CODE : 8 08 / 04 / 2012 Part – I : (PHYSICS)careerpoint.ac.in/studentparentzone/iitjee2012/IITPaper2Code8.pdf · PAPER-2 IIT-JEE 2012 EXAMINATION CAREER POINT, CP Tower, Road No.1,

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PAPER-2 IIT-JEE 2012 EXAMINATION

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

CAREER POINT

CODE : 8 08 / 04 / 2012

Part – I : (PHYSICS) SECTION – I (Single Correct Answer Type)

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which

ONLY ONE is correct.

Q.1 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 t = 0, 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 vr. In one time

period (T) of rotation of the discs, vr as a function of time is best represented by -

(A) (B)

(C) (D)

Ans. [A] Sol.

θ = ωtωR sin θ

ωR ωR cos θ

θ

ω

P

θ

ωR cos θωR

ωR sin θ

θ

θ

ω

θ = ωtQ

So, vr = 2ωR sin (ωt) At t = T/2, vr = 0 So two half cycles will take place.

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CAREER POINT

Q.2 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

a

y

x

(A) a2I k (B)

+

π 12

a2I k (C) –

+

π 12

a2I k (D) (2π + 1) a2I k

Ans. [B]

Sol. M = I × Area of loop k

= I × k424

aa2

2

×

×π

+ k12

aI 2

+π

×=

Q.3 An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform

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

as a function of the radial distance

r from the axis is best represented by -

(A) (B)

(C) (D)

Ans. [D] Sol.

R/2

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CAREER POINT

2Rr < ; B = 0

B at 2Rr =

B = 022

JR22

JR 00 =×

µ−

×µ

B at 2Rr >

B = 4

Rr2

J2JR 2

00 ×π

π×µ−

µ

−

µ=

r4Rr

2JB

20

if we put 2Rr =

B = 0 ∴ B is continuous at r = R /2

Q.4 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 ρc is the relative density of the material of the shell with respect to water,

then the correct statement is that the shell is -

(A) more than half-filled if ρc is less than 0.5

(B) more than half-filled if ρc is less than 1

(C) half-filled if ρc is more than 0.5

(D) less than half-filled if ρc is less than 0.5

Ans. [A]

Sol.

Vw

inside volume = V

container materialvolume = VC

mCg + mwg = FB

ρCVCg + 1 Vwg = 1

+2

V2V C g

Vw =

ρ−+ CC 21V

2V

if ρC < 21 ; Vw >

2V

correct ans. "A"

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CAREER POINT

Q.5 In the given circuit, a charge of +80 µC is given to the upper plate of the 4µF capacitor. Then in the steady

state, the charge on the upper plate of the 3µF capacitor is

(A) +32µC (B) +40µC (C) +48µC (D) +80µC

Ans. [C]

Sol.

+80–8080

q1 = 488053

=×

+48

–48–32

+32q2 = 3280

52

=× q2

2µF & 3µF are in parallel combination

Q.6 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 changed to

35ºC. The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K)

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

Ans. [D]

Sol. When balloon is expanding slowly. It is expanding against atmospheric pressure at constant pressure process. ∆Q = ∆W + ∆U = nR∆T + nCV∆T

∆Q =

+

2R3R2 × ∆T = 5 × 3.14 × 5 = 208 J

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CAREER POINT

Q.7 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 81 rotation (ii) their range

is less than half the disc radius and (iii) ω remains constant throughout. Then

(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. Ans. [C or D] Sol.

Rsinωt ωt

Rωt According to problem particle is to land on disc. If one consider a time 't' then x component of disc is Rωt Rsinωt < Rωt

This particle 'P' land on unshaded region. For "Q" x-component is very small and y-component equal to P it

will also land in unshaded region.

Now repeat same thing when right part is shaded then correct answer is "C" or "D"

Q.8 A student performing the experiment of Resonance Column. The diameter of the column tube is 4cm. The

frequency of the tuning fork is 512 Hz. The air temperature is 38ºC 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 the first

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

Ans. [B]

Sol.

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CAREER POINT

lλ/4

0.6 r

fλ = v

m512336

=λ ; l+×=λ r6.04

l+=××

2.11004512

336 ; l = 15.2 cm

SECTION – II (Paragraph Type)

This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has 4 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 of

mass 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) horizontally at 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 disc through 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 chase.

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

to x-y 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.

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CAREER POINT

Q.9 Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct ?

(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)

Ans. [A] Sol. In both the cases, the instantaneous axis will be along z-axis i.e. along vertical direction Q.10 Which of the following statements regarding the angular speed about the instantaneous axis (passing through

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

Ans. [D] Sol. w.r.t. centre of mass only pure rotation of disc will be seen. So in both the cases, angular speed about

instantaneous axis will be ‘‘ω’’

Paragraph for Questions 11 and 12

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 the kinetic 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. n → p + e– + ev , around 1930, Pauli

explained the observed electron energy spectrum. Assuming the anti neutrino ( )ev to be massless and

possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are

applied. From this calculation, the maximum kinetic energy of the electron is 0.8 × 106 eV. The kinetic

energy carried by the proton is only the recoil energy.

Q.11 If the anti-neutrino had a mass of 3eV/c2 (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) 0 ≤ K ≤ 0.8 × 106 eV (B) 3.0 eV ≤ K ≤ 0.8 × 106 eV

(C) 3.0 eV ≤ K < 0.8 × 106 eV (D) 0 ≤ K ≤ 0.8 × 106 eV

Ans. [D]

Sol. Total energy remain conserved. Energy is shared by antineutrino, proton and electron.

KE of e– is maximum when antineutrino does not share any KE.

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CAREER POINT

Now total energy is shared with proton & electron.

∴ K < 0.8 × 106 eV

Minimum KE of e– can be zero when total energy is shared by proton and antineutrino.

∴ 0 ≤ K ≤ 0.8 × 106 eV

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

(A) Zero (B) Much less than 0.8 × 106 eV

(C) Nearly 0.8 × 106 eV (D) Much larger than 0.8 × 106 eV.

Ans. [C]

Sol. When e– has zero kinetic energy total energy is shared by antineutrino and proton. This time energy of

antineutrino is its maximum possible kinetic energy.

As antineutrino is very light mass in comparison to proton so it will have almost contribution in total energy. ∴ its energy is almost 0.8 × 106 eV

Paragraph for Questions 13 and 14 Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring

material, then by Snell's law, 2

1

sinsin

θθ =

1

2

nn , 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, n =

vc = ± rrµε , where c is the speed of

electromagnetic waves in vacuum, v its speed in the medium, εr and µr are the relative permittivity and permeability of the medium respectively.

In normal materials, both εr and µr are positive, implying positive n for the medium. When both εr and µr 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 behaviour, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

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

(A)

θ1

θ2

Meta-material

Air

(B)

θ1

θ2Meta-material

Air

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CAREER POINT

(C)

θ1

θ2

Meta-material

Air

(D)

θ1

θ2Meta-material

Air

Ans. [C] Sol. As per snell's law, n1 sin θ1 = n2 sin θ2 So most probable answer is option ‘‘C’’. Q.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 v = |n|

c

(C) The speed of light in the meta-material is v = c (D) The wavelength of the light in the meta-material (λm) is given by λm = λair|n|, where λair is the

wavelength of the light in air Ans. [B]

Sol. The expression v = |n|

c is applicable.

SECTION – III (Multiple Correct Answer(s) Type)

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE or MORE are correct.

Q.15 In the given circuit, the AC source has ω = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)

(A) The current through the circuit, I is 0.3 A (B) The current through the circuit, I is 23.0 A

(C) The voltage across 100 Ω resistor = 210 V (D) The voltage across 50 Ω resistor = 10 V

Ans. [A, C] or [C]

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CAREER POINT

Sol. xL = ωL = 10 × 0.5 = 50 Ω

xC = C

1ω

= Ω=×× − 100

101001001

6

I1

100Ω 100Ω

Z1 = 2100 ; I1 = 2100

20 = 25

1

Vacross 100 Ω = 22

220100

251

×=× = 210 Ans. (C)

Phase diff. between I1 & V ⇒ cos φ1 = 1

1ZR =

2100100

φ1 = π/4

I1 lead V

I2

50Ω 50Ω

Z2 = 250 ; I2 = 250

20 = 25

2

Vrms across 50Ω = 2

2050250

2=× = 210 Ans.

φ2 = π/4

I2 lag V by π/4

I = I1 + I2

π/4

π/4

Ι2

Ι1

V

INet = 22

21 II +

I = 101

505

2251

2254

==×

+×

I = 0.316

As I is not exactly 0.3 therefore IIT give answer either C or (A,C)

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CAREER POINT

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

Given that K = 04

1πε 2L

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

+2q

+q –q

–2q

– q+q

E3

T

E2

O

A

BC

D

EF F

R

P

S

(A) The electric field at O is 6K along OD. (B) The potential at O is zero (C) The potential at all points on the line PR is same (D) The potential at all points on the line ST is same Ans. [A, B, C] Sol.

+2q

+q –q

–2q

– q+q

E3

E1

E2O

E2 = E3 = 2L

kq2

21 Lkq4E =

Eall = 321 EEErrr

++

= 20

2 Lq

416

Lkq6

πε×= = 6K (Along OD)

Vat O = 0 Vat line PR is zero because this line is equatorial axis for three dipole.

Q.17 Two spherical planets P and Q have the same uniform density ρ, masses MP and MQ, and surface areas A and 4A, respectively. A spherical planet R also has uniform density ρ and its mass is (MP + MQ). The escape velocities from the planets P, Q and R, are VP, VQ and VR respectively. Then.

(A) VQ > VR > VP (B) VR > VQ > VP (C) VR / VP = 3 (D) VP / VQ = 21

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CAREER POINT

Ans. [B, D]

Sol. Escape velocity, ve = RGM2

Surface area of P, AR4 2P =π

Surface area of Q, A4R4 2Q =π

⇒ 21

RR

Q

P =

⇒ RQ = 2RP

If MP = M = 3R34

πρ (Let R be the radius of P)

So, MQ = M8)R2(34 3 =πρ

So, MR = MP + MQ = 9M

For planet R, 9M = 3rR

34

πρ

So, Rr = (9)1/3R

Escape velocity from P, RGM2VP =

R2

)M8(G2VQ = = RGM22

VR = RGM2)9(

R)9()M9(G2 3/1

3/1 =

So, VR > VQ > VP & 21

VV

Q

P =

Q.18 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 2R rotating anti-clockwise with angular speed ω/2. 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,

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CAREER POINT

(A) the point O has a linear velocity 3Rω i

(B) the point P has a linear velocity iR4

11ω + kR

43

ω

(C) the point P has a linear velocity iR4

13ω – kR

43

ω

(D) the point P has a linear velocity kR41iR

433 ω+ω

−

Ans. [A, B]

Sol. )k30sinRi30cosR()j(2

iR3VP °+°×−ω

+ω=

= iR4

kR4

3iR3 ω−

ω+ω = kR

43iR

411

ω+ω

Q.19 Two solid cylinders P and Q of same mass and same radius start rolling down a flixed 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 than cylinder Q. (C) Both cylinders reach the ground with same translational kinetic energy (D) Cylinder Q reaches the ground with larger angular speed

Ans. [D] Sol.

f

mg

N

Translation motion : mg sin θ – f = macm ...........(i) Rotational motion fR = Icm α ..........(ii) Rolling without slipping αR = acm ........(iii) From (ii) & (iii)

f = 2cmcm

RaI

Put this in (i)

mg sin θ – 2cmcm

RaI = macm

acm =

+

θ

mRI

sinmg

2cm

As IP > IQ

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CAREER POINT

So aP < aQ So, vcm (Q) > vcm (P) Hence ωQ > ωP Q.20 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 induced 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 decreases at a steady rate

Ans. [A, C] Sol. Total flux associate with loop = 0 ∴ emf = 0 in any case

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CAREER POINT

Part – II : (Chemistry)

SECTION – I (Single Correct Answer Type)



Q.21 The major product H of the given reaction sequence is -

CH3 – CH2 – CO – CH3 →Θ CN G

HeatSOH%95 42 → H

(A) COOH

CH|

CCHCH

3

3 −=− (B) NC

CH|

CCHCH

3

3 −=−

(C) COOHCHCCHCHOH

3

|23

|−−− (D) 2

3

3 NHCO

CH|CCHCH −−=−

Ans. [A]

Sol. CH3–CH2–CO–CH3 →Θ

NC

CH3–CH2–C–CH3

OH

CN∆

→,SOH

%95

42

CH3–CH=C–C–OH

H3C O

Q.22 NiCl2P(C2H5)2(C6H5)2 exhibits temperature dependent magnetic behaviour (paramagnetic / diamagnetic).

The coordination geometries of Ni2+ 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 tetrahedral

Ans. [C]

Sol. 25602522

22)HC()HC(PClNi

−+

↓ ↓ negative neutral ligand ligand having –1 charge Ni ––––––––––––––––– 3d84s2 Ni+2 ––––––––––––––––3d84s0

3d 4s 4p

In paramagnetic state there will be no pairing therefore hybridization will be sp3 and geometry will be tetrahedral.

In diamagnetic state pairing of electron will take place.

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CAREER POINT

3d 4s 4p Since d-orbital is vacant so, hybridization will be dsp2 geometry is square planar.

Q.23 In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agent used are

(A) O2 and CO respectively. (B) O2 and Zn dust respectively.

(C) HNO3 and Zn dust respectively. (D) HNO3 and CO respectively.

Ans. [B]

Sol. 4Ag + 8NaCN + 2H2O + O2 → 4 NaAg (CN)2 + 4NaOH

2NaAg(CN)2 + Zn → Na2Zn (CN)4 + 2Ag

Q.24 The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus

containing compound. The reaction type ; the oxidation states of phosphorus phosphine and the other product

are respectively

(A) redox reaction ; –3 and –5 (B) redox reaction ; +3 and +5

(C) disproportionation reaction ; –3 and +5 (D) disproportionation reaction ; –3 and +3

Ans. [C]

Sol. (i) P4 + 3NaOH + 3H2O → 3NaH2 33

21

PHPO−+

+

(ii) NaH2PO2 + NaOH → 5 3

4PO+ −

(i) It depends upon the concentration of NaOH

(ii) The second step of reaction is possible when the concentration of NaOH is high. So the answer of this

question can be two.

Answer may be [C] or +1 and –3 because concentration of NaOH is not specified in the question.

So JEE has given zero marks to all.

Q.25 The shape of XeO2F2 molecule is -

(A) trigonal bipyramidal (B) square planar (C) tetrahedral (D) see-saw

Ans. [D]

Sol.

Xe

F

F OO

•• lp

see-saw

Q.26 For a dilute solution containing 2.5 g of a non-volatile non-electrolye solute in 100 g of water, the elevation

in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the

concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol–1)

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

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CAREER POINT

Ans. [A]

Sol. ∆Tb = Kb × m

2 = 1.0

M/5.276.0 ×

M = 9.5 Since solute is present in less amount

°

°P

P–P S = Nn

760

P–760 S = 18/100

5.9/5.2

Ps = 724

Q.27 The compound that undergoes decarboxylation most readily under mild condition is -

(A) (B)

(C) (D)

Ans. [B]

Sol.

COOH

O→∆

O

+ CO2↑

Q.28 Using the data provided, calculate the multiple bond energy (kJ mol–1) of a C≡C bond in C2H2. That energy

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

2C(s) + H2(g) → C2H2(g) ∆H = 225 kJ mol–1

2C(s) → 2C(g) ∆H = 1410 kJ mol–1

H2(g) → 2H(g) ∆H = 330 kJ mol–1

(A) 1165 (B) 837

(C) 865 (D) 815

Ans. [D]

Sol. ∆HR = (B.E)R – (B.E)P

225 = [1410 + 330] – [–∆HC≡C + 2 × 350]

∆HC≡C = 815 KJ/mol

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CAREER POINT

SECTION – II : Paragraph Type

This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph.

Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


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

I COONaCH

O)COCH(

3

23 → J 3

2

2

AlCl.anhyd)iii(SOCl)ii(

C/PdH)i( → K

J (C9 H8O2) gives effervescence on treatment with NaHCO3 and a positive Baeyer's test

Q.29 The compound I is -

(A) (B)

(C) (D)

Ans. [A]

Sol. CHO–Ph)I(

+ →NaOCO.CH

O)CO.CH(–

3

23

COOH–CHCH–Ph)J(

= 32

2

AlCl.Anhy)3(SOCl)2(

C/Pd,H)1( →

(K)

O Q.30 The compound K is -

(A)

(B)

(C) (D)

Ans. [C]

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CAREER POINT


The electrochemical cell shown below is a concentration cell.

M| M2+ (saturated solution of a sparingly soluble salt, MX2). || M2+ (0.001 mol dm–3) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf

of the cell at 298 K is 0.059 V.

Q.31 The value of ∆G(kJ mol–1) for the given cell is (take 1F = 96500 C mol–1) :

(A) – 5.7 (B) 5.7 (C) 11.4 (D) –11.4

Ans. [D] Sol. ∆G° = – °

cellnFE

= – 2 × 96500 × 1000

059.0

= – 11.387

~= – 11.4

Q.32 The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the given

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

(A) 1 × 10–15 (B) 4 × 10–15 (C) 1 × 10–12 (D) 4 × 10–12

Ans. [B]

Sol. Ecell = 0 – 2

0591. log 001.0

M 2+

M+2 = 1 × 10–5

MX2 M+2 + 2X–

S 2S

Ksp = S(2S)2

= 4S3

= 4 × (10–5)3

= 4 × 10–15


Bleaching powder and bleach solution are produced on a large scale and used in several house-hold products.

The effectiveness of bleach solution is often measured by iodometry.

Q.33 Bleaching power contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is

(A) Cl2O (B) Cl2O7 (C) ClO2 (D) Cl2O6

Ans. [A]

Sol. 2HOCl → Cl2O + H2O

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CAREER POINT

Q.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 Na2S2O3 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

Ans. [C]

Sol.

24.0252

25.048M =××

=2

25.048×

225.048×

CaOCl2 + 2KI + 2CH3COOH → CaCl2 + I2 + 2CH3COOK + H2O

n = 2 n = 1

0 12

–2

= 0.24

Section III : Multiple Correct Answer(s) Type


ONE or MORE is correct.

Q.35 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 ?

V

Padiabatic

isothermal

(P2, V2, T2)(P3, V2, T3)

(P1, V1, T1)

(A) T1 = T2 (B) T3 > T1 (C) wisothermal > wadabatic (D) ∆Uisothermal >

∆Uadiabatic

Ans. [A,D]

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

→ 42SOH.dil excess KI + K3[Fe(CN)6] brownish-yellow solution

ZnSO4

white precipitate + brownish-yellow filtrate

Na2S2O3

colourless solution

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CAREER POINT

(A) The first reaction is a redox reaction

(B) White precipitate is Zn3[Fe(CN)6]2

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

(D) White precipitate is soluble in NaOH solution

Ans. [A,C,D]

Sol. KI + K3Fe(CN)6 → K4Fe(CN)6 + KI3 Brownish yellow KI3 → 4ZnSO No reaction K4[Fe(CN)6] + ZnSO4 → K2Zn3 [Fe(CN)6]2 white ppt. KI3 + Na2SO3 → NaI + KI + Na2S4O6 starch blue K2Zn3 [Fe(CN)6]2 + NaOH ↓ [Fe(CN)6]4– + [Zn(OH)4]2–

Q.37 With reference to the scheme given, which of the given statement(s) about T, U, V and W is (are) correct ?

(A) T is soluble in hot aqueous NaOH

(B) U is optically active

(C) Molecular formula of W is C10H18O4

(D) V gives of effervescence on treatment with aqueous NaHCO3 Ans. [A,C,D]

Sol.

O

CH3

O

T

→ 4LiAlH

CH2OH

CH3 OH

U CrO3/H+

COOH

COOH

V

O)CO.CH(

excess

23

→

CH2–O–C–CH3

CH3O–C–CH3

W

||O

||O

22 / 35



CAREER POINT

Q.38 Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct ?

(A) M and N are non-mirror image stereoisomers

(B) M and O are identical

(C) M and P are enantiomers

(D) M and Q are identical

Ans. [A,B,C]

Q.39 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.

Ans. [B,D]

Q.40 The given graphs/data I, II, III and IV represent general trends observed for different physisorption and

chemisorption processes under mild conditions of temperature and pressure. Which of the following

choice(s) about I, II, III and IV is (are) correct ?

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CAREER POINT

(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 Ans. [A,C] Sol. I physiosorption

II chemisorption

III physiosorption

IV chemisorption

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CAREER POINT

Part – III : (Mathematics)




Q.41 Let a1, a2, a3, ....... be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for which

an < 0 is

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

Ans. [D]

Sol. a1 = 5 a20 = 25

T1 = 51 T20 =

251

T20 = 51 + 19D =

251

D =

−

51

251

191

= –)19)(25(5

20

Tn = 51 –

)19)(125()20)(1n( −

= )19)(125(

)20)(1n()19)(25( −− < 0

(25) (19) < (n – 1) (20)

n – 1 > )20(

)19)(25(

n > 4

)19(5 + 1

n > 4

95 + 1

n > 23.75 + 1

n > 24.75

n = 25

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CAREER POINT

Q.42 The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3

and at a distance 3

2 from the point (3, 1, – 1) is

(A) 5x – 11y + z = 17 (B) x2 + y = 23 – 1

(C) x + y + z = 3 (D) x – 2 y = 1 – 2

Ans. [A]

Sol. Equation of plane passing through intersecting of plane P1 & P2

is P1 + λP2 = 0

(1 + λ) x + (2 – λ) y + (3 + λ) z – 2 – 3λ = 0

distance of plane from pt (3, 1, –1) is 3

2

3

2 = 222 )3()2()1(

|323233|

λ−+λ−+λ+

λ−−λ−−λ−+λ+

on solving λ = –27

so equation of plane is

−

271 x +

+

272 y +

−

273 z – 2 +

221 = 0

5x – 11y + z = 17

Q.43 Let PQR be a triangle of area ∆ with a = 2, b = 27 and c =

25 , 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 P2sinPsin2P2sinPsin2

+− equals

(A) ∆43 (B)

∆445 (C)

2

43

∆ (D)

2

445

∆

Ans. [C]

Sol.

P

Q R

c

a

b

26 / 35



CAREER POINT

==

=

2/5c2/7b

2a ⇒ s = 4

PcosPsin2Psin2PcosPsin2Psin2

+− =

Pcos1Pcos1

+− ∆ =

23.

21.2.4 = 6

= tan2 2P

= 2

)as(s)cs)(bs(

−−−

= )2(423

21

= 323

=

6.169 =

2

43

∆

Q.44 If ar

and br

are vectors such that |ba|rr

+ = 29 and )k4j3i2(a ++×r

= b)k4j3i2(r

×++ , then a possible

value of )k3j2i7).(ba( ++−+rr

is

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

Ans. [C]

Sol. |ba|rr

+ = 29

)k4j3i2(a ++×r

= b)k4j3i2(r

×++

)ba(rr

+ × )k4j3i2( ++ = 0

barr

+ = λ )k4j3i2( ++

|ba|rr

+ = 222 1694 λ+λ+λ = |λ| 29

⇒ λ = 1, –1

barr

+ = ± )k4j3i2( ++

)ba(rr

+ . (–7 i + 2 j+ 3 k )

= ± )k4j3i2( ++ .(–7 i + 2 j+ 3 k ) = ± 4

27 / 35



CAREER POINT

Q.45 If P is a 3 × 3 matrix such that PT = 2P + I, where PT is the transpose of P and I is the 3 × 3 identity matrix,

then there exists a column matrix X =

zyx

≠

000

such that

(A) PX =

000

(B) PX = X (C) PX = 2X (D) PX = – X

Ans. [D]

Sol. PT = 2P + I

⇒ P = 2PT + I

⇒ P = 2(2P + I) + I

⇒ P = –I

PX = – I X

PX = – X

Q.46 Let α (a) and β(a) be the roots of the equation ( ) ( ) ( ) 01a1x1a1x1a1 623 =−++−++−+ where a > – 1.

Then +→0a

lim α(a) and +→0a

lim β(a) are

(A) – 25 and 1 (B) –

21 and – 1 (C) –

27 and 2 (D) –

29 and 3

Ans. [B]

Sol. (1 + a) = t6

(t2 –1) x2 + (t3 –1) x + (t –1) = 0

x = )1t(2

)1t()1t(4)1t()1t(2

2233

−

−−−−±−−

x = )1t()1t(2

)1t(4)1tt()1t()1t( 223

+−+−++−±−−

x = )1t(2

)1t(4)1tt()1tt( 222

++−++±++−

a → 0+ ⇒ t → 1+

x = )2(2

893 −±−⇒ x =

413 ±−

⇒ x = –1, – 21

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CAREER POINT

Q.47 Four fair dice D1, D2, D3 and D4, each having six faces numbered 1, 2, 3, 4, 5 and 6, are rolled

simultaneously. The probability that D4 shows a number appearing on one of D1, D2 and D3 is

(A) 21691 (B)

216108 (C)

216125 (D)

216127

Ans. [A]

Sol. required probability = 1 – P(D4 has diff.)

= 1 –

++4

23

63.4.5.64.5.1.6.C5.1.1.6

= 21691

Q.48 The value of the integral

∫π

π−

−π+π

+2/

2/

2

xxlnx cos x dx is

(A) 0 (B) 42

2−

π (C) 42

2+

π (D) 2

2π

Ans. [B]

Sol. ∫π

π

2/

2/–

2 dxxcosx + ∫π

π

+ππ

2/

2/–xx–nl cos x dx

↓ ↓ Even function Odd function

= 2 ∫π 2/

0

2x cos x dx + 0

= 2 2/0

2 ]xsin2–xcosx2xsinx[ π+ = 2

π 2–4

2

= 2

2π – 4

SECTION II : Paragraph Type

This section contains 6 multiple choice questions relating to three paragraphs with two questions on each

paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


A tangent PT is drawn to the circle x2 + y2 = 4 at the point P ( )1,3 . A straight line L, perpendicular to PT is

a tangent to the circle (x – 3)2 + y2 = 1.

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CAREER POINT

Q.49 A possible equation of L is

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

Ans. [A]

Sol.

O

30º (3, 0)

(3, 0), 1

0,

34

P( 3 , 1)

y

Slope of PT = tan (120º) = – 3

Slope of line L = 3

1

Line L ≡ x – 3 y + λ = 0

tangent to (x – 3)2 + y2 = 1

2

|3| λ+ = 1

λ + 3 = 2, – 2

λ = –1, –5

x – 3 y – 1 = 0

or x – 3 y –5 = 0

Q.50 A common tangent of the two circles is

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

Ans. [D]

Sol. Common tangent both circles

30 / 35



CAREER POINT

2 1

Q O (0, 0) (3, 0) P(6, 0)

So P ≡ (6, 0)

line through P

λx – y – 6λ = 0

tangent to circle 21

|3|

λ+

λ = 1

9λ2 = 1 + λ2 ⇒ λ2 =

81

λ = 22

1 , 221−

Equation of tangent x + 22 y = 6


Let f(x) = (1 – x)2 sin2x + x2 for all x ∈ IR, and let g(x) = ∫

−

+−

x

1

tln1t

)1t(2 f(t) dt for all x ∈ (1, ∞).

Q.51. Consider the statements :

P : There exists some x ∈ IR such that f(x) + 2x = 2(1 + x2)

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

Ans. [C]

Sol. (P): (sin2x) (1– x)2 + x2 + 2x = 2 + 2x2

(sin2x) (1 – x)2 = x2 – 2x + 2

sin2 x = 2

2

)x1(1)x1(

−+− , which is greater than 1 ⇒ no solution

⇒ P is false.

(Q) : 2 sin2x = 2

2

)x1(x−

– 1

0 ≤ 21)x1(

x2

2≤−

−.

This inequality is satisfied by some value of x.

31 / 35



CAREER POINT

Q.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, ∞)

Ans. [B]

Sol. g′(x) = 44 344 21

l

1xforve

xn1x

)1x(2

>−

−

+−

vealways

)x(f+

g′ (x) < 0 ⇒ g(x) ↓ on (1, ∞)


Paragraph for Questions 53 to 54

Let an 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 bn = the number of such n-digit integers ending with digit 1 and cn = the number of such n-digit integers ending with digit 0.

Q.53 The value of b6 is (A) 7 (B) 8

(C) 9 (D) 11 Ans. [B]

Sol. for bn

first and last place are fixed by 1

so case (1) if only one zero is used such cases = n–2C1

case (2) if two zero are used the two zeros are such that no two zeros are consecutive = n – 3C2

case (3) if three zeros are used then the positing of three zeros such that no two zeros are

consecutive = n – 4C3

So bn = n – 2C1 + n – .3C2 + n – 4C3 + n – 5C4 + n – 6C5 .......

for b6 = 4C1 + 3C2 + 1

when no zero is used

= 8

Q.54 Which of the following is correct?

(A) a17 = a16 + a15 (B) c17 ≠ c16 + c15

(C) b17 ≠ b16 + c16 (D) a17 = c17 + b16

32 / 35



CAREER POINT

Ans. [A]

Sol. bn = 1 + n – 2C1 + n – 3C2 + n – 4C3 + n – 5C4 + n – 6C5 + ……..

cn = 1 + n – 3C1 + n – 4C2 + n – 5C3 + n – 6C4 + ……..

an = 1 + n – 1C1 + n – 2C2 + n – 3C3 + n – 4C4 + ……..

a17 = 1 + 16C1 + 15C2 + 14C3 + 13C4 + 12C5 + 11C6 + 10C7 + 9C8 + 8C9

a16 = 1 + 15C1 + 14C2 + 13C3 + 12C4 + 11C5 + 10C6 + 9C7 + 8C8

a15 = 1 + 14C1 + 13C2 + 12C3 + 11C4 + 10C5 + 9C6 + 8C7

a17 = a16 + a15 so A is correct.

c17 = 1 + 14C1 + 13C2 + 12C3 + 11C4 + 10C5 + 9C6 + 8C7 + 7C8

c16 = 1 + 13C1 + 12C2 + 11C3 + 10C4 + 9C5 + 8C6 + 7C7

c15 = 1 + 12C1 + 11C2 + 10C3 + 9C4 + 8C5 + 7C6

c17 = c16 + c15

So, B is wrong.

b17 = 1 + 15C1 + 14C2 + 13C3 + 12C4 + 11C5 + 10C6 + 9C7 + 8C8

b16 = 1 + 14C1 + 13C2 + 12C3 + 11C4 + 10C5 + 9C6 + 8C7

c16 = 1 + 13C1 + 12C2 + 11C3 + 10C4 + 9C5 + 8C6 + 7C7

b17 = b16 + c17 so C is wrong.

a17 = 1 + 16C1 + 15C2 + 14C3 + 13C4 + 12C5 + 11C6 + 10C7 + 9C8 + 8C7

c17 = 1 + 14C1 + 13C2 + 12C3 + 11C4 + 10C5 + 9C6 + 8C7

b16 = 1 + 14C1 + 13C2 + 12C3 + 11C4 + 10C5 + 9C6 + 8C7

so D is wrong.

SECTION III : Multiple Correct Answer(s) Type

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of

which ONE or MORE are correct.

55. For every integer n, let an and bn be real numbers. Let function f: IR → IR be given by

f(x) =

∈π++∈π+

)n2,1–n2(xfor,xcosb]1n2,n2[xfor,xsina

n

n , for all integers n.

If f is continuous, then which of the following hold(s) for all n?

(A) an–1 – bn–1 = 0 (B) an – bn = 1

(C) an – bn+1 = 1 (D) an–1 – bn = –1

Ans. [B, D] Sol. At x = 2n x → 2n+ an + sin 2nπ = an x → 2n– bn + cos 2nπ = bn + 1 For continuous an = bn + 1

At x = 2n + 1

33 / 35



CAREER POINT

x → 2n + 1+ bn+1 + cosπ(2n + 1)= bn + 1–1 x → 2n + 1– an + sin π(2n + 1) = an for continuous an = bn+1 –1 an – bn+1 = – 1 for n = n – 1 an–1 – bn = – 1

Q.56 If the straight lines 2

1x − = k

1y + = 2z and

51x + =

21y + =

kz are coplanar, then the plane(s) containing these

two lines is (are) (A) y + 2z = –1 (B) y + z = –1 (C) y – z = –1 (D) y –2z = –1 Ans. [B, C] Sol. If these two lines are coplanar then shortest distance between them = 0

k252k2002

= 0

k = 2 or –2

so lines are

=+

=+

=+

=−

2z

21y

51xand

2z

21y

21x

set (i)

OR

−=

+=

+

=−+

=−

2z

21y

51xand

2z

21y

21x

set (ii)

the plane which contain these set of line should contain the points (1, –1, 0) and (–1, –1, 0) which is satisfied by all the four options and )k2j2i2( ++ & )k2j2i5( ++ OR

)k2j2i2( +− & )k2j2i5( −+ are perpendicular to normal of plane For first set option (C) is correct. For second set option (B) is correct.

Q.57 If the adjoint of a 3 × 3 matrix P is

311712441

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

(A) –2 (B) –1 (C) 1 (D) 2

Ans. [A, D]

Sol. adj (P) =

311712441

|adj P| = |P|n–1 = |P|2 |adj P| = 4 = |P|2 |P| = 2 or –2

34 / 35



CAREER POINT

Q.58 Let f : (–1, 1) → IR be such that f(cos 4θ) = θ− 2sec2

2 for θ ∈

π

4,0 ∪

ππ

2,

4. Then the value(s) of

f

31 is (are)

(A) 1 – 23 (B) 1 +

23 (C) 1 –

32 (D) 1 +

32

Ans. [A, B]

Sol. cos 4θ =31

2 cos2 2θ – 1 =31

cos 2θ = ±32

On solving

cos2θ = 2

3/21±

sec2θ = 3/21

2±

On putting this

f

31 = ±

23 + 1

Note : As f(1/3) is having two values, so this relation is not a function. So JEE has given zero marks to all.

Q.59 Let X and Y be two events such that P(X|Y) = 21 , P(Y|X) =

31 and P(X ∩Y) =

61 . Which of the following is

(are) correct?

(A) P(X ∪ Y) = 32 (B) X and Y are independent

(C) X and Y are not independent (D) P(XC ∩ Y) =31

Ans. [A, B]

YXP =

)Y(P)YX(P ∩ =

21 ⇒ P(Y) =

31

31

)X(P)XY(P

=∩ ⇒ P(X) =

21

P(X ∪Y) = 21 +

31 –

61 =

32

P(X ∩ Y) = P(X).P(Y) True

35 / 35



CAREER POINT

X Y

(XC ∩ Y)

P(XC ∩ Y) = P(Y) – P(X ∩ Y)

= 31 –

61 =

61

Q.60 If f(x) = ∫x

0

t2e (t – 2) (t –3) dt for all x ∈ (0, ∞), then

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

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

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

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

Ans. [A, B, C, D]

Sol. f(x) = ∫x

0

t2e (t – 2) (t –3) dt

f ′(x) = 2xe (x –2) (x –3)

+ + – 2 3

f ′(x) < 0 ∀ x ∈ (2, 3) so f(x) is decreasing on (2, 3)

also at x = 2, f ′(x) changes its sign from +ve to –ve. Hence x = 2 is point of maxima

At x = 3, f ′(x) changes its sign from –ve to +ve. Hence x = 3 is point of minima. Also f ′(2) = f ′(3) = 0 So from Rolle's Theorem there exist a point c such that f ′′(c) = 0

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CODE : 8 08 / 04 / 2012 Part – I : (PHYSICS)careerpoint.ac.in/studentparentzone/iitjee2012/IITPaper2Code8.pdf · PAPER-2 IIT-JEE 2012 EXAMINATION CAREER POINT, CP Tower, Road No.1,