Iit Jee 2012 Pet4 Solns p2

8/2/2019 Iit Jee 2012 Pet4 Solns p2

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Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns

BRILLIANTSPROGRESSIVE EVALUATION TEST

FOR STUDENTS OF

OUR ONE/TWO-YEAR POSTAL COURSES

TOWARDS

IIT-JOINT ENTRANCE EXAMINATION, 2012

SECTION I

1. (A)

= =

221 2

1 1 1R

n n

For first Lyman transition,

= = =

L 2 2

1 1 1 3R R

41 2

For first Paschen transition,

p2 2

1 1 1 7R R

1443 4

= = =

=PL3 7

: :4 144

73 :

36=

= 108 : 7

PART A: CHEMISTRY

IIT/ELITE 2012PET I/PET IV/CPM/P(II)/SOLNS

PAPER II SOLUTIONS

CHEMISTRY PHYSICS MATHEMATICS


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2. (D) Species Hybridization

4NH+ 5 4 1 8 4

2 2+ = = sp3

3NO

5 0 1 6

32 2

+ += = sp

2

2NO+

5 0 1 4

22 2

+ = = sp

3. (A) In the reaction,

H Li LiH+

Li undergoes oxidation and H undergoes reduction

( )H e H+ . Hence

hydrogen acts as an oxidising agent.

4. (A) Density of the solution,

= +

1 mol. wtd M

m 1000

1 98d 11.07

21.91 1000

= +

= 11.07 [0.0456 + 0.098]

= 11.07 0.1436 = 1.589 g/mL

5. (B) 2 2 31 3

N H NH

2 2

+

Partial pressure of 324.91

NH 100 24.91 atm100

= =

Pressure of (N2

+ H2) = 100 24.91 = 75.09 atm

Partial pressure of 23

H 75.09 56.32 atm4

= =

Partial pressure of 21

N 75.09 18.77 atm4

= =

( ) ( )= =

3

2 2

NH

p 1/2 3/2 1/ 2 3/ 2HN

P 24.91

K P P 18.77 56.32

24.91

4.3 421=

= 0.0136 atm

1


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6. (C) For gas A, A AA

A

d RTP

M=

For gas B, B BBB

d RTP

M=

A A A B

B B B A

P d T M

P d T M

=

[since M

A= M

B]

A

B

P 1 3 31

P 2 1 2

= =

or PA: PB = 3 : 2

7. (A) NO2

group exhibits I and M effect. These ve effects are greater for

NO2

group than for CN group. Also I effect is more when the group is

present in the ortho position. Thus

2CH

2NO

is the most stable carbanion

among them.

8. (C) I and M effects increase the strength of acid. Since NO2 group present in

para position exhibits M and I effect,

3NH

2NO

is most acidic.

SECTION II

9. (A), (C)

2 2 3 2 2 4 62Na S O I 2NaI Na S O+ +

(tetrathionate)


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10. (A), (B), (D)

NaI < NaCl < BaO < CaO (Lattice energy)BaCO

3> SrCO

3> CaCO

3> MgCO

3(Thermal stability)

Dipole moment of NH3

is greater than that of NF3.

Nuclear spins are in opposite direction in para hydrogen.

11. (A), (B), (D)

G = H TS

When H is ve and S is +ve, G becomes ve. So exothermic reactions arespontaneous at all temperatures.

When H is +ve, and S is ve, G is + ve. Hence such endothermic reactions

are not spontaneous at all temperatures.

When H is +ve and S is +ve, then G is ve only when H < TS i.e., G is

ve and spontaneous when temperature is high.

12. (A), (C)

Reductive ozonolysis

3 2 3CH C C CH CH3O

3 2 3CH C C CH CH

O

O O

2H O / Z n 3 2 3CH C C CH CH

O O

ZnO+

2 4

4

dil H SO3 2 3 3 2 31% HgSO

CH C C CH CH CH C C CH CH =

H OH

Rearrange

3 2 2 3CH CH C CH CH

O

4

cold alkaline3 2 3 KMnO

CH C C CH CH 3 2 3CH C C CH CH

OH

OH

OH

OH

22H O3 2 3CH C C CH CH

O O

3

2

(i) O3 2 3 3(ii) H O

CH C C CH CH CH COOH + 3 2CH CH COOH


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SECTION III

13. (5)

3CHOH

3CH

H

heat

+

3CH

2OH

3CH

2H OC H

3CH

3CH

Ring

expansion

3CH

3CH H

3CH

3CH+

14. (6)

All the six carbon atoms are in the sp2

hybrid state.

15. (2) + + CH CH 2 RMgX 2R H XMg C C MgX

Acetylene

16. (4) (g) (g) (g) (g)A 2B 2C D+ +

Initial 1 1.5 conc.

At (1 x) (1.5 2x) 2x xequilibriumconc.

At equilibrium since [A] = [D], (1 x) = x or x = 0.5

= =

2 2

c 2 2

[C] [D] (2 0.5) (0.5)K

[A] [B] (1 0.5) [1.5 1]

1 0.5 14

0.5 0.25 0.25

= = =


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17. (4)2 2

2 24 3C O 2H O 2CO 4H 2e 3 + + + +

2 24MnO 4H 3e MnO 2H O 2 + + + +

Add2 2

2 2 24 4 33C O 2MnO 2H O 6CO 2MnO 4H +

+ + + +

Since the reaction takes place in the basic medium,2 2

2 4 2 2 24 33C O 2MnO 2H O 4OH 6CO 2MnO 4H O

+ + + + +

or22

2 4 2 24 33C O 2MnO 4[OH ] 6CO 2MnO 2H O

+ + + +

18. (1) rmsT

vM

= =2

2

rms (H at 5 0 K )

rms (O at 800 K )

v 50 321

v 2 800

SECTION IV

19. (A) (p), (q); (B) (p), (r), (t); (C) (p), (s), (t); (D) (p), (q), (t)

(A) 50 ml of 0.1 M HNO3

+ 50 ml of 0.1 M KOH gives 50 10 3

0.1 mole of KNO3.

Complete neutralisation takes place. KNO3

does not undergo hydrolysis. It

simply ionises. Hence the pH of solution is 7.

(B) +3CH COOH NaOH

(moles) 3

50 0.2 10 350 0.2 10

+3 2CH COONa H O

gives 350 0.2 10 3moles of CH COONa

which is present in 100 ml of solution.

3

350 0.2 10

[CH COONa] 1000 moles / lit100

=

= 10 10 3

10 = 0.1 M

CH3COONa undergoes hydrolysis in solution

[ ]= + +w a1

pH pK pK log [salt]2

[ ]1

14 4.8 log 0.1

2

= + +

[ ]= + =1 1

14 4.8 1 (17.8)2 2

= 8.9


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(C) 4 4 23 3 3

HCl NH OH NH Cl H O

(moles) 50 0.1 10 50 0.1 10 50 0.1 10 in 100 ml

+ +

32

450 0.1 10

[NH Cl] 1000 5 10 M100

= =

NH4Cl undergoes salt hydrolysis.

[ ]= w b1

pH pK pK log (salt)2

( )21 14 4.8 log 5 102

=

= + 1

[14 4.8 2 0.6990]2

= 1

[11.2 0.6990]2

= 5.6 0.3495 = 5.25

(D) 5 10 3

moles of NH4OOC CH

3is formed in 100 ml.

Hence

= =

32

3 45 0.1 10 1000

[CH COONH ] 5 10 M10

This salt undergoes hydrolysis.

++3 4 3 4CH COONH CH COO NH

2H O3 3 44CH COO NH CH COOH NH OH

+++ +

= + w a b1

pH [pK pK pK ]2

1[14 4.8 4.8] 7

2= + =

20. (A) (p), (r), (t); (B) (p), (t); (C) (q); (D) (s)

(A) 34SF sp dii

(B)

3 3

6XeF sp d

ii

(C) CCl4 sp

3

(D) SF6 sp

3d

2


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SECTION I

21. (D) Let the initial velocity be iv along the +x direction and the final velocity fv

along the direction making an angle with iv .

Now = iv vi

= + fv v cos i v sin j

Change in velocity = ifv v

i.e., v v cos i v sin j vi = +

v (cos 1) i v sin j= +

2 2 2v v (cos 1) (vsin ) = +

2 2v cos 2cos 1 sin= + +

v 2 2 cos=

2v 4 sin2

=

2vsin2

=

average acceleration2vsinv 2a

t t

= =

22. (C) Since the rod is in rotational equilibrium, torque on it is zero.

N mg = 0

N = 39.2 N

Taking moments about O,

0.5 0.560 m sin N cos m

2 2 =

sin N

cos 60

=

39.260

=

= 0.65

= tan 1

(0.65)

iv

x

fv

y

F.B.D. for the rod

P

mg

O

N

PART B: PHYSICS


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23. (B) 1Elastic potential energyU stress strain

per unit volume 2

=

= 21

B (strain)2

, where B is bulk modulus

But strainy

(a coskx cos t)x x

= =

= ak sin kx cos t

= 21

U B ( ak sin kx cos t)2

2 2 2 21B a k sin kx cos t

2

=

The velocity of longitudinal wave is given by

Bc =

2B c=

Also ck

=

= 2 2 2 21

U a cos kx cos t2

= 2 2 21

U (x, 0) a cos kx2

In units of2 2a

2

U(x, 0) = cos2

kx

24. (A) For x-motion, x = vt (i)

But v 2gh= from Torricellis theorem

For y-motion, 21

(H h) gt2 = at the initial moment.

2(H h)t

g

=


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From (1),

2(H h)x 2ghg=

2 h(H h)=

1 3 32 m

4 4 2

= =

25. (A) P = k Fa

b

c

[ML2

T 3

] = [MLT 2

]a

[L]b

[ML 3

]c

a + c = 1

a + b 3c = 2

2a = 3

Solving, we get,

3 1a , b 1 and c

2 2= = =

=

3 112 2P kF

Let the power be P1

when weight, length and density are all altered. Then,

( ) ( )

=

1 1

3/221P k 2F 2

2

( ) =

3/ 2 11P 12 (2)2

12 2 2 4

2= =

26. (A) Centre of mass lies along the x-axis.

cm

x dmx

dm =

xL

L0

0

xLL

0

0

x e dx

e dx

=


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Let u = x and

x

Ldv e dx

=

x

Lv L e

=

L xx LLL

00

cm xLL

0

x L e L e dx

x

e dx

=

L

0

Lx

2 2L

0

x

L

L e L e

L e

=

2 2 2L e L e L

L (e 1)

+=

L L

(e 1) (2.718 1)= =

= 0.582 L

27. (C) Frequency of source = 1000 Hz

At t = 3 sec, velocity of detector v = a t

= 10 3 = 30 m/s

The frequency detected = 1100 Hz at t = 3 sec

c 301100 1000

c

+ =

, where c is the velocity of sound.

i.e., 11c = 10 c + 300

c = 300 m/s

28. (C) From conservation of momentum we have

m1

v1i

+ m2

v2i

= m1

v1f

+ m2

v2f

(1.60) (4.00) + (2.10) ( 2.50) = (1.60) (v1f

) + (2.10) ( 1.75)


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v1f

= 3.00 m/s

Again by law of conservation of energy

2 2 2 2 21 2 1 22i1i 1f 2f

1 1 1 1 1m v m v m v m v kx

2 2 2 2 2+ = + +

On substituting the numerical values and simplifying, we get compression inthe spring as

x = 0.17 m

SECTION II

29. (A), (B), (D)

=dQ dQ

AKdt dx

in the steady state, where K is coefficient of thermal conductivity.

Since dQdx

is constant,

dQ

Adt

. Hence rate of flow of heat decreases from A to C but increases from

C to B. The section at P at a distance x from A and the section at Q are at thesame distance from C. Since temperature gradient is uniform, temperature at

C is 50. Since the rod is symmetrical about C, rate of flow of heat throughP and Q is the same.

30. (A), (C), (D)

Since blocks move with uniform velocity, acceleration a = 0 for the system.

The F.B.D. for B is

T = N = mB

g (1)

The spring is stretched by tension T.By considering the F.B.D. for A, we have

T = mA

g (2)

From (1) and (2) we get mB

= 10 kg

For the spring we have

T = k = mA

g

2 9.8 = 1960

=1

100= 10

2m = 1 cm

Energy stored 2 41 1

kx 1960 102 2

= = = 9.8 10 2

J

Bgm

TNa


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31. (A), (B)

Potential energy for the system is

= + 2p1

U kx mgx sin 37 mg sin 372

For Up

to be a minimum

= >

2p p

2

U U0 , 0

x x

kx = mg sin 37

mg sin 37

x k

=

2 10 0.6

20

= = 0.6 m

Value of = + 2min1

V k (0.6) mg sin 37 mg (0.6) sin 372

21 20 (0.6) 2 10 0.4 0.62

= +

= 10 (0.6)2

+ 8 0.6 = 8.4 J

Maximum elastic potential energy occurs when

2 21 1mg sin 37 kx 20 x2 2

= =

2 10 (0.6) 1 = 10 x2

x = 1.2 m = 1.1 m

32. (A), (B), (C), (D)

Volume of the block = 125 cm3

Let h1 be the height of the block above mercury.

Then (5 h1) 25 13.6 g = 125 7.2 g

h = 5 2.65 = 2.35 cm


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45


Let h2

be the height of water required to just submerge the block. Then

wt. of block = wt. of (mercury + water) displaced

i.e., 900 = h2 2.5 1 + (5 h

2) 25 13.6

h2

= 2.54 cm

After pouring water, depth of mercury = 5 2.54 = 2.46 cm

SECTION III

33. (8) The kinetic energy of the system about BD is given by = K.E. about BC

=21

I K.E.2

of A about BD

21 mv2

=

22 Iv

m

=

22m( sin 60 )

m

=

2 23

4=

= 2 23

K.E. m8

Aliter

About BD about BC

2 21 1I mv2 2

= (K.E. of A)

=

22 2 2m

vm 4

= 2 23

4

2 21 3K.E. m2 4

=

A

B D C60


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34. (1) We have 1

0 1

q

4 r =

, where q is the charge on the metal sphere. The charge

induced on the shell is

q = 4 0

r1

1

Hence the potential acquired by the shell is2

2

q

r =

0 11

0 2

4 r

4 r

=

112

2

r

r

=

On comparison n = 1

35. (4) Average energy

T

0T

0

u dt

u

dt

=

T

0

1u dt

T=

But ( )2 2 21

u m A x2=

But x = A sin (t + )

( )T

2 2 2

0

1u mA cos t dt

2T = +

( )T

2 2 2

0

1mA cos t dt

2T= +

ButT

2

0

1cos ( t ) dt T

2 + =

2 21u mA4

=

n = 4


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36. (2) By the law of conservation of energy,

21 mghmvh2

1R

=

+

, where h is measured from the surface of the earth.

Given ev kv k 2gR= =

=

+

21 mg (r R)mk (2gR)r R2

1R

, where r is the distance measured from the centre

of the earth.

2 r Rk R 1 (r R)

R

+ =

i.e., k2

r = (r R)

2

Rr

1 k =

On comparison r = 2

37. (6) Due to extension produced in the cord, energy stored in it is converted into

kinetic energy K, when the stone flies away. Assuming that there is no loss of

energy in this process, the K.E. of the stone is given by21

K mv 4 J2

= = . This

must be equal to the work done in stretching the cord.

Therefore = =1

W F 4J2

4 2F

0.2

= = 40 N

Stress 6 22 3 2

F 40 401.415 10 Nm

A r (3 10 )

= = =

Strain20

0.47642

= =

Youngs modulus

= = = 6

6 2stress 1.415 10Y 2.97 10 Nm

strain 0.476

On comparison n = 6


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38. (4) The angular impulse about the centre of mass

J = Lf

Li

= I 0

But

=

J p

2

2mp

2 12

=

6p

m =

=

6p

2t m

n mt

48 p

=

On comparison n = 4

SECTION IV

39. (A) (p), (r), (t); (B) (p), (s), (t); (C) (p), (q), (t); (D) (p), (t)

Points x1

and x3

are points of zero displacement. Just to the left of x1, the

displacement is negative indicating that the gas molecules are displaced to the left,away from point x

1at some instant. Just to the right of x

1, the displacement is

positive, indicating that the molecules suffer displacement to the right which is

again away from x1. So at point x

1, the pressure is minimum.

At point x2, the pressure does not change because the gas molecules on both sides of

that point have equal displacements in the same direction. Hence pressure is

normal.At x3, the pressure is maximum, because the molecules on both sides of that point

are displaced towards point x3.

Since pressure p is proportional to density , pressure and density are in phase.

40. (A) (p), (q), (s); (B) (q), (r), (t); (C) (p), (q), (s); (D) (q), (r), (t)

Process AB is an isothermal process with T = constant. Therefore (P V) graph is

a hyperbola. (V T) graph is a straight line since T is constant. By ideal gas

equationPM

RT = . Therefore P. As T is a constant, ( T) graph is a straight

line.

Process BC is an isobaric process with P = constant. As P is constant, (P V) graph

is a straight line. In isobaric process, V T and hence (V T) graph is a straight

line.

When P is constant1

T . Hence ( T) graph is a hyperbola.

For isothermal process U = 0, but for isobaric process U 0. What applies toprocess AB, also applies to CD. Similarly what applies to BC, also applies to DA.


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49


SECTION I

41. (A) ax2

+ bx + c a (x ) (x ) p 1 p 2

a x xp p 1

+ +

+

Putting x = 1,1 1 a

a b c ap p 1 p(p 1)

+ + = =

+ +

( )2

2 2

2

b 4aca a ( ) 4 a b 4ac

a

= = + = =

42. (B)a

31 r

=

and3

3

a 108

131 r=

2

2

1 r r 13

41 2r r

+ + =

+

1r or 3

3 = . Here

1r

3= and a = 2

Now( )na 1 r 7281 r 243

>

n

11

72831 243

3

>

> < = >n n 51 728 1 1 11 n 5

729 7293 3 3

Minimum value of n = 6

43. (D)( ) ( )x 7 7log 2 log x log x5 x 2 24 + =

( )7 7 7log x log x log x5 2 2 24 2 4 + = = log7 x = 2 x = 72

= 49

44. (C) x + y + z = 16 and x, y, z 3

Let u = x 2, v = y 2, w = z 2

Then u + v + w = 10,

The number of positive integral solutions =9C

2= 36

PART C: MATHEMATICS


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50


45. (D) xA

= 1, xM

= 2, AP = 2

MAP 60 =

Required area = area QRPMQ

12 2 3

3=

43

3

=

46. (D) f ( ) 5 cos 3 cos cos 3 sin sin 33 3

= + +

( )13 3 3

cos sin 3 7 cos 32 2

= + = + + where =13

cos14

Range of f() = [3 7, 3 + 7] = [ 4, 10]

47. (B) P(a, a) must be outside the circle 2a2 4a 6 > 0

either a > 3 or a < 1

= = 21 1PT S 2a 4a 6 and 1CT r 2 2= =

< < < < >

1tan

3 6 2 2 2 3

2

2

2 2 1a 2a 3 2 3

32a 4a 6 >

Iit Jee 2012 Pet4 Solns p2

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