iit model paper answer 7

8/20/2019 Iit Model Paper Answer 7

1/19

Max Marks:360

KEY SHEET

Physics:

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

11) c 12) c 13) d 14) d 15) a 16) c 17) b 18) c 19) b 20) d

21) a 22) a 23) c 24) b 25) a 26) a 27) a 28) d 29) b 30) b

Mathematics:

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

41) d 42) d 43) a 44) a 45) b 46) a 47) d 48) a 49) a 50) b

51) b 52) b 53) b 54) a 55) b 56) c 57) a 58) b 59) b 60) d

Chemistry:

61) d 62) b 63) c 64) a 65) a 66) b 67) d 68) a 69) a 70) a

71) d 72) d 73) a 74) b 75) d 76) c 77) c 78) d 79) b 80) b

81) a 82) b 83) d 84) a 85) b 86) d 87) b 88) b 89) c 90) a

SOLUTIONS-:PHYSICS:-

1. (a)


2/19

2. (b)As we move away from the origin along X -axis, the area under p-V curve is continuously increasing

and hence work done by the system during the process is continuously increasing. As temperature is

constant the internal energy remains constant. From 1st

law, we can say that as W is continuously

increasing, the heat supplied is also continuously increasing as V is constant.

3. (a)The slope of x t graph at 0t is +ve i.e., initial velocity of the particle is +ve and x t graph isconcave down, so it means acceleration is –ve but whether it is constant or not it can’t be predictedfrom given information. As v and acceleration are in opposite directions, the motion is the retarded

one and finally particle stops.

4. (a)Here to solve this question we can use principle of superposition. The given structure can be

considered as combination of two as shown in figure.

E

at 0 (due to given structure)

0 0at at E due to I E due to II

2

0

4

dq E

Rπε towards dl

2 2 3

0 0

2

4 8

dldl R

R R

θθπ

πε π ε

.

5. (d)The charge of the capacitor means the charge appearing on the facing surface of the capacitor.

So, 2Q CV 2Q

V C

.

6. (a)

In steady state, capacitor can be treated as open circuit and hence circuit diagram would be asfollows:

Apply Kirchhoff’s voltage law to two loops 1 and 2

1 1 1 10 E V V E

2 1 2 2 1 20 E V V V V E

1 2 E E .Charge on 1,C 1 1 1 1 1Q V C C E

Charge on 2C , 2 2 2 2 1 2Q C V C E E .

7. (d)

As battery is connected across the resistor, thermal power developed in the circuit is,2

E H R

.

Here, E is constant, so1

H R

. As time increases, temperature of the resistor increases and therefore

its resistance and hence rate of production of thermal energy decreases.

8. (c)For refraction at the spherical surface,

1 21, , , , 2 R R u v Rµ µ µ 2 1 2 1

v u R

µ µ µ µ


3/19

or 11

22 R R

µ µ µµ

.

9. (c)This is a direct question based on concept of pure rolling. For no slipping relative velocity of point of

contact has to be zero.

Here velocity of the bottom-most point on cylinder is 0v Rω , and velocity of platform is v. So,

from definition of pure rolling, 0 0v R vω 0v v Rω .

10. (a)The circuit is steady state is shown in the figure. Capacitor would be open circuited while inductor

would be short-circuited.

Two 2R resistors have no role, whatever current is coming that goes to the inductor branch.

11. (c)

For 2 Li ion, ground state energy is 13.6 9 122.4 eV and on absorbing 91.8 eV energy electrontransits from 1

stenergy level to 2

ndenergy level.

So, increase in angular momentum,

3421.05 10

2 2 2

h h h L J s

π π π

.

12. (c)As medium remains same, the wave-speed remains same, but as frequency changes, new wavelength

is given by,

1

2 2 2

v v

f f

λλ

.

13. (d)Wavelength of the water waves is

2

5

vm

f λ .

Frequency as received by person in boat is,

velocity of wave relative to the person f

λ

10 2 12 530

2 f Hz

λ

.

14. (d)

Draw the free body diagram of block and platform. As M m , the acceleration of the platform,mg

a M

µ can be taken as zero and hence its velocity remains constant as 4m/s. Acceleration of the

block is,2

, 2 / block ground a g m sµ towards right.

Let velocity of block wrt ground at time t be v (towards right), then 2v t .Solve the question wrt platform frame of reference.

2

, 2 / block platforma m s

(towards right)

, 2 4 / block platformv t m s

(towards right)

Initial velocity of block wrt platform is 4 m/s towards left.

Let s be the required distance, then2 2

2 .v u a s

2

0 4 2 2 4s s m .

15. (a)


4/19

ˆ ˆU U F VU i j x y

ˆ ˆ7 24i j .

2 27 24 25F units

.

16. (c)

Let a be the acceleration of the centre of mass and α be the angular acceleration about an axispassing through the centre of mass.

22

,5

F Ma F h R MR α

For pure rolling a Rα

or 5

2

F h RF

M MR

or 5 7h R

1.4h R .

17. (b)

1st

method: Let us assume that the spheres are moving with velocities 1v and 2v when they are at a

separation of d .

Then from momentum conservation, 1 2mv Mv .From energy conservation,

2 2

1 2 0 02 2

mv Mv GMm

d

.

After solving above equation, we get

1 2

2G M m

v v d

.2

ndmethod: Using C frame or reduced mass concept.

Let v, be the relative when they are at a separation of d ,

then2

02

T mM v GMm

m M d

2T

G M mv

d

.

18. (c)

Elongation in the spring in equilibrium position is given by 0mg ky

0 10 y cm .The point mass is released from the location where elongation in spring is 15 cm.

So, amplitude of oscillation of point mass is 15 10 5cm .As Hooke’s law is valid, the point mass performs simple harmonic motion, let equation of simple

harmonic motion be,

sin y A t ω δ

where20 1

52000 10

A cm and ω .

At 0, 5t y cm so,

5 5sin δ

2

πδ .


5/19

So, equation of motion is, 5 sin10 2

t y cm

π

where t is in sec.

19. (b)

From,0 0

/

/

T A AlY T YA

l l l

where 0l is the length or natural length of string.

1 0

00

YA l l

T k l ll

where 1l is the length of string when tension in string is T .

1 04T N k a l

2 05T N k b l

3 03T N k l l

where l is the required length.Solving above equations, we get 5 4l b a .

20. (d)

21. (d)Time period is same so A Bω ω

Linear velocity A A A A

B B B B

v r r v r so

v r r

ωω

ω .

Force required is, 2F mr ω .

Now, masses of two particles can be different and hence ratio of the force required may not be, A

B

r

r .

22. (a)cosT mgα sinT α component would be balanced by force exerted on mirror by light.

2sin

IAT

cα

So,2

tan IA

cmgα for small angle tan α .

23. (c)Let V be the total volume of ice piece and V is the volume of ice piece which is outside the water,then

water iceV V g V gρ ρ

1000 900V V V

or 0.110

V V V .

So, fraction of ice piece outside the water is, 0.1V

f V

.

24. (b)

100 3 2 100dx da db dc

x a db c

.

For % age error we consider the worst case ---% age error is 3 % x in 2 %a error in %b error in c.

25. (a)The force acting on the ball are buoyancy force and gravitational force.


6/19

Net force on the ball will be zero when

BF mg

0v axg v gρ

0 x ρ

α , i.e., mean position is at a depth of 0

ρ

αfrom free surface.

But, it moves down from this mean position due to acquired speed. When it is at a depth of 0 xρ

α

from the free surface, net force acting on ball is 0 0 0F v x g v g v xgρ

ρα

.

[Towards mean position]

F x so simple harmonic motion.

Velocity of ball becomes zero when 02

x ρ

α from free surface,

i.e., amplitude of simple harmonic motion become 0ρ

α.

26. (a)

At a distance y from bottom end consider a string element of length dy.

Tension at this height is, m

T ygl

So, dy

dt velocity of wave

T gy

µ

2.45

0 0

t dydt

gy or 1t s .

27. (a)

For proper resolution, 1.22 y

D d

λ

where 1 , 500 , 3 y mm nm d mmλ .5 D m

max 5 D m .

28. (a)When multimeter shows the conduction on its connection with two legs of diode, it means the leg

which is connected to positive leads of multimeter is p-side of diode.

29. (d)Option (A) tells that transistor is not working, as there is no connection between base and collector

means there is some fault in this part of the transistor.

Option (B) tells that transistor is having no open circuit fault, i.e., it is having the continuity.

Transistor is faulty.

30. (b)Taking moments about the mid-point. 2 0.5 0.2g mg 5m kg .

:Mathematics:

31. (a)Since the planes are all parallel planes,

1 2 2 2

2 6 4 4

4 9 16 292 3 4 p

Equation of the plane 4 6 8 3 0 x y z can be written as 2 3 4 3/ 2 0 x y z


7/19

So 22 2 2

2 3/ 2 1

2 292 3 4 p

and 32 2 2

2 6 8

292 3 4 p

1 2 38 0 p p p

32. (b)

The equation of a line which passes through 2, 1 is 1 2 ....... y m x i

The equation (i) will touches the circle 2 2 1 x y if 2

2 1 41 0,

31

mm

m

4

1 2 4 3 5 03

i becomes y x x y

Now 5, 5 is a point on the above line. Its image by the line 1 y is 5,3

The equation of the incident ray is1 2

4 3 11 01 3 2 5

y x x y

33. (c)

We have to find the number of integral solutions if 1 2 3 4 5 6 x x x x x and that equals5 6 1 10

5 1 4C C

Thus Statement-1 is false.

Number of different ways of arranging 6 A’s and 4 B’s in a row

10

4

10

6 4C

= Number of different way the child can buy the six ice-creams.

Statement-2 is trueSo, Statement-1 is false, Statement-2 is true.

34. (a)

Mean x x ny

Mean deviation = x x

n

= 12 1

ny n

n

35. (d)

Here 2 2 2

OP OQ PQ

2 2

2 2 2 2 2 2

2 24a ab l l b l l

b b

2 22 2 2 2 2

2 20 3 0 3

3

a bl b a b a

b a

2 2 23 1a e a 2 2 21 / e b a

2 4 2

3 3e e

36. (d)

3let C x i y j k

2 2 21 1........ 1c x y z


8/19

. 0c a c a

0........... 2 x y z

0a b c

1 1 1

1 1 1 0 x y z

1 1 1 0 z y z x y x

0 z y z x y x

2 0 y z y z 2 x z

2 1 1, ,

6 6 6 x y z

12

6

c i j k

37. (c)

Let the sequence of 3n consecutive integers begins with m. Then 3n consecutive integers are m, m+2,

……..,m + (3n – 1)

3 integers from 3n can be selected in 3 3 ......n

C i

Now 3n integers can be divided into 3 groups.

1 :G n numbers of from 3p

2 :G n numbers of from 3 2 p The sum of 3integers chosen from 3n integers will be divisible by 3 if either all the three are from

same group or one integer from each group. The number of ways that the three integers are fromsame group is 3 3 3

n n nC C C and number of ways that the integers are from different groups is

1 1 1

n n nC C C

favourable cases 3 3 3 1 1 1n n n n n nC C C C C C

Required probability

3

3 1

3

3

3.n n

n

C C

C

23 3 2

3 1 3 2

n n

n n

38. (b)

3 4

3 4

x Let z

x

3 4

1

1 3 4 3 43 43 41 3 4 3 41

3 4

x

z x x x x z x x

x


9/19

6 6 3

8 8 4

x x x

4 1 4 4 6 6 10 22

3 1 3 1 3 1

z z z z f z

z z z

10 2 1 2 8 2

3 1 3 1 3

z

z z

8 2

3 1 3 f x

x

8 2

3 1 3 f x dx dx dx

x

2 2

log 13 3

x x c

39. (b)

Let Aα β

γ δ

We have

2

2 A

α βγ β α δ

γ α δ δ βγ

2

21 0

0 1 A I

α βγ β α δ

γ α δ δ βγ

giving 2 21α βγ δ βγ

and 0γ α δ β α δ

As 1, A A I , we have α δ

det1 1 0

0 11 A

βγ β

γ βγ

giving 2 21α βγ δ βγ

and 0γ α δ β α δ

As , A I A I , we have α δ

det1

1 11

Aβγ β

βγ βγ γ βγ

Statement-1 is therefore true.

tr 0 A α δ α δ

Statement-2 is false because 0tr A

40. (a)

21n

adj adjA A A

2 0 1

5 1 0 2 3 1 5 1

0 1 3

A

adj adjA A

41. (d)


10/19

Statement 1:

2 4 316 3 , y x y mx

m

2 22

1 11, 4 22 4

x y x m y m

21 1 1

2 1

4 ,

x

y mm m m

Now,

2 2

2

1

4 3 24

m m

2 2

2 2 2

1

48 2 244 4 2 2m m

m m m

4 22 24 0m m ………(1)

2 26 4 0 2m m m

Statement 2: If 4 3

y mx m is a common tangent to2

16 3 y x and ellipse2 2

2 4 x y , then

m satisfies 4 22 24 0m m From (1), statement 2 is a correct explanation for statement 1.

42. (d)

0 0 0

1 .n n n

n n n

r r r

r r r

r C r C C

1

1

0 0

. .n n

n n

r r

r r

nr C C

r

= 1 1.2 2 2 2n n nn n Thus Statement-1 is true.

Again 0 0

1 .n n

n r n r n r

r r r

r r

r C x r C x C x

= 0 0

1 .n n

n r n r n r

r r r

r r

n r C x r C x C x

1

1

0 0

n nn r n r

r r

r r

n C x C x

= 1

1 1n n

nx x x


11/19

43. (a)

At 2

0, sin 1 02 2

x f x x π π

At4

, 02 2 2

x f π π π

Equation has at least one root in4

0, 0

2 2

π π

44. (a)

sinsin

dx y y x dx x y

dy y dy y

1

log. .

dy y y I F e e y

sin cos 1 cos yx y ydy y y y dy c cos sin yx y y y c

cos sin y x y y c

45. (b)If the two A.P.’s are 2 .....a a d a d and 2 ..., A A D A D then

2 13 82

7 152 1

2

na n d

n

n n A n D

1

3 82

1 7 15

2

na d

n

n n A D

Substituting

111

2

n i.e. n = 23, we get

11 3 8

11 7 15

a d n

A D n

1

3 82

1 7 15

2

na d

n

n n A D

Substituting1

112

n i.e. n=23, we get

11 77

11 176

a d

A D

ratio of 12

thterms

7

16

46. (a)

Since, the person is allowed to select at most n coins out of 2 1n coins, therefore he can selectone, two, three, ……., n coins. Thus, if T is the total number of ways of selecting at least one coin,

then 2 1 2 1 2 11 2 ..... 255n n n

nT C C C …………. (i)

Using the bionomial thermo

2 12 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1

0 1 2 1 2 2 1..... ...... 1 1 2nn n n n n n n n

n n n nC C C C C C C

2 1 2 1 2 1 2 1 2 1 2 10 1 2 2 12 ..... 2n n n n n nn nC C C C C

2 11 2 1 2 nT


12/19

2 122

1 22

nn

T

2 2 81 255 2 2 2 4n n n 47. (d)

p q q

~ ~q p q (By Commutative law)

(~ ~q q p (By Commutative law)

~ ~ ~q q q p (By Distributive law)

~ ~q p p q 48. (a)

In , ABC Given

1

1 0

1

a b

c a

b c

21 0c ab a c a b b c

2 2 2 0a b c ab bc ca 2 2 22 2 2 2 2 2 0a b c ab bc ca

2 2 2 2 2 22 2 2 0a b ab b c bc c a ca

2 2 2

0a b b c c a

Here, sum of squares of three numbers can be zero, if an only if a b c . ABC is an equilateral triangle.

060 A B C 2 2 2 2 0 2 0 2 0sin sin sin sin 60 sin 60 sin 60 A B C

3 3 3 94 4 4 4

49. (a)

1 2 ..... 100ln f x ln x ln x ln x Differentiating

'1 1 1

.......1 2 100

f x

f x x x

Again differentiating

2" '

2

f x f x f x

f x

2 2 2

1 1 1...... 0

1 2 100 x x x

2

" ' 0 0 f x f x f x g x

0g x has no solution


13/19

50. (b)

Applying Leibnitz rule, we get

2 21

x f x x x f x f x

x

1 1 1

2 2

1/ 2 1/ 2 1/ 2

1 8ln

1 1 2 5

x x f x dx dx dx

x x

51. (b)

3 3,i k i jλ λ

and 2 sini j k λ λ λ

3

3

0 1

1 0 0

1 2 sin

λ

λ

λ λ λ

3 62 sin 0λ λ λ λ λ 7 3

2 sin 0 0λ λ λ λ λ

52. (b)

Let i z re θ . Then we have

2

2211

12 3

i ir e e r

θ θ

2 2 2cos

13 3

r r θ

2

2

33

cos 1

r

θ

max 3r i.e., max 3 z

53. (b)

S.D. between the lines x y z

l m n

α β γ

and' ' '

' ' '

x y z

l m n

α β γ is given by

S.D. =

' ' '

2

' '

' '

l m n mn m n

l m n

α α β β γ γ

2 2 2 2' ' 1 2 1 6mn m n

1 2 21

. . 2 3 4 66

3 4 5

S D


14/19

54. (a)

loge a is defined for a>0

2log 4 / 1 x x is defined for 24 / 1 0 x x

Now 2

4 2 2 x x x is realIf 2 2 x and 2 x

24 0 x when 2 x 2 2 x and 1 2 1 x x

Thus the domain of f x is 2, 1

To find the range

Let 24 / 1 x x u sin log , 2 1 f x u x

The function f x is continuous for 2 1 x

Also,

2 24

01 4

du x

dx x x

If 2 1 x Thus in the interval 2 1 x , u is an increasing functionAt 2, 0 x u and at 1, x u

0 u when 2 1 x when 2 1, log x u

when 2 1, 1 sin log 1 x f x u Thus the range of f x is 1,1

55. (b)3 2 4 2 3

30

1lim 1 .... 1 ..... ...

6 2 24 2 3 x

x x x x x L x x

x

=

0

1 1lim

6 3 x L

+terms

containing positive powers of 1/ 2 x 56. (c)

Given curve is sin y x x

Let at 1 1,P x y , tangent be parallel to x-axis, so 1 1, / 0 x ydy dx

1. 1 cos 0

2 sin x

x x

i.e., 1cos 1, x so 1sin 0 x .From (1) and (2), we have

2

1 1 1 1sin y x x x

Hence 1 1,P x y lies on2

y x , which is a parabola.


15/19

57. (a)

Odd prime on a dice are 3 and 5.

Let even A = one of the dice shows odd prime

And event B = remaining dice also shows odd prime

/

n A B p B A

n A

{ 3,1 , 3, 2 , 3,3 , 3,4 , 3,5 , 3,6 , 1,3 , 2,3 , 2,3 , 4,3 , 5,3 , 6,3 ,5,1 , 5,2 , 5,4 , 5,5 , 5,6 , 1,5 , 2,5 , 4,5 , 6,5 }

A

3,3 , 5,5 , 3,5 , 5,3 A B

/ 4 / 20 1/ 5 p B A 58. (d)

47, 45n M n C

53n T ,

80n M C T

' 100 80 20n M C T 59. (b)

Required area

= 1

2

0

2 2 log x x x x dx

=

1 13 2 2

2

0 0

2log

3 2 4

x x x x x

1 1 7

3 4 12

2

0

log log 0

x

x x

60. (d)

1 22 tan log 1 y x x x x

2

2 2

21

1 2 12

1 1

x

dy x

dx x x x

2 2

1 12

1 1 x x

=

2 2

22 2 1 11 x x

x

=2 2

2

1 2 10

1

x x x R

x

Hence y increases in ,


16/19

:Chemistry :61. (d)

1

2

a

H k

C

62. (b)(i) ; 21S aq AB aq AB H J

(ii) 2 2. ; 29.4S s AB xH O AB xH O H J

2. ; ?S aq AB xH O aq AB H

Equation (i) is equivalent to

2 2 1. ;S s AB xH O AB xH O H H

2 2. ;S aq AB xH O aq AB H H

1 2 21 H H

2

29.4 21 H 1

2 21 29.4 8.4 H J mol

63. (c)It is MA3 B3 type complex therefore it shows fac. & mer. isomers

&

64.

2 3CH O CH

O

Number of resonance less stable less acidic nature

||

||

O

O

S O

More stable more acidity65. (a)

Conceptual

66. (b)

Conceptual

67. (d)

(i) 132 2 11

9.1 102

CO CO O K

(ii) 122 2 2 21

7.1 102

H O H O K

object 2 2 2CO H CO H O equation (i) – (ii)

11

21.28

K K

K


17/19

68. (a)

2 21 12 2

d NO d NO d Or

dt dt dt

2 NO is reactant so (Z)Rate of disappearance of 2 NO rate of formation of NO

So NO is (X)69. (a)

OH OH

2CH OH

2CH Br

OH OH

3O C H OH

H Br

H Br

3CH Br

2 H O

70. (a)

3 H C 3CH

3 H C 3CH

2 Br C

C

C

C

3CH 3CH

3CH

3CH

3CH

Br

Br

Br

Br

3CH

Meso

(Not enantiomer’s)

3 H C

3 H C 2CH

2 Br C

3CH

2CH

2CH

3CH

Br *

Chiral center present

71. (d)

At equation 0 0G 7.5 1000

30025

H T K

S

72. (d)

2

2 2 MgCl Mg Cl

?

0.01 M2

2spK Q Mg OH

2

1210 0.01 OH


18/19

2 2 21 / 2 286

l

H O H O H kJ

6 323 6

X e F X e O H O H F

210

10OH 510OH

5 pOH 9 pH 73. (a)

(1)2 2

( ) ( )

H O H O

l g

44 H kJ

(2) 2 2 31

2 12733

B O B O kJ

(3) 2 6 2 2 3 23 3 B H O B O H O (g)

2035 H kJ (4)

Equation 2 4 3 1 3 3

1273 286 3 44 3 2035 36 H kJ 74. (b)

Conceptual

75. (d)

P = 2, Q = 3, R = 0, S = 1

76. (c)2 2

Zn Cu Zn Cu 2 22 2 2 2 Zn Cu Zn Cu

For 2 moles of Zn, n =40 0

4 96500 1.1 424.6CellG nFE kJ 77. (c)conceptual

78. (d)

According to Lagmuir curve

1

ap x

bp

a

p xb

0 p x p

79. (b)

3 3Sp d Hybridisation 3Sp Hybridisation

(distorted octahedral) (Pyramidal)


19/19

80. (b)2 2

4 2 4 2 216 2 5 2 10 8 H MnO C O Mn CO H O

2

4 2 4: 2 : 5 MnO C O

4 2 2 22 16 2 2 5 8K MnO HCl KCl MnCl Cl H O

4 : 2 :16 1:8 MnO HCl

81. (a)1 12 2

1 3

2 2

1 2 1 22 2

1 1 1 3 3213.6 / ;

1 1 27

1 2

E E eV atom

n n E

82. (b)

Conceptual

83. (d)

15 mL 0.1 N NaOH = 15 mL 0.1 N HCl

Vol. of HCl used with NH3 = 20-15 = 5 mL

% N =3

1.4 1.4 0.1 5

29.5 10

NV

w

70023.7

29.5

84. (a)

Conceptual

85. (b)

Conceptual

86. (d)

Lyophobic colloids have charge of adsorbed ions on surface whose repulsion stabilizes them.

Higher the zeta potential between fixed and diffused layers, higher is the stability.

87. (b)Conceptual

88. (b)

Conceptual

89. (c)

It is octahedral void

90. (a)

Conceptual

iit model paper answer 7

Documents