SECTION I (Single Correct Answer Type)

SECTION I(Single Correct Answer Type)

This section contains 13 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

1. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) us-ing a simple pendulum. They use different lengths of the pendulum and/or record time for differ-ent number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm Least count for time = 0.1 s

Student Length of the pendulum

(cm)

Number of oscillations

(n)

Total time for (n) oscillations

(s)

Time period

(s)I 64.0 8 128.0 16.0II 64.0 4 64.0 16.0III 20.0 6 36.0 6.0

If EI, EII and EIII are the percentage errors in g. i.e.

Dg

g¥Ê

ËÁˆ¯̃

100 for student I, II and III, respectively,

(a) EI = 0 (b) EI is minimum(c) EI = EII (d) EII is minimum

2. A block of mass m is held stationary against a wall by applying a horizontal force F on the block. Which of the following statements is false?(a) The frictional force acting on the block is

f = mg(b) The normal reaction force acting on the block

is N = F(c) No net torque acts on the block(d) N does not produce any torque.

Block

FO

Wall

3.M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P

3R4R

P

4R

(a) 2

74 2 5

GM

R( )- (b) - -2

74 2 5

GM

R( )

(c) GM

R4(d) 2

52 1

GM

R( )-

4. A block A of mass m is placed on a frictionless horizontal surface. Another block B of the same mass is kept on A and connected to the wall with the help of a spring of force constant k, as shown

blocks A and B is m. The blocks move together executing simple harmonic motion of amplitude a.The maximum value of frictional force between A and B is

A

Bmk

(a) ka (b) ka/2(c) zero (d) m mg

5. A bi-convex lens is formed with two thin plano-

index nsecond lens is 1.2. Both the curved surface are of the same radius of curvature R = 14 cm. For this

MODEL TEST PAPER—I

The questions in the practice papers are based on questions asked in previous years’ Physics Question Papers of IIT-JEE. Answer key and complete solutions of questions are provided at the end of each practice paper. Each paper contains 30 questions to be answered in 45 minutes.

MTPI.2 Comprehensive Physics—JEE Advanced

bi-convex lens, for an object distance of 40 cm,the image distance will be

n = 1.5 n = 1.2

R = 14 cm

(a) – 280.0 cm (b) 40.0 cm

6. Three very large plates of same area are kept parallel and close to each other. They are consid-ered as ideal black surfaces and have very high

maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is

(a) 65

2

1 4ÊËÁ

ˆ¯̃

/

T (b) 97

4

1 4ÊËÁ

ˆ¯̃

/

T

(c) 972

1 4ÊË

ˆ¯

/

T (d) (97)1/4 T

7. In the given circuit, a charge of + 80 mC is given to the upper plate of the 4 mF capacitor. Then in the steady state, the charge on the upper plate of the3 mF capacitor is

2 Fm 3 Fm

4 Fm

+80 Cm

(a) + 32 mC (b) + 40 mC(c) + 48 mC (d) + 80 mC

8. A student is performing the experiment of reso-nance column. The diameter of the column tube is

The air temperature is 38°C in which the speed of sound is 336 m/s. The zero of the meter scale co-

of the water level in the column is

(c) 16.4 cm (d) 17.6 cm

9. An RC circuit consists of a resistance R W.and a capacitance C = 1.0 mF connected in series with a battery. In how much time will the potential difference across the capacitor become 8 times that across the resistor? (Given loge (3) = 1.1)

(c) 44 s (d) 88 s10. A proton moving with a speed u along the positive

x-axis enters at y = 0 a region of uniform magnetic B = B0 k� which exists to the right of y-axis as

after some time with a speed v at co-ordinate y.Then

u+q x

B (out of page)

y

O

(a) v > u, y < 0 (b) v = u, y > 0(c) v > u, y > 0 (d) v = u, y < 0

11. A cylindrical conducting rod is kept with its axis along the x-axis. Also there exists a uniform

x-axis. The current induced in the cylinder is(a) clockwise as seen from the + x axis(b) anticlockwise as seen from the + x axis(c) along the axis towards – x direction(d) zero

12.

n0,n n n0 0 0

2 6 8, ,and

respectively. The angle of incidence q for which the beam just misses entering region IV is

Region I Region II Region III

n0

2

n0

6

n0

8

Region IV

q

n0

0 0.2 m 0.6 m

(a) sin- ÊËÁ

ˆ¯̃

1 3

4(b) sin- Ê

ËÁˆ¯̃

1 1

8

(c) sin- ÊËÁ

ˆ¯̃

1 1

4(d) sin- Ê

ËÁˆ¯̃

1 1

3

MTPI.3

13. Electrons with de-Broglie wavelength l fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is

16. A cubical region of side a has its centre at the ori-q at (0,

–a/4, 0), + 3q at (0, 0, 0) and –q at (0, +a/4, 0). Choose the correct options(s).

a

z

-q-q

3qy

x

x = +athe plane x = –a/2

y = +a/2

plane y = –a/2.

isq

e0

.

z = +athe plane x = +a/2.

17. A disc of mass M and radius R is rolling with an-gular speed w on a horizontal surface as shown in

the disc about the origin O is (here v is the linear velocity of the disc)

R

v

x

y

O

w

O

(a) l l0

22= mc

h(b) l0

2= h

mc

(c) l l0

2 2 2

2

2= m c

h(d) l0 = l

SECTION II(Multiple Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE than one choice/choices is/are correct.

14.inclined plane PQ which makes an angle q with the horizontal. A horizontal force of 1 N acts of the block through its centre of mass as shown in

g = 10 m/s2).

1 N

Q

O

qP

(a) q(b) q

towards P.(c) q

towards Q.(d) q

towards Q.15.

choose the correct option(s).

2 W

4 W

4 W

4 W

1 W 1 W

2 W

2 W

12 V

I1

I2P S

Q T

(a) The current through PQ is zero.(b) I1 = 3A(c) The potential at S is less than that at Q.(d) I2 = 2A


(a)3

22MR w (b) MR2w

(c) MRv (d)3

2MRv

18. The displacement x of a particle varies with time tas x – A sin2 wt + B cos2 wt + C sin wt cos wt. For what values of A, B and C is the motion simple harmonic?(a) All values of A, B and C with C π 0.(b) A = B, C = 2B(c) A = –B, C = 2B(d) A = B, C = 0

19. A beam of light consisting of two wavelengths

fringes in a Young’s double slit experiment. The separation between the slits is 1 mm and the dis-tance between the plane of the slits and the screen is 100 cm. The least distance from the central maxi-mum where the bright fringes due to both the wave-lengths coincide is ymin and y¢min is the correspond-ing distance where the dark fringes due to both the wavelengths coincide. Then

(a) ymin

(b) ymin = 2.0 mm

(c) y¢min(d) y¢min

20. off

voltage V0 and 1

l for three metals 1, 2 and 3, where

l is the wavelength of the incident radiation in nm.

0.001 0.002 0.004

1

l (nm )-1

V0

Metal 1 Metal 2 Metal 3

If W1, W2 and W3 are the work functions of metals 1, 2 and 3 respectively, then(a) W1 : W2 : W3 = 1 : 2 : 4(b) W1 : W2 : W3 = 4 : 2 : 1(c) The graphs for metals 1, 2 and 3 are parallel to

each other and the slope of each graph is hc/e,where h = Planck’s contant, c = speed of light and e = charge of an electron.

(d) Ultraviolet light will eject photoelectrons from metals 1 and 2 and not from metal 3.

Questions 21 and 22 are based on the following paragraph.

A light rod of length L having a body of mass M attached to its end hangs vertically. It is turned through 90° so that it is horizontal and then released.

21. The centripetal acceleration when the rod makes an angle q with the vertical is

(a) g cos q (b) 2g cos q(c) g sin q (d) 2g sin q

22. The tension in the rod when it makes an angle qwith the vertical is

(a) Mg cos q (b) 2 Mg cos q(c) 3 Mg cos q (d) zero

Questions 23 and 24 are based on the following passage.

A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of the same mass is attached to the other end of the rod. The two particles carry charges +q and –q. This

E such that the rod makes a small angle q

- q

B

O

q

+q

A

E

SECTION III(Linked Comprehension Type)

This section contains 4 questions based on a paragraph. Each question has four choices (a), (b), (c) and (d) out of which only ONE choice is correct.

MTPI.5

23. When the rod is released, it will rotate with an an-gular frequency w equal to

(a) qE

MLÊËÁ

ˆ¯̃

1 2/

(b) 2 1 2qE

MLÊËÁ

ˆ¯̃

/

(c) qE

ML2

1 2ÊËÁ

ˆ¯̃

/

(d) 1

2

1 2qE

MLÊËÁ

ˆ¯̃

/

24. The minimum time taken by the rod to align itself

given by

(a)p2 2

1 2ML

qEÊËÁ

ˆ¯̃

/

(b) 21 2

p ML

qEÊËÁ

ˆ¯̃

/

(c) 22

1 2

p ML

qEÊËÁ

ˆ¯̃

/

(d) 22

1 2

p ML

qEÊËÁ

ˆ¯̃

/

SECTION IV(Assertion-Reason Type)

This section contains 2 questions. In each question, Statement-1 is followed by Statement-2. Each question has the following four choices (a), (b), (c) and (d) out of which only ONE choice is correct.

(a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1.

(b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1.

(c) Statement-1 is true, Statement-2 is false.

(d) Statement-1 is false, Statement-1 is true.

25. Statement-1PV

T versus P graph for

a certain mass of oxygen gas at two temperatures T1and T2. It follows from the graph that T1 > T2.

PV

T

T1

T2

B

P

Statement-2 At higher temperature, real gas behaves more like

an ideal gas.26. Statement-1

A particle of mass M at rest decays into two par-ticles of masses m1 and m2 which move with ve-locities v1 and v2 respectively. Their respective de Broglie wavelengths are l1 and l2. If m1 > m2, then l1 > l2.Statement-2

The de Broglie wavelength of a particle having mo-mentum p is l = h/p.

SECTION V(Integer Answer Type)

This section contains 3 questions. The answer to each question is single digit integer, ranging from 0 to 9(both inclusive)

27. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a

uniform current density JP

is given by NaJ

12 0m , N.

a

OP

2a


28. A binary star consists of two stars A (mass 2.2 Ms)and B (mass 11 Ms), where Ms is the mass of the sun. They are separated by distance d and are rotat-ing about their centre of mass, which is stationary. Find the ratio of the total angular momentum of the binary star to the angular momentum of star Babout the centre of mass.

29. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature T1 and T2, respectively. The maximum intensity in the emission spectrum of A BConsidering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?

SECTION VI(Matrix Match Type)

This section contains 1 question. Each question has four statements (a, b, c and d) given in Column I and (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with ONE or MOREstatement(s) given in Column II. For example, if for a given question, statement B matches with the statements given

30. One mole of a monatomic gas is taken through a cycle ABCDA as shown in the P-V diagram.Column II gives the characteristics involved in the

Column I.P

AB

C D

3P

1P

1V 3V 9V V0

Column I Column II(a) Process A Æ B (p) Internal energy decreases

(b) Process B Æ C (q) Internal energy increases

(c) Process C Æ D (r)

(d) Process D Æ A (s)

(t) Work is done on the gas

AnswersSection-I

1. (b) 2. (d) 3. (b)4. (b) 5. (b) 6. (c)7. (c) 8. (b) 9. (b)

10. (d) 11. (d) 12. (b)13. (a)

Section-II14. (a, c) 15. (a, b, c, d) 16. (a, c, d)17. (a, d) 18. (a, b, c) 19. (a, d)20. (a, d)

Section-III21. (b) 22. (c) 23. (b)24. (a)

Section-IV

25. (a) 26. (d)

Section-V

27. 28. (6) 29. (9)

Section-VI

30. (a) Æ (p, r, t); (b) Æ (p, r), (c) Æ (q, s),(d) Æ (r, t).

Solutions

Section-I

1. TL

g= 2p

Time period Tt

n= , when n = number of oscilla-

tion and t is the total time for n oscillation. In terms of measured quantities,

t

n

L

gg

L n

t= fi =2

4 2 2

2p p

. Therefore,

D D Dg

g

L

L

t

t= + 2 (∵ D n = 0; there is no error

in counting the number of oscillation)

For student I, EI =Dg

g¥ = + ¥Ê

ËÁˆ¯̃ ¥100

0 1

64

2 0 1

128100

. .

= 5

16%

MTPI.7

For student II, EII =0 1

64

2 0 1

64 0100

. .

.+ ¥Ê

ËÁˆ¯̃ ¥

= 15

32%

For student III, EIII = 0 1

20 0

2 0 1

36100

.

.

.+ ¥ÊËÁ

ˆ¯̃ ¥

= 19

18%

Thus the percentage error in the measurement of g is minimum for student I.

2. Since the block is held stationary, it is in transla-

net force and no net torque acts on the block. No net force will act on the block if f = mg and N = F.No net torque will act on the block, if torque by frictional force f about centre O = counter torque by normal reaction N about centre Ochoice (d) is false.

3.mass from P Vp, where VP is the gravitational potential at PVP, we divide the disc into small elements, each of thickness dr. Consider one such element at a distance r from the centre of the disc, as shown

r

dr

r

R

2 +16

2

P

dm = M rdr

R R

( )

( ) ( )

2

4 32 2

pp p-

=2

7 2

M rdr

R

VP = -+

Ú3

4

2 216R

RG dm

r R

= -+Ú

2

7 1623

4

2 2 1 2

MG

R

rdr

r RR

R

( ) /

Putting r2 + 16 R2 = x2, we get 2r dr = 2x dx orrdr = x dx.

When r x = 9 16 52 2R R R+ =

When r = 4 R, x = 16 16 4 22 2R R R+ =

\ VP = - = - -Ú2

7

2

74 2 5

25

4 2

2

MG

Rdx

MG

RR

R

R

( )

VP = - -2

74 2 5

MG

R( ), which is choice (b).

4. The blocks will move together as long as the frictional force of block B = mass of block B ¥

f = mw2a

where w =k

m m

k

m( )+=

2

Thus f = mk

ma¥ ¥

2

= ka/2

5. 1 1 5 1

1

1

14

1 1

281

1

f= -Ê

ËÁˆ¯̃ -

•ÊËÁ

ˆ¯̃ = -.

cm

1 1 2 1

1

1 1

14

1

702

1

f= -Ê

ËÁˆ¯̃ •

--

ÊËÁ

ˆ¯̃

= -.cm

1 1 1 1

28

1

7020

1 2F f fF= + = + fi = cm

Now 1 1 1

v- =

u F

fi 1 1

40

1

2040

vv-

-= fi = cm

6. In the steady state, the rate at which the middle plate receives heat energy is equal to the rate at which heat energy is emitted by the other plates. Let A be the area of each plate and T0 be the steady state temperature of the middle plate. Since both sides of the middle plate receive heat energy, the total area of the middle plate receiving energy is 2A.

2T To 3T


From Stefan’s law

s (2A) (To)4 = sA (2T )4 + sA (3T )4

fi 2 To4 = 16 T 4 + 81 T 4 = 97 T 4

fi To = 97

2

1 4ÊËÁ

ˆ¯̃

/

T

7. Let q mC be the charge on the upper plate of 3 mFcapacitor. Then the charge on the upper plate of2 mF capacitor will be (80 – q) mC. Since poten-tial difference across 2 mF capacitor = potential difference across 3 mF capacitors,

80

2

- q =q

q3

48fi = mC.

8. End correction e = 0.3d = 0.3 ¥ 4 = 1.2 cm

Wavelength l = vn

= 336

512

Now L + e = l4

fi L = l4

– e = 65 6

4

. –1.2

9. At instant of time t, the charge on the capacitor is given by

q = q0 (1 – e–t/RC)

and the potential drop across the capacitor is given by (∵ V = q/C)

VC = V0 (1 – e–t/RC)

where V0 is the voltage of the battery. The poten-tial drop across the resistor is

VR = V0 – VC = V0 – V0 (1 – e–t/RC) = V0 e–t/RC

\V

VC

R=

11

- = --

-e

ee

t RC

t RCt RC

/

//

GivenV

VC

R

= 8. Therefore,

8 = et/RC – 1or et/RC = 9 = (3)2

t

RC= 2 loge (3)

or t = RC ¥ 2 loge (3)¥ 106) ¥ (1 ¥ 10–6) ¥ 2 ¥ 1.1

= 11 s

10. When the proton enters the region of the magnetic F given by

F = q (u ¥ B) where q is the charge of the proton. The force F

is perpendicular to both u and B. Since the force is perpendicular to the velocity of the particle, it

the velocity of the particle will remain unchanged;

v = u. Since u is perpendicular to B, the proton moves in a circular path. Since the charge of pro-ton is positive, u is along positive x-axis and B is directed out of the page, the proton will move in a circle in the x-y plane in the clockwise direction.

y coordinate will be negative, when it leaves the region. Thus the correct choice is (d).

11.

emf and current are zero. So the correct choice is (d).

12. The beam will not enter region IV if the angle refraction in region IV equals 90°. Apply Snell’s law at the interfaces, we have

q3 = 90°

q2

q2

q1

q1

q

I II III IV

n0 sin q =n0

12sin q

=n n0

20

6 890sin sinq = ∞

which gives sin q = 1

8.

13. de-Broglie wavelength is

l = h

mE2

where E is the kinetic energy of the electrons. The out-off wavelength is

MTPI.9

l0 = hc

E

From Eq. (1) E = h

m

2

22 l.

l0 = 2 2mc

h

l

Section II

14.

O P

Q

F cos q

mg sin q

mgmg cos q

F

q

q

q

Given F = 1 N, m = 0.1 kg and g = 10 ms–2. Let f be the frictional force between the block and the plane surface PQ.

The block will be stationary ifF cos q = mg sin qfi 1 ¥ cos q = 0.1 ¥ 10 ¥ sin qfi tan q = 1 fi q f = 0

If q q > cos q.mg sin q > F cos q (∵ F = mg = 1 N).

Therefore frictional force acts up the block to-wards Q.

If q q < cos qmg sin q < F cos q.

Therefore, in this case, frictional force f acts down in the plane towards P.

15.

2 W

4 W

4 W

4 W

1 W 1 W

2 W

2 W

12 V

I1

P S

Q T

A

a

b

V

( + )a b

B

a

( )a c-

c

c

b

( + )b c

U

Applying Kirchhoff’s loop rule to loops APQA,PSTQP and AQTBUVA, we get

2a + c – 4b = 0 (1) 2(a – c) – c – 4 (b + c) – c = 0fi a – 2b – 4c = 0 (2)and 4b + 4 (b + c) + 4b – 12 = 0fi 3b + c = 3 (3)Solving Eqs. (1), (2) and (3), we get a = 2 A, b = 1 A and c = 0Thus the current through PQ is zero. Also I1 = a + b= 3 A and I2 = a – c = 2 A.

Also VS – VQ = – c – 4 (b + c) = – 4b = – 4 ¥1 = – 4V (∵ c S is less than that at Q.

16.the cubical region is

f =q q q q qnet

e e e0 0 0

3= - - =

x = a/2 and the x = – a/2 is the same. Further, the positions of charges with respect to

x = a/2 and z = athrough the planes x = a/2 and z = a/2 is the same

y = a/2 and y = – a/2 is the same.17. The angular momentum about O is

�LO =

� � �L M RCM + ¥( )v

Its magnitude is (∵ � �R ^ v) and L = Iw

LO = Iw + MRv

=1

22MR MR R RÊ

ËÁˆ¯̃ + ¥ =w w w( )∵v

=3

22MR w

=3

2

3

22MR

RMR¥ Ê

ËÁˆ¯̃ =v

v

18. The displacement equation can be rewritten as

xA

tB

tC

t= - + + +2

1 22

1 22

2( cos ) ( cos ) sinw w w

or

x A B B A tC

t= + + - +1

2

1

22

22( ) ( )cos sinw w (1)

Choice (a): Equation (1) can be written asx = x0 + a cos 2wt + b sin 2wt (2)


where x0 =1

2

1

2( ), ( )A B a B A+ = -

and b =C

2.

Equation (2) can be recast asx = x0 + A0 sin(2w + f) (3)

where A0 = (a2 + b2)1/2 and tan f = a/b. Equa-tion (3) represents a simple harmonic motion of angular frequency 2w, amplitude = x0 + A0 and phase constant f.Choice (b): For A = B and C = 2B, Eq. (1) be-comes

x = B + B sin2 wt = B(1 + sin2 wt) This equation represents a simple harmonic mo-

tion of amplitude 2B and angular frequency 2w.Choice (c): For A = – B and C = 2B, Eq. (1) be-comes

x = B cos2 wt + B sin2 wt which represents a simple harmonic motion of

amplitude 2 B, angular frequency 2w and phase constant p/4.Choice (d): For A = B and C = 0, Eq. (1) reduces to

x = Awhich does not represent simple harmonic motion.

19. Let nth bright fringe of wavelength ln and the mth bright fringe of wavelength lm coincide at a distance y from the centre of the screen. Then

y =n D

d

m D

dn ml l

=

or n ln = m lm

orll

n

m

=m

n

or 750

450= m

n

or m

n= 5

3. The minimum integral values of m and

n that satisfy this equation are m n = 3.

Therefore, the minimum value of y is

ymin =n D

dnl

= ¥ ¥ ¥-

-3 750 10 1

10

9

3

¥ 10–3 m

For dark fringes to coincide, the condition is

y¢ = nD

dm

D

dn m-Ê

ËÁˆ¯̃ = -Ê

ËÁˆ¯̃

1

2

1

2

l l

fi750

450=

m

nn m

-ÊË

ˆ¯

-ÊË

ˆ¯

fi = +

1212

5 3 1

The minimum integral values which satisfy this condition are n = 2 and m

y ¢min = nD

dn-Ê

ËÁˆ¯̃

1

2

l

=2

12

750 10 1

10

9

3

-ÊË

ˆ¯ ¥ ¥ ¥-

¥ 10–3

The correct choices are (a) and (d).

20. Work function W = hn0 = hc

l0

, where l0 is the

W1 : W2 : W3 =hc hc hc

( ):

( ):

( )l l l0 1 0 2 0 3

=1 1 1

0 1 0 2 0 3( ):

( ):

( )l l l= 0.001 : 0.002 : 0.004= 1 : 2 : 4

-sion, the relation between V0 and l is given by

eV0 = h Whc

Wnl

- = -

or V0 =hc

e

W

e

1

lÊËÁ

ˆ¯̃ -

V0 and 1

l

is hc

e which is the same for all metals. Therefore,

choice (c) is correct. The threshold wavelength for the three metals are

1

0 1( )l = 0.001 nm–1, therefore (l0)l = 1000 nm

= 10,000 Å1

0 2( )l = 0.002 nm–1, therefore (l0)2

MTPI.11

1

0 3( )l = 0.004 nm–1, therefore (l0

For photoelectric emission, the wavelength of the incident radiation must be less than the threshold wavelength. Since the wavelength of ultraviolet light is about 1200 Å, it will eject photoelec-

choices are (a) and (d).Section III

21.

A

B C

O

qT

L

Mg cos qMg

L cos q

The loss of PE when the body falls from A to B= Mg ¥ OC = MgL cos q. If v is the velocity of the body at B, then

1

2M v2 = MgL cos q or v2 = 2gL cos q (1)

centripetal acceleration = v2

L

gL

L= 2 cos q

= 2g cos q, which is choice (b).

22. The centripetal force when the body is at B is

Fc = M

L

v2

Thus, we have

T – Mg cos q =M

L

v2

(2)

Using (1) in (2), we get

T – Mg cos q =M

LgL Mg¥ =2 2cos cosq q

or T = 3 Mg cos q Thus the correct choice is (c).

23. A non-conducting rigid rod having equal and op-posite charges at the ends is an electric dipole.

experiences a torque which tends to align it with F = qE each

acting at A and B constitute a couple whose torque is given by

t = force ¥ perpendicular distance

= F ¥ AC = F ¥ AB sin q = qEL sin q

So the correct choice is (a).

- q

B

O

q

qA

E

qEq

qE

Since q is small, sin q � q, where q is expressed in radian. Thus t = qELq\ t = –qELq (1)

If a is the angular acceleration of the rotatory motion,

t = Ia where I is the moment of inertia of the two masses

at A and B about an axis passing through the centre O and perpendicular to the rod. Since the rod is massless,

I = M ¥ (AO)2 + M ¥ (BO)2

= ML

ML ML¥ Ê

ËÁˆ¯̃ + ¥ Ê

ËÁˆ¯̃ =

2 2 2

2 2 2

Thus t = ML2

2

a (2)

Using Eq. (2) in Eq. (1), we get

a = -ÊËÁ

ˆ¯̃ = -2 2qE

MLq w q

where w = 2 1 2qE

MLÊË

ˆ¯

/

, which is choice (b).

24. The time period of oscillation is

T =2

22

1 2pw

p= ÊËÁ

ˆ¯̃

ML

qE

/

time taken by the rod to align itself parallel to

one-fourth of angular oscillation, i.e.

tmin =T ML

qE4 2 2

1 2

= ÊËÁ

ˆ¯̃

p /

So the correct choice is (a).

Section IV

25. The correct choice is (a). The line AB is parallel to the P-axis. This means that PV/T is a constant,

AB corre-sponds to an ideal gas for which PV/T = constant.


At higher temperatures, a real gas behave more T1 is greater than T2.

26. The correct choice is (d). The law of conservation of linear momentum gives

m1v1 + m2v2 = 0 or m

m2 2

1 1

1 0vv

= .

Since de Broglie wavelength l = h/mv, we will have

ll

1

2=

m

m2 2

1 1

1 0vv

= .

Section V

27. P due to complete cylinder is

B1 =m p

pp0

20

2 2

( )J a

a

Ja=

P due to cavity is

B2 =m p

p

m02

04

232

12

( )J aa

Ja/

ÊË

ˆ¯

=

\ P is

B = B BJa Ja Ja

1 20 0 0

2 12

5

12- = - =

m m m

which gives N28. Given M1 = 2.2 Ms and M2 = 11 Ms. Let R1 and

R2 be their respective distances from the centre of mass. The total angular momentum about the centre of mass is

Ltotal = (I1 + I2) w

and the angular momentum of B isL2 = I2 w

\L

Ltatal

2

=I I

I

I

I1 2

2

2

1

1+

= +

= 1 1 12

2 22

+M R

M R

= 1 1 12

2 22

+ =M

MR

v

vv( )∵ w

= 1 1 1

2 2

22

1

+ ÊËÁ

ˆ¯̃

¥M

M

M

M

vv

= 1 111

2 26+ ¥ =

.

(∵ M1 v1 = M2 v2)

29. From Wien’s displacement law lm T = constant, we have

lA TA = lllB B

A

B

B

A

TT

Tfi =

From Stefan’s law, E = sA ◊T 4 = s (4 pR2)T 4, we have

E

EA

B=

R

R

T

TA

B

A

B

ÊËÁ

ˆ¯̃

¥ ÊËÁ

ˆ¯̃

2 4

=R

RA

B

B

A

ÊËÁ

ˆ¯̃

¥ ÊËÁ

ˆ¯̃

2 4ll

=6

18

1500

50

2 4ÊËÁ

ˆ¯̃ ¥ Ê

ËÁˆ¯̃

= 9Section VI

30.(a) Process A Æ V μ T. There-fore TA > TB. DU = nCvDT = nCv(TB – TA). Since TB < TA, DU is negative, i.e. internal energy de-creases.DQ = nCpDTDW = 3P(VB – VA) = – 6PV. Which is negative.

\ (a) Æ (p, r, t)(b) Process B Æ P μ T. There-

fore TB > TC. DU = nCvDT = nCv (TC – TB) is negative, i.e. internal energy decreases.DW = PDV = 0

(∵ DV = 0)DQ = DU +

DW), DQ = DU. Since DU is negative, heat is lost.\ (b) Æ (p, r)

(c) Process C Æ D is isobaric, i.e. V μ TTD > TC. DU = nCv(TD – TCinternal energy increases.DQ = nCp(TD – TCgained by the gas.\ (c) Æ (q, s)

(d) In process D Æ A, the gas is returned to the ini-DU = 0. Therefore DQ = DW.

Since the gas is compressed, work is done on the gas, i.e. DW DQ is negative.

\ (d) Æ (r, t)Answer: (a) Æ (p, r, t) (b) Æ (p, r)

(c) Æ (q, s) (d) Æ (r, t)

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SECTION I (Single Correct Answer Type)