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Narayana IIT Academy INDIA

Sec: Jr. IIT_IZ CUT-16 Date: 04-12-17 Time: 07:30 AM to 10:30 AM 2012_P2 Max.Marks:198

KEY SHEET

PHYSICS 1 A 2 D 3 B 4 B 5 B

6 B 7 C 8 D 9 B 10 B

11 A 12 C 13 C 14 A 15 AD

16 ABCD 17 D 18 ABD 19 ABC 20 BD

CHEMISTRY

21 A 22 A 23 D 24 B 25 C

26 B 27 B 28 A 29 A 30 B

31 B 32 C 33 C 34 A 35 AB

36 ABC 37 BD 38 ABC 39 ABD 40 BD

MATHS

41 C 42 A 43 B 44 A 45 A

46 B 47 B 48 B 49 B 50 A

51 C 52 D 53 B 54 D 55 AB

56 ABC 57 AC 58 D 59 ABCD 60 ABD

Narayana IIT Academy 04-12-17_Jr. IIT_IZ_JEE-Adv_2012_P2_CUT-16_Key & Sol’s

Sec; Jr. IIT_IZ

SOLUTIONS PHYSICS

1. Considering the vertical motion, the time travel from A to B is 2 /t h g . In this time,

the particle must make an integral number of rotation, say n . Considering the

horizontal motion, distance covered 2 rn ut

2. When a particle undergoes normal collision with a floor of a wall, with coefficient of

restitution e , the speed after collision is e times the speed before collision. Thus in

this case, the change in momentum for the first impact is 1ep - -p p e , for the

second impact it is 1e ep ep ep e and so on. The total change in momentum is

21 1 .....p e e e e

3. 1 2 3, 3 / 2, / 2a g a a g

4. 2 2 2 /ca k rt v r or v krt

The tangential acceleration is 1dva krdt

The net tangential force on the particle t tma mkr F

Work is done on the particle only by tangential forces, as the radial forces are

perpendicular to v

the power delivered to the particle 2 21F v mkr krt mk r t

5. In sliding without friction and in rolling without slipping, no work is done against

friction Hence, kinetic energy = loss in gravitational potential energy. In rolling with

slipping, some energy is lost in doing work against friction and kinetic energy is less

than the loss in potential energy

6. 2ev Rg

At the surface of the planet, sGMV

R

At the centre of planet, 32cGMV

R

212 s cmv m V V

Or 2

2 32 12 2

evGM GMv RgR R


Sec; Jr. IIT_IZ

7. The air pressure is greater inside the smaller bubble 4 /S r . Hence, air flows

from the smaller to the larger bubble

8. As 2v r and the mass of the drop 3m r , its momentum 5p mv r . Here 52 32

9,10. The velocity of particle changes sign at 1sect

Distance from 0 2sect to t is

0 1

1 2

vdt v dt

0 1

3 2 3 2

1 2

3 3 32 2

t t t t m

Distance from 2

0

0 2sect to t is v dt

2

3 2

0

22

t t m

13.

2A j

2units

14.

3B j

3units

15.

u

v

V

L

As the block does not move, the ball moves along a circular path of radius l . The

centre of mass of the system always lies somewhere on the string

Let v speed of ball when the string makes an angle with the horizontal

2 21 1 sin2 2

mv mu mg

The horizontal component of 2sin 2 sinv V u gl


Sec; Jr. IIT_IZ

For V to be maximum 0dVd

, which gives 2

sin3ugl

16. Consider the vertical motion of the particle after entering the compartment. Let it

reach the floor in time t.

213.2 10 0.82

t or t s

Due to the velocity component u, which remains constant, it covers a distance of 2.4m

in 0.8s

2.4 3 /0.8

mu m ss

Due to the velocity v , point B must move forward by 16m in 0.8s

16 20 /0.8

mv m ss

17. In rolling without slipping, at constant speed, there is no force of friction between the

surfaces

Therefore removing the pin causes no change to the system

18. For a pulley with mass, the strings on its two sides do not have the same tension

when in motion

Also, the rotating pulley have some kinetic energy derived from the loss in potential

energy of the block B as it moves down

19. Tension in / 3BB T mg

Tension in 4 / 3A BA T T mg mg

4A BT T

Stress 2/T r S . Wire breaks when S = breaking stress

For , 4A B A Br r S S

A breaks before B

For 22 , BA B B

B

Tr r S

22 2

42

A B BA B

A BB

T T TS Sr rr

As the stresses are equal, either may break

20. Let 0 the angular velocity of the earth about its axis


Sec; Jr. IIT_IZ

00

2 22424

or

Let the angular velocity of the satellite

2 21.51.5

or

For a satellite rotating from west to east (the same as the earth), the relative angular

velocity , 1 0

Time period of rotation relative to the earth 1

2 1.6h

For a satellite from east to west (opposite to the earth), the relative angular angular

velocity 2 0

Time period of rotation relative to the earth 2

2 2417

h

CHEMISTRY

21. At A and D the temperature of the gas will be equal, so 0, 0E H

Now 0 0 0 0 02 ln 2AB BC CD oW W W W p V p V p V

0 02 ln 2p V

22.

2BaCl Carbon2 2 4 4KO S K SO BaSO BaS

(A) (B) (C)

HCl

H2S (D)

23. cos( 1)ml

l l

24. a)intermolecular hydrogen bonded (true)

b)structure of anions are different 23CO trigonal planar 2sp

O

C

OO

2-

2

4SO -trigonal pyramidal ( 3sp ) (false)


Sec; Jr. IIT_IZ

SOO

O

2-

c)

O=Xe

O

O

F

F no lone pair of electrons(false)

d)True 2 2374.4 ; 373.0D O k H O k

25. Meq of Cu+ = 0.72 100 1 = 72

Meq of 3NO = a 6 [nb of 3NO = 6; a = millimoles of 3NO ]

a 6 = 72 a = 12

Meq of 1 3

2 3NH OH NH

so millimoles of NH3 = 12 m moles

so volume of HCl = 120 ml

26.

ClBr/Na Ether

2Br hv.alc KOHBr

27. Conceptual

28.

2 3CHCH

2 2CHCHCl

C

Cl

H2Cl

hv

2 3CHCH

C

Cl

H

2 2CHCHCl

2 3CHCH

C

Cl

H2 3CHCH

C

Cl2CHCl CHCl

+

2 2CHCHCl+

Cl


Sec; Jr. IIT_IZ

C

Cl

HC

Cl

H+

2 2CHCHCl+

3CH CH

2 2CHCHCl

Cl2 2CHCHCl

Achiral I and II would formed as pair of diastereomers

29. A) 35f bE E kJ

155 35 120 /fE kJ mol

B) At 300 K, 1/2t for the forward reaction is 2 hrs

1

1/2

ln 2 0.693 0.16.93

k hrt

ln ln aff f

Ek A

RT

ln ln aff f

EA k

RT

log log2.303

off f

EA k

RT

120 10001 19.92.303 8.314 300

Or 19 18 10fA hr

Also 0E

RTk Ae

1f b

HE Ef f fRT RT

b b b

k A AK e e

k A A

ln ln f

b

A HKA RT

35000ln 0.16 ln8.314 300

f

b

AA

5ln 12.2 2 10f f

b b

A AA A

14 15 4 10

2 10f

b

AA h

31. Molar mass of albumin can be calculated using following relation :


Sec; Jr. IIT_IZ

22

w RTMV

----------------(i)

Given 1 12 1.08 , 0.0821w g R L atm K mol

5.85 50298 , ; 0.05760 1000

T K atm V L

Substituting these values in (i)

32. 35.85 101325 1 10 9.85760

h

27.958 10 7.958h m cm

33,34

Me

MeMe

Me Me

Me

2CH

3CH 3CH

3CH

3CH3CH

X (R) 03 0P 1Free radical

0Q 2Free radicalFree radical

Me

MeMe

MeMe

Me

2CH Cl

3CH 3CH

3CH

3CH3CHCl

Cl

Cl

H

3CH

3CH3CH

Cl

H

Dia

9 1=9 0 91 100 19.86%45.3

product

6 3.8 22.8 0 22.82 100 50.33%45.3

product

3 4.5 13.5 0 13.53 100 29.80%45.3

product

35. Conceptual

36. 2 3Trimetaborate ionBO

= all boron are 2sp

In 63 4Si O all the Si are in 3sp hybridization


Sec; Jr. IIT_IZ

37. Conceptual

38. (A), (B), (C) are optically active due to absence of plane of symmetry and centre of

symmetry

39. A, B : due to covalent nature LiCl have low m.p. than NaCl and dissolve more in

organic solvent.

C: Both LiCl and Nacl ionizes completely in water.

D: Fused LiCl have more conductivity (166) than Nacl (133) due to small size of Li-1

40. A) Bond angle

B)

Cl Cl

N

Cl

;Cl Cl

P

Cl

;Cl Cl

As

Cl

;Cl Cl

Sb

Cl0106.8ClNCl 0100.3ClPCl 097.7ClAsCl

097.2ClSbCl

C) In 2OF

angle O S F = 1130

angle F S F = 980

D)

O= N =O

0180

2NO

; O= N =O0134

2NO

O=N =O0115

2NO

;

MATHS 41. Taking x = y = 1, we get

f 1 f 1 f 1 2

2f 1 f 1 2 0 f 1 2 f 1 1 0

f 1 2 (as f(1) > 0)

Taking y = 1, we get

f(x). f(1) – f(x) = x + 1

f(x) = x + 1 1f (x) x 1

1 2f (x).f (x) x 1

(C) is the correct answer.


Sec; Jr. IIT_IZ

42. 2 3 2 31 10,1 0; 0 & . 1

1 1x xxx x x x

2 31 1 1 1

2 3

2 3

11 1 31tan tan tan tan11 41 .

11 not possible

xx x x

xx x xx x

x

43. = 0 i ≠ j and aij = (n – 1)2 + i i = j

Sum of all the element of 2n 1

2n

i 1

A [(n 1) i]

= (2n – 1) (n – 1)2 + (2n – 1)n = 2n3 – 3n2 + 3n – 1 = n3 + (n – 1)3

So, Tn = (–1)n [n3 + (n – 1)3] = (–1)n n3 – (–1)n-1 (n – 1)3 = Vn – Vn–1

102 102

3n n n 1 102 0

n 1 n 1T (V V ) V V (102)

102

nn 1

T2

520200

.

44. 2 21 1 1 1 2 2 2 24 4 0, 4 4 0D b a c D b a c

2 21 2 1 2 1 2a a c c b b the 2

3 1 2 1 2 1 24 0D b b a a c c 45. Let be a root of 0f x , so we have 0f and thus 0f f ,

0 0 0f .

We then have f x x x and thus 0, .

2f f x x x x x

We want such that 2x x has no real roots besides 0 and . We can easily find

that 0 4 .

46. an =

n ne

k 0

log 10 n!n! k! n k ! = ne nlog 10

2n!

= ne2 log 10

n!

Thus, a0 + a1 + a2 + . . . upto infinity is

= ne

n 0

2 log 10n!

= e2log 10e 100

(B) is the correct answer.


Sec; Jr. IIT_IZ

47.

2 2,a a lies of y x

2 2 0 1,2aa a

2 2 0 1, 2aa a

48. 1 4001

4000a a

= 10 a1a4001 = 400

a1 + a4001 = 50

(a1 – a4001)2 = (a1 + a4001)2 – 4a1 a4001

|a1 – a4001| = 30

49,50 Writing the equations of the roads such that constant terms are positive

-x + 2y + 4 = 0 , -2x + y + 4 = 0

a1 a2 + b1 b2 = 4 > 0. Origin lies in the obtuse angle.

Bisectors of acute angle is 2 2

2 4 2 4( 1) 4 ( 2) 1x y x y

3x – 3y – 8 = 0.

Solving x + y +2 = 0

3x – 3y – 8 = 0

P 1 7,3 3

51,52 3 2 21 45 0 3, 3,5

3 26 9 4 0

2 16 9 4 0A A I A

1 21 6 94

A A A I

53,54


Sec; Jr. IIT_IZ

A

B C

F

E

1

2

D

P 1

2

The position vectors of D and E are marked in figure.

The vector equation of CD and BE are

2 4 7 73

r r j k j k ....(i)

and 7 7 73

r i j k i j k ....(ii)

respectively.

CD and BE intersect at point P. At their point of intersection, we must have

2 4 7 7 7 7 73 3

i j k j k i j k i j k

7 7 71 1 , 2 13 3 3

and 74 1 73 3

6 / 7 and 3 / 7

Substituting the value of in (i) or that of in (ii), we obtain the position vector 1r of

point P as, 1 3 r i j k

Now, 1area of 2

ABC AB AC

1 3 2 22

i j k i j k 7 2 sq unit.2

Volume of the tetrahedron ABCF

13

(area of the base) height

1 7 2 723 2 3

cubic units

We have, 7 7 AB AC j k

Since, PF is parallel to

AB AC and 2PF units.


Sec; Jr. IIT_IZ

7 7

249 49

j kPF j k

. of

PV F j k

. of 3 4 P V F j k i j k i k

Vector equation of AF is,

2 2 4 2 2r i k i k j k

2 2 2 r i k i k .

55. When x 1 When x > 1

h 1 7f 1

3 1

p

Ltp

xf 1

p p

p

h x 7g xx x

x

P3x 17

x

h 1 77

4

g 17

7

h 1 21 g 1 49

g 1 h 1 28

g 1 h 1 70

56.

Y

XMo

( , ( ))x x ( , ( ))Q y y

( )y f xR

P

T

Take , ; ,p x x Q y y be any two points the curve y x

Let ‘R’ divides the line segment PQ in the ratio 2:1 then 22 ,3 3

x yx yR

Clearly TM > RM

223 3

x yx y


Sec; Jr. IIT_IZ

Equality holds iff x is a linear function. x ax b ' 0 1 1a 0 2 b = 2 2x x

57. By IVP,

f 0 f 1f c ,0 c 1

2 by

11 1

12 2

1 1LMVT,f 1/2 f 0 f c ,0 c2 2

1 1f 1 f 1/2 f c , c 12 2

Subtracting, we get,

1 12 1 112 1f c f c c c1f 1 f 0 2f f c 0

2 2 2

(UsingLMVT)

1f 1 f 0 2f

2

58. Let h x g x 4f x . Verify RT in 2,4

h 2 g 2 4f 2 32

h 4 g 4 4f 4 32

1h x 0 for at least one x 2,4 1 1g x 4f x for at least one x 2,4 59. 0, 2x are critical points

0, 1x are points of inflection

0x is the point where f x has cusp

60.

r x

c x

x

y

O

c x r x 3 3x

The number of phones per day 1268, 4732 for no loss

2 2dc dr xdx dx

Maximum profit when number of phones is 3414

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