 1 0 0 1

1 0 0 1 0 1

1 1 0 0 0 1

This is same as that of OR gate .

28. Conceptual29. Conceptual30. Conceptual

Chemistry

31.


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3

2 2 2 2

H

CH

CH C CH CH CH CH

|

O

3

2 2 2 2

H

CH O

CH C CH CH CH CH

|

|

1

3

2 2 2 2

SN

CH O H

CH C CH CH C H CH

| |

CH3O

H

H

CH2OH

O

H H

CH3

CH2OH

-H+

CH2OH

CH3

OH

32. Conceptual

33. Conceptual

34. p – nitrophenol (III) > phenol (I) > p – cresol (II).

35.22

2 3 2 3 24 2O

SiC NaOH Na SiO Na CO H O

36. Conceptual

37.

Cl THF Br Mg Cl Mg Br ClC

O M gBr

Cl C OH

O

H

1 mole X

C

O

O

O

38. Conceptual

39. Conceptual

40. - NO2 is a meta – directing group. As it is also a deactivating group so no chance of

introduction of second – Br atom.

41 2 2 32 6 2 42 2CH O CH CH

NaH B H NaB H


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42. Conceptual

43. A is CH3 – CHO, which then undergo aldol condensation to give compound B.

3 2 3 2

|| |

.

| |

NaOH

O OH

CH C H CH CHO CH CCH CHO

H H

44. Conceptual

45.1

1

100 1

1 200 4

X A

X A= ´

-1 1

8 1 A x A x= -

1 1

9 A

x =

1 8

100 2009 9Pm = ´ + ´ = 188.8 Torr46. As same charges are present at nearest position (less stable)

47. 4 3 3 444CuSO NH Cu NH SO 48.

1K 2B

2C2K

A Where1

2

1

2

K

K

2 2

2 0 0

A B C

Finally 0.51

2 1.53

0.5 2 0.5 2 0.5 1 2 3.549. Conjugate base is week conjugate acid strong50. Conceptual

51. Conceptual

52. 3 cis & 3 trans isomers

i)

C C

Cl

3CH

H

3CH ii)

C C 2

CH Cl

H

3CH

H

iii)

C C Cl

H

2 5C H

H

iv)

C

H

Cl2 3C H

3CH

C

H

Cl

3CH

2 3C H

iii

53. Conceptual

54. For isobaric process


6/12

2

1

logCPT

S nT

55.

2

0 5

NaNO HCl

C

/

NH2 2

N Cl

56. Conceptual

57. Conceptual58. Condensation polymerization

59. Conceptual

60. Conceptual

Mathematics61. R(b c) a bc

2R sin A bcb c

sinA2 bc

Now sin A 1b c

12 bc

2

b c 0 b c sin A 1 0

A 90 and b = cHence, triangle is right isosceles62. Since and are unit vectors and equally inclined to each other.

is are equilateral triangle and

Area of the base

If is the centroid of the then is a right angled triangle and


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Volume of the tetrahedron

63. Let be the required angle, then

64. Let be the of the line of the common perpendicular (or SD) to the

two given lines. Then, we have and

solving, these, we get or

of SD are

Also, is a point on first line and is a point on second line, then

and two lines are said to be skew lines or nonintersecting lines if they do not lie in the same plane.

65.


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66.

1 2 3

10 3 2 4

4s s s

No. of non negative integral solutions of 1 2 3 4s s s is 3 1 24 3 1 6c c

67. A-event that sum 5 occurs, B-sum 7 occurs

,probability that neither a sum 5 or 7 occur

68. Probability of equal number of W and L is

69.

f x kx c

0 1 1 f c ' 1 f x k

1 f x x

1 1 1

2 2 2

1 1 0

1 1 12

1 1 1 2

xdx dx dx

x x x

π

70.

2

2

2 2 2 2 2 2

1 1 1 1 1 1 1 1 1

1 1 1 1 1 5

1 1 1

α β α β α β α β α β

α β α β α β

71. Let equation of variable circles is 2 2

1 1 2 3 0 x y x yλ equation of common

chord of variable circle and given circle is

1 0 2 3 1 6 0S S x y xλ common chords always pass though point of

intersection of 1 – 6x = 0 & x + 2y – 3 = 0.

72. Let the numbers be 1,2,3,4,....., n and the erased number be x then 1 x n

Now,

172 35

1 17

n n x

n


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1 11

72 2351 17 1

7 235

2 17 2

7 2352 17 2

1470 2

17

69 70

n n n nn

n n

n n

n n

n n

n or

at n = 69 ;

69 x 7072 35 7

68 17

x

x

at n = 70 ; x I 73. We have to de- arrange 5 objects to 6 places

No. of ways = 44 44 4 9 9 3 2 2 2 1 1 = 256

74. Let 2 2

x be the point on ellipse 1

9 4

yP θ

Equation of Tangent,cos sin

13 2

x yθ θ --------- (1)

at 2 1 cos 2 1 cos

3 3,sin sin

x y T θ θ

θ θ

at 12 1 cos 2 1 cos3 3,sin sin x y T θ θ

θ θ

1T

P θT

3 x

3 x Circle with 1TT as diameter, 2 2 1 2 1 2 1 2 1 2 0 x y x x K y y y x x y y

2 2 5 0 x y yλ

It passes through 5, 075. Let 201 20 . 2

r r a b

r r T C x y kx y

20 1 11 20 . 2 ss a b

S sT C x y kx y

For given condition r – s=1

and1

120 .2 20 .2 13

s s

s sC C S

r=14 a=6

76. a,b,c are in.G.P 2b ac


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22 0ax bx c can be written as

2

0ax c

c x

a It is common root

Subsitute c

x

a

in given equation

77. Let ABCD be the given rectangle

and EFGH be the required reactangle

A = area of EFGH

2 2 2 2sin cos sin cos

1 1sin2

2 2

b a a b

ab a b ab a b

θ θ θ θ

θ

78. 6

24 2 5 4 3. 1 . 6 5 4g x x x x x x x dx

6

4 5 6 5 4 3. 6 5 4 x x x x x x dx Put 6 5 4t x x x

76

7

t t dt C

7

6 5 4

7

x x xC

But it passes through (0,0) 0C

731

7g

79. Draw the graphs,100

x y and y = sinx. By using this graph Find the no. of solutions

80. Let 3 3 1 f x x x

Then 1 23 1 f x x Let 1 0 1 f x x and f(1) f( – 1) < 0

81. Conceptual

82. f(x) is symmetric about y = x

Required area = 2

0

2x f x x dx

π

= 2

0

2 sin sin xd x dx

π π

π

= 32 2 2 2 3K

83.2

1

2

1

xdy

dx y

Tangent equation is 2 2 3 31 1 1 1 x x y y x y 2 2 3

1 1 x x y y a

Since, it passes through 2 2, x y 2 2 31 2 1 2 1 x x y y a

and 3 3 31 1 2 x y a


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3 3 32 2 3 x y a

By solving (1), (2), (3) we get result

84. : ( ) ( )cos sin cos sini i ie e i i ea b g a a b b+ = + + + =

Let , ,i i ia e b e c ea b g = = = then a b c+ = . Also 1 1 1a b c

+ = .

2ab cÞ = ( ) ( )2i ie ea b g +Þ = ( )sin sin 2a b g Þ + = .

85. The given expression is the coefficient of x4

in

4C0 (1 + x)

404 –

4C1 (1 + x)

303+

4C2 (1 + x)

202 –

4C3 (1 + x)

101+

4C4

i.e. in [(1 + x)101

– 1]4

i.e. in (101

C1x +101

C2 x2

+ ….. +101

C101 x101

)4

= (101)4

option (B) is correct.

86. ( ){ }2 2, : 4 3 0, , R x y x xy y x y N = - + = ÎLet x N Î , 2 24. . 3 0 x x x x- + =

( ), x x R\ Î \ R is reflexive .

we have ( ) ( )() ()2 2

3 4 3 1 3 1 9 12 3 0- + = - + = or ( )3,1 RÎ

Also ( )( ) ( )221 4 1 3 3 3 1 12 27 0- + = - + ¹ .

( )1,3 R\ Ï , R\ is not symmetric.

Again ( )9,3 RÎ because

( )( ) ( )229 4 9 3 3 3 108 108 0- + = - = .

and ( )3,1 RÎ because ( ) ( )() ()2 2

3 4 3 1 3 1 12 12 0- + = - =

and ( )9,1 RÎ if ( )( ) ( )22

9 4 9 1 3 1 0- + =

if 84-36=0 which is not possible.

( )( )9, 3 , 3,1 R\ Î and ( )9,1 RÏ R is not transitive\ .

\ Relation R is reflexive but neither symmetric nor transitive.87. As lines are perpendicular to each other ‘C’ moves on a circle with AB as diameter.

Now P, mid-point of AB (which is fixed) when joined with C is median.

Centroid is moving at a constant distance 1

PA3

from P.

Locus is a circle

A is point of intersection of x + 4y + 2 = 0 and x – y + 1 = 0 i.e.,6 1

,5 5

B is point of intersection of 4x – y + 6 = 0 and x + y + 3 = 0 i.e.,9 6

,5 5


12/12

3 7P ,

2 10

therefore locus is2 2

3 7 17x y

2 10 50

88.Let n be even , f x as x .

Let 3 2C i be the centre of the circle 3 2 4 z i . Join the origin to C and let it meet the circlein A and B ( see figure).

B

A

O

C

3

2

Least value of z OB

CB OC 2 24 3 2

4 13

Maximum value of 4 13 z OA

The absolute difference between the maximum and minimum values of z is 2 13

89. The inverse of p qÞ : is p qÞ: the contra positive of p qÞ: is q pÞ: .

[Contra positive of p qÞ is q pÞ: : ]90. Give that , mean 21= and median 22= .

we know, mod 3 2e median mean= - .( ) ( )mod e 3 22 2 21 66 42 24Þ = - = - =

iit model paper answer 5

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