IIT-JEE 2012 Solution of Paper - 2 - Momentum...

IIT-JEE 2012 Solution of Paper - 2

PPPPARARARART I :T I :T I :T I : PHY PHY PHY PHYSICSSICSSICSSICS

SECTION I : Single Correct Answer Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct.

1. An infinitely long hollow conducti9ng cylinder with inner radius / 2R and outer radius R carries a uniform

current density along its length. The magnitude of the magnetic field, | |B

as a function of the radial

distance r from the axis is best represented by :

(A)

R/2 Rr

B

(B)

R/2 Rr

B

(C)

R/2 Rr

B

(D)

R/2 Rr

B

1.(D) Field at P :

22

04

2

RJ x

Bx

µ π

π

−

=

Qx

PR/2

x

Field at Q is 0

2

IB

x

µπ

=

2. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in

water in half-submerged state. If C

ρ is the relative density of the material of the shell with respect to water,,

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then the correct statement is that the shell is :

(A) more than half-filled if C

ρ is less than 0.5

(B) more than half-filled if C

ρ is more than 1.0

(C) half-filled if C

ρ is more than 0.5

(D) less than half-filled if C

ρ is less than 0.5

2.(D)

h/2

h/2

3. In the given circuit, a charge of 80 Cµ+ is given to the upper plate of the 4 Fµ capacitor. Then in the

steady state, the charge on the upper plate of the 3 Fµ capacitor is :

+80 µC4 µF

3 µF2 µF

(A) 32 Cµ+

(B) 40 Cµ+

(C) 48 Cµ+

(D) 80 Cµ+

3.(C)80

3 2

q q−=

4µF

80µC

3µF2µFq(80 – ) q5 240q =

48q Cµ=

4. Two moles of ideal helium gas are in a rubber balloon at 30oC . The balloon is fully expandable and can be

assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly charged

to 35oC . The amount of heat required in raising the temperature is nearly

(take 8.31 / mol.R J K= )

(A) 62 J (B) 104 J (C) 124 J (D) 208 J

4.(D) 2n =

pQ nC T∆ = ∆

52 5 208

2R J= × × =

5. Consider a disc rotating in the horizontal plane with a constant angular speed ω about its centre O . The

disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown

in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously

projected at an angle towards R . The velocity of projection is in the y z− plane and is same for both

pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed




1

8rotation, (ii) their range is less than half the disc radius, and (iii) ω remains constant throughout. Then

Q

O

P

R

(A) P lands in the shaded region and Q in the unshaded region.

(B) P lands in the unshaded region and Q in the shaded region.

(C) Both P and Q land in the unshaded region.

(D) Both P and Q land in the shaded region.

5.(A)

6. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm.

The frequency of the tuning fork is 512 .Hz The air temperature is 038 C in which the speed of sound is

336 m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When thefirst resonance occurs, the reading of the water level in the column is(A) 14.0 cm (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm

6.(B) 4d cm= , 512f Hz= , 38o

t C= , 336 /v m s= , 0.6 2 0.6 1.2e r cm= = × =

4l e

λ+ =

0

e

l1.2

4

vl cm

f+ =

3360.164

4 512m= =

×

16.4 1.2 15.2l cm= − =

7. Two identical discs of same radius R are rotating about their axes in opposite directions with the same

constant angular speed ω . The discs are in the same horizontal plane. At time 0t = , the points P and Q

are facing each other as shown in the figure. The relative speed between the two points P and Q is v . In

one time period ( )T of rotaion of the discs, v as a function of time is best represented by

QP

R R




(A)

rv

O Tl

(B)

rv

O Tl

(C)

O Tl

(D)

O Tl

7.(A) ˆ ˆsin cosP

V v i v jθ θ= − −

ˆ ˆsin cosQ

V v i v jθ θ= −

ˆ2 sinQP

V v iθ=

| | | 2 sin | 2 |sin |QPV v v tθ ω= =

θ θv

θ θ

P Q

v

θ = ω t

8. A loop carrying current I lies in the x y− plane as shown in the figure. The unit vector k is coming out of

the plane of the paper. The magnetic moment of the current loop is

I

(A) 2 ˆa Ik (B) 21

2a I k

π +

(C) 21

2a I k

π − +

(D) ( ) 22 1 a I kπ +

8.(B)

2

2 14

2 4

aI A I aµ π

= = + ⋅

2 ˆ1

2I a k

π = +




SECTION II : Paragraph Type

This section contains 6 multiple choice questions relating to three paragraphs with two questions on eachparagraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 9 and 10

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre ofmass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontallyat its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin

on a horizontal frictionless plane with angular speed ω , the motion at any instant can be taken as a

combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the discthrough an instantaneous vertical axis passing through its centre of mass (as is seen from the changed

orientation of points P and Q ). Both these motions have the same angular speed ω in this case.

QP

x

z

P Qy

ω

Now consider two similar systems as shown in the figure : Case (A) the disc with its face vertical and

parallel to x z− plane; Case (B) the disc with its face making an angle of 45o with x y− plane and its

horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P , and the systems are

rotated with constant angular speed ω about the z-axis.

Case (A)

x

z

P

Q

y

ω

Case (B)

x

z

P

Q

y

ω

45o

9. Which of the following statements regarding the angular speed about the instantaneous axis (passingthrough the centre of mass) is correct?

(A) It is 2 ω for both the cases

(B) It is ω for case (A); and 2

ω for case (B)

(C) It is ω for case (A); and 2 ω for case(B)

(D) It is ω for both the cases

9.(D)

10. Which of the following statements about the instantaneous axis (passing through the centre of mass) iscorrect ?

(A) It is vertical for both the cases (A) and (B)

(B) It is vertical for case (A). and is at 45o to the x z− plane and lies in the plane of the disc for case (B)




(C) It is horizontal for case (A); and is at 45o to the x z− plane and is normal to the plane of the disc for

case (B)

(D) It is vertical for case (A); and is at 45o to the x z− plane and is normal to the plane of the disc for case

(B)10.(A)


The β -decay process, discovered around 1900 , is basically the decay of a neutron ( )n . In the laboratory,,

a proton ( )p and an electron ( )e− are observed as the decay products of the neutron. Therefore,

considering the decay of a neutron as a two-body decay process, it was predicted theoretically that thekinetic energy of the electron should be a constant. But experimentally, it was observed that the electron

kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. e

n p e v−→ + + ,

around 1930 , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ( )e

v to

be massless and possessing negligible energy, and the neutron to be at rest, momentum and energyconservation principles are applied. From this calculation, the maximum kinetic energy of the electron is

60.8 10 eV× . The kinetic energy carried by the proton is only the recoil energy..

11. If the anti-neutrino had a mass of 23 /eV c (where c is the speed of light) instead of zero mass, what

should be the range of the kinetic energy, K , of the electron ?

(A) 60 0.8 10K eV≤ ≤ × (B) 63.0 0.8 10eV K eV≤ ≤ ×

(C) 63.0 0.8 10eV K eV≤ < × (D) 60 0.8 10K eV≤ < ×

11.(D) 60.8 10Q eV×

If υ has mass, the mass defect will be less & energy released will be less, by 3eV

60 0.8 10K eV≤ < ×

12. What is the maximum energy of the anti-neutrino?

(A) Zero (B) Much less than 60.8 10 eV×

(C) Nearly 60.8 10 eV× (D) Much larger than 60.8 10 eV×12.(C)


Most materials have the refractive index, 1n > . So, when a light ray from air enters a naturally occurring

material, then by Snell’s law, 1 2

2 1

sin

sin

n

n

θθ

= , it is understood that the refracted ray bends towards the normal.

But it never emerges on the same side of the normal as the incident ray. According to electromagnetism,




the refractive index of the medium is given by the relation, r r

cn

vε µ

= = ± , where c is the speed of

electromagnetic waves in vacuum, v its speed in the medium, r

ε and r

µ are the relative permittivity and

permeability of the medium respectively.

In normal materials, both r

ε and r

µ are positive, implying positive n for the medium. When both r

ε and

rµ are negative, one must choose the negative root of n . Such negative refractive index materials can

now be artificially prepared and are called meta-materials. They exhibit significantly different optical behavior,

without violating any physical laws. Since n is negative, it results in a change in the direction of propagation

of the refracted light. However, simllar to normal materials, the frequency of light remains unchanged uponrefraction even in meta-materials.

13. For light incident from air on a meta-material, the appropriate ray diagram is :

(A)

θ 1

θ 2

Meta

material

Air(B)

θ1

θ 2

Meta

material

Air(C)

θ1

θ 2

Meta

material

Air(D)

θ 1

θ2

Meta

material

Air

13.(C)

14. Choose the correct statement :

(A) The speed of light in the meta-material is | |v c n=

(B) The speed of light in the meta-material is | |

cv

n=

(C) The speed of light in the meta-material is v c=

(D) The wavelength of the light in the meta-material ( )mλ is given by | |m air nλ λ= , where airλ is the

wavelength of the light in air.14.(B)

SECTION III : Multiple Correct Answer(s) Type

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE are correct.

15. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it.

The correct statement(s) is ( )are

(A) The emf induced in the loop is zero if the current is constant.

(B) The emf induced in the loop is finite if the current is constant.

(C) The emf in duced in the loop is zero if the current decreases at a steady rate.

(D) The emf induced in the loop is finite if the current desreases at a steady rate.

15.(A,C)

16. Six point charges are kept at the vertices of a regular haxagon of side L and centre O , as shown in the

figure. Given that 2

0

1

4

qK

Lπε= , which of the following statement (s) is ( )are correct?




q+F L E

D2q-

q+B

2q+S T

RC

P

O

(A) The electric field at O is 6k along OD .

(B) The potential at O the line PR is same.

(C) The potential at all points on the line PR is same.

(D) The potential at all points on the line ST is same.

16.(A, B, C)

q+

D2q-

q+B

2q+

C

O

k

kk

k

2k 2k

17. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from

the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while

Q has most of its mass concentrated near the axis. Which statement (s) is ( )are correct?

(A) Both cylinders P and Q reach the ground at the same time.

(B) Cylinder P has larger linear acceleration that cylinder Q .

(C) Both cylinders reach the ground with same translational kinetic energy.

(D) Cylinder. Q reaches the ground with larger angular speed.

17.(D)

2

sinMga

IM

R

θ=

+ P Q

I I>∵P Q

a a∴ <

18. To spherical planets P and Q have the same uniform density ρ , masses pM and Q

M , and surface

areas A and 4A , respectively. A spherical planet R also has uniform density ρ and its mass is

( )P QM M+ . The escape velocities from the planets P , Q and R , are ,

P QV V and

RV , respectively..

Then

(A) Q R PV V V> > (B) R Q P

V V V> > (C) / 3R P

V V = (D) 1

/2

P QV V =

18.(B,D)

A

P

a

MP

4A

Q

2a

MQ

8QM M=PM M=




9R

M M=

3 34 49

3 3b aπ ρ π ρ⋅ = ⋅ ⋅

( )1/ 39b a= ⋅

2e

GMv

R=∵

2P

GMv

a=

2 8 24

2Q

G M GMv

a a

⋅= = ⋅

2 / 3

1/ 3

2 9 29

9R

G M GMv

a a

⋅= = ⋅

⋅

1/ 32

9GM

a= ⋅

R Q Pv v v∴ > >

19. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on

a horizontal surface with angular speed ω and (ii) an inner disc of radius 2 R rotating anti-clockwise with

angular speed / 2ω . The ring and disc are separated by frictionless ball bearings. The system is in the

x z− plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle

of 30o with the horizontal. Then with respect to the horizontal surface,

30oP

x

z

ω/2

3R

2R

R

O

(A) the point O has a linear velocity ˆ3R iω

(B) the point P has a linear velocity 11 3 ˆˆ4 4

R i R kω ω+

(C) the point P has a linear velocity 13 3 ˆˆ4 4

R i R kω ω−




(D) the point P has a linear velocity 3 1 ˆˆ3

4 4R i R kω ω

− +

19.(A,B)

R

2

P

3 R

300

0 0 ˆˆ ˆ3 sin30 sin302 2

P

R RV R i i k

ω ωω

= + − +

1 3 ˆˆ34 4

R i R kω ω = − +

11 3 ˆˆ4 4

R i R kω ω= +

20. In the given circuit, the AC source has 100 / .rad sω = considering the inductor and capacitor to be ideal,

the correct choice (s) is ( )are

20 V

0.5 H 50

100100 F

(A) The current through the circuit, I is 0.3 .A

(B) The current through the circuit, I is 0.3 2 .A

(C) The voltage across 100 Ω resistor = 10 2 .V

(D) The voltage across 50 Ω resister = 10 .V

20.(A,C)

100 /rad sω =

2 2

1 50 50Z = + 50 2= [ ]LX Lω=

2 2

2 100 100Z = + 100 2=

6

1

100 100 10CX −

= × ×

R

Z

V

LX

1I

tan = 1

R

Z

2I

V

cX

tan =1




1 45oφ = & 2 45oφ =

1

20

50 2I∴ = 2

5=

2

20

100 2I =

1

5 2=

phase difference between 1I & 2I = 090

2 2

1 2I I I∴ = +

2 1

25 50= +

4 1

50

+=

1

10= 0.3A=

100 2

1x 100 10 2

5 2V I RΩ = = =

50 1

2x 50 10 2

5V I RΩ = = =

PPPPARARARART II :T II :T II :T II : CHEMISTR CHEMISTR CHEMISTR CHEMISTRYYYY


This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out

of which ONLY ONE is correct.

21. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are :

(A) 2O andCO respectively (B) 2O and Zn dust respectively

(C) 3HNO and Zn dust respectively (D) 3HNO and CO respectively

21.(C)

( )2 224 2Ag S NaCN NaAg CN Na S+ +

2 2 24 2Na S SO H O+ + →

2 42 4 2Na SO NaOH S+ +

( ) ( )22 42 2NaAg CN Zn Ag Na Zn CN+ → +

3Ag KNO Ag+ →

Anode - Impure Ag

Cathode - pure Ag plate

Electrolyte - 3 3AgNO HNO+

22. The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus

containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the otherproduct are respectively :

(A) redox reaction; 3− and 3+ (B) redox reaction ; 3+ and 5+(C) disproporationation reaction; 3− and 5+ (D) disproporationation reaction; 3− and 3+




22.(D) 4 24 2P NaOH H O+ + → 3 2 32 2PH Na HPO+-3 +3

23. The shape of 2 2XeO F molecule is :

(A) Trigonal bipyramidal (B) Square planer(C) tetrahedral (D) see-saw

23.(D) See-Saw but shape is see-saw because of lone pair.

Xe-O

O

F

F

24. For a diute solution containing 2.5g of a non-volatile non-electrolyte solute in 100g of water, the elevation inboiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentra-

tion of solvent, the vapour pressure (mm of Hg ) of the solution is (take 10.76

bK K kg mol

−= )

(A) 724 (B) 740 (C) 736 (D) 718

24.(A)2.5

2 0.760.1M

= ××

2.5 0.76

2M

×=

0

0

sP P n

P n N

−=

+

0

s

s

P P n

P N

−=

760 2.5 2 18

25 0.76 100

s

s

P

P

− × ×=

× ×

760 2 18 1

100 7.6 20

s

s

P

P

− ×= =

×

76020

ss

PP− =

760 20

21sP

×= 723.8= 724

25. The compound that undergoes decarboxylation most readily under mild condition is :

(A)

COOH

2CH COOH(B)

COOH

O

(C)

COOH

COOH(D)

2CH COOH

O

25.(B) β − keto acids are readily decarboxylate under mild condition because of stable carbanion is formed as an

intermediate (B) is correct :




α

O

C OH-

O

2CO

∆− ↑

→ O

H

26. Using the data provided, calculate the multiple bond energy ( 1kJmol − ) of C C≡ bond in 2 2C H . That

energy is (take the bond energy of a C H− bond as 350 1kJ mol− )

( ) ( ) ( )2 2 22C s H g C H g+ →

1225H kJ mol−∆ =

( ) ( )2 2C s C g→

11410H kJ mol−∆ =

( ) ( )2 2H g H g→

1330H kJ mol−∆ =(A) 1165 (B) 837 (C) 865 (D) 815

26.(D) ( ) ( ) ( )2 2 22C s H g C H g+ → ;

225H kJ∆ = ..1

( ) ( )2 2C s C g→ ; 1410H kJ∆ = ...2

( ) ( )2 2H g H g→ ; 330H kJ∆ = .. 3

This can be get by substrating 1, 2 & 3

( ) ( ) ( )2 22 2C g H g C H g+ → ;

1515H kJ∆ = −

2 52

C H C H C CH H H− ≡∆ = − ∆ − ∆

( )1515 2 350C C

H ≡+ = + + ∆

815C C

H ≡∆ =27. The major product H of the given reaction sequence is :

3 2 3CH CH CO CH− − − CNG

−

→

2 495%H SO

HeatH→

(A) 3CH CH C COOH- = -

3CH(B)

3CH CH C CN- = -

3CH

(C) 3 2CH CH C COOH- - -

3CH

OH

(D) 3 2CH CH C CO NH- = - -

3CH




27.(A)3 2CH CH C O- - -

3CH

CN

3 2CH CH C- -C N

O3

CHG

3 2CH CH C- -C N

OH3CH

3 2CH CH C- -C OH-

OH3CH

O

95%

2 4H SO( )2H O5%

Hydrolysis

β α

2H O-

Hydrolysis

3CH CH C C OH- = - -

O

3CH

28. ( ) ( )2 2 5 6 5 22 NiCl C H C H exhibits temperature dependent magnetic behaviour (paramagnetic/diamag-

netic). The coordination geometries of 2Ni + in the paramagnetic and diamagnetic states are respectively..

(A) tetrahedral and tetrahedral (B) square planar and square planar(C) tetrahedral and square planar (D) square planar and tetrahdral

28.(C) [ ]2 83Ni Ar d+ →

When it is paramagnetic

Hybridisation → 3sp

Shape → tetrahedral

When it is diamagnetic

Hybridisation → 2dsp

Shape →square planar






Paragraph : 1In the following reaction sequence, the compound J is an intermediate.

( ) ( )( )( )

3 2 2

3 2

3

, /

.

CH CO O i H Pd C

CH COONa ii SOCl

iii anhyd AlCl

I J K→ →

( )9 8 2J C H O gives effervescence on treatment with 3NaHCO and a positive Baeyer’s test.

29. The compound K is

(A) (B) O

(C)

O

(D) O

30. The compound I is

(A)

O H

(B)

OH

(C)

O 3CH

(D)

HH

H

29 & 30∵ J ( )9 8 2C H O gives effervescences with

3NaHCO & positive Baeyer’s test ∴ it has C OH- -

Ogroup & unsaturation.

2CHH

3 3CH C O C CH- - - -

3CH C ONa- -

O O

O

CH CH C OH= - -

O

Reduce onlydouble bond

2 | |H Pd C

2 2CH CH C O- - =

OH

2SOCl

NPSR→

2 2CH CH C O- - =

Cl




F.C.acylation ( )3 .AlCl Anhyd

2 2CH CH C O- - =

OK

29.(C)30.(A)Paragraph : 2

The electrochemical cell shown below is concentration cell.

2|M M+ (saturated solution of a sparingly soluble salt,

2MX ) || ( )2 30.001 |M mol dm M+ −

The emf of the cell depends on the difference in concentration of 2M

+ ions at the two electrodes.

The emf of the cell of 298 K is 0.059 V.

31. The value of ( )1G kJ mol

−∆ of the given cell is (take 11 96500F C mol−= )

(A) 5.7− (B) 5.7

(C) 11.4 (D) 11.4−31.(D) G nFE∆ = −

2 96500 0.059= − × × 111387J Mol−= −

111.387kJ Mol−= − 111.4 kJ mol

−−

32. The solubility product (3 9;spK mol dm ) of 2MX at 298 K based on the information available for the given

concentration cell is ( take 2.303 298 / 0.059R F V× × = )

(A) 151 10−× (B) 154 10−× (C) 121 10−× (D) 124 10−×

32.(D)2 2| |

C aM M

+ +→

0 0Cell

E = ( Assuming concentration cell)

2

2

|0.059log

2 |

a

C

mE

m

+

+

−=

( )1/3

4

3

/0.0590.059 log

2 10

spK

−

−=

1/3

9

0.0590.059 log

2 4 10

spK−

−= ×




( )32

910

4 10

spK−

−=

×154 10spK

−= ×

Paragraph : 3Bleaching powder and bleach solution are produced on a large scale and used in several house holdproducts. The effectiveness of bleach solution is often measured by iodometry.

33. Bleaching powder contians a salt of an oxoacid as one of its components. The anhydride of that oxoacid is:

(A) 2Cl O (B) 2 7Cl O (C) 2ClO (D) 2 6Cl O

33.(A) Bleaching powder → ( )Ca OCl Cl

acid of ClO− is HClO

and anhydride of that acid is 2Cl O

34. 25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4N acetic acid. In

the titration of the liberated iodine, 48 mL of 0.25 N 2 2 3Na S O was used to reach the end point. The molarity

of the household bleach solution is :(A) 0.48 M (B) 0.96 M (C) 0.24 M (D) 0.024 M

34.(C) 3 2. .B S CH COOH Cl+ →

2 2I Cl I Cl

− −+ → + ....... (1)

2 2

2 2 3 4 6I S O I S O

− − −+ → + ....... (2)

meq. of 2

2 3S O

−= meq. of 2I liberated

48 0.25 1× × = m. mols of 2 2I ×

m.mols of 2I = 24 0.25×

m. mols of 2I liberated = m. moles of Bleach sol.

24 0.25= ×So, 24 0.25 25 M× = ×

.24M = 0



35. The given graphs /data I, II, III and IV represent general trends observed for different physisorption andchemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s)about I, II, III and IV is (are) correct ?

Am

ount o

f gas

adso

rbed

P constantI

T

Am

ount o

f gas

adso

rbed

P constantII

T




Am

ount o

f gas

adso

rbed

III

Po

tential E

nerg

y

IV

0

1150ads

H KJmol-=

adsE

P

200k

250k

(A) I is physisorption and II is chemisorption(B) I is physisorption and III is chemisorption(C) IV is chemisorption and II is chemisorption(D)IV is chemisorption and III is chemisorption

35.(A,C)36. For the given aqueous reactions, which of the statement (s) is (are) true?

excess ( )3 6[ ]KI K Fe CN+

dilute 2 4H SO

4ZnSO

2 2 3Na S O

brownish-yellow solution

white precipitate + brownish -yellow filtrate

colourless solution

(A) The first reaction is a redox reaction

(B) white precipitate is ( )3 6 2Zn Fe CN

(C) Addition of filtrate to starch solution gives blue colour

(D) White precipitate is soluble in NaOH solution

36(A,C,D)

Excess ( )2 4

.

3 6

dil

H SOKI K Fe CN + →

( )4 36[ ]K Fe CN I

-+

+24ZnSO→

( )2 3 2 26

[ ]K Zn Fe CN I

white ppt Brownish yellow filtrate

+

−

NaOH 2 2 3Na S O

( )

( )

2

4

4

6

[ ]

[ ]

Zn OH

Fe CN

−

−

+ ( )

2

4 6I S O

Colourless

− −+




37. With respect to graphite and diamond, which of the statement(s) given below is (are) correct?(A) Graphite is harder than diamond(B) Graphite has higher electrical conductivity than diamond(C) Graphite has higher thermal conductivity than diamond

(D) Graphite has higher C C− bond order than diamond.

37. (C,D) diamond is harder than graphitedue to delocalisation graphite showconductivity.graphite has partial double bond character so greater bond order.

38. With reference to the scheme given, which of the given statment(s) about T, U, V and W is (are) correct?

O

T3H C

4LiAlH

3 /CrO H 3 2( )CH CO O

V U W

O

+

(A) T is soluble in hot aqueous NaOH(B) U is optically active

(C) Molecular formula of W is 10 18 4C H O

(D) V gives effervescence on treatment with aqueous 3

NaHCO

38.(C,D)

O

3H C T

4LiAlH

OH

3H C U

OH

Opticallyinactive

3CH C O- -

O

3CH C-

O

3O C CH- -

3H C W

O

3O C CH- -

O

10 18 4C H O

3 /CrO H+




3H CO

O

OO H

3H C

V

OH

O

OH

O

Give effervescences

on treatment with aq. 3NaHCO

∴ (C,D) both are correct

39. Which of the given statement(s) about N,O,P and Q with respect to M is (are) correct ?

3CH

HO

Cl

HO H

H

M

3CH

OHCl

HO H

H

N3CH

OH

Cl

HOH

H

O

3CH

ClHHO

H OH

P

3CH

ClHHO

H

Q

HO

(A) M and N are non-image stereoisomers(B) M and O are identical(C) M and P are enantiomers

(D) M and Q are identical

39.(A,B,C)

HO H

H

M

3H C

Cl

HO H

O 3CH

OHHO

H

HCl

HO H

M O=

3CH

Cl

HO H

.

3CH

ClHHO

H

P

OH

H

3CH

Cl

OHH

HO

H

3CH

Cl

H

OH

P

OH

P & M are non-super imposable

mirror image C




40. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure.Which of the following statement(s) is (are) correct ?

isotermal

adiabatic

( )2 2 2,P V T

( )3 2 3,P V T

P

V

(A) 1 2T T= (B) 3 1T T>

(C) isotermal adiabatic

W w> (D) isotermal adiabatic

U U∆ > ∆40.(A,C,D) In expansion temperature of adiabatic will decrease but of Isothermal reamains same

So 1 2T T= for Isothermal

2 3T T>Internal energy for adiabatic system will decrease

So, iso adi

V V∆ > ∆

| | | |iso adi

w w>

PPPPARARARART III :T III :T III :T III : MA MA MA MATHEMATHEMATHEMATHEMATICSTICSTICSTICS


This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct.

41. Let PQR be a triangle of area ∆ with 7

2,2

a b= = and 5

2c = , where ,a b and c are the lengths of the

sides of the triangle opposite to the angles at ,P Q and R respectively. Then 2sin sin 2

2sin sin 2

P P

P P

−+

equals

(A) 3

4∆(B)

45

4∆(C)

23

4

∆

(D)

245

4

∆

41.(C)

P

Q

2a =R

7 / 2b =5 / 2c =

2sin 2sin cos

2sin 2sin cos

p p p

p p p

−

+

1 cos

1 cos

p

p

−=

+




2 2 27 5 4cos

2.7.5p

+ −=

where cos 58 29

1 2.75 35

p= =

1 cos 35 29 3

1 cos 35 29 32

p

p

− −∴ = =

+ +

( )( )( )s s a s b s c∆ = − − −1 3

4 2 62 2

= × × × =

23 9 3

4 16 62 32

= = ∆ ×

42. If a

and b

are vectors such that 29a b+ =

and ( )ˆˆ ˆ2 3 4a i j k× + + ( )ˆˆ ˆ2 3 4i j k b= + + ×

, then a possible

value of ( ) ( )ˆˆ ˆ7 2 3a b i j k+ ⋅ − + +

is

(A) 0 (B) 3 (C) 4 (D) 8

42. (C) 0a v v b× − × =

0a v b v× + × =

( ) 0a b v+ × =

(2 3 4 )a b k i j k+ = + +

| | | | 29a b k+ =

29 | | 29 1k k= ⇒ = ± ( )( 7 2 3 )a b i j k∴ + − + +

(2 3 4 ).( 7 2 3 )i j k i j k= ± + + − + + ( 14 6 12) 4= ± − + + = ±

43. If P is a 3 3× matrix such that 2 ,TP P I= + where T

P is the transpose of P and I is the 3 3× identity

matrix, then there exists a column matrix

0

0

0

x

X y

z

= ≠

such that

(A)

0

0

0

PX

=

(B) PX X= (C) 2PX X= (D) PX X= −

43.(D) 2TP P I= +

( ) (2 )T T TP P I= +

2 TP P I= +1

2 ( )2

P I P I∴ + = −




4 2P I P I+ = − 3 3P I⇒ = −

P I= − 0P I∴ + =

( ) 0P I X∴ + = 0PX X⇒ + =

44. Let ( )aα and ( )aβ be the roots of the equation

( ) ( ) ( )23 61 1 1 1 1 1 0a x a x a+ − + + − + + − =

where 1a > − .

Then ( )0

lima

aα+→ and ( )

0lim

aaβ+→

are

(A) 5

2− and 1 (B)

1

2− and 1− (C)

7

2− and 2 (D)

9

2− and 3

44.(B) 3

1 1

1 1

a

aα β

+ −+ = − + −

6 1 1

3 1 1

a

aαβ

+ −= + −

( )( )

1/ 32 / 3(1 ) 1 1

1 1

a a

aα β

+ + + + ⇒ + = − + +

( )1

6 1 1aαβ =

+ +

( )0

3lim

2xα β

+→

−+ = ( )

0

1lim

2xαβ

+→=

45. Four fair dice 1 2 3, ,D D D and

4,D each having six faces numbered 1,2,3,4,5 and 6 , are rolled

simultaneously. The probabillity that 4

D shows a number appearing on one of 1 2,D D and

3D is

(A) 91

216(B)

108

216(C)

125

216(D)

127

216

45. (A) Favrable out comes 46 6 5 5 5= − × × ×

Total outcomes 46=

Probability

4

4

6 6 5 5 5 91

2166

− × × ×= =

46. The value of the integral

/ 2

2

/ 2

ln cosx

x x dxx

π

π

ππ−

+ + − ∫

(A) 0 (B)

2

42

π− (C)

2

42

π+ (D)

2

2

π




46.(B)

/ 22

/ 2

ln cosx

x xdxx

π

π

ππ

−

+ + − ∫

/ 22

/ 2

cosx x dx

π

π−

= ∫/ 2

2

0

2 cosx x dx

π

= ∫2

42

π= −

47. Let 1 2 3, , ,....a a a be in harmonic progression with

1 5a = and 20 25a = . The least positive integer n for

which 0na < is

(A) 22 (B) 23 (C) 24 (D) 25

47. (D)1

5a

=

125

19a d=

+

1 1 1

19 25 5d

∴ = −

1 20 4

19 25 5 19 25

− − = = × ×

1

1 4( 1)

5 19 25

na

n

=− + − ×

0na∴ <

1 4( 1) 0

5 19 25n

−⇒ + − <

×

1 4( 1)

5 19 25

n −<

×

19 251

4 5n

×⇒ + <

×

951

4n∴ > +

324

4n > 25n∴ =

48. The equation of a plane passing through the line of intersection of the planes 2 3 2x y z+ + = and

3x y z− + = and at a distance 2

3 from the point ( )3,1, 1− is

(A) 5 11 17x y z− + = (B) 2 3 2 1x y+ = − (C) 3x y z+ + = (D) 2 1 2x y− = −

48.(A)




2 3 ( 3) 0x y z x y zλ+ + + − + − =

(1 ) (2 ) (3 ) ( 2 3 ) 0x y zλ λ λ λ+ + − + + + − − =

2 2 2

| 3(1 ) (2 ) (3 ) ( 2 3 ) | 2

3(1 ) (2 ) (3 )

λ λ λ λ

λ λ λ

+ + − − + + − −=

+ + − + +

2

2

( 2 ) 4

33 4 14

λ

λ λ

−=

+ +2 23 3 4 14λ λ λ⇒ = + +

7 / 2λ = −

2( 2 3 2) 7( 3) 0x y z x y z∴− + + − + − + − =

5 11 17 0x y z− + − =



Paragraph for questions 49 and 50

A tangent PT is drawn to the circle 2 2 4x y+ = at the point ( )3,1P . A straight line L , perpendicular to

PT is a tangent to the circle ( )2 23 1x y− + = .

49. A common tangent of the two circles is

(A) 4x = (B) 2y = (C) 3 4x y+ = (D) 2 2 6x y+ =

49. (D) Meeting point of direct common tangent 2 3 1 0

,02 1

P× − × = −

(0,0) (3,0)

2x =

(6,0)=

If slope m=

0 ( 6)y m x− = −

is tangent to 2 2 4x y+ =

2 29 1m m= +

2 16 2 1

2 2m m m− = ± + ⇒ = ±




50. A possible equation of L is

(A) 3 1x y− = (B) 3 1x y+ = (C) 3 1x y− = − (D) 3 5x y+ =

50.(A) Equation of PT is

3 4 0x y+ − = ∴ Slope of 1

3L =

possible equation is

21 1

0 ( 3) 1 13 3

y x

− = − ± +

3 ( 3) 2y x= − ±

3 1x y− = or 5


Let ( ) ( )2 2 21 sinf x x x x= − + for all x IR∈ , and let ( ) ( ) ( )1

2 1ln

1

x tg x t f t dt

t

− = − + ∫ for all ( )1,x ∈ ∞ .

51. Consider the statements :

P : There exists some x IR∈ such that ( ) ( )22 2 1f x x x+ = +

Q : There exists some x IR∈ such that ( ) ( )2 1 2 1f x x x+ = +

Then

(A) both P and Q are true (B) P is true and Q is false

(C) P is false and Q is true (D) both P and Q are false

51. (C) 2 2 2( ) (1 ) sinf x x x x x R= − + ∀ ∈ 2 2 2 2:(1 )sin 2 2(1 )P x x x x x∴ − + + = +

2 2 2 2(sin )(1 ) 2 2 2x x x x x− = + − −

2 2 2(1 ) sin 2 2x x x x− = − +

2 2 2(1 ) sin ( 1) 1x x x− = − +

2 2(1 ) (sin 1) 1x x− − =

2

2

1cos

(1 )x

x

−=

−∴ no values of x exists so P false

Now, 2 2 22 (1 ) 1 2(1 ) sin 2x x x x x+ − = − +

2 22 1 2(1 ) sinx x x− = −

2

2

1 1sin

( 1) 2( 1)x

x x= +

− −

Now the value of R.H.S. will be from some large negative quantity to some large positive value, and value

of L.H.S. will be from 0 to 1. ∴ some value of x will exists.




52. Which of the following is true?

(A) g is increasing on ( )1, ∞ (B) g is decreasing on ( )1, ∞

(C) g is increasing on ( )1,2 and decreasing on ( )2,∞

(D) g is decreasing on ( )1,2 and increasing on ( )2,∞

52.(B) 2 2 2( ) (1 ) sinf x x x x x R= − + ∀ ∈

1

2( 1)( ) ln ( ) (1, )

1

x tg x t f t x

t

− = − ∀ ∈ ∞ + ∫

2( 1)'( ) ln ( ) (1, )

1

xg x x f x x

x

− = − ∀ ∈ ∞ +

( )f x always (+ve)

Let 2( 1)

( ) ln1

xh x x

x

−= −

+

2

4 1'( )

( 1)h x

xx= −

+

2

2

( 1)'( ) 0 (1, )

( 1)

xh x x

x x

− −= < ∀ ∈ ∞

+

( )h x⇒ is decreasing (1, )x∈ ∞ ( ) (1) (1, )h x h x⇒ < ∀ ∈ ∞

( ) 0h x⇒ < '( ) 0 0 (1, )g x x< ∀ ∈ ∞ ( )g x⇒ is decreasing (1, )x∀ ∈ ∞


Let n

a denote the number of all n -digit positive integers formed by the digits 0,1 or both such that no

consecutive digits in them are 0 . Let n

b = the number of such n -digit integers ending with digit 1 and n

c =

the number of such n - digit integers ending with digit 0 .

53. The value of 6

b is

(A) 7 (B) 8 (C) 9 (D) 11

53. (B) 46 2 (5 2 1) 8b = − + + =

54. Which of the follwing is correct?

(A) 17 16 15

a a a= + (B) 17 16 15

c c c≠ + (C) 17 16 16

b b c≠ + (D) 17 17 16

a c b= +

54. (A) 1 1a = 2 2a = 3 3a = 4 5a = 5 8a = 6 13a =

5 3 4a a a= +

6 4 5a a a= +............................................................

17 15 16a a a= +






55. If ( ) ( )( )2

02 3

xt

f x e t t dt= − −∫ for all ( )0,x ∈ ∞ , then

(A) f has a local maximum at 2x =

(B) f is decreasing on ( )2,3

(C) there exists some ( )0,c ∈ ∞ such that ( )" 0f c =

(D) f has a local minimum at 3x =

55.(A,B,C,D)2

'( ) ( 2)( 3) (0, )xf x e x x x− − − ∀ ∈ ∞

'( ) 0 (2,3)f x x< ∀ ∈

( )f x⇒ is increasing (2,3)x∀ ∈

sing of '( )f x

2 3

f⇒ has local maxima at 2x =

f⇒ has local minima at 2x =

'(2) '(3) 0f f= =∵

∴ from roolle’s Theorem ''( ) 0f c = for some ' 'in (0, )c ∞

56. If the straight line 1 1

2 2

x y z

k

− += = and

1 1

5 2

x y z

k

+ += = are coplanar, then the plane(s) containing these

two lines is (are)

(A) 2 1y z+ = − (B) 1y z+ = − (C) 1y z− = − (D) 2 1y z− = −

56. (B,C) lines are coplanar

1 1 1 1 0 0

2 2 0

5 2

k

k

+ − + −

=

22( 4) 0 2, 2k k− = ⇒ = −

0.( 1) 1( 1) 1( 0) 0x y z⇒ − + + − − = 1 0y z⇒ − + =

0.( 1) 1( 1) 1( 0) 0x y z− + + + − =

1 0y z+ + =




57. Let X and Y be two events such that 1 1

( | ) , ( | )2 3

P X Y P Y X= = and 1

( )6

P X Y∩ = which of the follow-

ing is (are) correct?

(A) 2

( )3

P X Y∪ = (B) X and Y are independent

(C) X and Y are not independent (D) 1

( )3

cP X Y∩ =

57. (A,B)

( ) 1 1( )

( ) 2 3

X P X YP P Y

Y P Y

∩ = = ⇒ =

( ) 1 1' ( )

( ) 3 2

Y P Y XP P X

X P X

∩ = = ⇒ =

( ) ( ) ( )P X Y P X P Y∩ =∵ &X Y∴ are independent

( ) ( ) ( ) ( )P X Y P X P Y P X Y∪ = + − ∩1 1 1 2

2 3 6 3= + − =

( ) ( ) ( )cP X Y P Y P X Y∩ = − ∩1 1 1

3 6 6= − =

58. If the adjoint of a 3 3× matrix P is

1 4 4

2 1 7

1 1 3

, then the possible value(s) of the determinant of P is (are)

(A) 2− (B) 1− (C) 1 (D) 2

58. (A,D)

1 4 4

2 1 7

1 1 3

adj P

=

2

1 4 4

| | 2 1 7 4 4 4 4

1 1 3

P = = − + + =

| | 2, 2P = −

59. Let : ( 1,1)f IR− → be such that 2

2(cos 4 )

2 secf θ

θ=

− for 0, ,

4 4 2

π π πθ ∈ ∪

Then the value(s) of

1

3f

is (are)

(A) 3

12

− (B) 3

12

+ (C) 2

13

− (D) 2

13

+




59. (A,B) 2

2(cos 4 )

2 secf θ

θ=

−

2

2

2cos(cos 4 )

2cos 1f

θθ

θ=

−

1 cos2(cos 4 )

cos2f

θθ

θ+

=

2 1(2cos 2 1) 1

cos2f θ

θ⇒ − = + 2 1

(2 1) 1f xx

⇒ − = +

2

3x = ±

1 11

3

2 / 3

f

⇒ = + ±

31

2= ±

60. For every integer n , let na and nb be real numbers. Let function :R IR IR→ be given by

sin , for [2 ,2 1]( ) ,

cos , for (2 1,2 )

n

n

a x x n nf x

b x x n n

π

π

+ ∈ +=

+ ∈ − for all integers n . If f is continuous, then which of the following

hold(s) for all n

(A) 1 1 0n na b− −− = (B) 1n na b− = (C) 1 1n na b +− = (D) 1 1n na b− − = −

60. (B,D) sin [2 ,2 1]

( )cos (2 1,2 )

n

n

a x x n nf x

b x x n n

π

π

+ ∀ ∈ +

+ ∀ ∈ −

Let 'f in continuous at 2x n= ⇒ L.H.L. = R.H.L (2 )f n=

2 2lim (2 ) lim (2 ) (2 )

h n h nf n h f n h f n

+ +→ →⇒ − = + = cos 2 sin 2 sin 2n n nb n a n a nπ π π⇒ + = + = +

1n nb a⇒ + = 1n na b⇒ − =

R.H.L of (2 1) cos (2 1)nn b nπ− = + − 1nb= −

In given function replasing ‘n’ by 1n −

[ ]1 sin 2 2,2 1na x x n nπ− + ∀ ∈ − −

[ ]1 cos 2 3,2 2nb x x n nπ− + ∀ ∈ − −

Now, L.H.L at 1(2 1) sin (2 1)nn a nπ−− = + − 1 0na −= +

1(2 1) nf n a −− =

11n nb a −− =

IIT-JEE 2012 Solution of Paper - 2 - Momentum...

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