Iit Model Paper Answer 12

8/20/2019 Iit Model Paper Answer 12

1/12

Max Marks:36

KEY SHEET

Physics :

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

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

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

Mathematics :

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

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

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

Chemistry :61)1 62) 1 63) 1 64) 4 65) 3 66) 2 67) 4 68) 2 69) 3 70) 1

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

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

SOLUTIONS

PHYSICS

1. Assuming that ionization occurs as a result of a completely inelastic collision, we can write 0 H mv m m u

Where m is the mass of incident particle H m the mass of hydrogen atom 0v the initial velocity of

incident particle and u the final common velocity of the particle after collision. Prior to collision the

KE of the incident particle was2

00

2

mv E

The total kinetic energy after collision

2 2 2

0

2 2

H

H

m m u m v E

m m

The decrease in kinetic energy must be equal to ionization energy.

1 0 0 H

H

m E E E E

m m

i.e, 1

0

H

H

E m

E m m

i.e, greater the mass m, the smaller the fraction of initial kinetic energy that be used for ionizartion.

2. 2PD V

10q C µ

3. In the first case


2/12

Point P is an antinode i,e the string is vibrating in its second harmonic. Let0 f be the fundamental

frequency. Then

02 100 f Hz

0 50 f Hz

Now P is an antinode (at length / 4l from one end) so centre should be a node. So, next higher

frequency will be sixth harmonic or0

6 f which is equal to 300 Hz as shown below:

4. Given that 1 2v v

Therefore, if 1v

makes an angle of α with horizontal than 2v

will make an

angle0

90 α 090 α with horizontal. Now horizontal component of velocity remains unchanged.Therefore

1 2cos sinv vα α

Or 1

2

3tan

4

v

vα

Or037α

Now minimum kinetic energy will be

2

2

min 1

1 1 4cos 2 3 5.76

2 2 5K m v J α

6. Wein’s displacement law is

mT bλ ( b Wein’s constant)62.88 10

2880m

b nm K

T K λ

1000nmλ

Energy distribution with wavelength will be as follows:

From the graph it is clear that

2 1U U (in fact 2U is maximum)


3/12

7.1 2

1 1 1

F f f

Or 1 21 1 1 1 1

1 1F R R

µ µ

1 2

R

µ µ Or

1 2

RF

µ µ

8. 0 0

0 0

mv vr

B q B α

3sin

2

x

r θ

060θ

06 3OA

T t

B

π

α

Therefore, x, co-ordinate of particle at any time0

3t

B

π

α will be

000

0 0

3cos60

2 3

v x v t

B B

π

α α

0 0

0 0

3

2 2 3

v vt

B B

π

α α

11. Free body diagram of the two bodies are as follows

Let acceleration of both the blocks towards left is a, then

2 20

2 4

f f a

Or 2 4 20 f f Or 8 f N

Maximum friction between the two blocks can be: 2m kg

max 0.5 2 10 10 f mg N µ

Now since max f f

Therefore, friction force between the two blocks is 8 N

12. In the circuit diode 1 D is forward biased, while 2 D is reverse biased. Therefore current i (through

1 D and 100 resistance) will be ;6

0.0250 100 150

i A

Here, 50 is the resistance of 1 D in forward biasing.

13. M iφ

d M diφ

Or 4

22

d di A

M

φ

Further d M diφ

2 1 2Wb

14. During fusion binding energy of daughter nucleus is always greater than the total binding energy of

the parent nuclei. The difference of binding energies is released. Hence


4/12

15. In perfectly inelastic collision between two particles linear momentum is conserved. Let θ be theangle between the velocities of the two particles before collision. Then

2 2 2

1 2 1 22 cos p p p p p θ

Or 2

2 22 2 cos

2

vm mv mv mv mv θ

Or 1 1 1 2cosθ Or1

cos 2θ Or0

120θ

16. 0 021 2

V V i and i

Z Z

;

Here,

2

2

1

1 Z R

C ω

And

2

2

2

1

2 Z R

C ω

As2 1 1 2 Z Z i i

17.2 2

tan 22

y

xθ

Or1

cot2

θ

tan 2tan

2 2

θφ

1cot

2θ

090φ θ

i.e, E

is along positive y-axis.

18. k k B A A B Z Z α αλ λ

cut of cut of A B A B

V V λ λ

19. Let nth minima of 400 nm coincides with mth minima of 560 nm, then

400 560

2 1 2 12 2

n m

Or2 1 7 14

.....2 1 5 10

n

m

i.e, 4

thminima of 400 nm coincides with 3

rdminima of 560 nm.

Location of this minima is

61

2 4 1 1000 400 1014

2 0.4

Y mm

Next 11th minima of 400 nm will coincide with 8th minima of 560 nmLocation of this minima is

62

2 11 1 1000 400 1042

2 0.1Y mm

Required distance 2 1 28Y Y mm

20. Let the initial amplitude decreases to1a to the other side i.e, after the first sweep;

Decreases in elastic potential energy Work done against friction

Or 2 21 11 1

2 2ka ka mg a aµ

Or 1 1 11

2k a a a a mg a aµ


5/12

Or 12 mg

a ak

µ

Similarly 1 12 mg

a ak

µ

……………………………….

1

2n n

mga a

k

µ

Adding all the above equations

2n

n mga a

k

µ

The block stops when

n n

mgmg ka or a

k

µµ

Substituting in the above equation we get:

2 1 mg

n ak

µ

Or

20 0.32 1 15

0.04 1 10

kan

mgµ

;

7n

21. Let l be the end correction. Given that, fundamental tone for a length 0.1m first overtone for the

length 0.35 m.

3

4 0.1 4 0.35

v v

l l

Solving this equation, we get

0.025 2.5l m cm 22. The average speed of molecules of an ideal gas is given by

8. ,

RT v i e v T

M α for same gas.

Since, temperature of A and C are same average speed of 2O , molecules will be equal in A and C i.e,

1v .

24. Linear acceleration of cylinder is zero i,e ,

sinmg θ Frictional force f upwards m massof cylinder

Angular acceleration about C is I

τα

Or2

2

1

2

fR f

mRmR

α R radius of cylinder Or2 sin 2 sinmg g

mR R

θ θα

For no slipping between cylinder and plank 2 sina R gα θ

25. In n-type semiconductors electrons are the majority charge carriers.

26. Current I can be independent of 6 R only when 1 2 3 4 6, , , R R R R and R form a balanced Wheatstone

bridge.

Therefore, 31 1 4 2 32 4

R Ror R R R R

R R

27. The average velocity in the first half of the distance v , while in the second half the averagevelocity is v. Therefore 1 2t t . The work done against gravity in both halves is / 2mgl .

28. The distribution of charge on the outer surface depends only on the charges outside and it

distribution itself such that the net, electric field inside the outer surface due to the charge on outer

surface and all the outer charges is zero. Similarly the distribution of charge on the inner surface

depends only on the charges inside the inner surface, and it distribution itself such that the net,


6/12

electric field outside the inner surface due to the charge on inner surface and all the inner charges is

zero.

Also the force on charge inside the cavity is due to the charge on the inner surface. Hence answer is

option.

29. 1 2

1 2

eq

YAK

K K LK YAK K

K L

YAK YA LK

2eq

mT

K π

2

m YA LK

YAK π

Note: Equivalent force constant for a wire is given by YA

K L

. Because in case of a wire,

YAF L

L

and in case of spring .F K x . Comparing these two we find K of wire

YA

L

30. 2 21

2 2

2

mR MRπ θ θ

4 22

M m mπ θ θ

8

4

m

m M

πθ

MATHEMATICS

31.2 3 4

z ,z , z 2 10

Form an equilateral triangle2 2 2

2 3 4 2 3 3 4 4 2z z z z z z z z z 2 2

2 3 2 3 2 3z z 2z 2z z z 6 0

32. Numbers in the form 7k 147k 1 15 7k 2 15 7k 3 14 7k 4 14 7k 5 14 7k 6 14

to get maximum number of elements, one element from 7k, and all elements of the from , 7k+1,7k+2, 7k+3


7/12


8/12

(1,2, a) lies on the plane 3x 4y 6z 7 0 a =3

& required plane, is nothing but plane, passing through (0,0,0) & the given line

x 1 4 2 z 91

2 3 1

is x+y+z=0

bαβ

cβγ

aγα abcαβγ 42. xp xq y(pxq) yp yq z(pxq) p (pxq) 0

2y xz y is Geometric mean of x & z

43.o o o

a 2 b c2R

sin 45 sin 60 sin 75

b 6 R 2 o3 31 absin 75

2 2

44.

2 21 x y x y

cos 1 1 02 3 4 9

2 2 29x 4y 12xy cos 36(1 cos )θ θ 2P 36sin θ

a) 0 0

p.d 18 1 cos 2 d 18

π π

θ θ θ π

b)

/2

20

6sin dI

6sin 36(1 sin )

πθ θ

θ θ

/2 /2

0 0

sin d cos d

sin cos sin cos

π πθ θ θ θ

θ θ θ θ

/2

0

2I 1d I4

π

πθ

45.1

k absin C2

1k bcsin A

2

1k acsin B

2 Given expression becomes

2 2 2a b c

2

46. In triangle APB

AP Bsin 2AB

In triangle AQC,C AQ

Sin2 AC

In triangle ABI,BI AB AI

A A B Csin Sin cos

2 2 2 2π

& In triangle ACI,CI AC

A BSin cos

2 2

AP AQ A

cotBI CI 2

47. w.k.t2 2 2

1 1 1

PA CA AD


9/12

Where D is the point of intersection of PC & AB

AD 4 AB 8

48.

1

ln(1 x )

x 0

1 tan 2 ln 1 xY l t

cos x(1 x)

ex 0

2 tan 2 ln 1 xlog y lim2 ln(1 x)

2y e

49. Centroid divides the line segment, joining the points circumcentre & orthocenter in 1:2 ratio circumcentre is P(2,3) & Image of orthocenter, with respect to any side of the triangle, lies on its

circumcircle (i.e) Q (8,5)

diameter = 2PQ 4 10

50. 2 21 1 2 2A 2t , t , B 2t , t P( 1,0) are collinear points

1 2 1 2t t 2t t & Centroid of triangle OAB

Is2 2

1 2 1 2

3t t &3 t t

2

αβ

2

1 2 1 2(t t ) 2t t 3β 29 3

34 2

α αβ

23x 2x 4y

51.cos x sin x cos x x sin x x x

I dx dx

x(cos x x) x(cos x x)

=1 (sin x 1)

dx dx log x log(cos x x) log cx cos x x

log cx log(cos x x)

Cxlog

cos x x

52. 2t tan x,dt sec x dt

5 3

3 1/2

9t 7t dt

(t t)

8 69 7 1/2

9t 7tdt

(t t )

9 7y t t

dy2 y c

y

53.

12 tan (sin x )

1 1 1

/2

1 e dxI

e tan (sin x) e tan (cos x)

π

ππ

2

1

/2

1 32I 1 dx

2

π

π

π

π


10/12

1

3I

4

1

1 1

5

sin (sin x )2

2 tan (sin x) tan (cosx)2

1 e dxI

e e

π

ππ

5

2

2 2

2

1 12I 1dx I4

π

ππ

I 1

54. 1sin2

α

1sin2

β

2n

2nn

1 (2sin )l t 2

(2sin )

αλ

β

3α

λ

β

55.3 5

4 4 4

0 0 0

I f (sin x)dx f (sin x)dx f (sin x)dx .....

π π π

to n terms = since4f (sin x) is periodic

function with period π ) /2

4 2 4

0 0

f (sin x)dx[1 3 5 ....to n terms] 2n f (sin x)dx

π π

56. Domain of 1y cos (2x 5) 1 is [-3,-2]

Domain of 1 11

y sin (2x) sin 2[2x]

Is 1 ,02 no solution57. Let A

1 2(Ct ,C / t ) B 2 2(Ct ,C / t )

Slope of AB = 1 21 2

1 2

C Ct t

1 t t 1Ct Ct

Equation of circle, AB as diameter is 2 2 1 2x y t t c[x y] 8 0 . 2C 4

2 2 2 2x y 8 (x y) 0 x y 0and x y 8 0λ circles pass through (2,2) & (-2,-2)

58. i ix 5 x 5n

n

2ix 400(x) 0 25n n

n=16.

59. Conceptual

60. Image of S(4,3), with respect to the tangent line x+y=3 is (0,-1) & the points (0,-1), (1,2) (6,a) are

collinear points a=17


11/12

CHEMISTRY

61. 4 2 4 10 2( ) ( )( ) A C B

P N O P O N

64.the electronic configuration of the given ion is as

2 2 2 6 2 6Ca 1 2 2 3 3s s p s p

2 2 2 6 2 6 51 2 2 3 3 3 Mn s s p s p d 2 2 2 6 2 6 101 2 2 3 3 3 Zn s s p s p d

Thus the size of 2 Zn ion is smallest and of 2Ca is largest among the three due to increase in effective

nuclear charge with increase in atomic number.Smaller the size of ion ,lighter is the hydration

enthalpy.Thus ,the order of hydration enthalpy is 1 1 1

2 2 2

2047 1841 1577kJ mol kJ mol kJ mol

Zn Mn Ca

.

67. 2 2 7 2 4 2 2 4 4 2' '

4 6 2 2 4 3redvapor p

K Cr O NaCl H SO CrO Cl KHSO NaHSO H O

2 2 2 42 yellowsolution

CrO Cl NaOH Na CrO HCl

2 4 3 2 4' '

2red solid Q

Na CrO AgNO Ag CrO

70. Conceptual72. A 2B C

t=0 Pi 0 0

at t Pi -x 2x x

if t 0 2Pi Pi

t iP P 2x

t iP Px2

and i iP

P 3p P 3

i

i

P2.303K log

t P x

73. The base in which lone pairs are more available for donation,forms most stable adduct with the lewis

acid, 3 3( ) B CH .

Presence of electron releasing groups like ,Me makes the electron more available for donation,where

as an electron withdrawing group decreases the electron releasing tendency of the base.

Thus N

CH3

forms the most stable adduct with 3 3( ) B CH .

N and

N

both are crowded molecules so do not form adduct with Lewis

Acids.

75. 3 23 4 4 2 4 4H PO PO H PO HPO

Initial milli moles : 20 0.1 20 0.1 - -

After reaction: - - 2 2

Buffer solution of 2 4H PO (acid) and 24HPO

(Conjugate base) is formed


12/12

a2

2

4KH

2 4

HPOP P log

H PO

= a2k

P

76. Due to lanthanide contraction, left to right atomic size decreases

78. 0.1 F Liberates 0.1 equivalents of 2O at anode and 0.1 equivalents of Ag at Cathode Total loss = (0.1)8 + (0.1) 108 = 11.6g Wt. of final solution = 108 11.6 96.4g

79. X is borazine.it is iso electronic with benzene.it reacts with hydrogen chloride to give an addition

product. 3 3 6 3 3 9 33 B N H HCl B N H Cl

81. f f T K .m

36 10.93 1.86x x

M 1.2

M 60 60

n 230 2 2m.f CH O

82.2

3 4 4 2 3 4[ ( ) ]( ) [ ( ) ]Cu NH ClO or Cu NH contains 2Cu ion

2 9[ ]3Cu Ar d

3d 4s 4p

2

3 4[ ( ) ] [ ]Cu NH Ar

3d 4s 4p

NH3

dsp2 square planar

3 NH being a strong field ligand transfer the unpaired electron to 4d orbital.In

3 4 4 3 4[ ( ) ]( ) [ ( ) ]Cu NH ClO or Cu NH

,Cu is present as Cu

ion.4s 4p

10

3 4[ ( ) ] [ ]3Cu NH Ar d

sp3 tetra hedral

Completely filled orbitals are highly stable, so 3 NH ligand occupy 4s and 4p orbitals.

85. Conceptual

87. ox

log tan(45 ) log p log 2m

x2p

m

4,at P 2atm

88. Due to emulsification surface tension decreases.

Iit Model Paper Answer 12

Documents

Transcript of Iit Model Paper Answer 12