iit jee2012phyiiquestionssolutions (1)

8/2/2019 IIT JEE2012PhyIIQuestionsSolutions (1)

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IIT-JEE - 2012 (08-Apr-12) Question & Solutions Paper-II - 1 - www. prernaclasses.com

Physics Paper-IIIIT-JEE 2012

R

Prerna Tower, Road No - 2, Contractors Area, Bistupur, Jamshedpur - 831001, Tel - (0657)2221892, www.prernaclasses.com

PART I - PHYSICS

SECTION - I : Single Correct Answer Type

This Section contains 8 multiple choice questions. Each question has four choices (A), (B),

( C) and (D) out of which ONLY ONE is correct.

1. A loop carrying current I lies in the xy plane as shown in the figure. The unit vectork is coming

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

I

y

xa

a

(A)kIa 2 (B) kIa

12

2

+

p(C) kIa

12

2

+

p- (D) (2p + 1) a2I

k

1. (B) ka.4

+

p== 2

2

( a / 2 )

I AI

2

= kIa12

+

p2


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

vertically in water in half-submerged state. Ifrcis the relative density of the material of the shellwith respect to water, then the correct statement is that the shell is

(A) More than half-filled ifrcis less than 0.5(B) More than half-filled ifrcis less than 1.0(C) Half-filled ifrcis more than 0.5

(D) Less than half-filled ifrcis less than 0.5

2. (A) [(V/ 2)+ (AL / 2)] rwg= rwAhg+ Vrwrcg [(V/ 2) + (AL / 2)] = Ah + rcV h = [(V/ 2A) + (L / 2) (rcV /A)]

= [(V/ 2) rcV) (1 /A) + (L / 2)ifrc> (1 / 2), h < (L / 2)ifrc< (1 / 2), h > (L / 2)Where A : internal cross sectional area.


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

uniform current density along its length. The magnitude of the magnetic field, Br

as a function of

the radial distance rfrom the axis is best represented by

(A)

|B|

rR/2 R

(B)

|B|

rR/2 R

(C)

|B|

rR/2 R

(D)

|B|

rR/2 R

3. (D) Amperes law

(1)R/2

R

(2)

(3)

loop(1) 00. == BdlB

loop (2)

-

-

=4

4

.2.2

2

22

0

Rx

RR

IxB p

p

mp

-=

4.

3

4.2.

22

2

0 RxR

IxB

mp

( )xR

RxIB

2

22

0

p

m

3

4/2 -=

loop(3)x

IBIdlBp

mm2

. 00 ==


4/13


4. Consider a disc rotating in the horizontal plane with a constant angular speed w about its centreO. 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

Pand Q are simultaneously projected at an angle towards R. The velocity of projection is in the

yzplane 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 / 8) rotation, (ii) their range is less than half the disc

radius, and (iii) w remains constant throughout. Then

QO

w

P

Ry

x

(A) Plands in the shaded region and Q in the unshaded region.

(B) Plands in the unshaded region and Q in the shaded region.

(C) Both Pand Q land in the unshaded region.

(D) Both Pand Q land in the shaded region.

4. (C)

y

x

QP'

Q'p/4

P

PV w.r.t. ground has also velocity wRalong x-axis while Q does not has any velocity along

x-axis. As given in figure Q1& P1are roughly landed position of pebble P& Q on the disc.

5. A student is performing the experiment of Resonance Column. The diameter of the column tubeis 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38 C in which the

speed of sound is 336 ms1. 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

5. (B)

d= 4 cm, f0= 512 Hz T= 273 + 38 = 311 C= 336 ms1

leff= (L + 0.3 D) = (l / 4) = (C/ 4f0) L = [336 / (4 512)] 0.3 102 4= (16.4 1.2) cm = 15.2 cm


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6. In the given circuit, a charge of +80 C is given to the upper plate of the 4 F capacitor. Then in

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

+80C 4F

3F2F

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

6. (C)

80F

4F

q1

q2

2F

3F

32

21 qq = 80 + q1+ q2= 0 8032

=+ 22 q

q q2= + 48 C


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

the same constant angular speedw. The discs are in the same horizontal plane. At time t= 0,the points Pand Q are facing each other as shown in the figure. The relative speed between

the two points Pand Q is vr. In one time period (T) of rotation of the discs, vras a function of time

is best represented by

w w

PR

QR

(A)

v

0T

t(B)

v

0T

t

(C)

v

0T

t(D)

v

0T

t

7. (A)

wt

P

wt

Q

y

x

( )PQr

VVtsinR2V -=ww= l

8. Two moles of ideal helium gas are in a rubber balloon at 30C. 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 35C. 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

8. (D) DQ = [2 . (5 / 2) 8.31 5] = 25 8.31= 207.75 208 J


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SECTION - II : Paragraph Type

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

question on each paragraph. Each question has four choices (A), (B), (C) and (D) out of

which ONE is correct.

Paragraph for Questions 9 and 10

The b-decay process, discovered around 1900, is basically the decay of a neutron (n). In thelaboratory, 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+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 energy conservation principles are applied.

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

energy carried by the proton is only the recoil energy.

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

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

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

9. (C) Considering electron at rest. Mass of anti-neutrino is negligible compared to mass of

proton. So using CLM, velocity of anti-neutrino is very large, almost entire energy is carried

away by the anti-neutrino.

10. If the anti-neutrino had a mass of 3 eV / c2(where cis 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 106eV (B) 3.0 eV K 0.8 106eV(C) 3.0 eV K


8/13



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

occurring material, then by Snells law, (sin q1/ sin q2) = (n2/ n1), it is understood that the refractedray 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,rr

e=

=

v

c

n , where cis the speed of electromagnetic waves in vacuum, vits

speed in the medium, erand rare the relative permittivity and permeability of the mediumrespectively. In normal materials, bothe erand rare positive, implying positive n for the medium.When both erand rare negative, one must choose the negative root ofn. Such negativerefractive index materials can now be artificially prepared and are called meta-materials. The

exhibit significantly different optical behaviour, without violating any physical laws. Since n is

negative, it results in a change in the directioin of propagation of the refracted light. However,

similar to normal materials, the frequency of light remains unchanged upon refraction even in

meta-materials.

11. 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|| n

c

v =

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

(D) The wavelength of the light in the meta-material (lm) is given by lm= lair|n|, where lair isthe wavelength of the light in air.

11. (B) |n| = (c / v) & speed is positive.

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

(A)

Air

Meta-material

q1

q2

(B)

Air

Meta-material

q1

q2

(C)

Air

Meta-material

q1

q2

(D)

Air

Meta-material

q1

q2

12. (C) n = (sinq1/ sinq2) sin q2= (sinq1/ n)


9/13



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 w, the motion at any instant can be taken as a comibnation of (i) a rotation of thecentre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous

vartical axis passing through its centre of mass (as is seen from the changed orientation of

points Pand Q). Both thesemotions have the same angular speed w in this case.

z

w

PQ

x

P Q y

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

and parallel tox-zplane Case (b) the disc with its face making an angle of 45 withx-yplane

and its horizontal diameter parallel tox-axis. In both the cases, the disc is welded at point P, and

the systems are rotated with constant angular speed w about the z-axis.

z

w

x

P

Q

y

case (a)

z

w

x

P

Q

y45

case(b)

13. Which of the following statements regarding the angular speed about the instantaneous axis

(passing through the centre of mass) is correct?

(A) It is 2w for both the cases.(B) It is w for case (a) and (w / 2) for case (b).(C) It is w for case (a) and 2w for case (b).(D) It is w for both the cases.

13. (D) wr

of the rigid body about any point remains same.


10/13

IIT-JEE - 2012 (08-Apr-12) Question & Solutions Paper-II - 1 0 - www. prernaclasses.com

14. 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 45 to the x-x plane and lies in the plane of the disc for

case (b.)

(C) It is horizontal for case (a) and is at 45 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 45 to thex-zplane and is normal to the plane of the disc

for case (b)

14. (A) Axis of Rotation is vertical in both the cases, as wr

is same in both the cases.


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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 may be correct.

15. Two solid cylinders Pand Q of same mass and same radius start rolling down a fixed inclinedplane from the same height at the same time. CylinderPhas most of its mass concentrated

near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is

(are) correce ?

(A) Bothe cylinders Pand Q reach the ground at the same time.

(B) Cylinder P has larger linear acceleration than cylinderQ.

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

(D) CylinderQ reaches the ground with larger angular speed.

15. (D) +

=

2MR

I

ga

1

sinq

QIP> IQ aP< aQ

Also, V= (2as) and t= (2S / a)\ VP< VQand t= (2S / a) and tQ< tP

and, V= wr \ wP< wQas VP< VQ

16. 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 it the current decreases at a steady rate.

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

16. (A,B,C,D) or(A,C)

Note:- [Option (ABCD) is correct provided zero is considered as finite else only (A,C)]

About the two sides of the diameter, the magnetic field due to infinite current carrying wire

are in opposite direction perpendicular to the plane of the circle.\ Net flux through the circle is always zero for constant as well as varying current.


12/13


17. In the given circuit, theACsource has w = 100 rad/s. Considering the inductor and capacitor tobe ideal, the correct choice(s) is (are)

~

100F 100W

0.5H 50WI

20V

(A) The current through the circuit, Iis 0.3 A.

(B) The current through the circuit, Iis 0.3 2A.(C) The voltage across 100 W resistor = 10 2V.(D) The voltage across 50 W resistor = 10 V.

17. (A,C) or(C)

Note:- [Option (AC) provided I= 0.316A 0.3 A is considered else only (C)]XC= (1 / wC) = 100 W ZC, R= 100 100 jXL= wL= 50 W ZL,R= 50 + 50 j

W=+

= 1020RC,RL,

RC,RL,

eqZZ

ZZZ

.AI 316.0

10

1

1020

20===

VZ

V 2101002100

20100

||

20=== ..100

RC

VZ

V 21005||

205 == .0

RC

18. Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shownin the figure. Given that K= (1 / 4pe0) (q / L2), which of the following statement(s) is (are)

correct?

L

A

ST D

+qF E

-q

-2q

B+q

O

R

P

C-q

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

(B) The potential at O is zero.

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

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

18. (A,B,C) 260cos.L

kq2

L

q2.xk2E

+=

220at

r= ( ) K6

L

kq624

L

kq==+

22 where, k= 1 / 4pe0

0=0atVr

(each +ve or ve changes are equidistant from 0)

Von line PR= ConstantVon line ST= Varying due to position of + 2q and 2q


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19. Two spherical planets Pand Q have the same uniform density r, masses MPand MQ, andsurface areas A and 4A, respectively. A spherical planet Ralso has uniform density r and itsmass is (MP+ MQ). The escape velocities from the planets P, Q and R, are VP, VQand VR,

respectively. Then

(A) VQ> VR> VP (B) VR> VQ> VP (C) VR/ VP= 3 (D) VP/ VQ= (1 / 2)

19. (BD) MP: MQ: MR= 1 : 8 : 9 also, ( )3

. RR34rM9 p= RR= 3

2/3. R

\ RP: RQ: RR= 1 : 2 : 32/3 Vescape= (2GM/ R) VeP: VeQ: VeR= 1 : 2 : 3

2/3

(VP/ VQ) = (1 / 2) and VR> VQ> VP

20. The figure shows a system consisting of (i) a ring of outer radius 3 Rrolling clockwise without

slipping on a horizontal surface with angular speed w and (ii) an inner disc of radius 2 Rrotatinganti-clockwise the angular speed w/2. The ring and disc are separated by frictionless ball bear-ings. The system is in thex-zplane. The point Pon the inner disc is at a distance Rfrom the

origin, where OPmakes an angle of 30 with the horizontal. Then with respect to the horizontal

surface,

3R

w/2 P

30R

2RO

z

w

x

(A) The point O has a linear velocityiR 3

(B) The point P has a linear velocity kR iR 4

33

11+

(C) The point Phas a linear velocity kR iR 4

34

13-

(D) The point Phas a linear velocity kR 4

1

iR 4

33 +

-

20. (AB) V= w(3R) since ICORat bottom point

\2

wr

rrr+= RVVP = kRiR

4

34

11ww +

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