iit model paper answer 9

8/20/2019 Iit Model Paper Answer 9

1/13

Max Marks:360

KEY SHEET

PHYSICS

1 4 2 2 3 2 4 1 5 2 6 3 7 3

8 3 9 3 10 3 11 2 12 3 13 3 14 3

15 3 16 3 17 1 18 4 19 1 20 4 21 2

22 2 23 4 24 2 25 4 26 1 27 1 28 2

29 2 30 3

MATHEMATICS

31 2 32 1 33 2 34 1 35 1 36 4 37 4

38 1 39 1 40 3 41 2 42 1 43 3 44 1

45 1 46 2 47 2 48 3 49 1 50 4 51 4

52 4 53 1 54 1 55 4 56 2 57 1 58 1

59 4 60 3

CHEMISTRY

61 4 62 4 63 2 64 4 65 3 66 4 67 1

68 4 69 1 70 3 71 3 72 4 73 3 74 3

75 2 76 4 77 2 78 4 79 3 80 1 81 2

82 1 83 1 84 4 85 1 86 4 87 4 88 2

89 3 90 2


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HINTS &SOLUTIONS

PHYSICS

12

0 0 0

0

1 2

2

I I E C or E

C ε

ε

0 12 8

2 4

8.8 10 3 10 E

155.5 NC 2 Conceptual

3 from input signals , we have

A B OUTPUT nand

GATE

0 0 1

1 0 1

0 0 1

1 1 0

0 0 1

The output signal is shown at (b)

4

A B Y1Y 2Y Y

0 0 1 1 1 0

0 1 1 1 0 1

1 0 1 0 1 11 1 0 1 1 0

5

61 1.5

4

t and drift t km

62

050cos37 v

R

2

40 ; 20 / 10

vv m s

7 conceptual

8


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9 30 20 5di di

l i t vedt dt

30 30 2sec; 1di

L at t i ampdt

10 Basic concept of phasor

11 This error occur in all the instruments which utilizes screw or nut-bolt system

12 L.C of vernier =M.S.D – V.S.D

45 4.9 10 0.01m mm L.C. of screw gauge =0.01mm

13

Let slope of line AC is 1 &m that of line DB is 2m , then

2 21 2

1 1

ln 2 lnm mnRT nRT m m

1

2

2T

T

14 The temperature of these objects being identical the cooling rate will depend upon surface area. The

circular pate has the largest area and hence will cool faster than cube and sphere, Sphere having less

surface area than a cube will cool at the lower rate

152

4 4 0.052 ; 10

2 10

S R cm P Pa

R

16


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0 3

tan378 4

r

6r m

20F P h g r ρ π

510 10 8000 10 36π 5 71.8 36 10 2 10π

17 since it is a conductor

0V = potential at surface=0

0 0

04 4 2

q Q

a aπε πε

2

Qq

18 A at surface B C OV V V V V

Charge is on the outer surface hence insideV remains constant

19 The potential at centre of sphere in which q charge is uniformly distributed throughout the volume is

0

1 3

4 2C

qV

Rπε

By symmetry the potential at centre due to half sphere will be half of the complete sphere

0

1 3 / 2

4 2C

qV

Rπε

0

1 3

4 2 2

Q qQ

Rπε

20 1 2 12

1 2

;

2

r

v v v A A v

v v

energy reflected

1

9energy incident

21

36 / 1.2

30

m sl m

0.3m


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22 Conservation of angular momentum2 2 2

0 '2 2 3

i f

mR mR ml L L ω ω

2

0

2 2

3'

3 2

R

R l

ωω

23 1 2 2 1 2(1 cos )cos 2 cos 2 cos 2 cos 2 E a f t f t a f t a f t f t π π π π π

2 1 2 1 21 1

cos 2 cos 2 cos 22 2

E a f t a f f t a f f t π π π

This is a complex vibration consists of harmonic vibrations of frequencies 2 1 2 1 2, f f f and f f

The highest is 1 2 max. f f So hv T φ

max 1 2T h f f φ

34

15 15

19

6.6 103.6 10 1.2 10 2.35

1.6 10

17.45ev

24 Apply conservation of momentum & energy

25 Let total sum of number of isotopes at the formation of the planet was 0 N and currently is N

0 1 0.22

t

X

N N e N

λ

….(i)

20 0.82

t

y

N N e N

λ

…..(ii)

Dividing equation (ii) by (i) 1 2 4

t e

λ λ

1 2ln 4t λ λ

9

ln 2 1 12ln2

10 2 4t

98 10t years26

At P, 8 3

,d d

x D

For 2nd

maxim,224

2 2d

x D

λ λ

27 0 0 0

2 2 2

I I I B i j k

R R R

µ µ µ

0 32

I B

R

µ


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28 In steady state the capacitor is fully charges and is treated as open circuit, so no current flows

through branch containing capacitor in steady state, So, the circuit can be redrawn as

Potential difference across capacitor in steady state

6 6V V V (-ve sign signifies that left hand plate is of negative polarity )

Charge 1 6 6CV C µ

29 In steady state current through capacitor is zero. For zero deflection A BV V

So, 11

q IR

C

And 1 22

2 2 1

q R C IR

C R C

30 4

3 2

02

4

T xU x x dx x Since there is no loss energy , so

42 2

1

13

4 2

x x E mv

4 24 12 0 x x

2 4 16 48

2 x

4 8

6, 22

2 x


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MATHEMATICS31 As z,iz 1 are collinear

1 1

0 1

1 1 1 1 1 1

z z z z

iz iz i z z i

[taking – I common from 2 R ]

1 1 1

0

0 0 1

z z

z i z i i

[Using 1 1 3 2 3C C C and C C ]

1 1 0 z z i z i z

0 z z z iz i z z i z z i

2 1 1 0 z z i z i z

1 10

2 2

i i z z z z

1 1

2 21

i z

Thus , z lies on a circle

32

As 1 2 3 , z z z circum centre of ABC is 0s z .Also, centriod of ABC is 0s z . Also

centroid of 1 2 31

3G ABCis z z z z . If H z denotes the orthocentres of ABC , then

1

23

G H S z z z

1 2 31 1

2 03 3

H z z z z

1 2 3 H z z z z

33 Let common root be α , then2 2

3 1 0 2 1 0 p and qα α α α Multiply the second expression by 3/2 and subtract from the first to obtain

3 1 1

2 2 2 3 p q

p qα α

Thus, 2

3 2 3 2 3 0 p p q p q

34 2 12n n na a a

/ 20

2cos 2 4 cos 2 cos 2 2

1 cos 2

n x nx n xdx

x

π

/ 2

0

2 cos 2 2 0n x dxπ


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35 30 4 27C

36 conceptual

37 We have

5 0 0 1 0 0

0 1 0 0 1 00 10 2 5 0 0 1

x

AB x x

1/ 5 x

38

1

1 0

1

a a

b b

c c

Applying 2 2 1 3 3 1C C C and C C C

And expand to obtain

1 1 1 1 1 1 0a b c b a c a b

Divide by 1 1 1a b c

39 Two tallest boy can be different groups in 2 2 11 1n nC C ways

40 , 22

α γ α γ δ β α δ γ β γ δ

413

2

04 13 | sin3

xt at t x

x

212 39 3sin / x x a x

212 39 3sin

a x x

x

2

6 3 sin 1a

x x

sin 1, 66

a x

42 From 2sin sin 1, x x we get 2sin cos x x . Now, the given expression is equal to

6 6 4 2cos cos 3cos 3cos 1 1 x x x x

3

6 2cos cos 1 1 x x

33

sin sin 1 1 x x

3

2sin sin 1 1 1 0 x x

43 We have 1 1 sin2S x β 2 1 2 cos2S x x β 3 1 2 3 cosS x x x β 4 1 2 3 4 sinS x x x x β

So that4

1 1 1 3

1 2 4

tan tan

1i

i

S S x

S S

1 sin 2 costan1 cos 2 sin

β β

β β


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1cos 2 sin 1

tansin 2sin 1

β β

β β

1 1tan cot tan tan / 2β π β / 2π β

44 33 87 y m x

1 87 33 33 872

m A mm

2

20 0

dA d A

dm dm

45 Perpendicular distance from centre to the line1

2

common difference

11

2 1 2 22

2 2 4

or or

46 Common tangent nearer P is x=1

directrix of ellipse is x=1e=1/2

1,1

2s

47 Perpendicular bisector of the base must meet the parabola at the vertex of the triangle.

48 Let P( a,b,c) be the foot of the perpendicular from the origin to the plane, then direction rations of

OP are

0, 0, 0a b c i.e, a,b,c

So the equation of the plane passing through P(a,b,c) the direction ratios of the normal to which are

a,b,c is

0a x a b y b c z c 2 2 2

ax by cz a b c

49 Since ' 2 cos 0 , f x x for all x R so f is one-to-one. Moreover,

f x as x and f x as x , hence the range of f is R. Therefore, f is onto aswell

50 Conceptual

51 0'9 limh f x h f x

f xh

0 0

0lim lim

0

h

h h

f x f h f x e g h

h h

0

lim 'h hh

e g h e g h

' 0 0 4g g

Hence 4 10 10 0 f x x c but f g

So 4 f x x

52 Conceptual


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53

1/ 1/

1 21n n

n n

f x x f o f x

x f x f x

Similarly

1/

1 3n

n

x f o f o f x

x

and likewise

1/

.....1

nn

x f o f o f n times x

nx

11/ 0

2

1/ . 11

nn

n n

nn

x x g x dx dx Put nx t

nx

We get the above integral as

111 1

1

nnt

dt t K n t n n

111

11

n nnx K n n

54 h x h I π π

Where sin 4

x

I t dt

π

π

Put ,t θ π so that

4

0sin

x

I d π θ θ

4

0sin

x

d h xθ θ h x h h xπ π

55 conceptual

56

2 2

2 ' ' 2 0d

f x g x f x f x g x g x f x g x f xdx

Hence 2 2 f x g x is constant . Thus

22 2 2 2 216 16 2 2 2 ' 2 16 16 32 f g f g f f

57 Let 1/

1 x

y x

2 3 41 1

log log 1 ...2 3 4

x x x y x x

x x

2 3

1 ..2 3 4

x x x

2 2

1 .. ..2 3 2 3

x x x xa

y e ee

2

2 211 ..... ..... ....

2 3 2! 2 3

x x x xe

2

21 1 1 10 0 .....2 3 2 2

y e ex ex x x

(0,(x) in terms containing x)

20

121 1 112lim3 8 24 x

y ex

e e x

58 Conceptual


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59 Conceptual

60 Vertex is (-2,1), length of the latus return =4a where a=1/2 latus rectum is

parallel to the tangent x+2=0 at the vertex, at a distance ½ from it. So its

equation is x=-3/2,1+1) and (-3/2, 1-1) i.e (-3/2,2) and (-3/2,0). The point not

lying on 2 3 0 3/ 2,2 x y is


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CHEMISTRY

61

4 2

1

f

f

T KCl i KCl

T X i x i x

20.5

4ix

for association of 3 molecules 1

1 1i xn

α

11 1 0.5

3α

0.75α

62 The boyle’s temperature of both 2 H and He is very low.

63 Along the body diagonal two B- ions and one A+ are removed

64 At high concentration of sulphide ions the cations of both II and IV groups are precipitated.

65 Follow CIP rules.

66 2 r n p l =2

12 nr n p l Þ =

67 Depends on stability of carbocations

68 (i) Molecules move faster for which T

M greater. obviously 2 H molecule move faster.

69 Cannizzarrows reaction

70 Rate of reaction depends on leaving ability of the group

71 Nucleophilic substitution

72 In the partial hydrolisys of 6 Xe F different products formed or 4 2 2 Xe O F and Xe O F

73 Tollen’s reagent oxidizes aldehyde group only

74 3 B PH = 2 2 A NaH PO=

75 Conductivity 6 6 64 2 10 1 10 1 10 BaSO

2 24 4

0 0 0

BaSO Ba SOλ λ λ Molar conductivities =

1 2 160 2 20 100ohm cm mol

In the case of sparingly soluble salt4

60 1000 1000 10

100 BaSO V

K

C C λ λ

3510

10 / 100

C moles litre

76 1C epimers are called anomers77 a) basic strength in vapour state depends on +I – effect

c) +R group increase basic strength at para and decrease the basic strength at meta due to – I.

d) Lesser the surface volume more the basic strength, hence NH3 is more basic. R – OH > H2O due to

+I – effect of ‘R’ group

78 Conceptual

79 Conceptual

80 3 B NH = 2C N = 3 2 D Mg N =

81 Conceptual

82


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83 Iodoform reaction84 Teflon, polystyrene and PVC are formed by addition polymerization

85 Conceptual

86 Ratio of concentration of reactants and products is same in all three.

87 0 t Kt C C = -

88 Follow the structure of dichromate ion.

89

2CH

2 H N

20

/

0.5.........

NaNO HCl

C

N N

2 N

H

or

H

90 Sod. Benzoate is metabolized to hippuric acid 6 5 2 .C H CO NH CH COOH

iit model paper answer 9

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