CHAPTER-5 CENTRE OF MASS€¦ · CENTRE OF MASS TRUE/FALSE 1. Two particles of mass 1 kg and 3 kg...

39

CHAPTER-5CENTRE OF MASS

TRUE/FALSE

1. Two particles of mass 1 kg and 3 kg move towards eachother under their mutual force of attraction. No otherforce acts on them. When the relative velocity ofapproach of the two particles is 2 m/s, their centre ofmass has a velocity of 0.5 m/s. When the relativevelocity of approach becomes 3 m/s, the velociy of thecentre of mass is 0.75 m/s. (1989; 2M)

OBJECTIVE QUESTIONS

Only One option is correct :1. Two particles A and B initially at rest, move towards

each other mutual force of attraction. At the instantwhen the speed of A is V and the speed of B is 2V,the speed of the centre of mass of the system is :

(1982; 2M)(a) 3V (b) V(c) 1.5 V (d) zero

2. A ball hits the floor and rebounds after an inelasticcollision. In this case : (1986; 2M)(a) the momentum of the ball just after the collision is

the same as that just before the collision.(b) the mechnical energy of the ball remains the same

in the collision(c) the total momentum of the ball and the earth is

conserved(d) the total energy of the ball and the earth is

conserved

3. A shell is fired from a cannon with a velocity v (m/s)at an angle θ with the horizontal direction. At thehighest point in its path it explodes into two pieces ofequal mass. One of the pieces retraces its path to thecannon and the speed (m/s) of the other pieceimmediately after the explosion is : (1986; 2M)(a) 3 v cos θ (b) 2 v cos θ

(c)32

v cos θ (d)32

v cos θ

4. An isolated particle of mass m is moving in horizontalplane (x-y), along the x-axis, at a certain height abovethe ground. It suddenly explodes into two fragment ofmasses m/4 and 3m/4. An instant later, the smallerfragment is at y = + 15 cm. The larger fragment at thisinstant is at : (1997 C; 1M)(a) y = – 5 cm (b) y = + 20 cm(c) y = + 20 cm (d) y = – 20 cm

5. Two particles of masses m1 and m2 in projectile motion

have velocities 21 vvrr

< respectively at time t = 0. They

collide at time t0. Their velocities become 1′vr

and 2′vr

at time 2t0 while still moving in air. The value of :

|)()(| 22112211 vmvmvmvmrrrr

+−′+′ (2001; S)

(a) zero (b) (m1 + m2)gt0

(c) 2 (m1 + m2)gt0 (d)12

(m1 + m2)gt0

FILL IN THE BLANKS

1. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments each of mass m arefound to move with a speed v each in mutually perpendicular directions. The total energy released in the processof explosion is ..... (1987; 2M)

2. The magnitude of the force (in newtons) acting on a body varies with time t (in microseconds) as shown in thefig. AB, BC and CD are straight line segment. The magnitude of the total impulse of the force on the body fromt = 4 µs to t = 16 µs is ...... N-s (1994; 2M)

800 –

600 –

400 –

200 –

0 2 4 6 8 10 12 14 16

A BE F D

C

Time (µs) →

Forc

e (N

) →

40

6. Two blocks of masses 10 kg and 4 kg are connectedby a spring of negligible mass and placed on africtionless horizontal surface. An impulse gives avelocity of 14 m/s to the heavier blocks in the directionof the lighter block. The velocity of the centre of massis : (2002; S)(a) 30 m/s (b) 20 m/s(c) 10 m/s (d) 5 m/s

7. A particle moves in the X-Y plane under the influenceof a force such that its linear momentum is

)],sin(ˆ)cos(ˆ[)( ktjktiAtp −=r

where A and k are

constants. The angle between the force and themomentum is: (2007; 3M)(a) 0° (b) 30°(c) 45° (d) 90°

8. Two small particles of equalmasses start moving in oppositedirections from a point A in ahorizontal circular orbit. Theirtangential velocities are v and 2v,respectively, as shown in the figure.

Av 2v

Between collisions, the particles move with constantspeeds. After making how many elastic collisions,other than that at A, these two particles will againreach the point A? (2009; M)(a) 4 (b) 3(c) 2 (d) 1

9. Look at the drawing given in thefigure which has been drawn withink of uniform line thickness. Themass of ink used to draw each ofthe two inner circles, and each ofthe two line segments is m.

x

y

The mass of the ink used to draw the outer circle is6m. The coordinates of the centres of the differentparts are: outer circle (0, 0) left inner circle (–a, a),right inner circle (a, a), vertical line (0, 0) and horizontalline (0, –a). The y-coordinate of the centre of mass ofthe ink in this drawing is (2009; M)

(a) 10a

(b) 8a

(c)12a

(d) 3a

OBJECTIVE QUESTIONS

More than one options are correct?

1. Two blocks A and B each of mass m, are connected bya massless spring of natural length L and spring

constant K. The blocks are initially resting on a smoothhorizontal floor with the spring at its natural length, asshown in fig. A third identical block C, also of mass m,moves on the floor with a speed v along the linejoining A and B, and collides elastically with A .Then : (1993; 2M)

C A B

v

L

(a) the kinetic energy of the A-B system, at maximumcompression of the spring, is zero

(b) the kinetic energy of the A-B system, at maximumcompression of the spring, is mv2/4

(c) the maximum compression of the spring is

( / )v m K

(d) the maximum compression of the spring is 2mvK

2. Two balls, having linear moments ipp1 =r

and ipp2 =r

,

undergo a collision in free space. There is no external

force acting on the balls. Let 1'pr

and 2'pr

be their final

momenta. The following option figure (s) is (are) NOTALLOWED for any non-zero value of p, a1, a2, b1, b2,c1 and c2. (2008; 4M)

(a) kji'p 1111 cba ++=r

ji'p 222 ba +=r

(b) k'p 11 c=r

j'p 22 b=r

(c) kji'p 1111 cba ++=r

kji'p 1222 cba −+=r

(d) ji'p 111 ba +=r

ji'p 122 ba +=r

SUBJECTIVE QUESTIONS

1. A circular plate of uniform thickness has a diameter of56 cm. A circular portion of diameter 42 cm is removedfrom one edge of the plate as shown in figure. Find theposition of the centre of mass of the remainingportion. (1980)

56 cm42cm

41

2. A body of mass 1 kg initially at rest, explodes andbreaks into three fragments of masses in the ratio1 : 1 : 3. The two pieces of equal mass fly-offperpendicular to each other with a speed of 30 m/seach. What is the velocity of the heavier fragment?

(1981; 3M)3. Three particles A, B and C

of equal mass move withequal speed V along themedians of an equilateraltriangle as shown infig.They collide at thecentroid G of the triangle.

A

B

G

CAfter the collision, A comes to rest, B retraces itspath with the speed V. What is the velocity of C?

(1982; 2M)

4. A block of mass M with a semicircular track of radiusR, rests on a horizontal frictionless surface. A uniformcylinder of radius r and mass m is released from restat the top point A (see Fig.). The cylinder slips on thesemicircular frictionless track. How far has the blockmoved when the cylinder reaches the bottom (point B)of the track? How fast is the block moving when hecylinder reaches the bottom of the track?(1983; 7M)

m R

M

B

r

5. Two bodies A and B of masses m and 2m respectivelyare placed on a smooth floor. They are connected bya spring. A third body C of mass m moves withvelocity v0 along the line joining A and B and collideselastically with A as shown in Fig. At a certain instantof time t0 after collision, it is found that theinstantaneous velocities of A and B are the same.Further at this instant the compression of the springis found to be x0. Determine (i) the common velocityof A and B at time t0 and (ii) the spring contant.

(1984; 6M)

C AB

6. A ball of mass 100 g is projected vertically upwards

from the ground with a velocity of 49 m/s. At the sametime another identical ball is dropped from a height of98 m to fall freely along the same path as that followedby the first ball. After some time the two balls collideand stick together and finally fall to the ground. Findthe time of flight of the masses. (1985; 8M)

7. A simple pendulum issuspended from a peg ona vertical wall. Thependulum is pulled awayfrom the wall to ahorizontal position Seefigure and released. The

L

ball hits the wall, the coefficient of restitution being

2

5. What is the minimum number of collisions after

which the amplitude of oscillations becomes less than60 degrees? (1987; 7M)

8. An object of mass 5 kg is projected with a velocity of20 m/s at an angle of 60° to the horizontal. At thehighest point of its path the projectile explodes andbreaks up into two fragments of masses 1 kg and 4 kg.The fragments separate horizontally after the explosion.The explosion released internal energy such that thekinetic energy of the system at the highest point isdoubled. Calculate the separation between the twofragments when they reach the ground.

(1990; 8M)

9. A block A of mass 2m is placed on another block B ofmass 4m which in turn is placed on a fixed table. Thetwo blocks have a same length 4d and they are placedas shown in figure. The coefficient of friction (bothstatic and kinetic) between the block B and table is µ.There is no friction between the two blocks. A smallobject of mass m moving horizontally along a linepassing through the centre of mass (CM) of the blockB and perpendicular to its face with a speed v collideselastically with the block B at a height d above thetable. (1991; 4+4M)

4d

d

B 4m2d

A 2m

P

vm

42

(a) What is the minimum value of v (call it v0) requiredto make the block A topple?

(b) If v = 2 v0 find the distance (from the point P inthe figure) at which the mass m falls on the tableafter collision (Ignore the role of friction duringthe collision). [1991; 4+4M]

10. A uniform thin rod of mass M and length L is standingvertically along the y-axis on a smooth horizontalsurface, with its lower end at the origin (0, 0). A slightdisturbance at r = 0 causes the lower end to slip onthe smooth surface along the positive x-axis, and therod starts falling. (1993; 1+ 5M)(i) What is the path followed by the centre of mass

of the rod during its fall?(ii) Find the equation of the trajectory of a point on

the rod located at a distance r from the lower end.What is the shape of the path of this point?

11. A small sphere of radius R is held against the innersurface of a larger sphere of radius 6R. The masses oflarger and small spheres are 4M and M respectively.This arrangement is placed on a horizontal table. Thereis no friction between any surfaces of contact. Thesmall sphere is now released. Find the co-ordinates ofthe centre of the larger sphere when the small spherereaches the other extreme position. (1996; 3M)

12. A wedge of mass m and triangular cross-section(AB = BC =CA =2R) is moving with a constant velocity

ivˆ− towards a sphere of radius R fixed on a smooth

horizontal table as shown in the figure. The wedgemakes an elastic collision with the fixed sphere andreturns along the same path without any rotation.Neglect all friction and suppose that the wedge remainsin contact with the sphere for a very short time ∆t

during which the sphere exerts a constant force Fr

onthe wedge. (1998; 8M)

z

y

v

A

R

x B C

(a) Find the force Fr

and also the normal force exertedby the table on the wedge during the time ∆t.

(b) Let h denote the perpendicular distance betweenthe centre of mass of the wedge and the line ofaction of F. Find the magnitude of the torque due

to the normal force Nr

about the centre of the

wedge during the interval ∆t.

13. A cylindrical solid of mass 10–2 kg and cross sectionalarea 10–4 m2 is moving parallel to its axis (the x-axis)with a uniform speed of 103 m/s in the positive direction.At t = 0, its front face passes the plane x = 0. Theregion to the right of this plane is filled with stationarydust particle of uniform density 10–3kg/m3. When adust particles collides with the face of the cylinder, itsticks to its surface. Assuming that the dimensions ofthe cylinder remains practically unchanged and thatthe dust sticks only to the front face of the x-coordinates of the front of the cylinder find the x-coordinate of the front of the cylinder at t = 150 s.

(1998; 5M)

14. Two blocks of mass 2 kg and M are at rest on aninclined plane and are separated by a distance of 6.0m as shown. The coefficient of friction between eachblock and the inclined plane is 0.25. The 2 kg block isgiven a velocity of 10.0 m/s up the inclined plane. Itcollides with M, comes back and has a velocity of 1.0m/s when it reaches its initial position. The other blockM after the collision moves 0.5 m up and comes to rest.Calculate the coefficient of restitution between theblocks and the mass of the block M. (1999; 10M)[Take sin θ ≈ tan θ = 0.05 and g = 10 m/s2]

θ

6.0m

2kg

M

15. A car P is moving with a uniform speed of 5 3 m/stowards a carriage of mass 9 kg at rest kept on the railsat a point B as shown in figure. The height AC is 120m. Cannon balls of 1 kg are fired from the car with aninitial velocity 100 m/s at an angle 30° with thehorizontal. The first cannon balls hits the stationarycarriage after a time t0 and sticks to it. Determine t0.At t0, the second cannon ball is fired. Assume that theresistive force between the rails and the carriage isconstant and ignore the vertical motion of the carriagethroughout. If the second ball also hits and sticks tothe carriage, what will be the horizontal velocity of thecarriage just after the second impact? (2001; 10M)

43

P

A

C

B

16. A particle of mass m, moving in a circular path ofradius R with a constant speed v2 is located at point(2R, 0) at time t = 0 and a man starts moving with avelocity v1 along the positive y-axis from origin at timet = 0. Calculate the linear momentum of the particlew.r.t. man as a function of time. (2003; 2M)

(0,0)

R

y

v1

v2

mx

17. Two point masses m1 and m2 are connected by aspring of natural length l0. The spring is compressedsuch that the two point masses touch each other andthen they are fastened by a string. Then the system ismoved with a velocity v0 along positive x-axis. Whenthe system reaches the origin the string breaks (t = 0).The position of the point mass m1 is given by x1 = v0t – A (1 – cos ω t) where A and ω are constants. Findthe position of the second block as a function of time.Also find the relation between A and l0. (2003; 4M)

18. There is a rectangular plate of mass M kg of dimensions(a × b). The plate is held in horizontal position bystriking n small balls each of mass m per unit area perunit time. These are striking in the shaded all region ofthe plate. The balls are colliding elastically with velocityv. What is v?

b

a

It is given n = 100, M = 3 kg, m = 0.01kg;b = 2m; a = 1m; g = 10 m/s2. (2006; 6M)

19. Two towers AB and CD are situated a distance d apart asshown in figure. AB is 20 m high and CD is 30 m high fromthe ground. An object of mass m is thrown from the topof AB horizontally with a velocity of 10m/s towards CD.

m

60°2m

C

DB

A

d

Simultaneously another object of mass 2m is thrownfrom the top of CD at an angle of 60° to the horizontaltowards AB with the same magnitude of initial velocityas that of the first object. The two objects move in thesame vertical plane, collide in mid air and stick to eachother. (1994; 6M)(i) Calculate the distance d between the towers.(ii) Find the position where the object hit the ground.

20. Three objects A, B and C are kept in a straight line ona frictionless horizontal surface. These have masses m,2m and m, respectively. The object A moves towardsB with a speed 9 m/s and makes an elastic collisionwith it. Thereafter, B makes completely inelastic collisionwith C. All motions occur on the same straight line.Find the final speed (in m/s) of the object C.

m 2m m

A B C

ASSERATION AND REASON

This question contains, statement I (assertion) andstatement II (reasons).

1. Statement-I : In an elastic collision between two bodiesthe relative speed of the bodies after collision is equalto the relative speed before the collison.

(2007; 3M)Because :Statement-II : In an elastic collision, the linearmomentum of the system is conserved.(a) Statement-I is true, statement -II is true, statement-

II is a correct explanation for statement-I.(b) statement-I is true, statement-II is true; statement-

II is NOT a correct explanaion for statement-I.(c) statement-I is true, statement-II is false.(d) statement-I is false, statement-II is true.

44

2. Statement-I : If there is no external torque on a bodyabout its centre of mass, then the velocity of the centreof mass remains constant. (2007; 3M)Because :Statement-II : The linear momentum of an isolatedsystem remains constant.(a) Statement-I is true, statement -II is true, statement-

II is a correct explanation for statement-I(b) statement-I is true, statement-II is true; statement-

II is NOT a correct explanaion for statement-I(c) statement-I is true, statement-II is false.(d) statement-I is false, statement-II is true.

PASSAGE BASED PORBLEM

Passage

A small block of mass M moves on a frictionlesssurface of an inclined plane, as shown in figure. Theangle of the incline suddenly changes from 60º at pointB. the block is initially at rest at A. Assume thatcollisions between the block and the incline are totallyinelastic (g = 10 m/s2)

B60º

30º

A

C

v

3 m 3 m3

1. The speed of the block at point B immediately after itstrikes the second incline is(2008; 4M)

(a) m/s60 (b) m/s45

(c) m/s30 (d) m/s15

2. The speed of the block at point C immediately beforeit leaves the second incline is(2008; 4M)

(a) m/s120 (b) m/s105

(c) m/s90 (d) m/s75

3. If collision between the block and the incline iscompletely elastic, then the vertical (upward) componentof the velocity of the block at point B, immediatelyafter it strikes the second incline is(2008; 4M)

(a) m/s30 (b) m/s15

(c) zero (d) m/s75

ANSWERS

FILL IN THE BLANKS

1. 232

mv 2. 5 × 10–3 N-s

TRUE/FALSE

1. F

OBJECTIVE QUESTION (ONLY ONE OPTION)

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

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)

1. (b, d) 2. (a), (d)

45


1. 9 cm from centre of bigger circle (leftwards) 2. 10 2 m/s at 45°

3. Velocity of C is V in a direction opposite to velocity of B.

4.( – ) 2 ( – )

,( )+ +

m R r g R rm

M m M M m 5. (i) 03

V (ii)

20

20

2

3

mV

x6. 6.53 s 7. 4 8. 44.25 m

9. (a) 5

64

µgd (b) 6 3d µ 10. (i) Straight line (ii) 2 2

21

–2

+ =

x yL rr

, ellipse

11. (L + 2R, 0) 12. (a) kmgt

mvkit

mv ˆ3

2),ˆˆ3(3

2

+

∆−

∆ (b) h

t

mv

∆3

4

13. 105 m 14. e = 0.84 M, = 15 kg 15. 12s, 15.75 m/s

16. jvtRv

vmitRv

mv ˆcosˆsin 12

22

2

−+

− 17. A

mmltA

mmtvx

+=−+= 1),?cos1(

2

10

2

102 18. 10 m/s


1. (b) 2.(d)

PASSAGE BASED PROBLEM

1. (b) 2. (b) 3. (c)

SOLUTIONS

FILL IN THE BLANKS

1. From conservation of linear momentum magnitude of

3Pr

should be 2mv in a direction opposite of 12Pr

(resultant of 1Pr

and 2Pr

).

∴ Speed of particle of mass 2m 22

)( 3 vm

Pv ==′

P3

P=mv2P =12

P = mv1

2 mv

∴ Total energy released is,

E ( )2 '21 12 2

2 2 = +

mv m v

22

2 = +

vmv m

232

= mv

2. Impulse ∫= Fdt = area under F-t graph

∴ Total impulse from t = 4 µs to t = 16 µs= Area EBCD= Area of trapezium EBCF +Area of triangle FCD

–6 –61 1(200 800)2 10 800 10 10

2 2= + × + × × ×

= 5 × 10–3 N-s

TRUE FALSE

1. Since, net force on the system is zero. Velocity ofcentre of mass will remain constant.

OBJECTIVE QUESTIONS (ONLY ONE OPTION)

1. Net force on system is zero. So 0=∆ COMVr

and

46

because initially centre of mass is at rest. Therefore,centre of mass always remains at rest.∴ (d)

2. In an inelastic collision only momentum of the systemremain conserved. Some energy is always lost in theform of heat, sound etc.∴ (c)

3. Let v' be the velocity of second fragment. Fromconservation of linear momentum,2m (v cos θ) = mv' – m (v cosθ)∴ v'= 3 v cos θ∴ (a)

4. Before explosion, particle was moving along x-axis i.e.,it has no y-component of velocity. Therefore, thecentre of mass will not move in y-direction or we cansay yCM = 0.

Now, m

ymm

mmymym

yCM

2

21

2211 4315

40×+×

=⇒++

=

or y2 = – 5cm∴ (a)

5. |)()(| 22112211 vmvmvmvmrrrr

+−′+′

= |)2()2(| 2211022011 vmvmgtvmgtvm −−−+−= 2 (m1 + m2) gt0∴ (c)

6. vCM 21

2211

mmvmvm

++

=rr

1014

14101410

041410 =×=+

×+×= m/s

B

v =14m/s1 v = 02

m = 4kg2m = 10kg1

7. jktkAiktkAdtPdF ˆ)cos(ˆ)sin( −−==rr

jktAiktAP ˆ)sin(ˆ)cos( −=r

Since, 0. =PFrr

∴ Angle between Fr

and Pr

should be 90°.∴ (c)

8. Let first collision be at an angle θ,

∴ vr

vr

2)2( θ−π=θ

θ−π=θ 22

°=π=θ 1203

2

Av2v

m2m1

B C

θθO

∴ After first collision at B, m2 will move back withspeed v and make collision with m1 at C.

(again at 32π=θ anticlockwise from OB)

Now, again m1 will move back with speed v and

meet m2 at A(at 32π=θ anticlockwise from OC)

∴ (c)

9. 106)()0()()()0(6 a

mmmmmammamammycm =

++++−++++=

x

y

(– ) ( , )a, a a a

(0, 0)(0, – )a

∴ (a)


1. After collisions between C and A, C stops while Amoves with speed of C i.e. v [in-head on elasticcollision, two equal masses exchange their velocities].At maximum compression, A and B will move withsame speed v/2 (from conservation of linear momentum).Let x be the maximum compression in this position.∴ KE of A-B system at maximum compression

21(2 )

2 2v

m = or, Kmax = mv2/4From conservation fo mechanical energy in twopositions shown in figure.

or, 222

21

41

21

kxmvmv +=

22

41

21

mvkx =

47

∴k

mvx2

=

∴ (b, d)

2. Initial momentum of the system 0pp 21 =+rr

∴ Final momentum 0'p'p 21 =+rr

should also be zero.

Option (b) is allowed because if we put 021 ≠−= cc

21 'p'prr

+ will be zero. Similary, we can check other

options.∴ correct options are (a) and (d).


1. comRr ( )2

2 2

427

[(56) –(42) ]

π= ×

π)ˆ( i−

= 9 cm

2. From conservation of linear momentum 3Puur

should be

equal and opposite to 12Puuur

(resultant of 1Puur

and 2Puur

).

So, let v' be velocity of third fragment, then

(3m)v' = 2mv

∴ v' = 2

3v

P3

P =mv2

P =12

P = mv1

2 mv

→

→

→

→45°45°

Here, v =30 m/s (given)

∴ v' =2 30

3×

= 210 m/sThis velocity is at 45° as shown in figure.

3. Before collision net momentum of the system was zero.No external force is acting on the system. Hencemomentum after collision should also be zero. A hascome to rest. Therefore B and C should have equaland opposite momentum or velocity of C should be Vin opposite direction of velocity of B.

4. (i) The centre of mass of M + m in this case will notmove in horizontal direction. For centre of mass not to

move along horizontal we should haveMx = m (R – r – x)

mMr)m(Rx

+−=

(ii) Let V1 be the speed of m towards right and v2 thespeed of M towards left. From conservation fo linearmomentum.

mv1 = Mv2 ...(1)From conservation of mechanical energy

mg (R – r) 2 21 2

1 12 2

mv Mv= + ...(2)

Solving these two equations, we get

)()(2

2 mMMrRg

mv+−

=

5. (i) Collision between A and C is elastic and mass ofboth the blocks is same. Therefore, they will exchangetheir velocities i.e., C will come to rest and A will bemoving will velocity vo. Let V be the common velocityof A and B, then from conservation of linear momentum,we have

C A B

(a)

v0

C A B

(b)

v0

At rest

C A B

(c)

V V

mAv0 = (mA + mB) Vor mv0 = (m + 2m) V

or V = 30v

(ii) Further, from conservation of energy we have,

202

1vmA = 2 2

01 1

( )2 2A Bm m V kx+ +

or2

021

mv =2

200

1 1(3 )

2 3 2V

m kx +

or 20

12

kx 203

1 mv=

or 20

20

32

xmvk =

6. For collision of both the balls,

22 8.921

988.921

49 ttt ×−=×−

48

∴ t = 2 sec., At this instant height from ground

= m 4.788.921

49 2 =×− tt

and using conservation of linear momentum, we get,

mvvmvm 22211 =− ,

Solving this, we get m/sec 9.4=v

∴ 2

21

atutS +=

⇒ 29.49.44.78 tt −=−⇒ .sec 53.4=tTotal time of flight = 4.53 + 2 = 6.53 sec.

7. As shown in figure initially when the bob is at A, itspotential energy is mgl. When the bob is released itstrikes the wall at B. If v be the velocity with which thebob strikes the wall, then

h

O l A

C

θn

B

mgl = 212

mv or glv 2= ...(1)

Speed of the bob after rebounding (first time)

Similarly, glev 21 = ..(2)

The speed after second rebound is v2 = e2 (2 )glIn general after n rebounds, the speed of the bob is

vn = en (2 )gl ..(3)

Let the bob rises to a height h after n rebounds.Applying the law of conservation of energy, we have

212 nmv = mgh

or h = leg

gleg

v nn

n .2

2.2

222

==

∴ h22 4

.55

n nl l

= =

...(4)

at ,60? °=

2)60cos1(

llh =°−=

∴2

60?l

h <⇒°<

or 21

54

254

<

⇒<

nn ll

∴ 3>n (here n is an integer)

8. Let v1 and v2 be the velocities after explosion in thedirections shown in figure. From conservation of linearmomentum, we have

5 (20 cos 60°) = 4v1 – 1 × v2or 4v1 – v2 = 50 ...(1)Further it is given that, kinetic energy after explosionbecomes two times. Therefore,

21

14

2v× × + 2

21

12

v× ×

=21

2 5 (20cos60 )2

× × °

or 2 21 24v v+ =1000 ...(2)

Solving Eqs. (1) and (2), we havev1 = 15 m/s, v2 = 10 m/s

or v1 = 5 m/s and v2 = – 30 m/sIn both the cases relative velocity of separation inhorizontal direction is 25 m/s.∴ x = 25t = distance between them when they

strike the ground

Here, t 2T

= = sin 20sin60

9.8u

gθ °

= = 1.77 s

∴ x = 25 × 1.77 = 44.25 m

9. If v1 and v2 are the velocities of object of mass m andblock of mass 4m, just after collision then byconservation of momentum,mv = mv1 + 4mv2, i.e. v = v1 + 4v2 ...(1)Further, as collision is elastic

2 2 21 2

1 1 14

2 2 2mv mv mv= + , i.e. 2 2 2

1 24v v v= + ...(2)

Solving, these two equations we get either

v2 = 0 or v225

v=

Therefore, v225

v=

Substituing in Eq. (1) v135

v=

when v2 = 0, v1 = v, but it is physically unacceptable.(a) Now, after collision the block B will start movingwith velociy v2 to the right. Since, there is no friction

49

between blocks A and B the upper block A wil stay atits positions and will topple if B moves a distance ssuch that

s > 2d ...(3)However, the motion of B is retarded by frictional forcef = µ (4m + 2m)g between table and its lower surface.So, the distance moved by B till it stops

22

60 – 2

4µmg

v sm

= i.e.,

22

3v

sµg

=

Substituting this value of s in Eq. (3), we find that fortoppling of A

22 6v µgd> ⇒

2 65

v µgd> ⇒ 56

2v µgd>

or gdvv µ625

0min ==

(b) if v = 2v0 5 6µgd= , the object will rebound with

speed.

v1 =3 3 65

v µgd=

and as time taken by it to fall down

t2 2h dg g

= = [as h = d]

The horizontal distance moved by it to the left of P inthis time

µ361 dtvx ==

Note :(a) Toppling will take place if line of action of weight does

not pass through the base area in contact.(b) v1 and v2 can obtained by using the equations of head

on elastic collision

11 2 uvv cmrrr

−=

22 2 uvv cmrrr

−=

10. (i) Since, only two forces are acting on the rod, itsweight Mg (vertically downwards) and a normalreaction N at point of contact B (vertically upwards).No horizontal force is acting on the rod (surface issmooth).

B

A

C

B

A

C

O M

P

yθ

x

(x, y)

(O, L/2 sin )θ

(a)(b)

Therefore, CM will fall vertically downwards towardsnegative y-axis i.e. the path of CM is a straight line.(ii) Refer figure (b). We have to find the trajectory ofa point P (x, y) at a distance r from end B. CB = L/2, OB = (L/2) cos θ, MB = r cos θ∴ x = OB – MB = cos θ (L/2 – r)

or )/(cos

rRLx

−=θ ...(1)

Similarly, y = r sin θ

or, sin θ = yr

...(2)

Squaring and adding Eqs. (1) and (2), we get

sin2θ + cos2θ = 2 2

2 2 2( /2) –

x y

L r r+

or, 1)2/( 2

2

2

2

=+− r

yrL

x...(3)

This is an equation of an ellipse. Hence, path of pointP is an ellipse whose equation is given by (3).

11. Since, all the surfaces are smooth, no external force isacting on the system in horizontal direction. Therefore,the centre of mass of the system in horizontal directionremains stationary.

yCC = 5R (in both cases)1 2

c1 c2

(L,0) (L + 5R,0)x

Initial

y

c1c2

(x,0)x

Finalc = (x – 5R, 0)2

x- coordinate of CM initially will be given by

x1 =1 1 2 2

1 2

m x m xm m

++

)(4

)5())(4(RL

MMRLMLM

+=+

++= ...(1)

Let (x, 0) be the coordinates of the centre of largesphere in final position. Then, x-coordinate of CMfinally will be

)(4

)5())(4(1 Rx

MMRxMxM

x −=+

−+= ...(2)

Equating Eqs. (1) and (2), we have x = L + 2RTherefore, coordinates of large sphere, when thesmaller sphere reaches the other extreme position, are(L + 2R, 0)

50

12. (a) (i) Since, the collision is elastic and sphere is fixed,

the wedge will return with velocity of ivˆz

y

x

–v i +v i

Fixed F

F sin 30°

Now, linear impulse in x-direction= change in momentum in x-direction.

∴ (F cos 30°)∆t=mv – (– mv) = 2mv

∴ F =2cos30

mvt∆ °

4

3

mv

t=

∆

F = 4

3

mv

t∆

∴ kFiFF ˆ)30sin(ˆ30cos °−°=r

or kt

mvit

mvF ˆ3

2ˆ2

∆−

∆

=r

(ii) Taking the equilibrium of wedge in vertical

F sin 30°

N

C

mg

z- direction during collision.N = mg + F sin 30°

N = 2

3

mvmg

t+

∆or in vector form

kt

mvmgN ˆ3

2

∆+=

r

(b) For rotational equilibrium of wedge [about (CM)]

N

C

mg

h

F

Magnitude of torque of N about CM = magnitude oftorque of F about CM

= F.h

ht

mvN

∆=

3

4|t|r

13. Given : m0 = 10–2 kg, A = 10–4 m2, v0 = 103 m/s

m0A vo

x=0

At t = 0

m

v

x

At t = t

and dustρ = ρ = 10 –3 kg/m3

m = m0 + mass of dust collected so far

= m0 + Ax dustρ

or m = m0 + Ax ρ

The linear momentum at t = 0 isP0 = m0v0

and momentum at t = t isPt = mv = (m0 + Ax ρ )v

From law of conservation of momentumP0 = Pt

∴ m0v0 = (m0 + Ax ρ ) v (but v dxdt

= )

∴ m0v0 = (m0 + Ax ρ )dxdt

=

or (m0 + A ρ x)dx = m0v0 dt

or

2

0∫ (m0 + A ρ x)dx =

150

0∫ m0v0dt

⇒ [ ]150000

0

2

0 2? tvm

xAxm

x

=

+

Hence,2

0 2x

m x A+ ρ = 150 m0v0

Solving this quadratic equation and substituting thevalues of m0, A, ρ and v0, we get, x = 105m.

14. Let v1 = velocity of block 2 kg just before collision.v2 = velocity of block 2 kg just after collision.and v3 = velocity of block M just after collision.Applying work energy theorem,(change in kinetic energy = work done by all the forcesat different stages as shown in figure.

51

θ

10 m/s

2kg

M

θ = 6 sin θh1

v1

6 m

Figure I.For 2 kg block between starting point and the momentjust before collision

∆KE = Wfriction + Wgravity

θ−θµ−=− sin6cos6)10(21 22

1 mgmgvm

Given, ,05.0?tan?sin,25.0µ =≈=

∴ 1?cos ≈

∴ ]05.010611025.06[210021 ××+×××−=−v

643610021 =−=v

∴ 81 =v m/s

Figure 2.For 2 kg block, just after collision and at the initialpoint


v21m/s

?sin6?cosµ6)])1[(21 2

22 mgmgvm +−=−

or 241 22 −=−v

∴ ,2522 =v or 52 =v m/s

Figure. 3.For block M, Just after collision and at the point ofstop,


?sin5.0?cos))(µ)(5.0(]0[21 2

3 MggMvM −−=−

05.0105.0)11025.05.0(23 ××+×××=v

35.1223 =×=v

75.13 =v m/s

θ

2kg

M

θ h= 0.5 sin 2 θ

V1

0.5

V3

(i) Coefficient of restitution (e)

=Relativevelocityof separationRelativevelocityof approach

= 2 3

1

v vv+

84.08

73.15 =+=

or e = 0.84(ii) Applying conservation of linear momentum beforeand after collision

2v1 = Mv3 – 2v2

∴ 73.126

73.1)58(2)(2

3

21 =+

=+

=v

vvM

M ≈ 15 kg

15. (i) Velocity of the ball relative to ground = 100 m/s

v = 5 3m/s

A

C

120m

uy

30°

u = 100m/s

uxy (horizontal)

y (vertical)

a = 0 andx

a = – g = – 10 m/sy2

Horizontal component of its velocity.

ux =u cos 30° = 50 3 m/s

and vertical component of its velocity,uy = u sin 30° = 50 m/s

Vertical displacement of the ball when it strikes thecarriage is –120 m or

Sy = uy 21

2 yt a t+ ⇒ –120 = (50 t)21

(–10)2

t + ⇒ t2 – 10t – 24 = 0∴ t = 12 s or – 2 sIgnoring the negative time, we have∴ t0 = 12 s

(ii) When it strikes the carriage, its horizontal component

of velocity is still 50 3 m/s. It stickes to the carriage.

52

Let V2 be the velocity of (carriage + ball) system aftercollision. Then, applying conservation of linearmomentum in horizontal direction :

∴ 2)91(3501 V+=×

∴ V2 = 5 3 m/sThe second ball is fired when the first ball strikes thecarriage i.e. after 12 s. In these 12 s the car will move

forward a distance of 12V1 or 60 3 m.The second ball also takes 12 s to travel a verticaldisplacement of – 120 m. This ball will strike thecarriage only when the carriage also covers the same

distance of 60 3 m in these 12s. This is possible only

when resistive forces are zero because velocity of car(V1) = Velocity of carriage after first collision (V2) =

5 3 m/s.

Hence, at the time of second collision :Let V be the by velocity of carriage after secondcollision. Conservation of linear momentum in horizontaldirection gives

1kg 5 3m/s

Before collision

µx = 50 3 m/s

11 V = (1) (50 3 ) + (10) (5 3 ) = 100 3

∴ 100 3

11V = m/s or V = 15.75 m/s

16. Angular speed of particle about centre of the circle

2 ,vR

ω = 2vt t

Rθ = ω =

θ

yv 2 v2

xt=0

R

v 1

(0, 0)

)ˆ?cosˆ?sin( 22 jvivv p +−=r

or

+−= jt

Rv

vitRv

vv pˆcosˆsin 2

22

2r

and jvvmˆ

1=r

∴ Linear momentum of particle w.r.t man as afunction of time is

)( mppm vvmprrr

−=

−+

−= jvtRv

vitRv

vm ˆcosˆsin 12

22

2

17. (i) x1 = v0 t – A (1 – cos ωt)

tvmm

xmxmxCM 0

21

2211 =++

=

(As there is no external force, so COM will movewith constant velocity v0)

∴ x21

02

(1–cos )m

v t A tm

= + ω

(ii) a1 = 2

212

–d x

dt= ω A cos ωt

The separation x2 – x1 between the two blocks will beequal to l0 and a1 = 0 or cos ωt = 0

x2 – x1 1

2

mm

= A (1 – cos ωt) + A (1 – cos ωt)

or, l0 = 1

21m A

m

+

(1 - cos ω t), (cos ω t = 0)

Thus, the relation between l0 and A is,

Amm

l

+= 1

2

10

18. (2 )2

p bF n a mv

t∆ = = × × × ∆

Equating the torque about hinge side, we have,

3(2 )

2 4 2b b b

n a mv Mg × × × × = × Substituting the given values we get,

v = 10 m/s

19. Acceleration of A and C both is 9.8 m/s2 downwards

53

30m

10m

20m

A

B D

C

Ed

V= V

– V

CA

C

A

→

→

→X

Y

V C

→

60°θ

θ vA

→

Let A be the origin, so

22

21 2

13510&

21

gttygty −−=−−=

For collision 21 yy =

⇒ t3510 =

sec3

2=t

⇒ m 32.173

30510 ==+= ttd

Horizontal (or x) component of initial linear momentumof projectile m and 2m,

PAx + PCx = 0)60cos(m2 =°− CA vmvi.e., x-component of momentum of combined mass aftercollision will also be zero i.e., the combined mass willhave momentum or velocity in vertical or y-directiononly.

d1d2

B F D

Cv = vcos60°cx c

v = vax A

Pt = 2

3s

A

V0 60°

Hence, the combined mass will fall at point F justbelow the point of collision P.

Here, d1 = VAxt = (10) × 3

2 = 11.55 m

∴ d2 = (d – d1)= (17.32 – 11.55)m = 5.77 m

Therefore, position from B is d1 i.e., 11.55 m and fromD is d2 or 5.77 m.

20. Let after collision velocity of block A and B be vA andvB respectively.

BCOMB uevevrrr

−+= )1(

m/sec 6039

)11( =−

+=

mm

CCOMC uevevrrr

−+= )1(

m/sec 403

62)1( =−

×

=m

m


1. In case of elastic collision, coefficient of restitution e = 1.or relative speed of approach= relative speed of separation.∴ (b)

2. If a force is applied at centre ofmass of a rigid body, its torqueabout centre of mass will be zerobut acceleration will be non-zero.Hence, velocity will change.

F→

C

∴ (d)

PASSAGE BASED PROBLEM

1. Between A and B, height fallen by block

m3º60tan31 ==h

∴ 111 ms6031022 −=××== ghv

In perfectly inelastic collision, component of v1perpendicular to BC will become zero, while componentof v1 parallel to BC will remain unchanged.∴ speed of block B immediately after it strikes thesecond incline is,

v2 = component of v1 along BC

= v1 cos 30º = 1ms)45(

23

)60( −=

54

∴ correct option is (b).

2. Height fallen by the block from B to C

m3º30tan332 ==hLet v3 be the speed of block, at point C, just before itleaves the second incline, then :

2223 2 ghvh +=

1ms1053245 −=×+=∴ correct option is (b).

3. In elastic collistion, component of v1 parallel to BC willremain unchanged, while component perpendicular toBC will remain unchanged in magnitude but its directionwill be reversed.

v11 = v1cos 30º = 1ms45

23

)60( −=

⊥v = v1 sin 30º = 1ms15

21)60( −=

Now vertical component of velocity of block :

0º60cosº30cos 11 =−= ⊥ vvv∴ (c)

81

CHAPTER-7GRAVITATION

FILL IN THE BLANKS1. The numerical value of the angular velocity of rotation of the earth shold be .... rad/s in order to make the effective

acceleration due to gravity at equator equal to zero. (1984, 2M)2. According to Kepler's second law, the radius vector to a planet from the sun sweeps out equal areas in equal

intervals of time. This law is a consequence of the conservation of ..... (1985, 2M)3. A geostationary satellite is orbiting the earth at a height of 6 R above the surface of the earth where R is the radius

of earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is .... hours.(1987, 2M)

4. The masses and radii of the Earth and the Moon are M1, R1 and M2, R2 respectively. Their centres are at distanced apart. The minimum speed with which a particle of mass m should be projected from a point midway between thetwo centres so as to escape to infinity is .... (1988, 2M)

5. The ratio of earth's orbital angular momentum (about the sun) to its mass is 4.4 × 1015 m2/s. The area enclosed byearth's orbit is approximately ....... m2. (1997C, 1M)

6. A particle is projected vertically upwards from the surface of earth (radius R) with a kinetic energy equal to halfof the minimum value needed for it to escape. The height to which it rises above the surface of earth is ....

(1997, 2M)TRUE/FALSE1. It is possible to put an artificial satellite into orbit in such a way that it will always remain directly over New Delhi.

(1984; 2M)

OBJECTIVE QUESTIONSOnly One option is correct :1. If the radius of the earth were to shrink by one per

cent, its mass remaining the same, the acceleration dueto gravity on the earth's surface would :(1981; 2M)(a) decrease (b) remain unchanged(c) increase (d) be zero

2. If g is the acceleration due to gravity on the earth'ssurface, the gain in the potential energy of an objectof mass m raised from the surface of the earth to aheight equal to the radius R of the earth, is :

(1983; 1M)

(a)12

mgR (b) 2 mgR

(c) mgR (d)14

mgR

3. Imagine a light planet revolving around a very massivestar in a circular orbit of radius R with a period ofrevolution T. If the gravitational force of attraction be-tween the planet and the star is proporional to R–5/2,then : (1989; 2M)(a) T2 is proportional to R2

(b) T2 is proportional to R7/2

(c) T2 is proportional to R3/2

(d) T2 is proportional to R3.75

4. If the distance between the earth and the sun were halfits present value, the number of days in a year wouldhave been : (1996; 2M)

(a) 64.5 (b) 129(c) 182.5 (d) 730

5. A statellite S is moving in an elliptical orbit around theearth. The mass of the satellite is very small comparedto the mass of the earth : (1998; 2M)(a) the acceleration of S always directed towards the

centre of the earth(b) the angular momentum of S about the centre of the

earth changes in direction, but its magnitude re-main constant

(c) the total mechanical energy of S varies periodi-cally with time

(d) the linear momentum of S remains constant inmagnitude

6. A simple pendulum has a time period T1 when on theearth's surface and T2 when taken to a height R abovethe earth' surface, where R is the radius of the earth.The value of T2/T1 is : (2001)

(a) 1 (b) 2(c) 4 (d) 2

7. A geostationary satellite orbits around the earth in acircular orbit of radius 36,000 km. Then, the time periodof a spy satellite orbiting a few hundred km above theearth's surface (Re = 6400 km) will approximately be :

(2002)(a) 1/2h (b) 1 h(c) 2 h (d) 4 h

82

8. A double star system consists of two stars A and Bwhich have time period TA and TB. Radius RA and RBand mass MA and MB. Choose the correct option.

(2006; 3M)(a) If TA > TB then RA > RB(b) If TA > TB then MA > MB

(c)

2 3A A

B B

T RT R

=

(d) TA = TB

9. A spherically symmetric gravitational system of par-

ticles has a mass density

>≤ρ

=ρRrRr

for0for0

where ρ0 is

a constant. A test mass can undergo circular motionunder the influence of the gravitational field of par-ticles. Its speed v as a function of distance r from thecentre of the system is represented by (2008; 3M)

OBJECTIVE QUESTIONSMore than one options are correct?1. A solid sphere of uniform density and radius 4 units

is located with its centre at the origin O of coordinates.Two spheres of equal radii 1 unit, with their centres atA (– 2, 0, 0) and B (2, 0, 0) respectively, are taken outof the solid leaving behind spherical cavities as shownin figure. Then : (1993; 2M)

A Bm

O

z

y

x

(a) the gravitational field due to this object at theorigin is zero.

(b) the gravitational field at the point B (2, 0, 0) is zero

(c) the gravitational potential is the same at all pointsof circle y2 + z2 = 36

(d) the gravitational potential is the same at all pointson the circle y2 + z2 = 4

2. The magnitude of the gravitational field at distance r1and r2 from the centre of a uniform sphere of radius Rand mass M and F1 and F2 respectively. Then :

(1994; 2M)

(a) 1 1

2 2

F rF r

= if r1 < R and r2 < R

(b) 2

1 22

2 1

F rF r

= if r1 > R and r2 > R

(c) 3

1 13

2 2

F rF r


(d) 2

1 12

2 2

F rF r


SUBJECTIVE QUESTIONS1. Two satellite S1 and S2 revolve round a planet in

coplanar circular orbits in the same sense. Their peri-ods of revolution are 1 h and 8 h respectively. Theradius of the orbit of S1 is 104 km. When S2 is closestto S1. Find : (1986; 6M)(i) the speed of S2 relative to S1.(ii) the angular speed of S2 as actually observed by an

astronaut in S1.

2. Three particles, each of mass m, are situated at thevertices of an equilateral triangle of side length a. Theonly forces acting on the particles are their mutualgravitational forces. It is desired that each particlemoves in a circle while maintaining the original mutualseparation a. Find the initial velocity that should begiven to each particle and also the time period of thecircular motion. (1988; 5M)

3. An artificial satellite is moving in a circular orbitaround the earth with a speed equal to half the mag-nitude of escape velocity from the earth (1990; 8M)(i) Determine the height of the satellite above the

earth's surface.(ii) If the satellit is stopped suddenly in its orbit and

allowed to fall freely onto the earth, find the speedwith which it hits the surface of the earth.

4. Distance between the centre of two stars is 10a. Themasses of these stars are M and 16 M and their radiia and 2a respectively. A body of mass m is firedstraight from the surface of the larger star towards thesurface of the smaller star. What should be its mini-mum initial speed to reach the surface of the smallerstar? Obtain the expression in terms of G, M and a.

(1996; 5M)

83

ANSWERS

FILL IN THE BLANKS

1. 1.24 × 10–3 2. angular momentum 3. 8.48 4. 1 22 ( )Gv M Md

= +

5. 6.94 × 1022 6. h = R

TRUE/FALSE1. F 2

OBJECTIVE QUESTION (ONLY ONE OPTION)1. (c) 2. (a) 3. (b) 4. (b) 5. (a) 6. (d) 7. (c)8. (d) 9. (c)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)1. (a, c, d) 2. (a, b)

SUBJECTIV QUESTIONS

1. (i) – π × 104km/h (ii) 3 × 10–4 rad/s 2.3

, 23

Gm av Ta Gm

= = π

3. (i) 6400 km (ii) 7.9 km/s

4. 3 5

2Gm

a5. 99.5R

ASSERTION AND REASON1. (a)

5. There is a crater of depth 100R

on the surface of the

moon (radius R). A projectile is fired vertically upwardfrom the crater with velocity, which is equal to theescape velocity v from the surface of the moon. Findthe maximum height attained by the projectile.


This question contains, statement I (assertion) andstatement II (reasons).

1. Statement-I : An astronaut in an orbiting space stationabove the earth experiences weightlessness.

(2008; 3M)

Because :Statement-II : An object moving around the earthunder the influence of earth's gravitational force is ina state of 'free-fall'.(a) Statement-I is true, statement -II is true, statement-

II is a correct explanation for statement-I.(b) statement-I is true, statement-II is true; statement-

II is NOT a correct explanaion for statement-I.(c) statement-I is true, statement-II is false.(d) statement-I is false, statement-II is true.

(2003; 4M)

84

FILL IN THE BLANKS

1. g' = g – Rω2 cos2φAt equator φ = 0∴ g' = g – Rw2

0 = g – Rw2

∴ ω =gR 3

9.8

6400 10=

×

= 1.24 × 10– 3 rad/s

2. 2dA Ldt M

= = constant because angular momentum of

planet (L) about the centre of sun is constant.Thus, this law comes from law of conservation ofangular momentum.

3. T ∝ r3/2

∴2

1

TT =

3 / 22

1

rr

or 2/3

1

2/3

1

22 7

5.3

=

=

RRT

rrT (24)h

= 8.48h

M1M2

m

d/2 d/2

4. Total mechanical energy of mass m at a point midwaybetween two centres is :

1 2– –/ 2 / 2

GM m GM mEd d

= 1 22– ( )Gm M M

d= +

Binding energy 1 22 ( )Gm M M

d= +

Kinetic energy required to escape the mass to infinityis,

212 emv = 1 2

2 ( )Gm M Md

+

∴ ve = 1 2( )2 G M Md+

5. Areal velocity of a planet round the sun is constantand is given by

dAdt = 2

Lm

where L = Angular momentum of planet (earth) aboutsun and m = mass of planet (earth).

Given : Lm = 4.4 × 1015 m2/s

∴ Area enclosed by earth in time T (365 = days) willbe

Area . .2

dA LT Tdt m

= =

13 214.4 10 365 24 3600

2m= × × × × ×

Area = 6.94 × 1022 m2.

6.

=

RGM

V2

21

From conservation of energy

)(42

21

hRGMm

RGMm

RGM

m+

−=−

)(4 hRGMm

RGMm

RGMm

+−=−

hRRR +−=−

1141

hRR +−=

− 14

41

RhR 4)(3 =+

3R

h =

TRUE FALSE1. New Delhi is not on the equatorial plane.


1. g = 2GM

R

or g ∝ 21

Rg will increase if R decreases.

2. ∆U =1

mghhR

+

Given, h = R

∆U =121

mgRmgRR

R

=+

SOLUTIONS

85

3.2mv

R∝ R–5/2

∴ v ∝ R–3/4

Now, T =2 R

vπ

or T2 ∝2R

v

or T2 ∝2

–3/4R

R

or T2 ∝ R7/2

4. From Kepler's third lawT2 ∝ r3 orT ∝ (r)3/2

∴2

1

TT =

3 / 22

1

rr

or T2 = T1

3 / 22

1

rr

=3 / 21

(365)2

T2 = 129 days

5. Force on satellite is always towards earth, therefore,acceleration of satellite S is always directed towardscentre of the earth. Net torque of this gravitationalforce F about centre of earth is zero.Therefore, angularmomentum (both in magnitude and direction) of Sabout centre of earh is constant throughout. Since, theforce F is conservative in nature, therefore mechanicalenergy of satellite remains constant. Speed of S ismaximum when it is nearest to earth and minimumwhen it is forthest.

6. T ∝ 1g

i.e. 2

1

1

2

gg

TT =

where g1 = acceleration due to gravity at a height h =R from earth's surface = g/4

22

1Using ( ) 2

/ 41

Tg gg h

T ghR

= = = +

7. Time period of a satellite very close to earth's surfaceis 84.6 min. Time period increases as the distance ofthe satellite from the surface of earth increase. So, timeperiod of spy satellite orbiting a few 100 km above theearth's surface should be slightly greater than 84.6 min.Therefore, the most appropriate option is (c) or 2 h.

8. In case of binary star system angular velocity andhence the time period of both the stars are equal.

9. For Rr ≤

rmv

rGmM 2

2=′

3

3

3

3

3434

Rr

R

r

MM =

ρπ

ρπ=′

3

2

3

32

RGMr

rRGMr

rGMv ==′= 3

3

RMrM =′

rR

GMv .3

=

rv ∝For Rr >

rmv

rGMm 2

2=

or, vr

GM =

∴ rv

1∝

So, (c) is correct.

OBJECTIVE QUESTIONS (ONLY ONE OPTION)1. (a) The gravitational field is zero at the centre of a solid

sphere. The small spheres can be considered as nega-tive mass m located at A and B. The gravitational forcedue to these masses on a mass at O is equal andopposite. Hence, the resultant force on mass at O iszero.(c and d) → are correct because plane of these circlesis y-z, i.e., perpendicular to x-axis i.e., potential at anypoint on these two circles will be equal due to thepositive mass M and negative masses – m and – m.

2. For r < R, F 3

.GM

rR

= or F ∝ r

1 1

2 2

F rF r

= and r1 < R and r2 < R

and for r > R, F 2

GM

r= or F ∝

21

r

i.e., 2

1 22

2 1

F rF r

= for r1 > R and r2 > R

86

SUBJECTIVE QUESTIONS1. T ∝ r3/2

or r ∝ T2/3

2

1

rr =

2 / 32

1

TT

( ) km 1041018 44

3/2

1

3/2

1

22 ×=×

=

= r

TTr

v2

v1

T2

T1

r1 r2

Now, v1 4

1

1

2 (2 )(10 )1

rTπ π

= = = 2π × 104km/h

v2 4

1

2

2 (2 )(4 10 )8

rTπ π ×

= = = (π × 104) km/h

(i) Seped of S2 relative to S1 = v2 – v2 = π × 104km/h(ii) Angular speed of S2 as observed by S1

ω =

4

1 27

2 1

510 m/s| – | 18| – | (3 10 m)

v vr r

π × × =×

= 0.3 × 10–3 rad/s= 3.0 × 10–4 rad/s

2. Centre should be at O and radius r.We can calculate r by figure (b).

30°30°

(a)

a

rO

aa

30°a/2

r

30°30°

Fnet

F

F

(b) (c)

/ 2ar

= cos30° 3

2=

∴ r =3

a

Further net force on any particle towards centreFnet = 2F cos 30°

=2

23

22

Gm

a

=2

23Gm

a

This net force should be equal to 2mv

r

∴2

23Gm

a=

2

/ 3mv

a

∴ v =Gma

Time period of circular motion

T =2 2 ( / 3)

/r a

v Gm aπ π=

=3

23aGm

π

3. (i) 2

2

)()( hRGMm

hRmv

+=

+

or, hRGMmmv

+=2

422

×==+veGM

mvGMmhR

844..2

2

2

2

2===

vegR

veR

RGM

∴ RV

gRhe

−=2

24

RRgR

gR=−=

24 2

(ii) From conservation of energy

2

21

2mv

RGMm

RGMm

+−=−

or, R

GMmR

GMmR

GMmmv222

2

=−=

RGMmmv

22

2

=

=2

4

87

gRR

GMV ==

310640010

8.9 ××= 4106498 ××=

km/s 9.7=

4. Le there are two stars 1 and 2 as shown below :

1 2

2a

C2C1

a

M 16M

r1P r2

Let P is a point between C1 and C2, where gravita-tional field strength is zero. Or at P field strength dueto star 1 is equal and opposite to the field strength dueto star 2. Hence,

21

GM

r = 22

(16 )G M

r

or2

14

rr

= also r1 + r2 = 10 a

∴ r2 =4

(10 ) 84 1

a a = + and r1 = 2aNow, the body of mass m is projected from the surfaceof larger star towards the smaller one. Between C2 andP it will be attracted towards. Therefore, the body shouldbe prjected to just cross point P because beyond that theparticle is attracted towards the smaller star itself.

From conservation of mechanical energy 2min2

1mv

= Potential energy of the body at P– Potential energy at the surface of large star

∴ 21min

2mv =

1 2

16– –GMm GMmr r

16– – –

10 – 2 2GMm GMma a a

16 8– – – –

2 8 8GMm GMm GMm GMm

a a a a =

or 2min

12

mV =458

GMma

∴ Vmin =3 5

2GM

a

5. Speed of particle at A, vA= escape velocity on the surface of moon

2GMR

=

At highest B, vB = 0Applying conservation of mechanical energy, decreasein kinetic energy= increase in gravitational potential energy

AVA

B

h

R100

V = 0B

or 212 Amv = UB – UA = m (VB – VA)

or2

2Av

= VB – VA

∴

−−−−

+−−=

22

3 1003

2RRR

RGM

hRGM

RGM

or 21 1 3 1 99 1

– – .2 2 100R R h R R

= + + Solving this equation, we get

h = 99.5R

ASSERTION AND REASON1. Force acting on astronaut is utilised in providing

necessary centripetal force, thus he feels weightless-ness, as he is in a state of free fall.∴ correct option is (a)

R=

28

CHAPTER-4WORK, POWER AND ENERGY

TRUE/FALSE1. A simple pendulum with a bob of mass m swings with an angular amplitude of 40°. When its angular displacement

is 20°, the tension in the string is greater than mg cos 20°. (1984; 2M)

positive x-axis to the point (a, 0) and then parallel tothe y-axis to the point (a, a). The toal work done bythe force F on the particle is : (1998; 2M)(a) – 2Ka2 (b) 2 Ka2

(c) – Ka2 (d) Ka2

7. A spring of force-sonstant k is cut into two piecessuch that one piece is double of the other. Then thelong piece will have a force-constant of :(1999; 2M)(a) (2/3)k (b) (3/2)k(c) 3 k (d) 6 k

8. A wind-powered generator converts wind energy intoelectric energy. Assume that the generator converts afixed fraction of the wind energy intercepted by itsblades into electrical energy. For wind speed v, theelectrical power output will be proportional to:

(2000; 2M)(a) v (b) v2

(c) v3 (d) v4

9. A particle, which is constrained to move along x-axis,is subjected to a force in the same direction whichvaries with the distance x of the particle from the originas F (x) = – kx + ax3. Here k and a are posiiveconstant. For x ≥ 0, the functional form of the potentialenergy U (x) of the particle is : (2002; 2M)

U (X)

X

U (X)

X

(a) (b)

U (X)

X

U (X)

X

(c) (d)

OBJECTIVE QUESTIONSOnly One option is correct :1. Two masses of 1 g and 4 g are moving with equal

kinetic energies. The ratio of the magnitudes of theirmomenta is : (1980; 2M)

(a) 4 : 1 (b) 2:1(c) 1 : 2 (d) 1 : 16

2. A body is moved along a straight line by a machinedelivering constant power. The distance moved by thebody in time t is proportional to : (1984; 2M)(a) t1/2 (b) t3/4

(c) t3/2 (d) t2

3. A uniform chain of length L and mass M is lying ona smooth table and one-third of its length is hangingvertically down over the edge of the table. If g isacceleration due to gravity, the work required to pullthe hanging part on to the table is : (1985; 2M)(a) MgL (b) MgL/3(c) MgL/9 (d) MgL/18

4. A particle of mass m is moving in a circular path ofconstant radius r such that its centripetal accelerationac is varying with time t as ac = k2 rt2, where k is aconstant. The power delivered to the particle by theforce acting on it is : (1994; 1M)(a) 2π mk 2 r2 (b) mk 2r2t

(c)4 2 5( )3

mk r t(d) zero

5. A stone tied to a string of length L is whirled in avertical circle with the other end of the string at thecentre. At a certain instant of time, the stone is at itslowest position, and has a speed u. The magnitude ofthe change in its velocity as it reaches a positionwhere the string is horizontal is : (1998; 2M)

(a) 2 – 2u gL (b) 2gL

(c) 2 –u gL (d) ( )22 –u gL

6. A force )ˆˆ( jxiyKF +−=r

(where K is a positive

constant) acts on a particle moving in the x-y plane.Starting from the origin, the particle is taken along the

29

10. An ideal spring with spring-constnat k is hung fromthe ceiling and a block of mass M is attached to itslower end. The mass is released with the springinitially unstretched. Then the maximum extension inthe spring is : (2002; 2M)

(a)4Mg

k (b)2Mg

k

(c)Mgk (d) 2

Mgk

11. A simple pendulum is oscillating without damping. Whenthe displacement of the bob is less than maximum, its

acceleration vector ar

is correctly shown in (2002; 2M)

a→

a→(a) (b)

→a →a(c) (d)

12. If W1, W2 and W3 represent the work done in movinga particle from A to B along different paths, 1, 2 and3 respectively (as shown) in the gravitational field ofa point mass m. Find the correct relation between W1,W2 and W3. (2003; 2M)

B

A

1 23

(a) W1 > W2 > W3 (b) W1 = W2 = W3(c) W1 < W2 < W3 (d) W2 > W1 > W3

13. A particle is placed at the origin and a force F = kx isacting on it (where k is a positive constant). IfU (0) = 0, the graph of U (x) versus x will be (whereU is the potential energy function : (2004; 2M)

U (x)

x

U (x)

x

(a) (b)

U (x)

x

U (x)

x

(c) (d)

14. A small block is shot into each of the four tracks asshown below. Each of the tracks rises to the sameheight. The speed with which the block enters the trackis the same in all cases. At the highest point of the track,the normal reaction is maximum in : (2001; 2M)

V V

(a) (b)

V V

(c) (d)

15. A bob of mass M is suspended by a massless stringof length L. The horizontal velocity v at position A isjust sufficient to make it reach the point B. The angleθ at which the speed of the bob is half of that at A,statisfies. (2008; 3M)

θ

v

B

L

A

(a)4π

=θ (b)24π

<θ<π

(c)42π3

<θ<π

(d) π<θ<π4

3

16. A block (B) is attached to two unstretched springs S1and S2 with spring constants k and 4k , respectively.The other ends are attached to two supports M1 andM2 not attached to the walls. The springs and supportshave negligible mass. There is no friction anywhere.The block B is displaced towards wall 1 by a small

30

distance x and released. The block returns and movesa maximum distance y towards wall 2. Displacements xand y are measured with respect to the equilibrium

position of the block B. The ratio xy

is (2008; 3M)

2M2 S2

M1S1

B

1

x2

M2 S2

M1S1

B

1

(a) 4 (b) 2

(c)21

(d)41

OBJECTIVE QUESTIONSMore than one options are correct?

1. A particle is acted upon by a force of constantmagnitude which is always perpendicular to the velocityof the particle. the motion of the particle takes place ina plane. It follows that : (1987; 2M)(a) its velocity is constant(b) its acceleration is constant(c) its kinetic energy is constant(d) it moves in a circular path

2. A simple pendulum of length L and mass (bob) M isoscillating in a plane about a vertical line betweenangular limits – φ and + φ. For an angular displacementθ (|θ| < φ), the tension in the string and the velocity ofthe bob are T and V respectively. The following relationshold good under the above conditions :

(1986; 2M)(a) T cos θ = Mg

(b) T – Mg cos θ 2MV

L=

(c) The magnitude of the tangential acceleration ofthe bob |aT| = g sin θ

(d) T = Mg cos θ

SUBJECTIVE QUESTIONS1. In the figure (a) and (b) AC, DG and GF are fixed

inclined planes, BC = EF = x and AB = DE = y. A smallblock of mass M is released from the point A. It slidesdown AC and reaches C with a speed VC. The sameblock is released from rest from the point D. It slidesdown DGF and reaches the point F with speed VF.The coefficients of kinetic frictions between the blockand both the surface AC and DGF are µ. Calculate VCand VF. (1980; 6M)

A

B C(a)

D

E F(b)

G

2. A body of mass 2 kg is being dragged with a uniformvelocity of 2 m/s on a rough horizontal plane. Thecoefficient of friction between the body and the surfaceis 0.20, J = 4.2 J/cal and g = 9.8 m/s2. Calculate theamount of heat generated in 5 s. (1990; 5M)

3. A lead bullet just melts when stopped by an obstacle.Assuming that 25 percent of the heat is absorbed bythe obstacle, find the velocity of the bullet if its initialtemperature is 27°C(Melting point of lead = 327°C, specific heat of lead =0.03 cal/g-C°, latent heat of fusion of lead = 6 cal/g°C,J = 4.2 J/cal). (1981; 3M)

4. Two blocks A and B are connected to each other bya string and a spring; the string passes over africtionless pulley as shown in the figure. Block Bslides over the horizontal top surface of a stationaryblock C and the block A slides along the vertical sideof C, both with the same uniform speed. The coefficientof friction between the surfaces of blocks is 0.2. Forceconstant of the spring is 1960 N/m. If mass of block Ais 2 kg. Calculate the mass of block B and the energystored in the spring. (1982; 5M)

B

AC

5. A 0.5 kg block slides from the point A (see Fig.) on ahorizontal track with an initial speed of 3 m/s towardsa weightless horizontal spring of length 1 m and forceconstant 2 N/m. The part AB of the track is frictionlessand the part BC has the coefficients of static andkinetic friction as 0.22 and 0.2 respectively. If thedistances AB and BD are 2 m and 2.14 m respectively,find the total distance through which the block movesbefore it comes to rest completely. (Take g = 10 m/s2)

(1983; 7M)

31

A B D C

3 m/s

6. A string, with one end fixed on a rigid wall, passingover a fixed fricionless pulley at a distance of 2 m fromthe wall, has a point mass M = 2 kg attached to it ata distance of 1 m from the wall. A mass m = 0.5 kgattached at the free end is held at rest so that thestring is horizontal between the wall and the pulleyand vertical beyond the pulley. What will be the speedwith which the mass M will hit the wall when the massm is released? (1985; 6M)

m

M

7. A bullet of mass M is fired with a velocity 50 m/s atan angle θ with the horizontal. At the highest point ofits trajectory, it collides head on with a bob of mass3M suspended by a massless string of length 10/3metres and gets embedded in the bob. After thecollision the string moves through an angle of 120°.Find :(i) the angle θ,(ii) the vertical and horizontal coordinates of the initial

position of the bob with respect to the point offiring of the bullet. (Take g = 10 m/s2)

(1988; 6M)

8. A particle is suspended verticallyfrom a point O by an inextensiblemassles string of length L. Avertical line AB is at a distance L/8 from O as shown in figure. Theobject is given a horizontal velocityu.

OL/8 A

L

uB

At some point, its motion ceasesto be circular and eventually theobject passes through the line AB.At the instant of crossing AB, itsvelocity is horizontal. Find u. (1999; 10M)

9. A spherical ball of mass m is kept at the highest pointin the space between two fixed, concentric spheres Aand B (see figure). The smaller sphere a has a radiusR and the space between the two spheres has a widthd. The ball has a diameter very slightly less then d. Allsurfaces are frictionless. The ball is given a gentlepush (towards the right in the figure). The angle madeby the radius vector of the ball with the upwardvertical is denoted by θ. (2002; 5M)

θ

OR

d

Sphere B

Sphere A

(a) Express the total normal reaction force exerted bythe spheres on the ball as a function of angle θ.

(b) Let NA and NB denote the magnitudes of thenormal reaction forces on the ball exerted by thespheres A and B, respectively. Sketch the variationsof NA and NB as function of cos θ in the range 0≤ θ ≤ π by drawing two separate graphs in youranswer book, taking cos θ on the horizontal axis.

10. A cart is moving along x-direction with a velocity of4 m/s. A person on the cart throws a stone with avelocity of 6 m/s relative to himself. In the frame ofreference of the cart the stone is thrown in y-z planemaking an angle of 30° with vertical z-axis. At thehighest point of its trajectory the stone hits an objectof equal mass hung vertically from branch of a tree bymeans of a sring of length L. A completely inelasticcollision occurs in which the stone gets embedded inthe object. Determine : (1997; 5M)(i) the speed of the combined mass immediately after

the collision with respect to an observer on theground.

(ii) the length L of the string such that tension in thestring becomes zero when the string becomeshorizontal during the subsequent motion of thecombined mass.

11. A light inextensible string that goesover a smooth fixed pulley as shownin the figure connects two blocks ofmasses 0.36 kg and 0.72 kg. Taking g= 10 m/s2, find the work done (inJoules) by the string on the block ofmass 0.36 kg during the first secondafter the system is released from rest.

32


This question contains, statement I (assertion) andstatement II (reason).

1. Statement-I : A block of mass m starts moving on arough horizontal surface with a velocity v. It stops dueto friction between the block and the surface aftermoving through a certain distance. The surface is nowtilted to an angle of 30° with the horizontal and thesame block is made to go up on the surface with thesame initial velocity v. The decrease in the meachnicalenergy in the second situation is smaller than that inthe first situation. (2007; 3M)

Because :Statement-II : The coefficient of friction between theblock and the surface decreases with the increase inthe angle of inclination(a) Statement-I is true, statement -II is true, statement-

II is a correct explanation for statmeent-I(b) statemen-I is true, statement-II is true; statement-

II is NOT a correct explanaion for statmeent-I(c) statement-I is true, statement-II is false(d) statement-I is false, statement-II is true

ANSWERS

TRUE/FALSE1. T

OBJECTIVE QUESTION (ONLY ONE OPTION)1. (c) 2. (c) 3. (d) 4. (b) 5. (d) 6. (c) 7. (b)8. (c) 9. (d) 10. (b) 11. (c) 12. (b) 13.(a) 14. (a)15. (d)16. (c)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)1. (c, d)2. (b, c)


1. VC = VF 2( – )= gy µgx 2. 9.33 cal 3. 409.8 m/s 4. 10 kg, 0.098 J

5. 4.24 m 6. 3.3 m/s

7. (i) θ = 30° (ii) The desired coordinates are (108.25m, 31.25m) 8.

+=

233

2gLu

9. (a) N = mg (3 cos θ – 2) (b) For θ ≤ cos–1 23

, NB 0, NA = mg (3cosθ – 2) and for θ ≥ cos –123

; NA = 0,

NB = mg (2 – 3 cos θ)


1. (c)

33

1. T - mg cos 20° 2

=mv

R

or T = mg cos 20° 2

+mv

R 20°

20°

T

mg

m

mgcos20°mgsin20°

∴T > mg cos 20° )0( ≠vQ∴ (T)


1. P = 2Km

or, 21

2

1

2

1 ==mm

PP

∴ (c)

2. constant=

= v

dtdv

mFv

dtvdv ∝

⇒ 23

21

tstdtds ∝⇒∝

∴ (c)

3. Considering reference for zero potential at the tablesurface,

−=

63L

gM

U i and 0=fU

∴ 18MgLUUW if =−=

∴ (d)

U = 0L3

2L/3

Before

After

4. krtvrtkr

vrtkac =⇒=⇒= 22

222

Therefore, tangential acceleration, at= =dv kr

dtTangential force, Ft = mat = mkrOnly tangential force does work.∴ Power =Ft v = (mkr) (krt)or Power = mk 2r2t∴ (b)

5. From energy conservationv2 = u2 – 2gL

Now, since the two velocity vectors shown in figureare mutually perpendicular, hence the magnitude of thechange of velocity will be given by

v

uA

BO

)(2)2(|| 22222 gLugLuuvuv −=−+=+=∆r

∴ (d)

6. )ˆˆ( jxiyKF +−=r

∫∫ −=+−=),(

)0,(

),(

)0,(

)()(aa

a

aa

a

xydKxdyydxKW

),()0,(][ aa

axyK−= = – Ka2

∴ (c)

7. l1 = 2l2 and l1 + l2 = l

∴ 123

=I l

Force constant, k ∝ )(springoflength1

l

∴ k1 = 32

k

∴ (b)

8. Power .= =ur rF v Fv

F =

dmv

dt 2?)?( AvAvv =×=

SOLUTIONS

TRUE/FALSE

34

where, ρ is density of wind A is area of cross-sectionof blades

∴ P = ρ Av3 ⇒ 3vP ∝∴ (c)

9. dU = – F. dx

∴ U (x) = – 3

0

(– )+∫x

kx ax dx

U (x) =2 4

–2 4

kx ax

U (x) = 0 at x = 0 and x 2= ka

U (x) = negative for x > 2ka

00 =⇒= xdxdU

and ,akx = therefore slope

of )(xU vs x curve is zero at hence points.

∴ (d)

10. Let x be the maximumextension of the spring. Fromconservation of mechanicalenergy : decrease ingravitational potential energy= increase in elastic potentialenergy

M

M

x

v = 0

v = 0

K

∴ Mg x 212

= kx

or, x = 2Mg

k∴ (b)

11. When the bob is betweenmean and extream position it

has both centripetal )( car

and

tangential )( tar

accelerations.

The net acceleration )(ar

is

vector sum of car

and tar

.O

at→

a→ac

→

∴ (c)12. Gravitational field is a conservative force field. In a

conservative force field work done is path independent.∴ W1 = W2 = W3∴ (b)

13. From F –= dUdx

( )

0 00

– – ( )= =∫ ∫ ∫U x

x xdU Fdx kx dx

∴ U (x)= – 2

2kx

as U (0) = 0∴ (a)

14. Since, the block rises to the same heights in all the fourcases, from conservation of energy, speed of the blockat highest point will be same in all four cases. Say itis V0.

V0

N+mg

Now,N + mg 2

0mVR

=

or2

0 –mV

N mgR

=

R (the radius of curvature) in first case is minimum.Therefore, normal reaction N will be maximum in firstcase.∴ (a)Note : In the questions it should be mentioned that all

the four tracks are frictionless. Otherwise oV will be

different in different tracks.

15. gLv 5= ...(i)

ghvv

22

22

−=

...(ii)

)cos1( θ−= Lh ...(iii)

Solving equations (i), (ii) and (iii), we get

87cos −=θ or °=

−=θ − 151

87

cos 1

∴ (d)

16. From energy conservation,

22 )4(21

21

ykkx =

∴ 21=

xy

∴ (c)

35


1. The given case is of uniform circular motion, in whichspeed or kinetic energy is constant. Direction ofvelocity and acceleraion keep on changing althoughits magnitude remains constant.∴ Correct options are (c) and (d).

2. From the FBD of bob,

T – mg cos θ L

mV 2

=

andMg sin θ = MaTor aT = g sin θ∴ (b) and (c) θ

θ

O

T

Mg sin θMg cos θMg

L

V

M


1. In both the cases work done by friction will be µ Mgx.

∴ 2 21 1–

2 2= =C FMV MV Mgy µMgx

∴ VC = VF = 2 – 2gy µgx

2. S = vt = 2 × 5 = 10 mQ = work done against friction = µmgs = 0.2 × 2 × 9.8 × 10 = 9.33 cal

3. Heat energy required to just melt the bullet.Q = Q1 + Q2

Here, Q1 = ms∆θ = (m × 103) (0.03 × 4.2) (327 – 27) = (3.78 × 104 m) J

Q2 = mL = (m × 103) (6 × 4.2) = (2.52 × 104m)∴ Q = (6.3 × 104)mOnly 75% of kinetic energy is utilized to melt the bullet

0.75 × 212

mv = Q

0.75 212

× × ×m v = 6.3 × 104m ⇒ v = 409.8 m/s

4. Normal reaction between blocks A and C will be zero.Therefore, there will be no friction between them. BothA and B are moving with uniform speed. Therefore, netforce on them shold be zero.

BT=kx

f=µmgB

B

m gA

T=kx

For equilibrium of A :

mAg = kx ⇒ x = Am gk

= (2)(9.8)

1960 = 0.01 m

For equilibrium of B :µmBg = T = kx = mAg

∴ mB = Am

µ2

0.2= = 10 kg

Energy stored in spring

U 212

= kx 12

= (1960) (0.01)2 = 0.098 J

5. From A to B, there will be no loss of energy. Now letblock compresses the spring by an amount x andcomes momentarily to rest. By work-energy theorem,

kx

fA B D v=0 C

x

ifsfNmg KKWWWW −=+++

22

21

21

)(00 mvkxxBDmg −=−+µ−+

Substituting the values

(0.2) (0.5) (10) (2.14 + x) 2 21 1(0.5)(3) – (2)( )

2 2= x

Solving this equation, we get, x = 0.1 mNow, spring exerts a force kx = 0.2N on the block. Butto stop the block from moving limiting static friction isµs mg = (0.22) (0.5) (10) = 1.1 N. Since, 1.1 N > 0.2 N,block will not move further and it will permanenly stopthere.Therefore, total distance covered before it comes torest permanently isd = AB + BD + x = 2 + 2.14 + 0.1 = 4.24 m

6. Let M strikes with v. Then, velocity of m at this instant

will be v cos θ or 2

5v. Further M will fall a distance

of 1 m while m will rise by ( )5 – 1 m. From

conservation of energy principle decrease in potenialenergy of M = increase in potenial energy of m +increase in kineic energy of both the blocks.

1 m

2 m

m

1 m

( 5–1) m√θ

v cos θ√5 m

θ

vθ

1

2

36

∴ (2) (9.8) (1)

= (0.5) (9.8) ( )5 – 1 212

2+ × × v

21 20.5

2 5

+ × ×

v

⇒ = 3.3 m/s

7. (a) At the highest point, velocity of bullet is 50 cos θ.So, by conservation of linear momentum

M (50 cos θ) = 4M VA

∴ VA = 504

cos θ ...(1)

At point, B, T = 0 but v ≠ 0

O

V= θ

VB

504

cos

Hence, 4 Mg cos 60° = 350

2)4( 2

2

==⇒ lgVlVM

BB

(as 103

=l m and g = 10 m/s2)

By conservation of energy between A and B

−=⇒−= lgVVghVV ABAB 2

322 2222

or, v2 = V2 – 2g 32

l

or, 100?cos4

503

50100

222 −

=

⇒−= AB VV

∴ cos θ = 0.86 or θ = 30°

(b) x = Range

2

21 sin22

θ =

ug

50 50 32 10 2× ×=× ×

= 108.25 m

y = H gu

2?sin22

= 50 50 12 10 4

× ×=× × = 31.25 m

Hence, the desired coordinates are (108.25 m, 31.25 m)

8. Let the string slacks at point Q as shown in figure. Pto Q path is circular and beyond Q path is parabolic.At point C, velocity of particle becomes horizontal,therefore, QD = half the range of the projectile.

θ

θQ

90° – θD

L

P u

L cos θ

L+L sin θ

mg

L8

v

We have following conditions,

(a) TQ = 0 Therefore, mg sin θ 2

=mv

L...(1)

(2) v2 = u2 – 2gh = u2 – 2gL (1 + sin θ) ...(2)

(3) QD 12

= (Range)

⇒

−

8?cos

LL

2 sin2(90 – )2

° θ=

vg

2 sin22

θ=

vg

−

81

?cos =2

vgL sin θ cos θ ...(3)

Substituting value of 2

vgL = sin θ from Eq. (1) we get

−

81

?cos = sin2θ. cos θ = (1 – cos2θ) cosθ

or, ?cos?cos81?cos 3−=−

∴ 81?cos3 =

or,cos θ = 12

or, θ = 60°From Eq. (1) v2 = gL sin θ = gL sin 60°

or v2 = 3

2gL

∴ Substituting this value of v2 in Eq. (2)u2=v2 + 2gL (1 + sin θ)

37

=3 32 1

2 2

+ +

gL gL

=3 3 3 32 2

2 2

+ = +

gL gL gL

u =3 32

2

+

gL

9. (a) h (1–cos )2

= + θ

dR

Velocity of ball at angle θ

θmg

θ v

h

v2 = 2gh ( )2 1–cos2

= + θ

dR g

Let N be the total normal reaction (away from centre)at angle θ. Then,

mg cos θ – N =2

2 +

mvd

R

Substituting value of v2 from Eq. (1) we getmg cos θ – N = 2 mg (1 – cos θ)

∴ N = mg (3 cos θ – 2)(b) The ball will lose contact with the inner spherewhen N = 0

or, 3 cos θ – 2 = 0 or θ = cos–123

After this it makes contact with outer sphere and normalreaction starts acting towards the centre. Thus for

θ ≤ cos–123

NB = 0 and NA = mg (3 cos θ – 2)

and for θ ≥ cos–123

NA = 0 and NB = mg (2 – 3 cos θ)The corresponding graphs are as follows

2/3 +1cos θ

mg

NA

–1 2/3 +1cos θ

2mg

5mg

NB

10. (i) Given iVcartˆ4=

r m/s

∴ =cartstoneV ,

r(6 sin30°) j$ + (6 cos 30°) $k

= (3 j$ + 3 3 $k ) m/s

∴ )ˆ33ˆ3ˆ4(, kjiVVV cartcartstonestone ++=+=rrr

m/s

At highest point of its trajectory, the vertical component(z) of its velocity will be zero, whereas the x and y-components will remain unchanged. Therefore, velocityof stone at highest point will be,

Vr

= (4 i$ + 3 j$ )m/s

or speed at highest point,

V ||Vr

= 2 2( (4) (3)= + m/s = 5m/s

Now, applying law of conservation of linear momentum,let V0 be the velocity of combined mass after collision.Then, mV = (2m)V0

∴ V0 5

2 2V

= = m/s = 2.5 m/s

(ii) Tension in the string becomes zero at horizontalposition. It implies that velocity of combined mass alsobecomes zero in horizontal position. Applyingconservation of energy, we have

V0 = 2.5m/s

L

T = 0V = 0

0 20 – 2V gL=

∴ L2

0

2V

g=

2(2.5)2(9.8)

= = 0.32m

∴ Length of the string is 0.32 m.

38

11. Let the two blocks move with acceleration a andtension in the string be T.

31010

36.072.036.072.0 =×

+−=a m/s2;

8.41036.072.0

36.072.02 =×+

××=T N;

mS35133.3

21 2 =××=

8=TW J


1. In statement-I : Decrease in mechanical energ in caseI will be

∆U1 = 212

mv

But decrease in mechanical energy in case II will be

∆U2 = 21–

2mv mgh

∴ ∆U2 < ∆U1or statement-I is correct.In statement-II : Coefficient of fricition will not changeor this statement is wrong.∴ Option (c) is correct

55

CHAPTER-6ROTATION

FILL IN THE BLANKS

1. A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal toone of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. The minimumvalue of F for which the cube begins to tip about the edge is ..... (Assume that the cube does not slide).

(1984; 2M)2. A smooth uniform rod of length L and mass M has two idenical beads of negligible size, each of mass m, which

can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating withan angular velocity ω0 about an axis perpendicular to the rod and passing through the mid-point of the rod (seefig.). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the systemis ...... (1988; 2M)

L/2 L/2

ω0

3. A cylinder of mass M and radius R is resting on a horizontal platform (which is parallel to the x-y plane) with itsaxis fixed along the y-axis and free to rotate about its axis.The platform is given a motion in the x-direction givenby x = A cos (ωt). There is no slipping between the cylinder and platform. The maximum torque acting on the cylinderduring its motion is ..... (1988; 2M)

4. A stone of mass m, tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due togravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centreof the circle constant. Then, the tension in the string is given by T = Arn where A is a constant, r is the instantaneousradius of the circle and n = ...... (1988; 2M)

5. A uniform disc of mass m and radius R is rolling up a rough inclined plane which makes an angle of 30° with thehorizontal. If the coefficient of static and kinetic friction are each equal of µ and the only forces acting aregravitational and frictional, then the magnitude of the frictional force acting on the disc is .... and its direction is.... (write up or down) the inclined plane. (1997C; 1M)

6. A rod of weight w is supported by two parallel knife edge A and B and is in equilibrium in a horizontal position.The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normalreaction on A is ..... and on B is .... (1997; 2M)

7. A symmetric lamina of Mass M consists of a square shape with a semicircular sectionover the edge of the square as shown in figure. the side of the square is 2a. The momentof inertia of the lamina about an axis thorugh is centre of mass and perpendicular to theplane is 1.6 Ma2. The moment of inertia of the lamina about the tangent AB in the planeof the lamina is .... (1997; 2M)

A B

O

2a

TRUE FALSE1. A triangular plate of uniform thickness and density is made to rotate about an

axis perpendicular to the plane of the paper and (a) passing through A, (b)passing through B, by the application of the same force F at C (mid-point ofAB) as shown in the figure. The angular acceleration in both the cases will bethe same. (1985; 3M) A C B

F

2. A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing throughits centre and perpendicular to its plane with an angular velocity ω. Another disc of the same dimensions but of

mass M/4 is placed gently on the first disc coaxially. The angular velocity of the system now is 2 / 5ω .

(1986; 3M)

56

OBJECTIVE QUESTIONSOnly One option is correct :1. A thin circular ring of mass M and radius r is rotating

about its axis with a constant angular velocity ω. Twoobjects, each of mass m, are attached gently to theopposite ends of a diameter of the ring the wheel nowrotates with angular velocity : (1983; 1M)(a) ωM/ (M + m) (b) ω(M – 2m) / (M + 2m)(c) ωM / (M + 2m) (d) ω(M + 2m)/ M

2. A particle of mass m is projected with a velocity vmaking an angle of 45° with the horizontal. Themagnitude of the angular momentum of the projectileabout the point of projection when the particle is at itsmaximum height h is : (1990; 2M)

(a) zero (b) 3 /(4 2 )mv g

(c) 3 /( 2 )mv g (d) 32m gh

3. A tube of length L is filled completely with anincompressible liquid of mass M and closed at boththe ends. The tube is then rotated in a horizontal planeabout one of its ends with a uniform angular velocityω. The force exerted by the liquid at the other endis : (1992; 2M)

(a)2

2M Lω (b) 2M Lω

(c)2

4M Lω (d)

2 2

2M Lω

4. Two point masses of 0.3 and 0.7 kg are fixed at theends of a rod of length 1.4 m and of negligible mass.the rod is set rotating about an axis perpendicular toits length with a uniform angular speed. The point onthe rod through which the axis should pass in orderthat the work required for rotation of the rod is minimum,is located at a distance of : (1995; S)(a) 0.42 m from mass of 0.3 kg(b) 0.70 m from mas sof 0.7 kg(c) 0.98 m from mass of 0.3 kg(d) 0.98 m from mass of 0.7 kg

5. A mass m is moving with a constant velocity along aline parallel to the x-axis, away from the origin. Itsangular momentum with respect to the origin:

(1997C; 1M)(a) is zero (b) remains constant(c) goes on increasing (d) goes on decreasing

6. Let I be the moment of inertia of a uniform square plateabout an axis AB that passes through its centre and isparallel to two of its sides. CD is a line in the plane ofthe plate that passes through the centre of the plateand makes an angle θ with AB. The moment of inertiaof the plate about the axis CD is then equal to :

(1998; 2M)(a) I (b) I sin2θ(c) I cos2 θ (d) I cos2 (θ/2)

7. A cubical block of side a moving with velocity v on ahorizontal smooth plane as shown. It hits a ridge atpoint O. The angular speed of the block after it hits Ois : (1999; 2M)

O

a

vM

(a) 3v/4a (b) 3v/2a

(c) 3 / 2a (d) zero

8. A smooth sphere A is moving on a frictionlesshorizontal plane with angular velocity ω and centre ofmass with velocity v. It collides elastically and head onwith an identical sphere B at rest. Neglect frictioneverywhere. After the collision their angular speedsare ωA and ωB respectively. Then : (1999; 2M)(a) ωA < ωB (b) ωA = ωB(c) ωA = ω (d) ωB = ω

9. A disc of mass M and radius R is rolling with angularspeed ω on a horizontal plane as shown. Themagnitude of angular momentum of the disc about theorigin O is : (1999; 2M)

O x

y

ωM

(a)21

2MR ω

(b) MR2 ω

(c)23

2MR ω

(d) 2MR2 ω

3. A ring of mass 0.3 kg and radius 0.1 m and a solid cylinder of mass 0.4 kg and of the same radius are given thesame kinetic energy and released simultaneously on a flat horizontal surface such that they begin to roll as soonas released towards a wall which is at the same distance from the ring and the cylinder. The rolling friction in bothcases is negligible. The cylinder will reach the wall first. (1989; 2M)

57

10. An equilateral triangle ABC formed from a uniform wirehas two small identical beads initially located at A. Thetriangle is set rotating about the vertical axis AO. Thenthe beads are released from rest simultaneously andallowed to slide down, one along AB and other alongAC as shown. Neglecting frictional effects, thequantities that are conserved as beads slides downare :` (2000)

A

B O C

g

(a) angular velocity and total energy (kinetic andpotential)

(b) total angular momentum and total energy(c) angular velocity and moment of inertia about the

axis of rotation(d) total angular momentum and moment of inertia

about the axis of rotation.

11. A cubical block of side L rests on a rough horizontalsurface with coefficient of friction µ. A horizontal forceF is applied on the block as shown. If the coefficientof friction is sufficiently high, so that the block doesnot slide before toppling, the minimum force requiredto topple the blocks is : (2000)

F

L

(a) infinitesimal (b) mg/4(c) mg/2 (d) mg (1 – µ)

12. A thin wire of length L and uniform linear mass densityρ is bent into a circular loop with centre at O asshown. The moment of inertia of the loop about theaxis XX' is : (2000)

O

90°X'X

(a)3

28

Lρ

π(b)

3

216

Lρ

π

(c)3

25

16

Lρ

π(d)

3

23

8

Lρ

π

13. One quarter section is cut from auniform circular disc of radius R.This section has a mass M. It ismade to rotate about a lineperpendicular to its plane andpassing through the centre of theoriginal disc. Its moment of inertiaabout the axis of rotation is : (2001)

(a) 212

MR (b) 214

MR

(c)21

8MR (d) 22MR

14. A cylinder rolls up an inclined plane, reaches someheight and then rolls down (without slipping throughoutthese motions). The directions of the frictional forceacting on the cylinder are : (2002)(a) up the incline while ascending and down the

incline while descending(b) up the incline while ascending as well as

descending(c) down the incline while ascending and up the

incline while descending(d) down the incline while ascending as well as

descending

15. A circular platform is free to rotate in a horizontal planeabout a vertical axis passing through its centre. Atortoise is sitting at the edge of the platform. Now theplatform is given an angular velocity ω0. When thetortoise move along a chord of the platform with aconstant velocity (with respect to the platform). Theangular velocity of the platform ω (t) will vary withtime t as : (2002)

ω ( )t

ω0

t(b)

ω0

t(a)

ω ( )t

ω ( )t

ω0

t(c)

ω ( )t

ω0

t(d)

58

16. Consider a body, shown in figure, consisting of twoidentical balls, each of mass M connected by a lightrigid rod. If an impulse J = Mv is imparted to the bodyat one of its end, what would be its angular velocity?

(2003)L

M

J = Mv

M

(a) v/L (b) 2v/L(c) v/3L (d) v/4L

17. A particle undergoes uniform circular motion. Aboutwhich point on the plane of the circle, will the angularmomentum of the particle remain conserved? (2003)(a) Centre of circle(b) On the circumference of the circle(c) Inside the circle (d) Outside the circle

18. A disc is rolling (with slipping) on a horizontal surfaceC is its centre and Q and P are two points equidistantfrom C. Let vP, vQ and vC be the magnitude of velocitiesof points P, Q and C respectively, then :

(2003)Q

C

P

(a) vQ > vC > vP (b) vQ < vC < vP

(c) vQ = vP, vC 12 Pv=

(d) vQ < vC < vP

19. A child is standing with folded hands at the centre ofa platform rotating about its central axis. The kineticenergy of the system is K. the child now stretches hisarms so that the moment of inertia of the systemdoubles. The kinetic energy of the system now is :

(2004)

(a) 2K (b)2K

(c)4K

(d) 4K

20. A particle moves in a circular path with decreasingspeed. Choose the correct statement : (2005)(a) Angular momentum remains constant

(b) Acceleration ( )ar

is towards the centre

(c) Particle moves in a spiral path with decreasingradius

(d) The direction of angular momentum remainsconstant

21. From a circular disc of radius R and mass 9M, a smalldisc of radius R/3 is removed from the disc. Themoment of inertia of the remaining disc about an axisperpendicular to the plane of the disc and passingthrough O is : (2005)

R/3

2R/3

OR

(a) 4MR2 (b)240

9MR

(c) 10MR2 (d)237

9MR

22. A solid sphere of radius R has moment of inertia Iabout its geometrical axis. If it is melted into a disc ofradius r and thickness t. If it's moment of inertia aboutthe tangential axis (which is perpendicular to plane ofthe disc), is also equal to I, then the value of r is equalto : (2006; 3M)

l

r

(a)2

15R (b)

2

5R

(c)3

15R (d)

315

R

23. A ball moves over a fixed track as shown in the figure.From A to B the ball rolls without slipping. If surfaceBC is frictionless and KA, KB and KC are kineticenergies of the ball at A, B and C respectively, then :

(2006; 5M)

59

A C

hchA

B(a) hA > hC ; KB > KC(b) hA > hC ; KC > KA(c) hA = hC ; KB = KC(d) hA < hC ; KB > KC

24. A small object of uniform density rolls up a curvedsurface with an initial velocity v. It reaches up to a

maximum height of 23

4vg

with respect to the initial

position. The object is : (2007; 3M)

v

(a) ring (b) solid sphere(c) hollow sphere (d) disc

25. A long horizontal rod has a bead which can slide alongits length and is initially placed at a distance L fromone end A of the rod. The rod is set in angular motionabout A with a constant angular acceleration α. If thecoefficient of friction between the rod and bead is µ,and gravity is neglected, then the time after which thebead starts slipping is : (2000; 2M)

(a)µa

(b)µ

α

(c)1µα (d) infinitesimal

26. A block of mass m is at rest under the action of forceF against a wall as shown in figure. Which of thefollowing statement is incorrect? (2005)

a

F

(a) f = mg [where f is the frictional force](b) F = N [where N is the normal force](c) F will not produce torque(d) N will not produce torque

27. A block of base 10 cm × 10 cm and height 15 cm is kepton an inclined plane. The coefficient of friction between

them is 3 . The inclination θ of this inclined plane

from the horizontal plane is gradually increased from0°. Then (2009; M)(a) at θ = 30°, the block will start sliding down the

plane(b) the block will remain at rest on the plane up to

certain θ and then it will topple(c) at θ = 60°, the block will start sliding down the

plane and continue to do so at higher angles(d) at θ = 60°, the block will start sliding down the

plane and on further increasing θ, it will topple atcertain θ

OBJECTIVE QUESTIONSMore than one options are correct?1. A uniform bar of length 6 a and mass 8 m lies on a

smooth horizontal table. Two point masses m and 2mmoving in the same horizontal plane with speed 2v andv respectively, strike the bar [as shown in the Fig] andstick to the bar after collision. Denoting angular velocity(about the centre of mass), total energy and centre ofmass velocity by ω, E and VC respectively, we haveafter collision : (1991; 2M)

2a

m

2v

c

2m

v

a

(a) VC = 0 (b)35

va

ω =

(c) 5va

ω = (d)2

35

mvE =

2. The moment of inertia of a thin square plate ABCD, ofuniform thickness about an axis passing through thecentre O and perpendicular to the plane of the plate is

(1992; 2M)

60

A4

B 1

3

2CD

O

(a) I1 + I2 (b) I3 + I4(c) I1 + I3 (d) I1 + I2 + I3 + I4where I1, I2, I3 and I4 are respectively moments ofinertia about axes 1, 2, 3 and 4 which are in the planeof the plate.

3. The torque τr

on a body about a given point is found

to be equal to Aur

× Lur

where Aur

is a constant vector

and Lur

is the angular momentum of the body aboutthat point. From this it follows that : (1998; 2M)

(a)dtdLuuur

is perpendicular to Lur

at all instants of time

(b) the component of Lur

in the direction of Aur

doesnot change with time

(c) the magnitude of Lur

does not change with time

(d) Lur

does not change with time

4. A solid sphere is in pure rolling motion on an inclinedsurface having inclination θ. (2006; 5M)

θ

(a) frictional force acting on sphere is f = µ mg cos θ(b) f is dissipative force(c) friction will increase its angular velocity and

decreases its linear velocity(d) If θ decreases, friction will decrease

5. If the resultant of all the external forces acting on asystem of particles is zero, then from an inertial frame,one can surely say that (2009; M)(a) linear momentum of the system does not change

in time(b) kinetic energy of the system does not change in

time(c) angular momentum of the system does not change

in time(d) potential energy of the system does not change in

time

6. A sphere is rolling without slipping on a fixed horizontalplane surface. In the figure, A is the point of contact,B is the centre of the sphere and C is its topmostpoint. Then, (2009; M)

A

B

C

(a) )(2 CBAC VVVVrrrr

−=−

(b) ABBC VVVVrrrr

−=−

(c) ||2|| CBAC VVVVrrrr

−=−

(d) ||4|| BAC VVVrrr

=−

SUBJECTIVE QUESTIONS1. A particle is projected at time t = 0 from a point P on

the ground with a speed v0, at an angle of 45° to thehorizontal. Find the magnitude and direction of theangular momentum of the particle about P at timet = v0/g. (1984; 6M)

2. A small sphere rolls down without slipping from thetop of a track in a vertical plane. The track has anelevated section and a horizontal part. The horizontalpart is 1.0 m above the ground level and the top of thetrack is 2.6 m above the ground. Find the distance onthe ground with respect to the point B (which isvertically below the end of the track as shown in fig.)where the sphere lands. During its flight as a projectile,does the sphere continue to rotate about its centre ofmass? Explain. (1987; 7M)

2.6m

1.0m

A

B

3. A thin uniform bar lies on a frictionless horizontalsurface and is free to move in any way on the surface.

Its mass is 0.16 kg and length is 3 m. Two particles,

each of mass 0.08 kg are moving on the samesurface and towards thebar in a directionperpendicular to the barone with a velocity of 10m/s, and the other with 6m/s, as shown in Fig.

10m/s

6m/sB

A

61

The first particle strikes the bar at point A and theother at point B. Points A and B are at a distance of0.5 m from the the centre of the bar. The particles strikethe bar at the same instant of time and stick to the baron collision. Calculate the loss of kinetic energy of thesystem in the above collision process. (1989; 8M)

4. A carpet of mass M made of inextensible material isrolled along its length in the form of a cylinder ofradius R and is kept on a rough floor. The carpet startsunrolling without sliding on the floor when a negligiblysmall push is given to it. Calculate the horizontalvelocity of the axis of the cylindrical part of the carpetwhen its radius reduces to R/2. (1990; 8M)

5. A homogeneous rod AB of length L = 1.8 m and massM is pivoted at the centre O in such a way that it canrotate freely in the vertical plane (Fig.). The rod is initiallyin the horizontal position. An insect S of the same massM falls vertically with speed v on the point C, midwaybetween the points O and B. Immediately after falling,the insect move towards the end B such that the rodrotates with a constant angular velocity ω.(1992; 8M)

OL/2 L/4 L/4

C

v

S

A B⊗

(a) Determine the angular velocity ω in terms of v and L.(b) If the insect reaches the end B when the rod has

turned through an angle of 90°, determine v.

6. A block X of mass 0.5 kg is held by a long masslessstring on a frictionless inclined plane of inclination 30°to the horizontal. The string is wound on a uniformsolid cylindrical drum Y of mass 2 kg and of radius0.2 m as shown in figure. The drum is given an initialangular velocity such that the block X starts movingup the plane. (1994; 6M)

30°

X

Y

(i) Find the tension in the string during the motion.(ii) At a certain instant of time the magnitude of the

angular velocity of Y is 10 rad s–1. Calculate thedistance travelled by X from that instant of timeuntil it comes to rest.

7. Two uniform rods A and B of length 0.6 meach and of masses 0.01 kg and 0.02 kgrespectively are rigidly joined end to end.The combination is pivoted at the lighterend P as shown in figure. Such that it canfreely rotate about point P in a verticalplane. A small object of mass 0.05 kg, movinghorizontally, hits the lower end of thecombination and sticks to it. What shouldbe the velocity of the object, so that thesystem could just be raised to the horizontalposition. (1994; 6M)

B

A

P

8. A rectangular rigid fixed block has a long horizontaledge. A solid homogeneous cylinder of radius R isplaced horizontally at rest with its length parallel to theedge such that the axis of the cylinder and the edgeof the block are in the same vertical plane as shownin figure. There is sufficient friction preseent at theedge, so that a very small displacement causes thecylinder to roll of the edge without slipping Determine:

(1995; 10M)

R

(a) the angle θ, through which the cylinder rotatesbefore it leaves contact with the edge,

(b) the speed of the centre of mass of the cylinderbefore leaving contact with the edge and

(c) the ratio of the translational to rotational kineticenergies of the cylinder when its centre of mass isin horizontal line with the edge.

9. Two thin circular discs of mass 2 kg and radius 10 cmeach are joined by a rigid massless rod of length 20 cm.The axis of the rod isalong theperpendicular to theplanes of the discthrough their centres.

O

20cm

This object is kept on a truck in such a way that theaxis of the object is horizontal and perpendicular to thedirection of motion of the truck. Its friction with the

62

floor of the truck is large enough, so that the objectcan roll on the truck without slipping. Take x-axis asthe direction of motion of the truck and z-axis as thevertically upwards direction. If the truck has anacceleration 9 m/s2. calculate : (1997; 5M)(i) the force of friction on each disc and(ii) the magnitude and direction of the frictional torque

acting on each disc about the centre of mass O ofthe object. Express the torque in the vector form

in terms of unit vectors, i$ , j$ and $k in x, y andz-direction.

10. A uniform disc of mass m and radius R is projectedhorizontally with velocity v0 on a rough horizontalfloor, so that it starts off with a purely sliding motionat t = 0. After t0 seconds, it acquires a purely rollingmotion as shown in figure. (1997 C; 5M)

t = t0t = 0

v0

(i) Calculate the velocity of the centre of mass of thedisc at t0.

(ii) Assuming the coefficient of friction to be µ,calculate t0. Also calculate the work done by thefrictional force as a function of time and the totalwork done by it over a time t much longer than t0.

11. A uniform circulardisc has radius R andmass m. A particle,also of mass m, isfixed at a point A onthe edge of the disc

QR/4

P

R

A

C

as shown in the figure. The disc can rotate freelyabout a horizontal chord PQ that is at a distance R/4from the centre C of the disc. The line AC isperpendicular to PQ. Initially the disc is held verticallywith the point A at its highest position. It is thenallowed to fall, so that it starts rotation about PQ. Findthe linear speed of the particle as it reaches its lowestposition. (1998; 8M)

12. A man pushes a cylinder of mass m1 with the help ofa plank of mass m2 as shown. There is no slipping atany contact.The horizontal component of the forceapplied by the man is F. Find : (1999; 10M)

F m2

m1

(a) the accelerations of the plank and the centre ofmass of the cylidner and

(b) the magnitudes and directions of frictional forcesat contact points.

13. A rod AB of mass M and length L is lying on ahorizontal frictionless surface. A particle of mass mtravelling along the surface hits the end A of the rodwith a velocity v0 in a direction perpendicular to AB.The collision is elastic. After the collision the particlecomes to rest. (2000)(a) Find the ratio m/M.(b) A point P on the rod is at rest immediately after

collision. Find the distance AP.(c) Find the linear speed of the point P after a time

πL/3V0 after the collision.

14. Two heavy metallic plates are joined together at 90° toeach other. A laminar sheet of mass 30 kg is hinged atthe line AB joining the two heavy metallic plates. Thehinges are frictionless. The moment of inertia of thelaminar sheet about an axis parallel to AB and passingthrough its centre of mass is 1.2 kg - m2. Two rubberobstacles P and Q are fixed, one on each metallic plateat a distance 0.5 m from the line AB. This distance ischosen, so that the reaction due to the hinges on thelaminar sheet is zero during the impact. Initially thelaminar sheet hits one of the obstacles with an angularvelocity 1 rad/s and turns back. If the impulse on thesheet due to each obstacles is 6N-s. (2001; 10M)

BP

A

Q

(a) Find the location of the centre of mass of thelaminar sheet from AB.

(b) At what angular velocity does the laminar sheetcome back after the first impact?

(c) After how many impacts, does the laminar sheetcome to rest?

15. Three particles A, B and C, each of mass m, areconnected to each other by three massless rigid rodsto form a rigid, equilateral triangular body of side l.This body is placed on a horizontal frictionless table(x-y plane) and is hinged to it at the point A, so thatit can move without friction about the vertical axisthrough A (see figure). The body is set into rotationalmotion on the table about A with a constant angularvelocity ω. (2002; 5M)

63

BF

l

ω

xA

y

C

(a) Find the magnitude of the horizontal force exertedby the hinge on the body.

(b) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (asshown). Obtain the x-component and the y-component of the force exerted by the hinge onthe body, immediately after time T.

16. A rod of length L and mass M is hinged at point O.A small bullet of mass m hits the rod as shown in thefigure. The bullet gets embedded in the rod. Findangular velocity of the system just after impact.

(2005; 2M)O

L, M

mv

17. A solid cylinder rolls without slipping on an inclined planeinclined at an angle θ. Find the linear acceleration of thecylinder. Mass of the cylinder is M. (2005; 4M)

COMPREHENSION

PassageTwo discs A and B are mounted coaxially on a verticalaxle. The discs have moment of inertia I a n d 2Irespectively about the common axis. Disc A is impartedan initial angular velocity 2ω using the entire potentialenergy of a spring compressed by a distance x1. DiscB is imparted an angular velocity ω by a spring havingthe same spring constant and compressed by a distancex2. Both the discs rotate in the clockwise direction.

1. The ratio 1

2

xx is : (2007; 4M)

(a) 2 (b)12

(c) 2 (d)1

2

2. When disc B is brought in contact with disc A, theyacquire a common angular velocity in time t. Theaverage frictional torque on one disc by the otherduring this period is : (2007; 4M)

(a)23Itω

(b)92Itω

(c)94Itω

(d)32Itω

3. The loss of kinetic energy during the above processis : (2007; 4M)

(a)2

Iω(b)

2

3Iω

(c)2

4Iω

(d)2

6Iω

PassageA uniform thin cylindrical disk of mass M and radiusR is attached to two identical massless springs ofspring constant k which are fixed to the wall as shownin the figure. The springs are attached to the axle ofthe disk diammetrically on either side at a distance dfrom its centre. The axle is massless and both thesprings and the axle are in a horizontal plane. Theunstretched length of each spring is L. The disk isinitially at its equilibrium position with its centre ofmass (CM) at a distance L from the wall. The disk rolls

without slipping with velocity ivv ˆ0

rr= . The coefficient

of friction is µ .

4. The net external force acting on the disk when itscentre of mass is at displacement x with respect to itsequilibrium position is (2008; 4M)(a) – kx (b) – 2kx

(c) 32kx− (d) 3

4kx−

5. The centre of mass of the disk undergoes simpleharmonic motion with angular frequency ω equal to

(2008; 4M)

(a)Mk

(b)M

k2

(c) Mk

32

(d) Mk

34

64

6. The maximum value of v0 for which the disk will rollwithout slipping is (2008; 4M)

(a)Mkgµ (b) k

Mg

2µ

(c) kM

g3

µ (d) kM

g25

µ


This question contains, statement-I (assertion) andstatement II (reason).

1. Statement-I : Two cylinders, one hollow (metal) andthe other solid (wood) with the same mass and identical

dimensions are simultaneously allowed to roll withoutslipping down an inclined plane from the same height,The hollow cylinder will reach the bottom of theinclined plane first. (2008; 3M)Because :Statement-II : By the principle of conservation ofenergy, the total kinetic energies of both the cylindersare identical when they reach the bottom of the incline.(a) Statement-I is true, Statement -II is true, Statement-

II is a correct explanation for Statmeent-I(b) Statement-I is true, Statement-II is true; Statement-

II is NOT a correct explanaion for Statmeent-I(c) Statement-I is true, Statement-II is false(d) Statement-I is false, Statement-II is true

ANSWERS

FILL IN THE BLANKS

1.23

mg 2.06

MM m

ω+ 3.

21 43

MR ω 4. – 3 5. ,6

mg up

6. dxw

wd

xd,

−

7. 28.4 ma

TRUE/FALSE1. F 2. F 3. F

OBJECTIVE QUESTION (ONLY ONE OPTION)1. (c) 2. (b) 3. (a) 4. (c) 5. (b) 6. (a) 7. (a)8. (c) 9. (c) 10. (b) 11. (c) 12. (d) 13. (a) 14. (b)15. (c) 16. (a) 17. (a) 18. (a) 19. (b) 20. (d) 21. (a)22. (a) 23. (a) 24. (d) 25. (a) 26. (d) 27. (b)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)1. (a, c, d) 2. (a, b, c) 3. (a, b, c) 4. (c, d) 5. (a) 6. (b, c)


1.30

2 2mv

g in a direction perpendicular to paper inwards 2. 2.13m, yes 3. 2.72 J

4. 14

3Rgv = 5. (a)

127

VL (b) 3.5 ms –1 6. (i) 1.63 N (ii) 1.22 ln 7. 6.3 m/s

8. (a) θ = cos–1

74

(b) 4

7gR

(c) 6

9. (i) 6i$ (ii) 0.6 ($k – j$ ), 0.6 (– j$ – $k ), 0.85 N-m 10. (i) 2

03V (ii) [ ]

20 0

0–

, 3 – 2 ,3 2 6V mVmµgt

µgt Vµg

11. 5gR 12. aCM 1 2

43 8

Fm m

=+ , aplank

1 2

83 8

Fm m

=+ = 2aCM (b)

1 1

1 2 1 2

3,

3 8 3 8Fm Fm

m m m m+ +

65

FILL IN THE BLANKS1. Taking moment about point O :

N

Omg

f

F

3a4

a2

Moment of N (normal reaction) and f (force of friction)are zero. As block just starts tipping about O, Nreaction will pass through O. To tip about the edge,moment of F should be greater than moment of mg. Or,

34a

F

> (mg) 2a

⇒ F > 23

mg

∴ mgF32

min =

2. By principle of conservation of angular momentumI1ω1 = I2ω2

∴ ω2 = 11

2.

II

ω ⇒ 2

02 2( /12)

[( /12) 2 ( /2) ]

ML

ML m Lω

+

or ω2 06M

M m = ω +

3. Considering the motion of cylinder from the frame ofplatform. The torque on cylinder will be maximum when

the acceleration of platform is maximum i.e. Aa 2max ω=

αF = m Ap ω2

f

a = Amax ω2fPlatform

Now, mafAm =−ω2

α= IfR

Ra α=Solving above three equation we get,

Amf 2

31 ω=

fR=τmax

ARm 2

31 ω=

4. mvr = k (constant) kv

mr⇒ =

rmv

T2

= = Ar–3 2

where,k

Am

=

∴ n = – 3

5. Under the given conditions only possibility is thatfriction is upwards and it accelerates downwards asshown below :

13. (a) 14

(b) 23

L (c) 0

2 2

v14. (a) 0.1 m (b) 1 rad/s (c) sheet will never come to rest

15. (a) 23 ωml (b) net–

( ) ,4xF

F = 23)( ω= mlF ynet

16. 3

(3 )mv

L m M+ 17.2 sin3

g θ

COMPREHENSION

1. (c) 2.(a) 3. (b)4. (d) 5.(d) 6. (c)


1. (d)SOLUTIONS

66

mg sin θ

θ= 30°

mg cos θ

N f i

a

α

The equations of motion are :

mafmg =−°30sin ...(1)

α=2

2mRfR ...(2)

For rolling (no slipping)a = Rα ...(3)

Solving (1), (2) and (3), we get

+

θ==

22

21

sin

mRIR

IgRIa

f

Q θµ= cosmax mgf mgµ=23

Q maxff < (upwards)

∴ 6mgf =

6. Net torque of all the forces about CM should be zero.

x d-x

A CM

W

B

NBNA

About CM, )( xdNxN BA −= ...(1)

For vertical equilibrium of rod —NA + NB = W ...(2)

Solving (1) and (2), we get

NB Wdx

=

or, Wd

xdN A

−

=

7. Assuming the lamina to be in x-y plane.Ix + Iy = Iz

Q Ix = Iy (by symmetry)and Iz = 1.6 Ma2

∴ 28.02

MaI

I zx ==

Now, from the parallel axis theorem,

A B

2a

x

y

2aO

IAB = Ix + M (2a)2

= 0.8 Ma2 + 4Ma2 = 4.8 ma2

TRUE FALSE

1.Iτ

α =

Fr⊥=τ : Torque is same in both the cases but

moment of inertia depends on distribution of massfrom the axis. Distribution of mass in both the casesis different. Therefore, moment of inertia will be differentor the angular acceleration α will be different.

2. I1ω1 = I1ω2

∴ ω2 = 11

2.

II

ω

=

2

2 2425

.2 4 2

MR

MR M R

ω = ω

+

3. Total energy of the ring

22

21

21

mvI +ω=

2222

21

21

ω+ω= mrmr

22ω= mr ...(1)

Total K.E. of the cylinder

22

21

21

′+′ω′= MVI

222222

43

21

21

21

′ω=′ω+′ω

= MrMrMr ...(2)

Equating (i) and (ii),

2222

43

′ω=ω Mrmr

67

14.033.04

34

2

2=

××==

ω′ω

Mm

ω=′ωor both will reach simultaneously.


1. I1ω1 = I1ω2

∴ ω2 = 1

2.

II

ω

2

2 22

Mr

Mr mr

= ω +

2M

M m = ω +

2.2

vL m r⊥=

y

h

x

v cos 45° = v2

Here 2 2sin 45

2v

r hg⊥

°= =

∴ L =2

42

v vm

g

=3

4 2mv

g

3. Mass of the element dx is m M

dxL

=

This element needs centripetal force for rotation.

dx x=L

x=0

ω

F F+dF

∴ dF = mxω2 =2M

x dxL

ω

∴2

0

ω== ∫ LM

dFFL 2

02

L M Lxdx

ω=∫

Thiis is the force exerted by the liquid at the other end.

4. Work done 212

W I= ω

If x is the distance of mass 0.3 kg from the centre ofmass, we will have,

I = (0.3)x2 + (0.7) (1.4 – x)2

For work to be minimum, the moment of inertia (I)should be minimum, or

dIdx = 0

or, 2 (0.3x) – 2(0.7) (1.4 – x) = 0or, (0.3)x = (0.7) (1.4 – x)

⇒ x =(0.7)(1.4) 0.980.3 0.7

m=+

5. |Vur

| = V = constant and |rr

| = r (say)Angular momentum of the particle about origin O willbe given by

( )m= × = ×L r p r Vur r r r ur

r

mV

h

x

y

O

or | |Lur

= L = mrV sinθ = mV (r sin θ) = mVhNow, m, V and h all are constants.Therefore, angular momentum of particle about origin

will remain constant. The direction of ×r Vr ur

also remainsthe same (negative z).

6. A'B' ⊥ AB and C'D' ⊥ CDFrm symmetry IAB = IA 'B 'and ICD = IC'D 'From theorem of perpendicular axes,IZZ = IA'B' + IAB = ICD + IC'D' = 2IAB = 2ICD

IAB = ICD

68

B

A′D

D′

Β′C

A

C′

7. 22a

r = or 2

22a

r =

M M

C ω

Net torque about O is zero.Therefore, angular moment (L) about O will beconserved,or, Li = Lf

ω=

02I

aMv

= (ICM + Mr2)ω

2 222

6 2 3Ma a

M Ma = + ω = ω

34

va

ω =

8. Since, it is head on elastic collision between twoidentical balls, they will exchange their linear velocitiesi.e., A comes to rest and B starts moving with linearvelocity v. As there is no friction anywhere, torque onboth the spheres about their centre of mass is zero andtheir angular velocities remain unchanged.Therefore,

ωA = ω and ωB = 0.

9. Angular momentum about O

O x

y

M V= Rω

(a)

= OCOMOCOMCOM VRML , ,

rrr×+

ω=+ω= 22

23

21

MRMRVMR (Q )ω= RV

10. Net external torque on the system is zero. Therefore,angular momentum is conserved. Force acting on thesystem are only conservative. Therefore, totalmechanical energy of the system is also conserved.

11. At the critical condition, normal reaction N will passthrough point P. In this conditionthe block will topple when net torque about G is clockwise, so

0<τ+τ+τ+τ fFmgN

⇒ 022

02

<−−+flFlNl

⇒ fFN +<

⇒ FFmg +< ,( Ff =Q for translational

equilibrium)

⇒ F minimum = 2

mg

F

f

G

L2

mg

L

N

12. Mass of the ring M = ρ LLet R be the radius of the ring, then

L = 2π R or 2L=π

Moment of inertia about an axis passing through Oand parallel to XX' will be

I0 = 212

MR

Therefore, moment of inertia about XX' (from parallelaxis theorm) will be given by

222

23

21

MRMRMRI XX =+=′

Substituting values of M and R

69

2

3

2

2

8

3

4)(

23

π

ρ=

πρ=′

LLLI XX

13. (i) 2dmrdI ==

2dArAM

∫=

r

∫π

π= 2

2 24 rdrrRM

22 2

0

32

MRdrr

R

MR

== ∫

14. As cylinder rolls up, due to θsinmg its acceleration

is downward,∴ For ,α= Ra friction should be upward

& Same way when cylinder rolls down, to maintainsame condition ,α= Ra friction should act inupward direction only.

15. Since, there is no external torque, angular momentumwill remain conserved. The moment of inertia will firstdecrease till the tortoise moves from A to C and thenincrease as it moves from C to D. Therefore, ω willinitially increase and then decrease.

O

r

CB DA

vt

Let R be the radius of platform, m the mass of disc andM is the mass of platform.Moment of inertia when the tortoise is at A.

I1 = mR2 + 2

2MR

and moment of intertia when the tortoise is at B.

I1 = mr2 + 2

2MR

Here, r2 = a2 + [ 2 2 2– – ]R a vtFrom conservation of angular momentum

ω0I1 = ω (t) I2

16. Let ω be the angular velocity of the rod. Applyingangular impulse = change in angular momentum aboutcentre of mass of the system

J.2

= ωCL

I

∴ ( )2

LMv =

2(2)

4

ω

ML

∴ ω =vL

17. In uniform circular motion the only force acting on theparticle is cenripetal (towards centre). Torque of thisforce about the centre is zero. Hence, angularmomentum about centre remain conserved.

18.

Q

C

P

V

V

VRω

Rω

∴ PCQ vvv >>

19. Using 2211 ω=ω II ⇒ 21 2 ω=ω II

2,21

2

2

1 ==ωω

II

∴ 24212

21

222

211

2

1 =×=ω

×ω=I

IKK

21

2K

K =

20. As Torque of tangential force will not be zero. Soangular momentum will not be conserved about centreof circle.

Here TC aaarrr

+=

But as direction of VRrr

× remain same.

So direction of anuglar momentum remains constant.

21. Iremaining = Iwhole – Iremoved

or

+

−=

222

32

321

))(9(21 R

mR

mRMI ..(1)

Here, m = 2

29

3 ×π = π

M RM

RSubstituting in Eq. (1) we have

I = 4MR2

∴ (a)

70

22.2 2 22 1

5 2= +MR Mr Mr

or, 2 22 35 2

=MR Mr ∴2

15=I R

23. If ball is released from A, then

0. =AEK

At B, 22

2222

21

52

21

21

21.. mv

RvmRmvIEK +×=+ω=

2

107 mv=

At C, 2

2

222

51

52

21

21

.. mVRv

mRIEK =

=ω=

,0( =τQ so ω will remain constant)

∴ ACB EKEKEK ...... >>& Applying mechanical energy conservation

between A and C, we get

CCA KmvImghmgh ==ω=− 22

51

21

∴ mgK

hh CCA +=

∴ CA hh >∴ (a)

24.

=

+

gv

mgRv

Imv43

21

21 22

2

∴ I = 212

mR

Body is disc.∴ (d)

25. Tangential force (Ft) of thebead will be given by thenormal reaction (N), whilecentripetal force (FC) isprovided by friction (fr). Thebead starts sliding when thecentripetal force is just equalto the limiting friction.

A

ω

f

N

F = C m Lω2

F = t m Lα

From the frame of end A,N = mαLf = µN = mω2L∴ µmαL = mω2L

µα=ω

Angular velocity at time t is

ω = αt ⇒ αµ=t

∴ (a)

26. For translational equilibriumΣF = 0∴ F = N and f = mgFor rotational equilibrium,

Στc = 0

mg

N

f

FC

∴ 0N fτ + τ =uuur uur

Since, 0fτ ≠uur

∴ 0Nτ ≠uuur

∴ (d)

27. The block will start sliding down (if it does not topple)at angle of reposei.e. µ = tan θ ∴ θs = 60°

θmg sin θ

15 cm

mg cos θ

NFs

The block will start toppling (if it does not slide) atangle θ if

θ=

θ

210

cos2

15sin mgmg

⇒

=θ −

32

tan 1t

∴ As θt < θs block will first topple before it slides.∴ (b)

71


1. Pi = 0 ∴ Pf = 0 or Vc = 0Li = Lf or (2mV) a + (2mV) (2a) = Iω ...(1)

Here, I 2

2 2 2(8 )(6 )(2 ) (2 )( ) 30

12= + + =

m am a m a ma

Substituting in Eq. (1) we get

ω =5Va

Further, 2 2

2 21 1 3(30 )

2 2 5 5 = ω = × =

V mVE I ma

a

∴ (a), (c) and (d).

2. Since, it is a square laminaI3 = I4 and I1 = I2 (by symmetry)

From perpendicular axes theorem,Moment of inertia about an axis perpendicular tosquare plate and passing from O is

I0 = 1 + I2 = I3 + I4or I0 = 2I2 = 2I3Hence, I2 = I3Rather we can say I1 = I2 = I3 = I4Therefore, I0 can be obtained by adding any two i.e.,I0 =I1 + I2 = I1 + I3 = I1 + I4 = I2 + I3= I2 + I4 = I3 + I4

3. (a) = ×A Lr ur urτ

i.e. = ×dL A Luuur ur ur

dt

The relation implies that dLuuur

dt is perpendicular to both

Aur

and Lur

. Therefore, option (a) is correct.

(c) Here, 2. =L Lur ur

L

Differentiating w.r.t time, we get

. . . 2+ =dL dLL Luuur uuurur ur dLLdt dt dt

⇒ 2 . 2=dLLuuurur dLLdt dt

But since, ⊥ dLLuuurur

dt

∴ . 0=dLLuuurur

dt

Therefore, from Eq. (1) 0=dLdt

or magnitude of Lur

does not change with time,(b) So far we are confirmed about two points

L

L

L

L

(1) τr

or ⊥dL Luuur ur

dt and

(2) |Lur

| = L is not changing with time, therefore, it is

a case when direction of Lur

is changing but its

magnitude is constant and τr

is perpendicular to Lur

atall points. This can be written as :

If ( cos ) ( sin )= θ + θL i jur r r

a aHere, a = positive constant

Then ( sin ) – ( cos )τ = θ θi jr r r

a a

So, that 0τ =Lur r

and ⊥ τLur r

Now, Aur

is constant vector and it is always

perpendicular to τr

. Thus, Aur

can be written as

Aur

= $kA

we can see that Lur

. Aur

= 0

i.e., Lur

⊥ Aur

also.

Thus, we can say that component of Lur

along Aur

is

zero or component of Lur

along Aur

is always constant.

Finally we conclude, τr

, Aur

and Lur

are always mutuallyperpendicular.

4. In case of pure rolling.

f =2

sin

1

θ

+

mg

mRI

(upwards)

∴ f ∝ sin θTherefore, as θ decrease force of friction will alsodecrease.As net work done by f is zero in case of rolling, so fis not a dissipative force.

5. ∴ If 0=extFr

∴ =Pr

constant

72

But extτr

may be non-zero ∴ ≠Lr

constant

For example in case of a non-rigid body, (K.E)system ≠constant∴ (a)

6. If VVB

rr= then VVC

rr2= and 0

rr=AV

∴ (b, c)


1. )( vrmLrrr

×=

θ== ⊥ cos0maxvmhvmr

g

mvg

mvv

gmv

2422

12

cos2sin 3

030

0

220 ==θ

θ=

2. h = 2.6 – 1.0 = 1.6 mDuring pure rolling mechanical energy remainsconserved.By conservation of mechanical energy, we get

22

21

21

)6.1( mvImg +ω=

22

22

21

52

21

)6.1( mvRv

mRmg +

=

2

107

)6.1( mvmg =

716 g

v =

gght 122 ×==

∴ m13.27

3227

16 ====g

gvtx

hv

ω

B C

In air, during its flight as a projectile only mg is acting

on the sphere which passes through its centre ofmass. Therefore, net torque about centre of mass iszero. Or angular velocity will remain constant.

3. Let v be the velocity of centre of mass (at c) of rod andtwo particles and ω the angular velocity of the system.From conservation of linear momentum.(0.08) (10 + 6) = [0.08 + 0.08 + 0.16]v

6m/s

10m/sA

C

B

C

B

A

∴ v = 4 m/sAC = CB = 0.5 mSimilarly conservation of angular momentum aboutpoint C.(0.08) (10) (0.5) – (0.08) (6) (0.5) = Iω ..(1)Here, I = Irod + ITwo particles

( )2

2(1.6) 3

2(0.08)(0.5)12

= +

= 0.08 kg-m2

Substituing in Eq. (1), we geω = 2 rad/s

Loss of kinetic energy

2 21 1(0.08)(10) (0.08)(6)

2 2= +

2 21 1– (0.08 0.08 0.16)(4) – (0.08)(2)

2 2+ +

= 4 + 1.44 – 2.56 – 0.16 = 2.72J

4. Let M' be the mass of unwound carpet. Then,

M' =2

2 2 4 π =

π

M R M

R

M'

R/2v

M

R

From conservation of mechanical energy :

2 21 1– '

2 2 4 2 = + ω

R MMgR M g v I

or, 22 21 1–

4 2 8 2 2 4 4 / 2

= + × ×

M R Mv M R vMgR gR

73

or78

MgR =23

16Mv

∴ v = 14

3Rg

5. In this problem we will write K for the angular momentumbecause L has been used for length of the rod.(a) Angular momentum of the system (rod + insect)about the centre of the rod O will remain conservedjust before collision and after collison i.e., Ki = Kf.

OL/2 L/4 L/4

C

v

M

A B

Just before collision

O CA B

Just after collision

ω

or ω

+=ω=

22

4124L

MML

Il

Mv

or ω= 2

487

4MLLMv

i.e.., Lv

712=ω ...(1)

(b) Due to the torque of weight of insect about O, angularmomentum of the system will not remain conseved(although angular velocity ω is constant). As theinsect moves towards B, moment of inertia of thesystem increases, hence, the angular momentum of thesystem will increase.Let at time t1 the insect be at a distance x from O andby then the rod has rotated through an angle θ. then,angular momentum at that moment,

L =2

212

+ ω

MLMx

Hence, dtdL

= 2 ω dxM xdt (ω = constant)

⇒ τ = 2Mωx.

τ=

dtdL

dtdx

⇒ Mgx cos θ = 2M ω xdxdt

⇒ dx = cos2

ω ω

gt dt (θ = ωt)

At time t = 0, x = L/4 and at time t = T/4 or π/2ω,x = L/2.Substituing these limits in Eq. (2) we get

/ 2

/ 4∫L

Ldx =

/ 2

0(cos )

2π ω

ωω ∫g t dt

[ ] / 2/ 4

LLx =

ωπωω

2/02

][sin2

tg

⇒ –2 4

L L=

°−

π

ω0sin

2sin

2 2

g

4L =

22ω

g or 2ω = gL

Substituting in Eq. (1), we get

2gL

= LV

712

or V =7

212

gL =7

2 10 1.8 3.5 /12

× × = m s

Hence, v = 3.5m/s6. Given mass of block X, m = 0.5 kg

30°

X

Y

θ=30°

X

Y

Mass of drum Y, M = 2kgRadius of drum, R = 0.2 mAngle of inclined plane, θ = 30°Let a be the linear retardation of block X and α theangular retardation of drum Y. Then,

a = Rαmg sin 30° – Τ =ma ...(1)

or –2

mgT = ma ...(2)

α =21

2

τ=

TRI MR

or α =2TMR

...(3)

Solving Eqs. (1), (2) and (3) for T, we get

T =12 2+

M mgM m

Substituting the value, we get

74

NT 63.1)2)(5.0(2)8.9)(5.0)(2(

21 =

+

=

T = 1.63N(ii) From Eq. (3), angular retardation of drum

2α =

TMR

=(2)(1.63)(2)(0.2) = 8.15 rad/s2

or linear retardation of blocka = Rα = (0.2) (8.15) = 1.63 m/s2

At the moment when angular velocity of drum isω0 = 10 rad/s

The linear velocity of block will bev0 = ω0 R = (10) (0.2) = 2m/s

Now, the distance (s) travelled by the block until itcomes to rest will be given by

20

2= vs

a [Using v2 = 2

0v – 2as with v = 0]

23.163.12

)2( 2=

×= m

7. System is free to rotate but not free to translate.During collision, net torque on the system (rod A + rodB + mass m) about point P is zero.Therefore, angular momentum of system before collision= Angular momentum of system just after collision.(About P). Let ω be the angular velocity of system justafter collision, then

Li = Lf⇒ mv (2l) = IωHere, I = moment of inertia of system about P

= m (2l)2 + mA (l2/3) + m3

++

22

212l

ll

Given : l = 0.6 m, m = 0.05 kg, mA = 0.01 kg and mB= 0.02 kgSubstituting the values, we get

I = 0.09 kg-m2

Therefore, from Eq. (1)

ω =I

mvl2 (2)(0.05)( )(0.6)0.09

= v

ω = 0.67v ..(2)Now, after collision, mechanical energy will beconserved.

ω

ω–0

Therefore, decrease in rotational KE = increase ingravitational PE

or, 212

ωI

++

+=

22)2(

llgm

lgmlmg BA

or, 2 (4 3 )+ +ω = A Bgl m m m

I

(9.8)(0.6)(4 0.05 0.01 3 0.02)0.09

× + + ×=

= 17.64 (rad/s)2

∴ ω = 4.2 rad/s ...(3)Equating Eqs. (2) and (3), we get

v =4.2

0.67 m/s

or, v = 6.3 m/s

8. (a) The cylinder rotates about the point of contact.Hence, the mechanical energy of the cylinder will beconserved

R

θR

R cos θ

V'

ω

∴ (PE + KE)1 = (PE + KE)2

∴ mgR + 0 = mgR cos θ 2 21 12 2

+ ω +I mv

but ω = v/R (No slipping at point of contact)

and I = 212

mR

Therefore,

mgR = mgR cos θ 2

2 22

1 1 12 2 2

+ +

vmR mv

R

or 234

v = gR (1 – cos θ)

or v2 =4 (1–cos )3

θgR

75

or2v

R=

4 (1–cos )3

θg ...(1)

θV

mgN=0mg s

in θ

mgcos θ

At the time of leaving contact, normal reaction N = 0and θ = θc. Hence

mg cos θ =2mv

R

or2v

R= g cos θ (2)

From Eqs. (1) and (2),

(1–cos )34 θcg = g cos θ

or7

cos4

θc = 1

or cosθc = 4/7or θc = cos–1 (4/7)

(b) 4 (1–cos )3

= θv gR [From Eq. (1)]

At the time of losing contactcosθ = cos θc = 4/7

∴ v =4 4

1 –3 7

gR

or v =47

gR

Therefore, speed of CM of cylinder just before losing

contact is 47

gR

(c) At the moment, when cylidner loses contact

v =47

gR

Therefore, rotational kinetic energy, KR 212

= ωI

or KB =2

22

1 12 2

vmRR

=21 1 4

4 4 7 =

mv m gR

or KR = 7mgR

Now, once the cylinder loses its contact, N = 0, i.e.,the frictional force, whcih is responsible for its rotation,also vanishes. Hence, its rotational kinetic energynow becomes contant, while its translational kineticenergy increase.Applying conservation of energy at 1 and 3.Decrease in gravitation PE = Gain in rotational KE +translation KE∴ Translational KE (KT)

= Decrease in gravitational PE – KR

or KT = (mgR) – 6

7 7=mgR mgR

From Eqs. (3) and (4)

T

R

KK =

67

7

mgR

mgR

orT

R

KK = 6

9. (i) In the frame of truck, a pseudo frame ma is appliedon each disc on the centre of mass.∴ Torque acting on the disc is

α= IfR

f

ma ma

f

yxa = 9 m/s2

z

Let a′ be acceleration of C.M. of disc in frame of truck.For pure rolling

Ra α=′

2RIa

f′

=

By Newton's law in x-direction

′=− mafma ∴mf

aa −=′

2

21

mRI =

76

∴

−=

mf

aR

mRf 2

2

21

fmaf −=2 ⇒3

maf = = 6 N

(in + x –direction)

)ˆ6( if = N

(ii) τ = ×r fr

$ $

Here, ( 6 )N=f ir $ (for both the discs)

$1 –0.1 – 0.1p = =r r j k

uur ur $

and $2 0.1–0 .1Q = =r r j k

uur uur $

Therefore, frictional torque on disk 1 about point O(centre of mass).

mNikjfr −×−−=×=τ )ˆ6()ˆ1.0ˆ1.0(11

rrr

$(0.6 –0.6 )= k j$

or, $1 0.6( – )τ = k juur $ N-m

and 2 21| | (0.6) (0.6)τ = +

uur= 0.85 N-m

Similarly, 2τ = ×2r fuur uur r

= 0.6 $(– – )j k$

and 1 2| | | |τ = τuur uur

= 0.85 N-m

10. By conserving AM about bottom most point, we get

ω+=

2

2

0mR

mvRRmv

mRvRvmR

mvRRmv23

2

2

0 =+=

32 0vv =

After, ,0tt > it will have pure rolling,

20

22

21

21

21

.. mvImvEKW f −ω+=∆=

202

222

21

21

21

21

mvRv

mRmv −

+=

6294

43

21

43 2

020

202

02 mvmvvmmvmv −=−=−=

11. Initial and final positions are shown below.Decrease in potential energy of mass m

= mg 5 5

24 2R mgR × =

Decrease in potential energy of disc

= 24 2R mgR

mg × =

P Q

C

R 5R4

3R4

R/4

P QC

m

m

3R4

R 5R4

R/4

ω

Therefore, total decrease in potential energy of system

mgRmgRmgR

322

5=+=

Gain in kinetic energy of system 212

I= ω

where I = moment of inertia of system(disc + mass) about axis PQ= moment of inertia of disc.+ moment of inertia of mass

2 22 54 4 4

mR R Rm m

= + +

I =215

8mR

From conservation of mechanical energy,Decrease is potential energy = Gain in kinetic energy

∴ 3mgR =2

21 152 8

mR ω

⇒ ω =165

gR

Therefore, linear speed of particle at its lowest point

v =5 5 164 4 5R R g

R ω =

v = 5gR

12. We can choose any arbitary directions of frictionalforces at different contacts.In the final answer the negative values will show theopposite directions.Let f1 = friction between plank and cylinder

77

f2 = friction between cylinder and grounda1 = acceleration of planka2 = acceleration of centre of mass of cylinderand α = angular acceleration of cylinder about its CM.Directions of f1 and f2 are as shown hereSince, there is no slipping anywhere= a1 = 2a2 ...(1)(Acceleration of plank = acceleration of top point ofcylinder)

f2

f1

a2m1

a1 =1

2

–F fm ...(2)

a2 =1 2

1

f fm+

...(3)

IRff )( 21 −

=α ...(4)

(I = moment of inertia of cylinder above CM)

a = 2a1 2

a2

1 2

21

( – )12

f f R

m R=

1 2

1

2( – )f fm R

α =

1 22

1

2( – )f fa R

m= α = ...(5)

(Acceleration of bottom most point of cylinder = 0)(a) Solving Eqs. (1), (2), (3) and (5), we get

Rα a2

a1 =1 2

83 8

Fm m+

and a2 =1 2

43 8

Fm m+

(b) f1 =1

1 2

33 8

m Fm m+

f2 = 1

1 23 8m F

m m+

13. (a) Let just after collision, velocity of CM of rod is vand angular velocity about CM is ω. Applyingfollowing three laws :

CM CM v x

v0

L2

L2

ω

mm

Before collision Before collision

(i) External force on the system (rod + mass) inhorizontal plane along x-axis is zero.∴ Applying conservation of linear momentum in x-direction.

mv0 = Mv ...(1)(ii) Net torque on the system about CM of rod is zero∴ Applying conservation of angular momentum about

CM of rod, we get 0 2L

mv I = ω

or 0 2L

mv =2

12ML

ω

or mv0 = 6MLω

...(2)

(iii) Since, the collision is elestic,

0

21v

lve

ω+== ...(3)

From Eqs. (1), (2) and (3) we get the following results

mM

=14

0mvv

M= and 06mv

MLω =

(b) Point P will be at rest if x ω = v

78

orvx =ω =

0

0

/ '6 /mv Mmv ML

or x = L/6

CM V

ωL2

P

A

x∴ AP = 2 6

L L+

or AP =23

L

(c) After time t =03Lv

π

angle rotated by rod, θ = ωt 0

0

6.3

mv LML v

π=

=

π=

π=

41

41

22Mm

Mm

∴ θ =2π

Therefore, situation will be as shown below :∴ Resultant velocity of point P will be

| |Pvr

00 0

22 24 2 2

vmv v vM

= = = =

AP

V

or | |Pvr

= 0

2 2

v

14. Let r be the perpendicular distance of CM from the lineAB and ω the angular velocity of the sheet just aftercolliding with rubber obstacle for the first time.Obviously the linear velocity of CM before and aftercollision will be v1 = (r) (1 rad/s) = r and vf = rω

ivuur

and fvuuur

will be in opposite direction

Now, linear impulse on CM= change in linear momentum of CMor 6 = m (vf + vi) = 30 (r + rω)

or r (1 + ω) =15

Similarly, angular impulse about AB = change in angularmomentum about ABAngular impulse = Linear impulse × perpendiculardistance of impulse from ABHence, 6 (0.5) = IAB (ω + 1)[Initial angular velociy = 1 rad/s)or 3 = [ICM + Mr2] (I + ω)or 3 = [1.2 + 30r2] (1 + ω) ...(2)Solving Eqs. (1) and (2) for r, we get

r = 0.4 m and r = 0.1 mBut r = 0.4m, ω comes out to be negative (–0.5 rad/s)which is not acceptable. Therefore,(a) r = distance of CM from AB = 0.1 m(b) Substituting r = 0.1 m in Eq. (1), we get ω = 1 rad/s i.e., the angular velocity with which sheet comesback after the first impact is 1 rad/s.(c) Since the sheet returns with same angular velocityof 1 rad/s, the sheet will never come to rest.

15. (a) The distance of centre of mass (CM) of the systemabout point A will be :

3

lr =

Therefore, the magnitude of horizontal force exerted bythe hinge on the body is

CM

ll

FB C

A y

xω α,

3/2l

F = centripetal forceor F = (3m) rω2

or2

3)3( ω

= lmF

or, 23 ω= mlF(b) Angular acceleration of system about point A is

α =

( )

2

32

2A

A

F l

I ml

τ =

=3

4F

mlNow, acceleration of CM along x-axis is

ax = rα 3

43l F

ml

= or ax 4

Fm

=

Now, let Fx be te force applied by the hinge along x-axis. Then,

Fx + F = (3m)ax

or Fx + F = (3m) 4Fm

or Fx + F = 34

F or Fx = –4F

79

Further if Fy be the force applied by the hinge alongy-axis. Then,

Fy = centripetal force

or 23 ω= mlFy

16. Angular momentum of the system about point O willremain conserved.

Li = Lf

∴ mvL = Iω 2

23

MLmL

= + ω

∴ ω =3

(3 )mv

L m M+

17. For rolling without slipping, we haveUsing formula

21

sin

MRI

ga

+

θ=

2

21

MRI =

∴ θ= sin32

ga

θ

M

α

a

Mg sin θ

f

a = Rα

COMPREHENSION

1. 21(2 )

2I ω = 2

112

kx ...(i)

( )21(2 )

2I ω = 2

212

kx ...(i)

From Eqs. (i) and (ii), we have

1

2

xx = 2

2. Let ω' be the common velocity. Then from conservationof angular momentum, we have(I + 2I)ω' = I (2ω) + 2I (ω)

ω' =43

ω

From the question,

angular impulse = change in angular momentum, forany of the disc, we have

τ .t = ( ) 4 22 –

3 3I

I Iω ω ω =

∴ τ =23Itω

3. Loss of kinetic energy = Ki – Kf

2 21 1(2 ) (2 )( )

2 2I I = ω + ω

221 4

– (3 )2 3 3

I I1 ω = ω

4.

f

kxkx a

∴ Mafkx =−2

α=2

2MRfR

α= Ra (Because it is having a pure rolling motion)

∴2

Maf =

MakxMaMa

kx23

22

2 =⇒=−

Mkxa

34=

∴ 32

34

2kx

MkxM

f =

=

∴ kxMafnet 34==

∴ (d)

5. α= IfR

RIaRkx =

32

xa

IkR =3

2 2

Mk

Mk

xa

MRkR

34

34

322 22

2=ω⇒=ω⇒=

××

80

6. MkxxvAv

34

maxmax0max =ω=⇒ω=

∴kMvx4

30max =

mgkxf µ==3

2 maxmax

⇒ mgkMvk µ=4

332

0

Mk

kmgv

34

23

0 ×µ=

kMg 3µ=

∴ (c)


1. In case of pure rolling on inclined plane,

2/1sin

mRIga

+θ=

hollowsolid II <

∴ hollowsolid aa >Kinetic energy of solid cylinder when it reaches ground

22

21

21

mVI +ω=

222

22

43

21

21

21

mvmvRV

mR =+

=

And 3sin22

1

sin22

2

2 lg

mRI

lgasv

θ×=

+

θ==

∴ lmglgmEK solid θ=θ×= sin3sin4

43..

For Hollow cylinder, K.E. = 22

21

21

mvI +ω

222

21

21

mvmvmv =+=

& lglg

asv θ=+

θ== sin

11sin2

22

∴ lmgmv θ= sin2

∴ Both are having identical K.E.

88

CHAPTER-8SIMPLE HARMONIC MOTION

FILL IN THE BLANKS

1. An object of mass 0.2 kg executes simple harmonic motion along the x-axis with a frequency of (25/π) Hz. At theposition x = 0.04, the object has kinetic energy of 0.5 J and potential energy 0.4 J. the amplitude of oscillationsis ..... m. (1994, 2M)

OBJECTIVE QUESTIONSOnly One option is correct :

1. A particle executes simple harmonic motion with afrequency f. The frequency with which its kineticenergy oscillates is : (1987; 2M)(a) f/2 (b) f(c) 2f (d) 4f

2. Two bodies M and N of equal masses are suspendedfrom two separate massless springs of spring constantsk1 and k2 respectively. If the two bodies oscillatevertically such that their maximum velocities are equal,the ratio of the one amplitude of vibration of M to thatof N is : (1988; 1M)

(a) k1/k2 (b) 2 1/k k

(c) k2/k1 (d) 1 2/k k

3. A uniform cylinder of length L and mass M havingcross-sectional area A is suspended, with its lengthvertical, from a fixed point by a massless spring, suchthat it is half-submerged in a liquid of density ρ atequilibrium position. When the cylinder is given asmall downward push and released, it starts oscillatingvertically with a small amplitude. If the force constantof the spring is k , the frequency of oscillation of thecylinder is : (1990; 2M)

(a)1 / 21 –

2k A g

Mρ

π (b)

1 / 212

k A gM

+ ρ π

(c)

1/3212

k gLM

+ ρ π

(d)1 / 21

2k A g

A g + ρ π ρ

4. A highly rigid cubical block A of small mass M andside L is fixed rigidly on to another cubical block B ofthe same dimensions and of low modulus of rigidity ηsuch that the lower face of A completely covers theupper face of B. The lower face of B is rigidly held ona horizontal surface. A small force F is appliedperpendicular to one of the sides faces of A. After theforce is withdrawn, block A executes small oscillations,the time period of which is given by : (1992; 2M)

(a) 2 M Lπ η (b) 2 MLηπ

(c) 2ML

πη (d) 2

ML

πη

5. One end of a long metallic wire of length L is tied tothe celling. The other end is tied to a massless springof spring constant K. A mass m hangs freely from thefree end of the spring. The area of cross-section andthe Young's modulus of the wire are A and Yrespectively. If the mass is slightly pulled down andreleased, it will oscillate with a time period T equal to:

(1993; 2M)

(a) 2π (m/K)1/2 (b)( )2 m YA KL

YAK+π

(c) 2π [(mYA/KL)1/2 (d) 2π (mL/KA)1/2

6. A particle of mass m is executing oscillation about theorigin on the x-axis. Its potential energy is U (x) = k|x|3, where k is a positive constant. If the amplitude ofoscillation is a, then its time period T is : (2001)

(a) proportional to a/1(b) independent of a

(c) proportional to a(d) proportional to a3/2

7. A spring of force constant k is cut into two piecessuch that one piece is double the length of the other.Then the long piece will have a force constant of :

(1999; 2M)(a) 2/3k (b) 3/2k(c) 3k (d) 6k

8. A particle free to move along the x-axis has potentialenergy given by U (x) = k (1 – exp (–x2)] For–∞ ≤ x ≤ + ∞, where k is a positive constant ofappropriate dimensions. Then (1999; 2M)

89

(a) At points away from the origin, the particle is inunstable equilibrium.

(b) for any finite non-zero value of x, there is a forcedirected away from the origin

(c) if its total mechanical energy is k/2, it has itsminimum kinetic energy at the origin

(d) for small displacement from x = 0, then motion issimple harmonic

9. The period of oscillation of simple pendulum of lengthL suspended from the roof of the vehicle which moveswithout friction, down an inclined plane of inclinationα, is given by : (2000; 2M)

(a) 2cosL

gπ

θ (b) asinp2

gl

(c) 2Lg

π (d) 2tanL

gπ

α

10. A particle executes simple harmonic motion betweenx = – A and x = + A. The time taken for it to go fromO to A/2 is T1 and to go from A/2 to A is T2, then

(2000; 2M)(a) T1 < T2 (b) T1 > T2(c) T1 = T2 (d) T1 = 2T2

11. For a particle executing SHM, the displacement x isgiven by x = A cos ωt. Identify the graph whichrepresents the variation of potential energy (PE) as afunction of time t and displacement x : (2003; 2M)

t x

IV

III PE PE

I II

(a) I, III (b) II, IV(c) II, III (d) I, IV

12. A simple pendulum has time period T1. The point ofsuspension is now moved upward according to therelation y = Kt2, (K = 1 m/2) where y is the verticaldisplacement. The time period now becomes T2. The

ratio of 2

12

2

T

T is : (g = 10 m/s2) (2005; 2M)

(a)65 (b)

56

(c) 1 (d)45

13. A block P of mass m is placed on a horizontal frictionlessplane. A second block of same mass m is placed on itand is connected to a spring of spring constant k , thetwo blocks are pulled by distance A. Block Q oscillateswithout slipping. What is the maximum value of frictionalforce between the two blocks? (2004; 2M)

kQ

P

µs

(a) kA/2 (b) kA(c) µs mg (d) Zero

14. The x-t graph of a particle undergoing simple harmonicmotion is shown below. The acceleration of the particleat t = 4/3 s is (2009; M)

t(s)40

–1

1

8 12

(a)2

323 π cm/s2 (b)

32

2π− cm/s2

(c)32

2π cm/s2 (d)

2

323 π− cm/s2

15. A uniform rod of length L and mass M is pivoted atthe centre. Its two ends are attached to the springs ofequal spring constants k. The springs are fixed to rigidsupports as shown in the figure, and the rod is free tooscillate in the horizontal plane. The rod is gentlypushed through a small angle θ in one direction andreleased. The frequency of oscillation is (2009; M)

(a)Mk2

p21

(b)Mk

p21

90

(c)Mk6

p21

(d)M

k24p2

1

16. The mass M shown in the figure oscillates in simpleharmonic motion with amplitude A. The amplitude ofthe point P is (2009; M)

P

k1 k2

M

(a)2

1

kAk

(b)1

2

kAk

(c)21

1

kkAk

+ (d)21

2

kkAk

+

OBJECTIVE QUESTIONSMore than one options are correct:

1. A linear harmonic oscillator of force constant 2 × 106

N/m and amplitude 0.01 m has a total mechanicalenergy of 160 J. Its : (1989; 2M)(a) maximum potential energy is 100 J(b) maximum kinetic energy is 100 J(c) maximum potential energy is 160 J(d) maximum potential energy is zero

2. Three simple harmonic motions in the same directionhaving the same amplitude and same period aresuperposed. If each differ in phase from the next by45°, then : (1999; 3M)

(a) the resultant amplitude is ( )1 2 a+

(b) the phase of the resultant motion relative to thefirst is 90°.

(c) the energy associated with the resulting motion is

( )3 2 2+ times the energy associated with any

single motion(d) the resulting motion is not simple harmonic

3. Function x = A sin2 ωt + B cos2ωt + C sin ωt cos ωtrepresents SHM : (1999; 3M)(a) for any value of A, B and C (except C = 0)

(b) If A = – B, C = 2B, amplitude = | 2 |B(c) If A = B; C = 0(d) If A = B; C = 2B; amplitude = |B|


1. Two masses m1 and m2 are suspended together by amassless spring of spring constant k (Fig.). When the

masses are in equilibrium m1 is removed withoutdisturbing the system. Find the angular frequency andamplitude of oscillation of m2. (1981; 3M)

m1

m2

2. An ideal gas is enclosed in a vertical cylindricalcontainer and supports a freely moving piston of massM. The piston and the cylinder have equal cross-secional area A. Atmospheric pressure is P0, the volumeof the gas is V0. The piston is now displaced slightlyfrom its equilibrium position. Assuming that the systemis completely isolated from is surroundings, show thatthe piston executes simple harmonic motion and findthe frequency of oscillation (1981; 6M)

3. A thin fixed ring of radius 1 m has a positive charge1 × 10–5C uniformly distributed over it. A particle ofmass 0.9 g and having a negative charge of 1 × 10–6

C is placed on the axis at a distance of 1 cm from thecentre of the ring. Show that the motion of thenegatively charged particle is approximately simpleharmonic. Calculate the time period of oscillations.(1982; 5M)

4. Two light springs of force constant k1 and k2 and ablock of mass m are in one line AB on a smoothhorizontal table such that one end of each spring isfixed on rigid supports and the other end is free asshown in the figure. The distance CD between the freeends of the spring is 60 cm. If the block moves alongAB with a velocity 120 cm/s in between the springs,calculate the period of oscillation of the block. (k1 = 1.8N/m, k2 = 3.2 N/m, m = 200 g) (1985; 6M)

k1k2v

60cm

DCA B

m

5. A point particle of mass M attached to one end of amassless rigid non conducting rod of length L. Anotherpoint particle of the same mass is attached to the otherend of the rod. The two particles carry charges + q and– q respectively. This arrangement is held in a regionof a uniform electric field E such that the rod makes asmall angle θ (say of about 5 degree) with the fielddirection as shown in figure. Find an expression for theminimum time needed for the rod to become parallel tothe field after it is set free. (1989; 8M)

91

+q Eθ–q

6. Two non-viscous incompressible andimmiscible liquids of densities ρ and1.5 ρ are poured into the two limbsof a circular tube of radius R andsmall cross section kept fixed in avertical plane as shown in figure.

RO

θ

Each liquid occupies one-fourth the circumference ofthe tube. (1991; 4+4M)(a) Find the angle θ that the radius to the interface

makes with the vertical in equilibrium position.(b) If the whole liquid column is given a small

displacement from its equilibrium position, showthat the resulting oscillations are simpleharmonic.Find the time period of these oscillations.

7. Two identical balls A andB, each of mass 0.1 kg,are attached to twoidentical masslesssprings. The spring-masssystem is constrained to

A

P

B

Q 0.06m

π/6 π/6

move inside a rigid smooth pipe bent in the form of acircle as shown in figure. The pipe is fixed in a circleof radius 0.06 m. Each springs has a natural length of0.06 metre and spring constant 0.1 N/m. Initially, boththe balls are displaced by an angle θ = π/6 radian withrespect to the diameter PQ of the circle (as shown infig.) and released from rest. (1993; 6M)(i) Calculate the frequency of oscillation of ball B.(ii) Find the speed of ball A when A and B are at the

two ends of the diameter PQ.

MATCH THE COLUMN

1. Column 1 describes some situations in which a small object moves. Column II describes some characteristics ofthese motions. Match the situations in Column I with the characteristics in Column II. (2007; 6M)

Column I Column I(A) The object moves on the x-axis under a conservative (p) The object executes a simple harmonic motion.

force in such a way that its 'speed' and 'position'

satisfy 21 2 –v c c x= , where c1 and c2 are positive

constants.(B) The object moves on the x-axis in such a way that (q) The object does not change its direction.

its velocity and its displacement from the originsatisfy v = – kx, where k is a positive constant.

(C) The object is attached to one end of a massless (r) The kinetic energy of the object keeps onspring of constant. The other end of the spring is decreasing

(iii) What is the total energy of the system?

8. A thin rod of length L anduniform cross-section ispivoted at its lowest pointP inside a stationaryhomogeneous and non-viscous liquid. The rod isfree to rotate in a verticalplane about a horizontalaxis passing through P.

d2d1

PThe density d1 of the material of the rod is smaller thanthe density d2 of the liquid. The rod is displaced bysmall angle θ from its equilibrium position and thenreleased. Show that the motion of the rod is simpleharmonic and determine its angular frequency in termsof the given parameters.(1996; 5M)

9. A solid sphere of radius R is floating in a liquid ofdensity ρ with half of its volume submerged. If thesphere is slightly pushed and released, it startsperforming simple harmonic motion. Find the frequencyof these oscillations. (2004; 4M)

10. A mass m is undergoing SHM in thevertical direction about the meanposition y0 with amplitude A andangular frequency ω. At a distancey from the mean position, the massdetaches from the spring.

ym

Assume that the spring contracts anddoes not obstruct the motion of m.Find the distance y (measured fromthe mean position) such that theheight h attained by the block ismaximum. (Aω2 > g). (2005)

92

attached to the ceiling of an elevator. Initiallyeverything is at rest. The elevator starts goingupwards with a constant acceleration a. The motionof the object is observed from the elevator duringthe period it maintains this acceleraion.

(D) The object is projected from the earth's surface (s) The object can change its direction only once.

vertically upwards with a speed 2 e

e

GMR , where

Me is the mass of the earth and Re is the radiusof the earth. Neglect forces from objects other thanthe earth.

2. Column 1 give a list of possible set of parameters measured in some experiments. The variations of the parametersin the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphsgiven in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 X 4 matrix given in theORS. (2008; 7M)

Column I Column I

(A) Potential energy of a simple pendulum (y-axis) as a (p)function of displacement (x-axis)

(B) Displacement (y-axis) as a function of time (x-axis) (q)the for a one dimensional motion at zero or constantacceleration when the body is moving along thepositive x-direction.

(C) Range of a projectile (y-axis) as a function of its (r)velocity (x-axis) when projected at a fixed angle.

(D) The square of the time period (y-axis) of a simple (s)pendulum as a function of its length (x-axis).

ANSWERS

FILL IN THE BLANKS

1. 0.06


1. (c) 2. (b) 3. (b) 4. (d) 5. (b) 6. (a) 7. (b) 8.(d) 9. (a) 10. (a) 11. (a) 12. (a) 13. (a) 14. (d) 15. (c)16. (d)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)1. (b, c) 2. (a, c) 3. (b, d)


1.1

2, m gk A

m kω = = 2. )(

p22

0

0

MgAAPr

MV

+ 3. 0.628 s 4. 2.82s

93

1. Since, v 1

2km

=π

25π =

12 0.2

kπ

or k = 50 × 50 × 0.2 = 500 N/mIf A is the amplitude of oscillation

Total energy = KE + PE

212

kA = 0.5 + 0.4

A =2 0.9

500×

= 0.06 m


1. In SHM frequency with which kinetic energy oscillatesis two times the frequency of oscillation of diplacement.

2. (vM)max = (vN)max∴ ωMAM = ωNAN

orNM

N M

AA

ω=

ω =2

1

kk

km

ω =

∴ (b)

3. When cylinder is displaced by an amount x from itsmeans position, spring force and upthrust both willincrease. Hence,Net restoring force = extra spring force + extra upthrust

k

or F = Net restoring force = (kx + Ax ρ g)

or a = –k Ag

xM

+ρ

Now,1 | |2

afx

=π

=12

k AgM

+ ρπ

∴ (b)4. Modulus of rigidity = F/Aθ

Here, A = L2 and θ xL

=

Therefore, restoring force,F = η Aθ = – η Lx

or acceleration, a FM

= –L

xMη

=

Since, | a | ∝ x and direction of a is opposite to x,hence oscillations are simple harmonic in nature, timeperiod of which is given by

T = 2π displacementacceleration

= 2π xa

2ML

= πη

5.1 2

1 2eq

K KK

K K=

+

YAKL

YA KL

=+

M

M

K

Y, A, L

K = K2

K =1

YAL

5. 2 2MLqE

π6. (a)

–1 1tan5

θ = (b) 26.11

Rπ 7. (i) 1π Hz (ii) 0.0628 m/s (iii) 3.9 × 10–4J

8. 2 1

1

3 ( – )2

g d dd L

ω = 9.1 32 2

gRπ

10. 2g

ω

MATCH THE COLUMN

1. A-p, B-q, C-p, D-q, r2. A-p, B-q,s, C–s, D-q

SOLUTIONS

FILL IN THE BLANKS

94

YAKYA LK

=+

∴ T = 2eq

mK

π

=( )2 m YA LK

YAK+π

6. 3||)( xkxU =

∴ [k] = [ ] 2 –2

3 3[ML T ]

L

U

x=

Now, time period may depend onzyx kT )()amplitude()mass(∝

[M0 L0 T] = [M]x [L]y [ML–1T–2]z

= [Mx+zLy–zT–2z]Equating the powers, we get

– 2z = 1 or z = – 1/2y – z = 0 ory = z = – 1/2

Hence, T ∝ (amplitude)–1/2

∝ (a)–1/2

or T ∝ 1

a

7. l1 = 2l2

Using lll =+ 21

We get, l1 = 23

l

Force constant k ∝ 1

lengthofspring

∴ k1 = 32

k

8.2–( ) (1– )xU x k e=

It is an exponentially increasing graph of potentialenergy (U) with x2. Therefore, U versus x graph will beas shown.From the graph it is clear that at origin,

K

x

U

Potential energy is minimum at x = 0.∴ x = 0 is the state of stable equilibriumNow if we displaced the particle from x = 0 then forsmall displacements the particle tends to regain the

position x = 0 with force 2 2 xekxF −=

For x to be small, xF ∝ .

9. Whenever point of suspension is accelerating

α

g

g sin α

90°+α

Take T = 2eff

lg

π

where aggeffrrr

−=

ar = Acceleration of point of suspension

In this question ar = g sin α (down the plane)

∴ effgag =− ||rr

)a90cos()asin)((2)asin( 22 +°++= gggg

= g cos α.

10.In SHM, time taken to go from x = 0 to x = A/2 isT/12 by using a equation A/2 = Asinωtand Time taken to move from x = A/2 to x = A isT/4 – T/12 = T/6.

11. Potential energy is minimum (in this case zero) at meanposition (x = 0) and maximum at extreme positions(x = +A).At time t = 0, x = A. Hence PE should be maximum.Therefore, graph I is correct. Further in graph III, PEis minimum at x = 0. Hence, this is also correct.

12. y = Kt2

2

2d y

dt = 2K

or ay = 2m/s2 (as K = 1 m/s2)

glT π= 21

and T2 = 2y

lg a

π+

95

∴2

12

2

T

T =

yg a

g

+

=10 2 6

10 5+ =

∴ (a)

13. Angular, frequency of the system,

2k k

m m mω = =

+

mkAAa2

2max =ω=

This acceleration to the lower block is provided byfriction.

∴fmax = mamax 2kAmm

= 2

kA=

∴ (a)

14. sTtAx 8;sin =ω= and A = 1 cm

ππ

−=ωω−=34

82

sin8

41sin 2

22 tAa

323

23

16

22 π−=×π−= cm/s2

∴ (d)

15. a12

22

?2

2MLLLk =×

a12

?2

22 ML

Lk

−=

?6

aM

k−=

⇒k

MT6

p2= ⇒Mkf 6

p21=

∴ (c)

16. Let 1x and 2x are the amplitudes of points P and Q

respectively.

Axx =+ 21

2211 xkxk = Pk1 k2

MQ

⇒21

21 kk

Akx

+=

∴ (d)


1. (a) KEmax = 2 6 –2 21 12 10 (10 ) 100

2 2KA J= × × × =

So Umin = 160 - 100 = 60 J KEmin = 0So Umax = 160 J

2. From superposition principle.y = y1 + y2 + y3= a sin ωt + a sin (ωt + 45°) + a sin (ωt + 90°)= a [sin ωt + sin (ωt + 90°)] + a sin (ωt + 45°)= 2a sin (ωt + 45°) cos 45° + a sin (ωt + 45°)

= ( )2 1+ a sin (ωt + 45°)

= A sin (ωt + 45°)Therefore, resultant motion is simple harmonic ofAmplitude.

A = ( )2 1 a+

and which differ in phase by 45° relative to the first.Energy in SHM ∝ (amplitude)2

[E = 2 212

mA ω ]

∴ resultant

single

EE

=2A

a

( )22 1 (3 2 2)= + = +

∴ Eresultant = ( )3 2 2+ Esingle

3. For A = – B and C = 2BX = Bcos 2ωt + B sin 2ωt

+=

4p

?2sin2 tB

If A = B, C = 2B, then tBBx ?sin 2+= ,

This is also equation of SHM about the point X = Bof amplitude B.


1. (i) When m1 is removed, only m2 is left. Therefore,angular frequency :

ω = 2

km

(ii) Let x1 be the extension when only m2 is left. Then,kx1 = m2g

96

or x1 = 2m gk ....(1)

Similarly, let x2 be extension in equilibrium when bothm1 and m2 are suspended. Then,

(m1 + m2)g = kx2

x2 = 1 2( )m m g

k+

...(2)

From Eqs. (1) and (2), amplitude of oscillation :

A = x2 – x1 1m gk

=

2. In equilibrium, P = 0Mg

PA

+ ...(1)

When piston is displacedslightly by an amount x ,change in volume,

dV = – Ax x

Since, the cylinder is isolatedfrom the surroundings,process, is adiabatic is nature.In adiabatic process.

dPdV = – P

Vγ

or increase in pressure inside the cylinder,

( )– ( )PdP dVVγ= = –

0

0( . )

MgPA A x

V

+ γ

This increase in pressure when multiplied with area ofcross-section A will give a net upward force (or therestoring force). Hence

F = – (dP) A = –2

0

0

P A MgAx

V

+γ

or,F

aM

= =2

0

0–

P A MgAx

V M

+γ

Since, a ∝ – x, motion of the piston is simple harmonicin nature. Frequency of this oscillation is given by :

f =12

axπ

= MVMgAAP

0

20 )(

21 +γπ

Hence,)(

22

0

0

MgAAP

MVT

+γπ=

3. Given Q = 10–5Cq = 10–6CR = 1 m andm = 9 × 10–4kg

x m–q

E

+Q

R

Electric field at a distance x from the centre on the axisof a ring is given by :

E =2 2 3 / 2

0

14 ( )

Qx

R xπε +

If x < < R, R2 + x2 = R2

and E = 304

Qx

Rπε

Net force on negatively charged particle would be qEand towards the centre of ring. Hence, we can write

F = 30

–(4 )

Qqx

Rπε

or acceleration Fam

=3

0–

(4 )

Qqx

mR=

πε

as | a | ∝ – x and nature is restoring motion of theparticle is simple harmonic in nature. Time period ofwhich will be given by :

T = 2xa

π

or T =3

0(4 )2

mRQq

πεπ

Substituting the values, we get

T =( )3–4

9 –5 –6

(9 10 ) 12

(9 10 )(10 )(10 )

×π

×

= 0.628 s

4. Between C and D block will move with constant speedof 120 cm/s. Therefore, period of oscillation will be(starting from C).

T = 2 12 2CD DCT T

t t+ + +

Here, 11

2 mTk

= π and 22

2 mTk

= π

and tCD = tDC 60

120= = 0.5s

∴ T =2 0.2 2 0.20.5 0.52 3.2 2 1.8π π+ + +

97

(m = 200g = 0.2 kg)T = 2.82 s

5. A torque will act on the rod, which tries to align therod in the direction of electric field. This torque willbe of restoring nature and has a magnitude PE sin θ.Therefore, we can write

τ = –PE sin θor Iα = –PE sin θ ...(1)

Here2

22L

I M = =

2

2ML

and P = qL

Further since θ is small, we can write, sin θ ≡ θSubstituting these values in Eq. (1) we have

2

2ML

α = – (qL) (E) θ

or α =2

–qE

ML θ

As α is porportional to – θ and its nature is restoringmotion of the rod is simple harmonic in nature, timeperiod of which is given by:

T = 2 22MLqE

θπ = π

α

The desired time will be,

T = 4 2 2T ML

qEπ

=

6. (a) In equilibrium, pressure of same liquid at same levelwill be same.Therefore, P1 = P2or P + (1.5 ρgh1) = P + (ρgh2)(P = pressure of gas in empty part of the tube)

θ

θ

θ

1.5

h1

h2

21ρ

ρ

∴ 1.5h1 = h21.5 [R cos θ – R sin θ] = ρ (R cos θ + R sin θ)or 3 cos θ – 3 sin θ = 2 cos θ + 2 sin θor 5 tan θ = 1

θ = tan–115

(b) When liquids are slightly disturbed by an angle β.Net restoring pressure ∆P = 1.5 ρgh + ρgh = 2.5 ρghThis pressure will be equal at all sections of the liquid.Therefore, net restoring torque on the whole liquid.

h

h

β

β

τ = – (∆P)(A) (R)or, τ = – 2.5 ρgh AR = – 2.5 ρgAR [R sin (θ + β) – R sin θ] = – 2.5 ρg AR2 [sin θ cos β + sin β cos θ – sinθ]Assuming cos β = 1 and sin β ≈ β (as β is small)∴ τ = – (2.5 ρAgR2 cos θ)βor Iα = – (2.5 ρAgR2 cos θ)β ...(1)Here, I = (m1 + m2) R2

2)5.1(22

RAR

AR

ρ

π

+ρ

π

=

= (1.25πR3ρ)A

and cosθ =5

0.9826

=

Substituting in Eq. (1), we have

α = –(6.11)

Rβ

As angular acceleration is proportional to – β, motionis simple harmonic in nature.

11.622

RT π=

αβ

π=

7. Given - Mass of each block A and B, m = 0.1 kgRadius of circle, R = 0.06 m

A B

θ θθ π = /6

x=Rθx=Rθ

Natural length of spring l0 = 0.06 π = πR (Half circle) and spring constant, k = 0.1 N/mIn the stretched position elongation in each spring

x = Rθ.By energy method of calculation of angular frequencyof oscillation

98

constant21

2)2(21

2 22 =×+θ× mvRk

By differentiating this with respect to time, we get

04 2 =+θω mvakR ) and ( α=ω= RaRvQor (mR2)α = – 4kR2θ

or α = –k

m4 θ

∴ Frequence of oscillation f1

2=

πaccelerationdisplacement

=12

απ θ

f =1 42

kmπ

Substituting the values, we have

f =1 4 0.1 12 0.1

× =π π

Hz

(ii) In stretched position, potential energy of the systemis

PE = 2 212 2 4

2k x kx =

and in mean position, both the blocks have kineticenergy only. Hnce,

KE = 2 212

2mv mv =

From energy conservation

–∆PE = ∆KE∴ 4kx2 = mv2

∴ v = 2 2k kx Rm m

= θ


v = 2 (0.06) (π/6) 0.10.1

or v = 0.0628 m/s(iii) Total energy of the system E = PE in stretchedpositionor = KE in mean position

E = mv2 = (0.1) (0.0628)2Jor E = 3.9 × 10–4J

8. Let S be the area of cross-section of the rod.In the displaced position, as shown in figure. weight(W) and upthrust (FB) both pass through its centre ofgravity G.

Q

Fa

r 1G

WP

Given that d1 < d2. Therefore, W < FB.Therefore, net force acting at G will be :F = FB – W = (SLg) (d2 – d1) upwards Restoring torqueof this force about point P is

τl = F × r = (SLg) (d2 – d1) (Q G)

or τ = – (SLg)(d2 – d2) sin2L θ

Here, negative sign shows the restoring nature oftorque

or τ =2

2 1( – )–

2SL g d d θ

sin θ ≈ θ for small values of θFrom Eq. (1), we see that

| τ | ∝ θHence motion of the rod will be simple harmonic.Rewriting Eq. (1) as —

2

2dIdt

θ= –

22 1( – )

2SL g d d θ

...(2)

Here, I = moment of inertia of rod about an axispassing through P.

2

3MLI = =

21( )3

SLd L

Substituting this value of I in Eq. (1), we have

2

2d

dt

θ=

2 1

1

( – )3–2

g d dd L

θ

Comparing this equation with standard differentialequation of SHM, i.e.,

2

2d

dt

θ= – ω2θ

The angular frequency of oscillation is —

ω =2 1

1

3 ( – )2

g d dd L

9. Half of the volume of sphere is submerged.For equilibrium of sphere,weight = upthrust

99

∴ Vρsg = ( )( )2 LV

gρ

ρs =2Lρ

...(1)

When slightly pushed downwards by x, weight willremain as it is while upthrust will increase. Theincreased upthrust will become the net restoring force(upwards).F = – (extra upthrust)= – (extra volume immersed) (ρL) (g)or ma = (πR2)xρLg (a = acceleration)

∴34

3 2R a

ρ π = (πR2ρLg)x

∴ a = –32

gx

R

as | a | ∝ x and its nature is restoring motion is simpleharmonic

f =12

axπ

=1 32 2

gRπ

10. At distance y above the mean position velocity of theblock.

v = 2 2–A yω

After detaching from the spring net downwardacceleration of the block will be g.Therefore, total height attained by the block above themean position,

h =2

2v

yg

+

=2 2 2( – )

2A y

yg

ω+

For h to be maximum 0dhdy

=

Putting 0dhdy

= , we get

y = 2g

ω

MATCH THE COLUMN

1. (A) As 221 xccv −= is similar to 22 xAv −ω=

(B) At x = 0, v = 0 and particle comes to rest.(C) As in the lift’s reference frame, a pseudo force acts on

its, which is constant in nature. Hence the motion isstill simple harmonic.

(D) As speed is 2 times the escape speed, so the object

does not change its direction.

101

CHAPTER-9PROPERTIES OF MATTER

FILL IN THE BLANKS

1. A wire of length L and cross sectional area A is made of a material of Young's modulus Y. If the wire is stretchedby an amount x, the work done is .... (1987, 2M)

2. A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container.A massless piston of area A floats on the surface of the liquid. When a mass M is placed on the piston to compressthe liquid the fractional change in the radius of the sphere, δR/R, is .... (1988, 2M)

3. A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are γ1 and γ2respectively. If the temperature of both mercury and the metal are increased by an amount ∆T, the fraction of thevolume of the metal submerged in mercury changes by the factor ...... (1991, 2M)

4. A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross-sectional areais 10 cm2, the water velocity is 1ms –1 and the pressure is 2000 Pa. The pressure of water at another point where thecross-sectional area is 5 cm2, is .... Pa. (Density of water = 103kg/m3) (1994, 2M)

5. A uniform rod of length L and density P is being pulled along a smooth floor with a horizontal acceleration a (seefig.) The magnitude of the stress at the transverse cross-section passing through the mid point of the rod is ....

(1983, 1M)L

a

TRUE/FALSE

1. A barometer made of a very narrow tube (see fig) is placed at normal temperatureand pressure. The coefficient of volume expansion of mercury is 0.00018/°C andthat of the tube is negligible. The temperature of mercury in the barometer is nowraised by 1°C but the temperature of the atmosphere does not change. Then, themercury height in the tube remains unchanged. (1983; 2M)

Vaccum

Hg

2. Water in a closed tube (see fig.) is heated wih one arm vertically placed above a lamp. Water will begin to circulatealong the tube in counter-clockwise direction. (1983; 2M)

A B

3. A block of ice with a lead shot embedded in it is floating on water contained in a vessel. The temperature of thesystem is maintained at 0°C as the ice melts. When the ice melts completely the level of water in the vessel rises.(1986; 3M)

102

OBJECTIVE QUESTIONS

Only One option is correct :

1. A metal ball immersed in alcohol weighs W1 at 0°C andW2 at 50°C. The coefficient of cubical expansion of themetal is less than that of the alcohol. Assuming thatthe density of the metal is large compared to that ofalcohol, it can be shown that : (1987; 2M)(a) W1 > W2 (b) W1 = W2(c) W1 < W2 (d) all of these

2. A vessel containing water is given a constantacceleration a towards the right along a straighthorizontal path. Which of the following diagramrepresents the surface of the liquid? (1981; 2M)

(a) a (b) a

(c) a (d) None of these

3. The following four wires are made of the same material.Which of these will have the largest extension whenthe same tension is applied? (1981; 3M)(a) Length = 50 cm, diameter = 0.5 mm(b) Length = 100 cm, diameter = 1 mm(c) Length = 200 cm, diameter = 2 mm(d) Length = 300 cm, diameter = 3 mm

4. A body floats in a liquid contained in a beaker. Thewhole system as shown in figure falls freely undergravity. the upthrust on the body is : (1982;3M)

(a) zero(b) equal to the weight of liquid displaced(c) equal to the weight of the body in air(d) equal to the weight of the immersed portion of the

body

5. A U-tube of uniform cross-section is partially filledwith a liquid I. Another liquid II which does not mixwith liquid I is poured into one side. It is found thatthe liquid levels of the two sides of the tube are thesame, while the level of liquid 1 has risen by 2 cm. Ifthe specific gravity of liquid I is 1.1, the specificgravity of liquid II must be : (1983; 1M)(a) 1.12 (b) 1.1(c) 1.05 (d) 1.0

6. A vessel contains oil (density = 0.8 g/cm3) over mercury(density = 13.6 g/cm3). A homogeneous sphere floatswith half of its volume immersed in mercury and theother half in oil. The density of the material of thesphere in g/cm3 is : (1988; 2M)(a) 3.3 (b) 6.4(c) 7.2 (d) 12.8

7. A homogeneous solid cylinder of length L (L < H/2)cross-sectional area A/5 is immersed such that it floatswith its axis vertical at the liquid-liquid interface withlength L/4 in the denser liquid as shown in the figure.The lower density liquid is open to atmosphere havingpressure P0. Then density D of solid is given by :

(1995; 2M)

d

2d

(a)54

d (b)45

d

(c) 4d (d) 5d

8. Water from a tap emerges vertically downwards withan initial speed of 1.0 m/sec. The cross-sectional areaof tap is 10–4m2. Assume that the pressure is constantthroughout the stream of water and that the flow issteady, the cross-sectional area of stream 0.15 m belowthe tap is : (1998; 2M)(a) 5.0 × 10–4 m2 (b) 1.0 × 10–4 m2

(c) 5.0 × 10–5 m2 (d) 2.0 × 10–5 m2

9. A given quantity of an ideal gas is at pressure P andabsolute temperature T. the isothermal bulk modulus ofthe gas is : (1998; 2M)

(a)23

P (b) P

(c)32

P (d) 2P

10. A closed compartment containing gas is moving withsome acceleration in horizontal direction. Neglect effectof gravity. Then, the pressure in the compartment is :(1999; 2M)(a) same everywhere(b) lower in front side(c) lower in rear side(d) lower in upper side

11. A large open tank has two holes in the wall. One is a

103

square hole of side L at a depth y from the top and theother is a circular hole of radius R at a depth 4y fromthe top. When the tank is completely filled with water,the quantities of water flowing out per second fromboth holes are the same. Then, R is equal to :

(2000; 2M)

(a) L/ 2π (b) 2πL(c) L (d) L/2π

12. A hemispherical portion of radius R is removed fromthe bottom of a cylinder of radius R. The volume of theremaining cylinder is V and mass M. It is suspendedby a string in a liquid of density ρ, where it staysvertical. The upper surface of the cylinder is at a depthh below the liquid surface. The force on the bottom ofthe cylinder by the liquid is : (2001; 2M)

h

2R

ρ

(a) Mg (b) Mg - Vρg(c) Mg + πR2hρg (d) ρg (V + πR2h)

13. When a block of iron floats in mercury at 0°C, fractionk1 of its volume is submerged, while at the temperature60°C, a fraction k2 is seen to be submerged. If thecoefficient of volume expansion of iron is γFe and thatof mercury is γHg, then the ratio of k1/k2 can beexpressed as : (2001; 2M)

(a)Fe

Hg

1 601 60

+ γ+ γ (b)

Fe

Hg

1–601 60

γ+ γ

(c)Fe

Hg

1 601–60

+ γγ (d)

Hg

Fe

1 60

1 60

+ γ

+ γ

14. A wooden block, with a coin placed on its top, floatsin water as shown in figure. The distance l and h areshown there. After some time, the coin falls into thewater. Then : (2002; 2M)

l

h

Coin

(a) l decreases and h increases(b) l increases and h decreases(c) both l and h increase(d) both l and h decrease

15. The adjacent graph shows the extension (∆l) of a wireof length 1 m suspended from the top of a roof at oneend and with a load W connected to the other end. Ifthe cross sectional area of the wire is 10–6m2, calculatethe Young's modulus of the material of the wire.

(2003; 2M)

4

3

2

1

20 40 60 80

∆l (×10 m)–4

W (N)

(a) 2 × 1011 N/m2 (b) 2 × 10–11 N/m2

(c) 3 × 1012 N/m2 (d) 2 × 1013 N/m2

16. The pressure of a medium is changed from 1.01 × 105

Pa to 1.165 × 105 Pa and change in volume is 10%keeping temperature constant. The bulk modulus ofthe medium is : (2005; 2M)(a) 204.8 × 105 Pa (b) 102.4 × 105 Pa(c) 51.2 × 105 Pa (d) 1.55 × 105 Pa

17. Water is filled in a cylindrical container to a height of3 m. The ratio of the cross-sectional area of the orificeand the beaker is 0.1. The square of the speed of theliquid coming out from the orifice is (g = 10 m/s2) :

(2005; 2M)

3m

52.5cm

(a) 50 m2/s2 (b) 50.5 m2/s2

(c) 51 m2/s2 (d) 52 m2/s2

18. Water is filled up to aheight h in a beaker ofradius R as shown inthe figure. The densityof water is ρ, thesurface tension ofwater is T and theatmospheric pressureis P0. Consider avertical section ABCDof the water columnthrough a diameter ofthe beaker. The force

B

A

C

D

2R

h

104

on water on one side of this section by water on theother side of this section has magnitude. (2007; 2M)(a) |2 P0 Rh + ρR2gh – 2RT|(b) |2 P0 Rh + Rρgh2 – 2RT|(c) |P0 πR2 + Rρgh2 – 2RT|(d) |P0 πR2 + Rρgh2 + 2RT|

19. A glass tube of uniform internal radius (r) has a valveseparating the two identical ends. Initially,< the valveis in a tightly closed position. End I has a hemispherialsoap bubble of radius r. End 2 has sub-hemisphericalsoap bubble as shown in figure.

Just after opening the value. (2008; 3M)(a) air from end 1 flows towards end 2. No change in

the volume of the soap bubbles(b) air from end 1 flows towards end 2. Volume of the

soap bubble at end 1 decreases(c) no change occurs(d) air from end 2 flows towards end 1. Volume of the

soap bubble at end 1 increases

OBJECTIVE QUESTIONS

More than one options are correct?1. The spring balance A read 2

kg with a block m suspendedfrom it. Balance B reads 5 kgwhen a beaker with liquid isput on the pan of the balance.The two balances are now soarranged that the hangingmass is inside the liquid in thebeaker as shown in the figure.In thissituation : (1985; 2M)

m

B

A

(a) the balance A will read more than 2 kg(b) the balance B will read more than 5 kg(c) the balance A will read less than 2 kg and B will

read more than 5 kg(d) the balances A and B will read 2 kg and 5 kg

respectively


1. Two identical cylindrical vessels with their bases atthe same level each contain a liquid of density ρ. Theheight of the liquid in one vessel is h1 and in the otheris h2. The area of either base is A. What is the workdone by gravity in equalising the levels when the twovessels are connected? (1981; 4M)

2. A wooden plank of length 1 m and uniform crosssection is hinged at one end to the bottom of a tankas shown in figure. The tank is filled with water uptoa height 0.5 m. The specific gravity of the plank is 0.5.Find the angle θ that the plank makes with the verticalin the equilibrium position. (exclude the case θ = 0°)

(1984; 8M)

θ

3. A ball of density d is dropped on to a horizontal solidsurface. It bounces elastically from the surface andreturns to its original position in a time t1. Next the ballis released and it falls through the same height beforestriking the surface of a liquid of density dL.

(1992; 8M)(a) If d < dL, obtain an expression (in terms of d, t1 and

dL) for the time t2 the ball takes to come back tothe position from which it was released.

(b) Is the motion of the ball simple harmonic?(c) If d = dL, how does the speed of the ball depend

on its depth inside the liquid? Neglect all frictionaland other dissipative forces. Assume the depth ofthe liquid to be large.

4. A container of large uniform cross-sectional area Aresting on a horizontal surface, holds two immiscible,non-viscous and incompressible liquids of densities dand 2d, each of height H/2 as shown in figure. Thelower density liquid is open to the atmosphere havingpressure P0. (1995; 5+5)(a) A homogeneous solid cylinder of length

L (L < H/2), cross-sectional area A/5 is immersedsuch that it floats with its axis vertical at theliquid-liquid interface with length L/4 in the denserliquid. Determine:

d

2d

H/2

H/2

x

h

(i) the density D of the solid.

105

(ii) the total pressure at the bottom of the container.(b) The cylinder is removed and the original

arrangment is restored. A tiny hole of areas (s < < A) is punched on the vertical side of thecontainer at a height h (h < H/2). Determine :

(i) the initial speed of efflux of the liquid at the hole,(ii) the horizontal distance x travelled by the liquid

initially, and(iii) the height hm at which the hole should be punched

so that the liquid travels the maximum distance xminitially. Also calculate xm.

(Neglect the air resistance in these calculations)

5. A large open top container of negligile mass anduniform cross sectional area A has a small hole ofcross sectional area A/100 in its side wall near thebottom. The container is kept on a smooth horizontalfloor and contains a liquid of density ρ and mass m0.Assuming that the liquid starts flowing out horizontallythrough the hole at t = 0. Calculate : (1997C; 5M)(i) the acceleration of the container and(ii) its velocity when 75% of the liquid has drained

out.

6. A non-viscous liquid ofconstant density 1000kg/m3 flows in streamlinemotion along a tube ofvariable cross section.The tube is kept inclinedin the vertical plane asshown in the figure.

P

2m

Q

5m

The area of cross section of the tube at two points Pand Q at heights of 2 m and 5 m are respectively 4 ×10–3m2 and 8 × 10–3m2. The velocity of the liquid atpoint P is 1 m/s. Find the work done per unit volumeby the pressure and the gravity forces as the fluidflows from point P and Q. (1997; 5M)

7. A wooden stick of length L, radius R and density ρ hasa small metal piece of mass m (of negligible volume)attached to its one end. Find the minimum value for themass m (in terms of given parameters) that would makethe stick float vertically in equilibrium in a liquid ofdensity σ (> ρ). (1999; 10M)

8. A uniform solid cylinder of density 0.8 g/cm3 floats inequilibrium in a combination of two non-mixing liquidsA and B with its axis vertical. The densities of theliquids A and B are 0.7 g/cm3 and 1.2 g/cm3, respectively.The height of liquid A is hA = 1.2 cm. The length of thepart of the cylinder immersed in liquid B is hB = 0.8 cm.

(2002; 5M)

Air

A

B

h

hA

hB

(a) Find the total force exerted by liquid A on thecylinder.

(b) Find h, the length of the part of the cylinder in air.(c) The cylinder is depressed in such a way that its

top surface is just below the upper surface ofliquid A and is then released. Find the accelerationof the cylinder immediately after it is released.

9. A soap bubble is being blown at the end of verynarrow tube of radius b. Air (density ρ) moves with avelocity v inside the tube and comes to rest inside thebubble. The surface tension of the soap solution is T.After sometimes the bubble, having grown to radius Rseparates from the tube. Find the value of R. Assumethat R > > b so, that you can consider the air tobe falling normally on the bubble's surface.

(2003; 4M)10. A liquid of density 900

kg/m3 is filled in acylindrical tank of upperradius 0.9 m and lowerradius 0.3 m. A capillarytube of length l is attachedat the bottom of the tankas shown in the figure.The capillary has outerradius 0.002 m and innerradius a. When pressureP is applied at the top ofthe tank, volume flow rate

0.9m

0.3m

l

H

of the liquid is 8 × 10–6m3/s and if capillary tube isdetached, the liquid comes out from the tank with avelocity 10 m/s. Determine the coefficient of viscosityof the liquid. (2003; 4M)[Given ; πa2 = 10–6m2 and a2/l =2 × 10–6m]

11. A container of width 2a isfilled with a liquid. A thinwire of mass per unit length λis gently placed over theliquid surface in the middle ofthe surface as shown in thefigure. As a result, the liquidsurface is depressed by a

y

2a

106

distance y (y < < a). Determine the surface tension ofthe liquid. (2004; 2M)

12. A small sphere falls from rest in a viscous liquid. Dueto friction, heat is produced. Find the relation betweenthe rate of production of heat and the radius of thesphere at terminal velocity. (2004; 2M)

13. Consider a horizontally oriented syringe containingwater located at a height of 1.25 m above the ground.The diameter of the plunger is 8 mm and the diameterof the nozzle is 2 mm. The plunger is pushed with aconstant speed of 0.25 m/s. Find the horizontal rangeof water stream on the ground.(Take g = 10m/s2). (2004; 2M)

D = 8 mm D = 2 mm

1.25m

14. A U-shaped tube contains a liquid of density ρ and itis rotated about the line as shown in the figure. Findthe difference in the levels of liquid column.

(2005; 2M)

H

ω

L

15. Two soap bubbles A and B are kept in a closedchamber where the air is maintained at pressure8 N/m2. The radii of bubbles A and B are 2 cm and 4cm, respectively. Surface tension of the soap-waterused to make bubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air inbubbles A and B, respectively. [Neglect the effect ofgravity.] [2008]

16. A cylindrical vessel of height 500 mm has an orifice(small hole) at its bottom. The orifice is initially closedand water is filled in it up to height H. Now the top iscompletely sealed with a cap and the orifice at thebottom is opened. Some water comes out from theorifice and the water level in the vessel becomessteady with height of water column being 200 mm. Find

the fall in height (in mm) of water level due to openingof the orifice.[Take atmospheric pressure = 1.0 × 105 N/m2, densityof water = 1000 kg/m3 and g = 10 m/s2. Neglect anyeffect of surface tension.] [2008]

COMPREHENSION

PassageA wooden cylinder of diameter 4r, height h and density

ρ/3 is kept on a hole of diameter 2r of a tank, filled withliquid of density ρ as shown in the figure. The height ofthe base of cylinder from the base of tank is H.1. Now level of the liquid starts decreasing slowly when

the level of liquid is at a height h1 above the cylinderthe block just starts moving up. At what value of h1,will the block rise : (2006; 5M)

2r

ρ/3 h

ρ

h1

(a) 4h/9 (b) 5h/9

(c)53h

(d) remains same

2. The block in the above question is maintained at theposition by external means and the level of liquid islowered. The height h2 when this external force reducesto zero is : (2006; 5M)

ρ/3

ρh 2

(a)49h

(b)59h

(c) remains same (d)23h

3. If height h2 of water level is further decreased, then:(2006; 5M)

(a) cylinder will not up and remains at its originalposition

(b) for h2 = h/3, cylinder again starts moving up(c) for h2 = h/4, cylinder again starts moving up(d) for h2 = h/5 cylinder again starts moving up

Passage

A small spherical monoatomic ideal gas bubble

=γ

35

107

is trapped inside a liquid of density ρ l (see figure).Assume that the bubble does not exchange any heat withthe liquid. The bubble contains n moles of gas. Thetemperature of the gas when the bubble is at the bottomis T0, the height of the liquid is H and the atmosphericpressure is p0 (Neglect surface tension)

Hy

p0 Liquid

4. As the bubble moves upwards, besides the buoyancyforce the following forces are acting on it

(2008; 4M)(a) Only the force of gravity(b) The force due to gravity and the force due to the

pressure of the liquid(c) The force due to gravity, the force due to the

pressure of hte luiquid and the force due toviscosity of the liquid

(d) The force due to gravity and the force due toviscosity of the liquid

5. When the gas bubble is at a height y from the bottom,its temperature is (2008; 4M)

(a)

5/2

0

00

ρ+ρ+

gypgHpT

l

l(b)

5/2

0

00

)(

ρ+

−ρ+gHp

yHgpTl

l

(c)

5/3

0

00

ρ+ρ+

gypgHpTl

l(d)

5/3

0

00

)(

ρ+

−ρ+gHp

yHgpTl

l

6. The buoyancy force acting on the gas bubble is(Assume R is the universal gas constant)

(2008; 4M)

(a) 5/70

5/20

0 )()(

gypgHpnRgTl

ll ρ+

ρ+ρ

(b) 5/30

5/20

0

)]([)( yHgpgHpnRgT

ll

l

−ρ+ρ+ρ

(c) 5/80

5/30

0 )()(

gypgHpnRgTl

ll ρ+

ρ+ρ

(d) 5/20

5/30

0

)]([)( yHgpgHpnRgT

ll

l

−ρ+ρ+ρ


This question contains, statement-I (assertion) andstatement-II (reasons).

1. Statement-I : The stream of water flowing at highspeed from a garden hose pipe tends to spread like afountain when held vertically up, but tends to narrowdown when held vertically down.

(2008; 3M)Because :Statement-II : In any steady flow of an incompressiblefluid, the volume flow rate of the fluid remains constant.(a) Statement-I is true, statement -II is true, statement-

II is a correct explanation for statement-I(b) statement-I is true, statement-II is true; statement-

II is NOT a correct explanation for statement-I(c) statement-I is true, statement-II is false.(d) statement-I is false, statement-II is true.

108

MATCH THE COLUMN

*1. Column II shows five systems in whcih two objects are labelled as X and Y. Also in each case a point P is shwon.Column I gives some statements about X and/or Y. Match these statements to the appropriate system(s) fromColumn II.

Column - I Column - II(A) The force exerted (p) Block Y of mass M left on a fixed inclined plane X, slides,

by X on Y has on it with a constant velocity.magnitude Mg.

(B) The gravitational (q) Two ring magnets Y and Z, each of mass M, are kept inpotential energy frictionless vertical plastic stand so that they repel eachinnermost other. Y rests on the base X and Z hangs in air incontinuously equilibrium. P is the topmost point of the stand on theincreasing. common axis of the two rings. The whole system is in a

lift that is going up with a constant velocity.

(C) Mechanical (r) A pulley Y of mass m0 is fixed to a tablethrough a clampenergy of the X. A block of mass M hangs from a string that goes oversystem X + Y the pulley and is fixed at point P of the table. The wholeis continuously system is kept in a lift that is going down with a constantdecreasing. velocity.

(D) The torque of (s) A sphere Y of mass M is put in a nonviscous liquid Xthe weight of Y kept in a container at rest. The sphere is released and itabout point P moves down in the liquid.

(t) A sphere Y of mass M is falling with its terminal velocityin a viscous liquid X kept in a container.

ANSWERS

FILL IN THE BLANKS

1.21

2YA

xL

2. 3MgAK 3. T

T∆+

∆−

1

21

?1)??(

4. 500 5. 2ρ αL

TRUE/FALSE1. F 2. F 3. F


1. (c) 2. (c) 3. (a) 4. (a) 5. (b) 6. (c) 7. (a)8. (c) 9. (b) 10. (b) 11. (a) 12. (d) 13. (a) 14. (d)15. (a) 16. (d) 17. (a) 18. (b) 19. (b)


1. (b, c)


1. 21 2( – )

Agh h

rρ

2.45° 3. (a) 1

–L

L

t dd d (b) no (c) remains same

YX

P

PZY

X

PY

X

PX

Y

PX

Y

109

SOLUTIONS

FILL IN THE BLANKS

1. W = 212

Kx

Here, YA

KL

=

∴ W = 21

2YA

xL

2. ∆P = MgA

VV∆

= P Mg

K AK∆

=

Now as, V34

3R= π or V ∝ R3

∴ 3V R

V R∆ ∆ =

or 13

R VR V

∆ ∆ = ...(1)

∴ From Eq. (1)

RR

∆= 3

MgAK

3. The condition of floating is,Weight = Upthrust

4. (i) 54d

(ii) P = 0(6 )

4dg H L

P P+

= + (b) (i) (3 – 4 )2gH h (ii) (3 – 4 )h H h (iii) At

3 3;8 4Hh H=

5. (i) 50g

(ii) 0

2gm

Aρ 6. 29025 J/m3, 29400 J/m3

7. πR2 L ( )–ρσ ρ 8. (a) zero (b) 0.25 cm (c) g/6 9. 24T

vρ 10. 1

720 Ns/m2

11. 2ay

λ12.

5dQ rdt

α 13. 2m 14. H 2 2

2Lg

ω=

COMPREHENSION

1. (c) 2.(a) 3.(a)

ASSERTION AND REASON

1. (a)

MATCH THE COLUMN

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

Vρ1g = Viρ2g (ρ1 = density of metal, ρ2 = density ofmercury)

∴iV

V = 1

2

ρρ

⇒ fraction of volume of metal submerged in mercury= x (say)Now, when the temperature is increased by ∆T.

ρ′1 T∆γ+ρ

=1

1

1 and ρ′2 T∆γ+ρ

=2

2

1

∴1 2 1 2

1 2 2 1

1 1'1 1

T TxT T

ρ + γ ∆ ρ + γ ∆= = + γ ∆ ρ ρ + γ ∆

TT

xx

TT

xx

∆γ+∆γ+

−=′

−⇒∆γ+∆γ+

=′

1

2

1

2

11

1111

)?1()??(

1

21

TT

∆+∆−

=

4. From continuity equationA1v1 = A2v2

∴ v2 1

12

A vA

=

=

10(1)

5

= 2 m/s

Applying Bernoulli's theorem at 1 and 2

110

v1

1 2

v2

222

211 ?

21

?21

vPvP +=+

∴ P2 = ( )2 21 1 2

1–

2P v v+ ρ

= 312000 10 (1–4)

2 + ×

∴ P2 = 500 Pa

5. Let A be the area of cross-section of the rod. From theFBD of rod at mid-point

L/2

m T

Mass (m) = volume × density ρ××= AL2

∴ T = ma aLAρ=21

∴ Stress aLAT

ρ==21

TRUE/FALSE

1. .atm121 ghPP ρ=== On changing the temperature,

g will not change. So, =ρh constant. On increasing

the temperature of mercury, its density will decrease.Hence, level of mercury in barometer tube will increase.

2. When tube is heated the density of water at A willdecrease, hence the water will rise up or it will circulatein clockwise direction.

3. When ice melts, level of water does not change. Incase of lead, it was initially floating i.e., it would haddisplaced the water equal to the weight of lead. So,volume of water displaced would be,

V1 w

m=

ρ (m = mass of lead)

Now, when ice melts, lead will sink and it woulddisplace the water equal to volume of lead itself. So,volume of water displaced in this case would be,

l

mV

ρ=2

Now, as ρ l > ρw, V2 < V1 or level will fall.


1. Wapp = Wactual – UpthrustUpthrust F = VSρLgHere, VS = volume of solid, ρL = density of liquidAt higher temperature

F' =' 'S LV gρ

∴' ''

.S L

S L

VFF V

ρ=

ρ = (1 )(1 )

S

L

+ γ ∆θ+ γ ∆θ

Since, γS < γL (given)∴ F′ < F

or 12 WW >

2. Net force on the free surface of the liquid in equilibrium(from accelerated frame) should be perpendicular to it.Forces on a water particle P on the free surface havebeen shown in the figure. In the figure, ma is thepseudo force.

90° FP ma

Fnet mg

3.

Yd

FlAYFl

l

π==∆

4

2

or (∆l) ∝ 2l

d

Now, 2dl

is maximum in (a).

4. In a freely falling system geff = 0 and since, Upthrust= ViρLgeff(Vi = immersed volume, ρL = density of liquid)∴ Upthrust = 0

111

5. P1 = P2

h

1 2

P0 + ρ1gh = P0 + ρII gh∴ ρ1 = ρII

6. Let density of material of sphere (in g/cm3) be ρ.Applying the condition of floatation,

Weight = Upthrust

or oil Hg2 2V V

V g g gρ = ρ + ρ

or .8 13.6

2 2 2 2Hgoil ρρ 0

ρ = + = + = 7.2 g/cm3

7.

3L/4

L/4

D

d

2d

(i) Considering vertical equilbrium of cylinder —Weight of cylinder = Upthrust due to upper liquid +upthrust due to lower liquid∴ (A/5) (L) Dg =(A/5) (3L/4) (d) g + (A/5) (L/4) (2d) (g)

∴ D =3 1

(2 )4 4

d d +

D = 54

d

8. From conservation of energy22v = 2

1 2v gh+ ...(1)

[can also be found by applying Bernoulli's theorembetween 1 and 2]From continuity equation

A1v1 = A2v2

v2 =1

12

A vA

...(2)

Substituting value of v2 from Eq. (2) in Eq. (1)

22112

2.

Av

A = 2

1 2v gh+

or 22A =

2 21 1

21 2

A v

v gh+

∴ A2 =1 1

21 2

A v

v gh=

+

Substituting the given values

h

v 2

A2

A1

v 1

1

2A2 =

–4 2

2

(10 )(10 / )

1.0 / ) 2(10)(0.15)

m m s

m s +

A2 = 5.0 × 10–5m2

9. In isothermal processPV = constant

∴ PdV + VdP = 0

ordVdV

= –PV

∴ Bulk modulus, B –/

dPdV V

= –

dPV

dV =

∴ B =– –P

V PV

= ∴ B = PNote : Adiabatic bulk modulus is given by B = γP

10. If a fluid (gas of liquid) is accelerated in positive x-direction, then pressure decrease in positive x-direction.Change in pressure has following differential equation.

a

y

x

dPdx = – ρ a

where ρ is the density of the fluid. Therefore, pressureis lower in front side.

11. Velocity of efflux at a depth h is given by 2v gh=Volume of water flowing out per second from both theholes are equal.

112

∴ a1v1 = a2v2

or (L2) 2 ( )g y = 2 2 (4 )R g yπ

or R =2

L

π12. F2 – F1 = upthrust

∴ F2 = F1 + upthrustF2 = ρgh (πR2) + Vρg

or F2 = ρg (V + πR2h)

13. mgFB =

gVdgdV SLi =

CL

si

CL

s

C

i

dd

VV

kdd

VV

k°°°

′′

=

′′

=

=

=60

200

1 ,

∆γ+∆γ+=

TT

dd

s

L

L

s

11

Fe

Hg

kk

γ+

γ+=

601

601

1

2

14. l will decrease because the block moves up, h willdecrease because the coin will displace the volume ofwater (V1) equal to its own volume when it is in thewater whereas when it is on the block it will displacethe volume of water (V2) whose weight is equal toweight of coin and since density of coin is greater thanthe density of water, V1 < V2.

15. lAWlY

∆=

Al

lW

∆=

Al

=

slope1

= 2 × 1011 N/m2

16. From the definition of bulk modulus,

B =( / )

dPdV V


B 5(1.165–1.01) 10

(10/100)×

= Pa = 1.55 × 105Pa

17. Applying continuity equation at 1 and 2, we haveA1v1 = A2v2 ...(1)

Further applying Bernoulli's equation at these twopoints, we have

h

1

2

P0 + ρgh + 21

12

vρ = P0 + 0 + 22

12

vρ ...(2)

Solving Eqs. (1) and (2), we have

22v =

222

1

2

1–

gh

A

A


22v =

22 10 2.475

1–(0.1)

× × = 50m2/s2

18. Force from right hand side liquid on left hand sideliquid.(i) Due to surface tension force

= 2RT (towards right)(ii) Due to liquid pressure force

= 00

( )(2 . )x h

x

P gh R x d x=

=

+ ρ∫

=(2P0Rh + Rρgh2) (towards left)∴ Net force is |2P0Rh+ Rρgh2 – 2RT|

19.2

21

14

and4

rT

prT

p =∆=∆

2121 , pprr ∆>∆∴<∴ Air will flow from 1 to 2 and volume of bubble

at end-1 will decrease.Therefore correct option is (b).


1. Liquid will apply an upthrust on m An equal force willbe exerted (from Newton's third law) on the liquid.Hence A will read less than 2 kg and B more than 5 kg.Therefore, the correct options are (b) and (c).

SUBJECTIVE QUESTIONS1. Let h be the same level on both the tubes in equilibrium.

Equating the volumes, we have

113

hh1

⇒

h

Ah1 + Ah2 =2Ah

∴ h = 1 22

h h+

Work done by gravity = Ui – Uf

W 1 21 2 1 2– ( )

2 2 2h h h

m g m g m m g = + +

[ ]1 1 2 2 1 21 2–

2 2 4Ah gh Ah gh h h

Ah Ah gρ ρ + = + ρ+ ρ

Simplifying this, we get

W = 21 2( – )

4Ag

h hρ

2. Submerged length = 0.5secθ, F = Upthrust,W = Weight.(i) Weight which will act at centre of plank.

θ θ

OW

F

0.5m

(ii) Upthrust which will act at centre of submergedportion.Taking moments of all forces about point O. Momentof hinge force will be zero.

∴ Moment of W (clockwise) = Moment of F (anti-clockwise)

∴ θ

θ

ρθ=θρ sin2sec5.0

))()(sec5.0(sin2

)( gAl

Alg w

∴ cos2θ = 2

2(0.5) (1)

( )( )l ρ =

2(0.5) 1(0.5) 2

= (As l = 1m)

∴ cosθ1

2=

or θ = 45°

3. In elastic collision with the surface, direction of velocityis reversed but its magnitude remains the same.Therefore time of fall = time of rise.

or time of fall = 21t

Hence, velocity of the ball just before it collides withliquid is

v = 12t

g ...(2)

Retardation inside the liquid

upthrust–weightamass

=

– –L LVd g Vdg d dg

Vd d = =

(V=Volume of ball) ...(2)

Time taken to come to rest under this retardation willbe

1 1 1–2 2( – )2 L L

gt gt dtvt

d da a d dgd

= = = =

Same will be the time to come back on the liquidsurface. Therefore,(a) t2 = time the ball takes to came back to the positionfrom where it was released

= t1 + 2t

=1

1 –L

dtt

d d+

= 1 1–L

dtd d

+

or t2 = 1

–L

L

t dd d

(b) The motion of the ball is periodic but not simpleharmonic because the acceleration of he ball is g in air

and –Ld d

gd

inside the liquid which is not

proportional to the displacement, which is necessaryand sufficient condition for SHM.(c) When dL = d, retardation or acceleration inside theliquid becomes zero (upthrust = weight). Therefore, theall will continue to move wih constant velocity v = gt1/2 inside the liquid.

4. (a) (i) Considering vertical equilibrium of cylinder :Weight of cylinder = upthrust due to upper liquid +upthrust due to lower liquid

Note that h1 and h2 ≠ 2H

∴ 3

( ) . ( ) (2 )( )5 5 4 5 4A A L A L

L D g d g d g = +

114

∴ D = 3 1

(2 )4 4

d d +

D = 54

d

(ii) Considering vertical equilibrium of two liquids and thecylinder.Total pressure at the bottom of the cylinder =Atmospheric pressure + Pressure due to liquid ofdensity d + Pressure due to liquid of density 2d +Pressure due to cylinder

A

gd

LA

dgH

dgH

P×××

+×++= 45

52220

or P = P0 + (6 )

4dg H L+

(b) (i) Applying Bernoulli's theorem at 1 and 2

0 2 –2 2H H

P dg dg h + + 2

01

(2 )2

P d v= +

Here, v is velocity of efflux at 2.Solving this, we get

P

1

H/2

x

hV

– hH2 2

v = (3 – 4 )2gH h

(ii) Time taken o reach the liquid to the bottom will be

t = 2 /h g∴ Horizontal distance x travelled by the liquid is

x = vt = 23 – 4 )

2g hH h

g

x = (3 – 4 )h H h(iii) For x to be maximum

dxdh = 0

or1

(3 – 8 ) 02 (3 – 4 )

H hh H h

=

or h = 38H

Therefore, x will be maximum at h 38H=

The maximum value of x will be

xm =3 3

3 – 48 8H H

H

xm = 4

H3

5. (i) Mass of water = (Volume)(density)

F

H

A/100 V

A

∴ m0 = (AH)ρ

∴ H = 0m

Aρ ...(1)

Velocity of efflux, 02 2m

V gH gA

= =ρ

02m gA

=ρ

Thrust force on the container due to draining out ofliquid from the bottom is given by.F = ρaV2

F = ρ (A/100)V2 = ρ (A/100) 02m g

A ρ

F = 0

50m g

∴ Acceleration of the containder, a = F/m0 = g/50a = g/50

(ii) Velocity of efflux when 75% liquid has been drainedout i.e., height of liquid.

ρ==

AmH

h44

0

V 2gh=V

h=H4

02

4m

gA

= ρ

2h

115

02m g

VA

=ρ

6. Given : A1 = 4 × 10–3m2, A2 = 8 × 10–3m2

h1 = 2m, h2 = 5mv1 = 1 m/s and ρ = 103kg/m3

From continuity equaion, we have

A1v1 = A2v2 or v2 1

12

A vA

=

or v2 =–3

–34 10

8 10

× ×

(1m/s)

v2 =12

m/s

A1

1

2

h1

h2

A2

v2

v1

Applying Bernoulli's equation at section 1 and 2

2 21 1 1 2 2 2

1 12 2

P v gh P v gh+ ρ + ρ = + ρ + ρ

or ( )2 21 2 2 1 2 1

1– ( – ) –

2P P g h h v v= ρ + ρ ...(2)

(i) Work done per unit volume by the pressure as thefluid flows from P to Q.W1 = P1 – P2

2 22 1 2 1

1( – ) ( – )

2g h h v v= ρ + ρ [From eq. (1)]

2 22 1 2 1

1( – ) ( – )

2g h h v v= ρ + ρ J/m3

= [29400 – 375) J/m3

= 29025 J/m3

(ii) Work done per unit volume by the gravity as fluidflows from P to Q.W2 = ρg (h2 – h1) = (103) (9.8) (5 – 2)Jm3

or W2 = 29400 J/m3

7. Let M = Mass of stick = πR2 ρL

GBC

L/2l/2

YCM

Ll

B

(M + m)g

C

F

(1) (2)

l = Immersed length of the rodG = CM of rodB = Centre of buoyant force (F)C = CM of (rod + mass (m))YCM = Distance of C from bottom of the rod

B

(M + m)g

C

(3)

m

F

BG

Mass m should be attached to the lower end becauseotherwise B will be below G and C will be above G andthe torque of the couple of two equal and oppositeforces F and (M + m)g will be counter clockwise anddisplacing the rod leftwards. Therefore, the rod cannotbe in rotational equilibrium. See the figure (3).Now, refer figures (1) and (2).For vertical equilibrium

Mg + mg = F (upthrust)or (πR2L)ρg + mg = (πR2l) σg

∴ l =2

2R L m

R

π ρ+ π σ

...(1)

Position of CM (of rod + m) from bottom

YCM =.2LM

M m+

( )( )

2

22Lr L

R L m

π ρ=

π ρ +...(2)

Centre of buoyancy (B) is at a height of 2l

from the

bottomWe can see from figure (2) that for rotational equilibriumof the rod, B should either lie above C or at the samelevel of B.

h

116

Therefore,12

≥ YCM

or,2

22

R L m

R

π ρ +

π σ≥

2

2

( )2

( )

LR L

R L m

π ρ

π ρ +

or m + ρR2 Lρ ≥ πR2L ρσ

or m ≥ πR2 L ( – )ρσ ρ

∴ Minimum value of m is πR2L ( – )ρσ ρ .

8. As the pressure exerted by liquid A on the cylinder isradial and symmetric.So the force due to this pressure cancels out and netvalue is zero.

Air

A

B

h

hA

hB

(b) In equilibrium :Weight of cylinder = Net upthrust on the cylinderLet s be the area of cross-section of the cylinder, thenweight = (s) (h + hA + hB) ρcylindergupthrust on the cylinder = upthrust due to liquidA + upthrust due to liquid B= shAρAg + shBρBgEquating these two

s (h + hA + hB) ρcylinderg = gshgsh BBAA ?? +or (h + hA + hB) ρcylinder = hAρA + hBρBSubstituting,hA = 1.02cm, hB = 0.8 cm and ρA = 0.7 g/cm3

ρB = 1.02 g/cm3 and ρcylinder = 0.8 g/cm3

In the above equation, we geth = 0.25 cm

(c) Net acceleration M

MgFa B −

=

ghhh

hhhhhh

BA

BABBAA

++

++−++=

)(?

?)()(??

cylinder

cylinder

Substituting the values of h, hA, hB,ρB and ρcylinder we get

a = 6g

(upwards)

9. The bubble will separate from the tube when thrustforce due to striking air at B is equal to the force dueto excess pressure.

r

vB

Tdl

θ

θ

Rb

bTAv =θπθ=ρ cos ,2cos22

(A = Area of bubble at B where air strikes)

∴ ART

Av

=ρ

42

R = 24T

v

ρ

10. When the tube is not there.

P + P0 + 2 21 2 0

1 12 2

v gH v Pρ + ρ = ρ +

∴ P + ρgH = 2 22 1

1( – )

2v vρ

A1v1 = A2v2 or v1 2 2

1

A vA

=

∴ P + ρgH =

22 22 2

1

1–

2A

v vA

×ρ

= 22

22 2

1 (0.3)1–

2 (0.9)v

π ×ρ× π

= 21 1

(10) 1 –2 81

×ρ×

= 34 10

81× ρ

= 34 10 90081

× ×

= 54 10

9× N/m2

This is also the excess pressure ∆P.By Poiseuille's equation, the rate of flow of liquid inthe capillary tube

Q =4( )

8P a

lπ ∆

η

117

∴ 8 × 10–6 = 2 2( )( )8

a P al

π ∆ η

∴ η =

( )( )2

2

–68 8 10

aa pl

π ∆

× ×

Substituting the value, we have

η = ( )–6 3 –6

–6

(10 ) 10 2 109

8 8 10

4 × ×

× ×

=1

720 N–s/m2

11. Free body diagram of the wire is as shown in figure.

T Tθ θ

λ = mg/T= surface tension

l

a

yθ

Considering the equilibrium of wire in vertical direction.2Tlcosθ = mg ...(1)

For y << a, cos θ=ya

Substituting the values in Eq. (1), we ge

yag

T2

λ=

12. Terminal velocity vT ( )22 –

9 s Lr g= ρ ρη

and viscous force F = 6 Trvπη

TdQ Fvdt

= = ( 6 Trvπη ) (vT) = 26 Trvπη

22)??(

?92p?6

−= Lsgrr

= ( )2

2 58 –27 s L

g rπ ρ ρη

ordQdt ∝ r5

13. Equation of continuity (Av = constant)

2(8) (0.25)4π

= 2(2) ( )4

vπ

...(1)

Here, v is the velocity of water with which water comesout of the syringe (Horizontally).Solving Eq. (1), we get

v = 4m/sThe path of water after leaving the syringe will beparabola. Substituting proper values in equation oftrajectory.

ghtgth 2

21 2 =⇒=

we have,

m 210

25.124 =×== vtx

(x = horizontal range)

14. For circular motion of small elements dx, we havedF = (dm) x ω2

∴ (dP)A = (ρAdx). x ω2

or dP = ρω2 x.dx

∴2

1

P

PdP∫ = 2

0

Lxdxρω ∫

∴ P2 – P1= 2 2

2Lρω

ω

dx

x

∴ ρgH2

22Lρω=

∴ H = 2 2

2Lg

ω

15. For bubble A, 210004.04

84

0××

+=+=A

A RS

PP

= 16 N/m2

For bubble B, 410004.04

84

0××

+=+=B

B RS

PP

= 12 N/m2

118

68166412

p34

p34

3

3

=×

×==

AA

BB

A

B

RP

RP

nn

16. 0? PghP =+

5101000200101000 =××+P

45 108.9200010 ×=−=P N/m2

if VPPV 0=

P

500 mm

200 mmH

⇒ 1000)500(1010

1000300108.9 44 HAA −×××=×××

H = 206 mmHeight fallen = 206 - 200 = 6 mm

COMPREHENSION

1.

( P + gh)4 r0 1ρ π 2

(P+ g(h+h))3 r0 1ρ π 2 ρ3

4 r hgπ 2

P r0π2

Equating the net downward forces and net upwardforces we get

210

20

2210 3))((4

34)( rhhgPrPhgrrghP π+ρ++π=π

ρ+πρ+

hh35

1 =

2 Again equating the forces, we get

P 4 r0 π 2

(P+ gh)3 r0 2ρ π 2 ρ3

4 r hgπ 2

P r0π2

2

202

022

0 3)(43

4)( rghPrPhgrrP πρ++π=πρ+π

Solving this we get,

94

2hh =

3. For h2 < 49h

buoyant force will further decrease.

hence, the cylinder remains at its original position.

4. (d)

5. For adiabatic process =−γ

γ1

. PT constant

( ) ( )[ ] 5/20

5/200 .. −− −+=+ yHgPTgHPT ll ρρ

( )

5/2

0

00

−

−+

+=

yHgPgHP

TTl

l

ρρ ( ) 5/2

0

00

+

−+=

gHPyHgP

Tl

l

ρρ

∴ (b)

6. PnRT

V =

⇒ ( )[ ]( ) 5/2

0

0

0

0

+

−+−+

=gHP

yHgPyHgP

nRTV

l

l

l ρρ

ρ

Buoyant force

( ) ( )[ ] 5/30

5/20

0

yHgPgHP

nRgTgV

−++==

ll

ll

ρρ

ρρ

∴ (b)


1. From continuity equation, Av = constant

or vA 1∝

∴ correct option is (a)

184

CHAPTER-12OPTICS

FILL IN THE BLANKS

1. A light wave of frequency 5 × 1014 Hz enters a medium of refractive index 1.5. In the medium the velocity of thelight wave is .... and its wavelength is .... (1983, 2M)

2. A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis witha distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then d is equal to..... cm. (1985, 2M)

3. A monochromatic beam of light of wavelength 6000Å in vacuum enters a medium of refractive index 1.5. In themedium its wavelength is ..... its frequency is .... (1985, 2M)

4. In Young's double slit experiment, the two slits act as coherent sources of equal amplitude A and of wavelengthof λ. In another experiment with the same set up the two slits are sources of equal amplitude A and wavelengthλ, but are incoherent. The ratio of the intensity of light at the mid point of the screen in the first case to that inthe second case is .... (1986, 2M)

5. A thin lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a medium of refractiveindex. 4/3, its focal length will become ... cm. (1987, 2M)

6. A point source emits sound equally in all directions in a non-absorbing mediuim. Two points P and Q are at adistance 9 m and 25 m respectively from the source. The ratio of amplitudes of the waves at P and Q is .....

(1989, 2M)7. A slab of material of refractive index 2 shown in figure has a curved surface APB of radius of curvature 10 cm and

a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given inthe figure. An object O is placed at a distance of 15 cm from the pole P as shown. The distance of the final imageof O from P, as viewed from the left is .... (1991, 2M)

An = 1.01

P

15cmB

20cm

OE

Cn = 2.02

D

n =3

43

8. A thin rod of length f/3 placed along the optic axis of a concave mirror of focal length f such that its image whichis real and elongated, just touches the rod. The magnification is .... (1991, 1M)

9. A ray of light undergoes deviation of 30° when incident on a equilateral prism of refractive index 2 . The angle

made by the ray inside the prism with the base of the prism is .... (1992; 1M)10. The resolving power of electron microsocope is higher than that of an optical microscope because the wavelength

of electrons is .... than the wavelength of visible light. (1992, 1M)11. If ε0 and µ0 are, respectively, the electric permittivity and magnetic permeability of free space, ε and µ the

corresponding quantitites in a medium, the index of refraction of the medium in terms of the above parameter is ....(1992, 1M)

12. A slit of width d is placed in front of a lens of focal length 0.5 m and is illuminated normally with light of wavelength5.89 × 10–7 m. The first diffraction minima on either side of the central diffraction maximum are separated by2 × 10–3m. The width d of the slit is .... m. (1997, 1M)

13. Two thin lenses, when in contact, produce a combination of power + 10 diopters. When they are 0.25 m apart, thepower reduces to + 6 diopters. The focal lengh of the lenses are ..... m and m. (1997, 2M)

14. A ray of light is incident normally on one of the faces of a prism of apex angle 30° and refractive index 2 . The

angle of deviation of the ray is ..... degrees. (1997, 2M)

185

OBJECTIVE QUESTIONSOnly One option is correct :

1. When a ray of light enters a glass slab from air :(1980; 1M)

(a) its wavelength decrease(b) its wavelength increases(c) its frequency increases(d) neither its wavelength nor its frequency changes

2. In Young's double slit experiment, the separationbetween the slits is halved and the distance betweenthe slits and the screen is doubled. The fringe widthis : (1981; 2M)(a) unchanged (b) halved(c) doubled (d) quadrupled

3. A glass prism of refractive index 1.5 is immersed inwater (refractive index 4/3). A light beam incidentnormally on the face AB is totally reflected to reach theface BC if : (1981; 3M)

B A

C

θ

(a) sin θ > 8/9 (b) 2/3 < sin θ < 8/9(c) sin θ < 2/3 (d) none of these

4. A convex lens of focal length 40 cm is in contact witha concave lens of focal length 25 cm. The power of hecombination is : (1982; 3M)(a) – 1.5 D (b) – 6.5D(c) + 6.5D (d) + 6.67 D

5. A ray of light from a denser medium strikes a rarermedium at an angle of incidence i (see figure). Thereflected and refracted rays make an angle of 90° with

eahc other. The angles of reflection and refraction arer and r'. The critical angle is : (1983; 1M)

i r

r'

(a) sin–1 (tan r) (b) sin–1 (cot i)(c) sin–1 (tanr') (d) tan–1 (sin i)

6. Two coherent monochromatic light beams of intensitiesI and 4I are superposed. The maximum and minimumpossble intensities in the resulting beam are :

(1988; 1M)(a) 5I and I (b) 5I and 3I(c) 9I and I (d) 9I and 3I

7. A short linear object of length b lies along the axis ofa concave mirror of focal length f at a distance u fromthe pole of the mirror. The size of the image isapproximately equal to : (1988; 2M)

(a)1 / 2–u f

bf

(b)1 / 2

–f

bu f

(c)–u ff

(d)

2

− fufb

8. A beam of light consisting of red, green and bluecolours is incident on a right angled prism. The refractiveindices of the material of the prism for the above red,green and blue wavelengths are 1.39, 1.44 and 1.47respectively. The prism will : (1989; 2M)

45°

TRUE/FALSE

1. The intensity of light at a distance r from the axis of a long cylindrical source is inversely proportional or r.(1981; 2M)

2. A convex lens of focal length 1m and a concave lens of focal length 0.25 m are kept 0.75 m apart. A parallel beamof light first passes through the convex lens, then through the concave lens and comes to a focus 0.5 m awayfrom the concave lens. (1982; 2M)

3. A beam of white light passing through a hollow prism give no spectrum.(1983; 2M)

4. Two slits in a Youn g's double slit experiment are illuminated by two different sodium lamps emitting light of thesame wavelength. No interference pattern will be observed on the screen. (1984; 2M)

5. In a Young's douuble slit experiment performed with a source of while light, only black and white fringes areobserved. (1987; 2M)

6. A parallel beam of white light fall on a combination of a concave and a convex lens, both of same material. Theirfocal lengths are 15 cm and 30 cm respectively for the mean wavelength in white light. On the same side of thelens system, one sees coloured patterns with violet colour nearer to the lens. (1988; 2M)

186

(a) separate the red colour from the green and bluecolours

(b) separate the blue colour from the red and greencolours

(c) separate all the three colours from the one another(d) not separate even partially any colour from the

other two colours

9. An astronomical telescope has an angular magnificationof magnitude 5 for distance objects. The separationbetween the objective and the eyepiece is 36 cm andthe final image is formed at infinity. The focal lengthf0 of the objective and the focal length fe of theeyepiece are : (1995; 2M)(a) f0 = 45 cm and fe = – 9 cm(b) f0 = 50 cm and fe = 10 cm(c) f0 = 7.2 cm and fe = 5 cm(d) f0 = 30 cm and fe = 6 cm

10. A thin prism P1 with angle 4° and made from glass ofrefractive index 1.54 is combined with another thinprism P2 made from glass of refractive index 1.72 toproduce dispersion without dviation. The angle of theprism P2 is : (1990; 2M)(a) 5.33° (b) 4°(c) 3° (d) 2.6°

11. Two thin convex lenses of focal length f1 and f2 areseparated by a horizontal distance d (where d < f1 , d< f2) and their centres are displaced by a verticalseparation ∆ as shown in the figure:

O

y

d

∆x

Taking the origin of coordinates O, at the centre of thefirst lens, the x and y-coordinates of the focal point ofthis lens system, for a parallel beam of rays comingfrom the left, are given by : (1993; 2M)

(a)1 2

1 2,

f fx y

f f= = ∆

+

(b)1 2

1 2 1 2

( ),

–f f d

x yf f d f f

+ ∆= =

+ +

(c) dffdf

ydff

dfdffx

−+−∆

=−+

−+=

21

1

21

121 )(,

)(

(d)1 2 1

1 2

( – ), 0

–f f d f d

x yf f d

+= =

+

12. Spherical aberration in a thin lens can be reduced by:(1994; 2M)

(a) using a monochromatic light(b) using a double combination(c) using a circular annular mark over the lens(d) increasing the size of the lens

13. A narrow slit of width 1 mm is illuminated bymonochromatic light of wavelength 600 nm. Thedistance between the first minima on either side of ascreen at a distance of 2 m is : (1994; 1M)(a) 1.2 cm (b) 1.2 mm(c) 2.4 cm (d) 2.4 cm

14. An isosceles prism ofangle 120° has a refractiveindex 1.44. Two parallel ofmonochromatic light enterthe prism parallel to eachother in air as shown. Therays emerge from theopposite face :(1995; 2M)

120°

rays

(a) are parallel to each other(b) are diverging(c) make an angle 2 [sin–1 (0.72) – 30°] with each other(d) make an angle 2 sin–1 (0.72) with each other

15. The focal lengths of the objective and the eyepiece ofa compound microscope are 2.0 cm and 3.0 cmrespectively. The distance between the objective andthe eyepiece at 15.0 cm. The final image formed by theeyepiece is at infinity. The two lenses are thin. Thedistance in cm of the object and the image producedby the objective, measured from the objective lens,and respectively : (1995; 2M)(a) 2.4 and 12.0 (b) 2.4 and 15.0(c) 2.0 and 12.0 (d) 2.0 and 3.0

16. A diminished image of an object is to be obtained ona screen 1.0 m from it. This can be achieved byappropriate placing : (1995; 2M)(a) a concave mirror of suitable focal length(b) a convex mirror of suitable focal length(c) a convex lens of focal length less than 0.25 m(d) a convex lens of suitable focal length

17. Consider Fraunhoffer diffraction pattern obtained witha single slit illuminated at normal incidence. At theangular position of the first diffraction minimum thephase difference (in radian) between the wavelets fromthe oppoosite edges of the slit is : (1995; S)

(a)4π

(b)2π

(c) 2π (d) π

187

19. A real image of a distant object is formed by aplanoconvex lens on its principal axis. Sphericalaberration : (1998; 2M)(a) is absent(b) is smaller if the curved surface of the lens faces

the object(c) is smaller if the plane surface of the lens faces the

object(d) is the same whichever side of the lens faces the

object

20. A parallel monochromatic beam of light is incidentnormally on a narrow slit. A diffraction pattern isformed on a screen placed perpendicular to the directionof the incident beam. At the first minimum of thediffraction pattern, the phase difference between therays coming from the two edges of the slit is :

(1998; 2M)(a) zero (b) π/2(c) π (d) 2π

21. A concave mirror is placed on a horizontal table withits axis directed vertically upwards. Let O be the poleof the mirror and C its centre of curvature. A pointobject is placed at C. It has a real image also locatedat C. If the mirror is now filled with water, the imagewill be : (1998; 2M)(a) real and will remain at C(b) real and located at a point between C and ∞.(c) virtual and located at a point between C and O(d) real and located at a point between C and O

22. A spherical surface of radius of curvature R, separatesair (refractive index 1.0) from glass (refractive index1.5). The centre of curvature is in the glass. A pointobject P placed in air is found to have a real image Qin the glass. The line PQ cuts the surface at a pointO and PO = OQ. The distance PO is equal to :

(1998; 2M)(a) 5 R (b) 3 R(c) 2 R (d) 1.5 R

23. Yellow light is used in a single slit diffraction experimentwith slit width of 0.6 mm. If yellow light is replaced byX-rays, then the observed pattern will reveal :(1998;2M)(a) that the central maximum is narrower(b) more number of fringes(c) less number of fringes(d) no diffraction pattern

24. A thin slice is cut out of a glass cylinder along a planeparallel to its axis. The slice is placed on a flat planeas shown. The observed interference fringes from this

combination shall be : (1999; 2M)

(a) straight (b) circular(c) equally spaced(d) having fringe spacing which increases as we go

outwards

25. A concave lens of glass, refractive index 1.5 has bothsurfaces of same radius of curvature R. On immersionin a medium of refractive index 1.75, it will behave asa : (1999; 2M)(a) convergent lens of focal length 3.5 R(b) convergent lens of focal length 3.0 R(c) divergent lens of focal length 3.5 R(d) divergent lens of focal length 3.0 R

26. In a compound microscope, the intermediate image is:(2000; 2M)

(a) virtual, erect and magnified(b) real, erect and magnified(c) real, inverted and magnified(d) virtual, erect and reduced

27. A hollow double concave lens is made of very thintransparent material. It can be filled with air of eitherof two liquids L1 or L2 having refracting indices n1 andn2 respectively (n2 > n1 > 1). The lens will diverge aparallel beam of light if it is filled with : (2000; 2M)(a) air and placed in air(b) air and immersed in L1(c) L1 and immersed in L2(d) L2 and immersed in L1

28. A diverging beam of light from a point source S havingdivergence angle α falls symmetrically on a glass slabas shown. The angles of incidence of the two extremerays are equal. If the thickness of the glass slab is tand its refractive index is n, then the divergence angleof the emergent beam is : (2000; 2M)

n

iα

S

i

t

(a) zero (b) α(c) sin–1(l/n) (d) 2 sin–1(l/n)

29. In a double slit experiment instead of taking slits ofequal widths, one slit is made twice as wide as the

188

other, then in the interference pattern : (2001; 2M)(a) the intensities of both the maxima and the minima

increases(b) the intenseity of the maxima increase and the

minima has zero intensity(c) the intensity of maxima decreases and that of

minima increases(d) the intensity of maxima decreases and the minima

has zero intensity

30. A point source of light B, placed at a distance L infront of the centre of a plane mirror of width d, hangingvertically on a wall. A man walks in front of the mirroalong a line parallel to the mirror at a distance 2L fromit as shown. The greatest distance over which he cansee the image of the light source in the mirror is :

(2000; 2M)

d B

L

2L

(a) d/2 (b) d(c) 2d (d) 3d

31. A rectangular glass slab ABCD of refractive index n1,is immersed in water of refractive index n2(n1> n2). Aray of light is incident at the surface AB of the slab asshown. The maximum value of the angle of incidenceαmax, such that the ray comes out only from the othersurface CD, is given by : (2000; 2M)

αmax

B

A D

C

n1 n2

(a)–1 –11 2

2 1sin cos sinn n

n n

(b)–1 –1

12

1sin cos sinnn

(c)–1 1

2sin n

n

(d)–1 2

1sin n

n

32. Two beams of light having intensities I and 4I interfereto produce a fringe pattern on a screen. The phasedifference between the beams is π/2 at point A and πat point B. Then the difference resultant intensities atA and B is : (2001; 2M)(a) 2I (b) 4 I(c) 5 I (d) 7 I

33. In a young's double slit experiment, 12 fringes areobserved to be formed in a certain segment of thescreen when light of wavelength 600 nm is used. If thewavelength of light is changed to 400 nm, number offringes observed in the same segment of the screen isgiven by : (2001; 2M)(a) 12 (b) 18(c) 24 (d) 30

34. A ray of light passes through four transparent mediawith refractive indices µ1, µ2, µ3 and µ4 as shown in thefigure. The surfaces of all medcia are parallel. If theemergent ray CD is parallel to the incident ray AB, wemust have : (2002; 2M)

µ1 µ2 µ3 µ4

A

D

B C

(a) µ1 = µ2 (b) µ2 = µ3(c) µ3 = µ4 (d) µ4 = µ1

35. A given ray of light suffers minimumd eviation in anequilateral prism P. Additional prism Q and R o fidentical shape and of the same mateiral as P are nowadded as shown in the figure. The ray will suffer :

(2001; 2M)

Q

P R

(a) greatier deviation(b) no deviation(c) same devaition as before(d) total internal reflection

36. An observer can see through a pin-hole the top endof a thin rod of height h, placed as shown in the figure.The beaker height is 3h and its radius h. When thebeaker is filled with a liquid up to a height 2h, he cansee the lower end of the rod. Then the refractive indexof the liquid is : (2002; 2M)

189

3h

h

2h

(a)52

(b)52

(c)32

(d)32

37. Which one of the following spherical lenses does notexhibit dispersion? The radii of curvature of the surfaceof the lenses are as given in the diagrams:

(2002; 2M)

(a) R2R1 (b) ∝R1

(c) RR (d) R ∝

38. In the ideal double-slit experiment, when a glass-plate(refractive index 1.5) of thickness t is introduced in thepath of one of the interfering beams (wavelength λ),the intensity at the position where the central maximumoccurred previously remains unchange.d The minimumthickness of the glass-plate is : (2002; 2M)

(a) 2λ (b)23λ

(c) 3λ

(d) λ

39. Two plane mirrors A and B are aligned parallel to eachother, as shown oin the figure. A light ray is incidentat an angle 30° at a point just inside one end of A. Theplane of incidence coincides with the plane of thefigure. The maximum number of times the ray undergoesreflections (including the first one) before it emergesout is : (2002; 2M)

B

A

0.2m

2 3µ

30°

(a) 28 (b) 30(c) 32 (d) 34

40. In the adjacent diagram cP represeent a wavefront andAO and BP, the corresponding two rays. Find thecondition on θ for constructive interference at Pbetween the ray BP and reflected ray OP.(2003; 2M)

O R

P

B

A

C d

θ θ

(a) cos θ 32d

λ= (b) cos θ 4dλ=

(c) sec θ – cos θ dλ= (d) sec θ – cos θ

4dλ=

41. The size of the image of an object, which is at infinity,as formed by a convex lens of focal length 30 cm is 2cm. If a concave lens of focal length 20 cm is placedbetween the convex lens and the image at a distanceof 26 cm from the convex lens, calculate the new sizeof the image. (2003; 2M)(a) 1.25 cm (b) 2.5 cm(c) 1.05 cm (d) 2 cm

42. A ray of light is incident at the glass-water interfaceat an angle i, it emerges finally parlalel to the surfaceof water, then the vlaue of µg would be :(2003; 2M)

rµ = w 4/3

i

r

Air

WaterGlass

(a) (4/3) sin i (b) 1/sin i(c) 4/3 (d) 1

43. White light is incident on the interface of glass and airas shown in the figure. If green light is jsut totalyinternally reflected then the emerging ray in air contains.

(2004; 2M)

Glass Water

Air

(a) yellow, orange, red(b) violet, indigo, blue(c) all colours (d) all colours except green

190

44. A ray fo light is incident on an equilaterla glass prismplaced on a horizontal table. For minimum deviationwhich of the following is true ? (2004; 2M)

P

Q R

S

(a) PQ is horizontal (b) QR is horizontal(c) RS is horizontal(d) Either PQ or RS is horizontal

45. A point object is placed at the centre of a glass sphereof radius 6 cm and refractive index 1.5. The distance ofthe virtual image from the surface of the sphere is :

(2004; 2M)(a) 2 cm (b) 4 cm(c) 6 cm (d) 12 cm

46. In a YDSE bi-chromatic light of wavelength 400 nm and560 nm are used. The distance between the slits is 0.1mm and the distance between the plane of the slits andthe screen is 1 m. The minimum distance between twosuccessive regions of complete darkness is :

(2004; 2M)(a) 4 mm (b) 5.6 mm(c) 14 mm (d) 28 mm

47. In Young's double slit experiment intensity at a pointis 1/4th of the maximum intensity. Angular position ofthis point is : (2005; 2M)

(a) sin–1 dλ

(b) sin–1 2dλ

(c) sin–1 3dλ

(d) sin–1 4dλ

48. A container is filled with water (µ = 1.33) upto a heightof 33.25 cm. A concave mirror is placed 15 cm abovethe water level and the image of an object placed at thebottom is formed 25 cm below the water level. Thefocal length of the mirror is : (2004; 2M)

1

O

25 cm

15 c

m33

.25

cm

(a) 10 cm (b) 15 cm(c) 20 cm (d) 25 cm

49. A convex lens is in contact with concave lens. Themagnitude of the ratio of their focal length is 2/3. Theirequivalent focal length is 30 cm. What are theirindividual focal lengths? (2005; 2M)(a) – 70, 50 (b) –10, 15(c) 75, 50 (d) – 15, 10

50. A point object is placed atdistance of 20 cm from a thinplanoconvex lens of focallength 15 cm. The plane surfaceof the lens is now silvered. 20 cmThe image created by the system is at : (2006; 3M)(a) 60 cm to the left of the system(b) 60 cm to the right of the system(c) 12 cm to the left of the system(d) 60 cm to the right of the system

51. The graph between object distnace u and image distancev for a lens is given below. The focal length of the lensis : (2006; 3M)

+11

+10

+9 –9 –10 –11

u

v

(a) 5 + 0.1 (b) 5 + 0.05(c) 0.5 + 0.1 (d) 0.5 + 0.05

52. A biconvex lens of focal length f forms a circular imageof radius r of sun in focal plane. Then which optionis correct? (2006; 3M)(a) πr2 ∝ f (b)πr2 ∝ f2

(c) If lower half part is covered by black sheet, thenarea of the iamge is equal to πr2/2

(d) If f is doubled, intensity will increase

53. A ray of light travelling in water is incident on itssurface open to air. The angle of incidence is θ, whichis less than the cirtical angle. Then there will be :

(2007; 3M)(a) only a reflected ray and no refracted ray(b) only a reflected ray and no reflected ray(c) a reflected ray and a refracted ray and the angle

between them would be less than 180° – 2θ(d) a reflected ray and a refracted ray and the angle

between them would be greater than 180° – 2θ

54. In an experiment to determine the focal length (f) of aconcave mirror by the u-v method, a student places theobject pin A on the principal axis at a distance x from

191

the pole P. The student looks at the pin and itsinverted image from a distance keeping his/her eye inline with PA. When the student shifts his/her eyetowards left, the image appears to the right of theobject pin. Then : (2007; 3M)(a) x < f (b) f < x < 2f(c) x = 2f (d) x > 2f

55. Two beams of red and violet colours are made to passseparately through a prism (angle of the prism is 60º).In the position of minimum deviation, the angle ofrefraction will be (2008; 3M)(a) 30º for both the colours(b) greater for the violet colour(c) greater for the red colour(d) equal but not 30º for both the colours

56. A light beam is travelling from Region I to Region IV(Refer Figure). The refractive index in Regions I, II, III

and IV are ,8

and6

,2

, 0200

nnnn respectively. The angle

of incidence θ for which the beam just misses enteringRegion IV is (2008; 3M)Region I Region II Region III Region IV

20n

60n

80n

(a)

−

43

sin 1(b)

−

81

sin 1

(b)

−

41

sin 1(d)

−

31

sin 1

57. A ball is dropped from a height of 20 m above thesurface of water in a lake. The refractive index of wateris 4/3. A fish inside the lake, in the line of fall of theball, is looking at the ball. At an instant, when the ballis 12.8 m above the water surface, the fish sees thespeed of ball as [Take g = 10 ms/s2.] (2009; M)(a) 9 m/s (b) 12 m/s(c) 16 m/s (d) 21.33 m/s

OBJECTIVE QUESTIONSMore than one options are correct?

1. In the Young's double slit experiment, the intereferncepattern is found to have an intensity ratio between thebright and dark fringes as 9. This implies that :

(1982; 2M)(a) the intensities at the screen due to the two slits

are 5 units and 4 units respectively(b) the intensities at the screen due to the two slits

are 4 units and 1 unit respectively.(c) the amplitude ratio is 3(d) the amplitude ratio is 2

2. White light is used to illuminate the two slits in a Young's double slit experiment. The separation between theslits is b and the screen is at a distance d (>> b) fromthe slits. At a point on the screen directly infront ofone of the slits, certain wavelengths are missing. Someof these missing wavelengths are : (1984; 2M)(a) λ = b2/d (b) λ = 2b2/d(c) λ = b2/3d (d) λ= 2b2/3d

3. A converging lens is used to form an image on ascreen. When the upper half of the lens is covered byan opaque screen : (1986; 2M)(a) half of the iamge will disappear(b) complete image will be formed(c) intensity of the image will increase(d) intensity of the image will decrease

4. A planet is observed by an astronomical refractingtelescope having an objective of focal length 16 m andan eyepiece of focal length 2 cm : (1992; 2M)(a) the distance between the objective and the eyepiece

is 16.02 m(b) the angular magnification fo the planet is – 800(c) the image of the planet is inverted(d) the objective is larger than the eyepiece

5. In an interfernce arrangement similar to Young's double-slit experiment, the slits S1 and S2 are illuminated withcoherent microwave sources, each of frequency 106 Hz.The sources are synchronized to have zero phasedifference. The slits are separated by a distanced = 150 M. The intensity I (θ) is measured as a functionof θ, where Q is defined as shown. If I0 is the maximumintensity, then I (θ) for 0 ≤ θ ≤ 90° is given by :

(1995; 2M)

θ

S1

S2

d/2

d/2

(a) I (θ) = I0/2 for θ= 30°(b) I (θ) = I0/4 for θ= 90°(c) I (θ) = I0 for θ = 0°(d) I (θ) is constant for all values of θ

6. Which of the following form (s) a virtual and erectimage for all position of the object? (1998; 2M)(a) Convex lens(b) Concave lens(c) Convex mirror(d) Concave mirror

192

7. A ray of light travelling in a transparent medium fallson a surface separating the medium from air at an angleof incidence 45°. The ray undergoes total internalreflection. If n is the refractive idnex of the mediumwith respect to air, select the possible value (s) of nfrom the following : (1998; 2M)(a) 1.3 (b) 1.4(c) 1.5 (d) 1.6

8. In a Young's double slit experiment, the separationbetween the two slits is d and the wavelength of thelight is λ. The intensity of light falling on slit 1 is fourtimes the intensity of light falling on slit 2. Choose thecorrect choice (s). (2007; 4M)(a) If d = λ, the screen will contain only one maximum(b) If λ < d < 2λ, at least one more maximum (besides

the central maximum) will be observed on thescreen

(c) If the intensity of light falling on slit 1 is reducedso that it becomes equal to that of slit 2, theintensities of the observed dark and bright fringeswill increase

(d) If the intensity of light falling on slit 2 is increasedso that it becomes equal to that of slit 1, theintensities of the observed dark and bright fringeswill increase


1. The convex surface of a thin concave-convex lens ofglass of refractive index 1.5 has a radius of curvature20 cm. The concave surface has a radius of curvature60 cm. The convex side is silvered and placed on ahorizontal surface. (1981; 2M)

(i) Where should a pin be placed on the optic axissuch that is image is formed at the same place?

(ii) If the concave part is filled with water of refractiveindex 4/3, find the distance through which the pinshould be moved, so that the image of the pinagain coincides with the pin.

2. Screen S is illuminated by two point sources A and B.Another source C sends a parallel beam of lighttowards point P on the screen (see figure). Line AP isnormal to the screen and the lines AP, BP and PC arein one plane. The distances AP, BP and CP are 3 m,1.5 m and 1.5 m respectively. The radiant powers ofsources A and B are 90 W and 180 W respectively. The

beam from C is of intensity 20 W/m2. Calculate intensityat P on the screen. (1982; 5M)

A 60°

60° P

S B

C

3. In Young's double slit experiment usingmonorochormatic light the fringe pattern shifts by acertain distance on the screen when a mica sheet ofrefractive index 1.6 and thickness 1.964 microns isintroduced in the path of one of the interfering waves.The mica sheet is then removed and the distancebetween the slits and the screen is doubled. It is foundthat the distance between successive maxima (orminima) now is the same as the observed fringe shiftupon the introduction of the mica sheet. Calculate thewavelength of the monochromatic light used in theexperiment. (1983; 6M)

4. A plano-convex lens has a thickness of 4 cm. Whenpalced on a horizontal table, with the curved surfacein contact with it, the apparaent depth of the bottommost point of the lens is found to be 3 cm. If the lensis inverted such that the plane face is in contact withthe table, the apparent depth of the centre of the planeface is found to be 25/8 cm. Find the focal length ofthe lens. Assume thickness to be negligible whilefinding its focal length. (1984; 6M)

5. A beam of light consisting of two wavelength, 6500Åand 5200 Å is used to obtain interference fringe in aYoung's double slit experiment. (1985; 6M)(i) Find the distance of the third bright fringe on the

screen from the central maximum for wavelengh6500Å

(ii) What is the least distance from the central maximumwhere the bright fringes due to both thewavelengths coincide?

The distance between the slits is 2 mm and the distancebetween the plane of the slits and the screen is 120 cm.

6. Monochromaic light is incident on a plane interface ABbetween two media of refractive indices n1 and n2(n2 > n1) at an angle of incidence θ as shown in thefigure. The angle θ is infinitesimally greater than thecritical angle of the two media so that total internalreflection takes place. Now if a transparent slab DEFGof uniform thickness and of refractive index n3 isintroduced on the interface (as shown in the figure),show tha for any value of n3 all light will ultimately be

193

reflected back again into medium II. Consider separatelythe cases: (1988; 6M)(i) n3 < n1 and n3 > n1

θ

A B

Medium III( n )3

DG

EF

Medium I( n )1

Medium II( n )2

7. A right prism is to be made by selecting a propermaterial and the angles A and B (B ≤ A), as shown infigure. It is desired that a ray of light incident on theface AB emerges parallel to the incident direction aftertwo internal reflections. (1987; 7M)

A

C

B

(i) What should be the minimum refractive index n forthis to be possible?

(ii) For n 53

= is it possible to achieve this with the

angle B equal to 30 degrees?

8. A parallel beam of light travelling in water (refractiveindex = 4/3) is refracted by a spherical air bubble ofradius 2 mm situated in water. Assuming the light raysto be paraxial. (1988; 6M)(i) Find the position of the image due to refraction at

the first surface and the position of the finalimage.

(ii) Draw a ray diagram showing the positions of boththe images.

9. In a modified Young's double slit experiment, amonochromatic uniform and parallel beam of light ofwavelength 6000Å and intensity (10/π) Wm–2 is incidentnormally on two apartues A and B of radii 0.001 m and0.002 m respectively. A perfectly transparent film ofthickness 2000 Å and refractive index 1.5 for thewavelength of 6000 Å is placed in front of aperture A(see figure.) Calculate the power (in W) received at thefocal spot F of the lens. The lens is symmetricallyplaced with respect to the apertures. Assume that 10%of the power received by each aperture goes in theoriginal direction and is brought to the focal spot.

(1989; 8M)

F

A

B

10. A narrow monochromatic beam of light of intensity Iis incident on a glass plate as shown in figure. Anotheridentical glass plate is kept close to the first one-andparallel to it. Each glass plate reflects 25 per cent of thelight incident on it and transmits the remaining. Findthe ratio of the minimum and maximum intensities inthe interference pattern formed by the two beamsobtained after one reflection at each plate.

(1990; 7M)

1 2

1

11. Two parallel beams of light P and Q (separation d)containing radiations of wavelengths 4000Å and 5000Å(which are mutually coherent in each wavelengthseparately) are incident normally on a prism as showninf igure. The refractive index of the prism as a functionof wavelength is given by the relation, µ (λ) = 1.20 +

2b

λ where 2

b

λ is in Å and b is positive constant. The

value of b is such that the condition for total reflectionat the face C is just satisfied for one wavelength andis not satisfied for the other. (1991; 2 + 2 + 4M)

d

P

QB

90°

A

sin = 0.8θ

C

θ

(a) Find the value of b.(b) Find the deviation of the beams transmitted through

the face AC.(c) A convergent lens is used to bring these

transmitted beams into focus. If the intensities ofthe upper and the lower beams immediately aftertransmission from the face AC, are 4I and Irespectively, find the resultant intensity at thefocus.

194

12. Light is incident at an angle α on one planar end ofa transparent cylindrical rod of refractive index n.Determine the least value of n so that the light enteringthe rod does not emerge from the curved surface of therod irrespective of the value of α. (1992; 8M)

α

13. In given figure, S is a monochromatic point sourceemitting light of wavelength λ= 500 nm. A thin lens ofcircular shape and focal length 0.10 m is cut into twoidentical halves L1 and L2 by a plane passing througha diameter. The two halves are placed symmetricallyabout the central axis is SO with a gap of 0.5 mm. Thedistance along the axis from S to L1 and L2 is 0.15 mwhile that from L1 and L2 to O is 1.30 m. The screenat O is normal to SO. (1993; 5 + 1 M)

0.15m 1.30m

Screen

O

A

0.5mmL1

L2

S

(i) If the third intensity maximum occurs at the pointA on the screen, find the distance OA.

(ii) If the gap between L1 and L2 is reduced from itsoriginal value of 0.5 mm, will the distance OAincreases, decrease or remain the same.

14. An image Y is formed of point object X by a lenswhose optic axis is AB as shown in figure. Draw a raydiagram to locate the lens and its focus. If the imageY of the object X is formed by a concave mirror (havingthe same optic axis as AB) instead of lens, drawanother ray diagram to locate the mirror and its focus.Write down the steps of construction of the raydiagrams. (1994; 6M)

A B

X

Y

15. A ray of light travelling in air is incident at grazingangle (incident angle = 90°) on a long rectangular slabof a transparent medium of thickness t = 1.0 m. The

point of incidence is the origin A (0,0). The medium hasa variable index of refraction n (y) is given byn (y) = [ky3/2 + 1]1/2 where k = 1.0 (meter)–3/2.The refractive index of air is 1.0 (1995; 10M)

A (0, 0) Air xθ

B (x, y)Mediumt =

1.0

m

Airy

P (x , y )1 1

(a) Obtain a relation between the slope of the trajectoryof the ray at a point B (x, y) in the medium and theincident angle at the point.

(b) Obtain an equation for the trajectory y (x) of theray in the medium.

(c) Determine the coordinaes (x1, y1) of the point P,where the ray intersects thre upper surface of theslab-air boundary.

(d) Indicate the path of the ray subsequently.

16. Angular width of central maximum in the Fraunhoferdiffraction pattern of a slit is measure.d The slit isilluminated by light of wavelength 6000Å, When theslit is illuminated by light of another wavelength, thewidth decreases by 30%. Calculate the wavelength ofthis light. The same decreases in the angular width ofcentral maximum is obtained when the originalapparatus is immersed in a liquid. Find refractive indexof the liquid. (1996; 2M)

17. A right angle prism (45°-90°-45°) of refractive index nhas a plane of refractive index n1 (n1 < n) cemented toits diagonal face. The assembly is in air. The ray isincident on AB. (1996; 3M)

n1n

B C

A

(i) Calculate the angle of incidence at AB for which theray strikes the diagonal face at the critical angle.

(ii) Assuming n = 1.352, calculate the angle ofincidence at AB for which the refracted ray passesthrough the diagonal face undeviated.

18. A double slit apparatus is immersd in a liquid ofrefractive index 1.33. It has slit separation of 1 mm anddistance between the plane of slits and screen is1.33 m. The slits are illuminated by a parallel beam oflight whose wavelength in air is 6300Å. (1996; 3M)

195

(i) Calculate the fringe width.(ii) One of the slits of the apparatus is covered by a

thin glass sheet of refractive index 1.53. Find thesmallest thickness of the sheet to bring the adjecentminimum as the axis.

19. A thin plano-convex lens of focal length f is split intotwo halves. One of the halves is shifted along theoptical axis. The separation between object and imageplanes is 1.8 m. The magnification of the image formedby one of the half lens is 2. Find the focal length ofthe lens and separation between the halves. Draw theray diagram for image formation. (1996; 5M)

1.8m

O

20. In Young's experiment, the source is red light ofwavelength 7 × 10–7 m. When a thin glass plate ofrefractive index 1.5 at this wavelength is put in thepath of one of the interfering beams, the central brightfringe shifts by 10–3 m to the position previouslyoccupied by the 5th bright fringe. Find the thicknessof the plate. When the source is now changed to greenlight of wavelength 5 × 10–7 m, the central fringe shiftsto a position initially occupied by the 6th bright fringedue to red light. Find the refractive index of glass forgreen light. Also estimate the change in fringe widthdue to the change in wavelength. (1997C; 5M)

21. A thin equiconvex lens of glass of refractive index µ= 3/2 and of focal length 0.3 m in air is sealed into anopening at one end of a tank filled with water µ = 4/3. On the oppoosite side of the lens, a mirror is placedinside the tank on the tank wall perpendicular to thelens axis, as shown in figure. The separation betweenthe lens and the mirror is 0.8 m. A small object ispalced outside the tank in front of. Find the position(relative to the lens) of the image of the object formedby the system (1997C; 5M)

0.9m

0.8m

22. In a Young's experiment, the upper slit is covered bya thin glass plate of refractive index 1.4, while the lowerslit is covered by another glass plate, having the same

thickness as the first one but having refractive index1.7. Interference pattern is observed using light ofwavelength 5400 Å. It is found that the point P on thescreen, where the central maximum (n = 0) fall beforethe glass plates were inserted, now has 3/4 the originalintensity. It is further observed that what used to bethe fifth maximum earlier lies below the point P whilethe sixth minima lies above P. Calculate the thicknessof glass plate. (Absorption of light by glass plate maybe neglected). (1997; 5M)

23. A prism of refractive index n1 and another prism ofrefractive index n2 are stuck togehter with a gap asshown in the figure. The angle of the prism are asshown n1 and n2 depend on –l, the wavelength of light

according to : n1 4

210.8 101.20 ×= +

λ and n 2

4

21.80 101.45 ×= +

λ where λ is in nm. (1998; 8M)

A60°

n1

n2

CD

B

20°

40°

70°

(a) Calcualte the wavelength λ0 for which rays incidentat any angle on the interface BC pass throughwithout bending at that interface.

(b) For light of wavelength λ0, find the angle ofincidence i on the face AC such that the deviationproduced by the combination of prisms is minimum.

24. A coherent parallel beam of microwaves of wavelengthλ = 0.5 mm falls on a Young's double slit apparatus.The separation between the slits is 1.0 mm. Theintensity of microwaves is measured on a screen placedparallel to the plane of the slits at a distance of 1.0 mfrom it as shown in the figure. (1998; 8M)

D = 1.0m

Screen

d = 1.0mm x

y

30°

(a) If the incident beam falls normally on the doubleslit apparatus, find the y0-coordinates of all theinterference minima on the screen.

(b) If the incident beam makes an angle of 30° wit the

196

x-axis (as in the dotted arrow shown in figure),find the y-coordinates of the first minima on eitherside of the central maximum.

25. The Young's double slit experiment is done in a mediumof refractive index 4/3. A light of 600 nm wavelengthis falling on the slits having 0.45 mm separation. Thelower slit S2 is covered by a thin glass sheet ofthickness 10.4 µm and refractive index 1.5 Theinterfernce pattern is observed on a screen placed 1.5m from the slits as shown in the figure.

(1999; 10M)

S 1

S2

O

y

S

(a) Find the location of central maximum (bright fringewith zero path difference) on the y-axis.

(b) Find the light intensity of point O relative to themaximum fringe intensity.

(c) Now, if 600 nm light is replaced by white light ofrange 400 to 700 nm, find the wavelengths of thelight that form maxima exactly at point O.

[All wavelengths in the problem are for the givenmedium of refractive index 4/3. Ignore dispersion]

26. The x-y plane is the boundary between two transparent

media. Medium–1 with z ≥ 0 has a refractive index 2

and medium –2 with z ≤ 0 has a refractive index 3 .

A ray of light in medium –1 given by vector

$6 3 8 3 –10= +A i j kur $ $ is incident on the plane of

separation. Find the unit vector in the direction of therefracted ray in medium-2. (1999; 10M)

27. A quarter cylinder of radius R and refractive index 1.5is placed on a table. A point object P is kept at adistance of mR from it. Find the value of m for whicha ray from P will emerge parallel to the table as shownin figure. (1999; 5M)

P A

RmR

+ve

28. A convex lens of focal length 15 cm and a concavemirror of focal length 30 cm are kept with their opticsaxis PQ and RS parallel but separated in verticaldirection by 0.6 cm as shown. The distance betweenthe lens and mirror is 30 cm. An upright object AB ofheight 1.2 cm is placed on the optic axis PQ of the lensat a distance of 20 cm from the lens. If A'B′ is the imageafter refraction from the lens and the reflection fromthe mirror, find the distance of A'B' from the pole of themirror and obtain its magnification. Also locate positionsof A' and B' with respect to the optic axis RS.

(2000; 6M)

R

P

A

B QS0.6cm

30 cm

29. A glass plate of refractive index 12.5 is coated withathin layer of thickess t and refractive index 1.8. Lightof wavelength λ travelling in air is incident normally onthe layer. It is partly reflected at the upper and thelower surfaces of the layer and the two reflected raysinterfere. Write the condition for their constructiveinteference. If λ = 648 nm, obtain the least value of tfor which the rays interfere constructively.

(2000; 4M)

30. The refractive indices of the crown glass for blue andred light are 1.51 and 1.498 respectively and those ofthe flint glass are 1.77 and 1.73 respectively. Anisosceles prism of angle 6° is made of crown glass. Abeam of white light is incident at a small angle on thisprism. The other flint glass isosceles prism is combinedwith the crown glass prism such that there is nodeviation of the incident light. (2001; 5M)(i) Determine the angle of the flint glass prism.(ii) Calculate the net dispersion of the combined

system.

31. A vessels ABCD of 10 cm width has two small slits S1and S2 sealed with identical glass plates of equalthickness. The distance between the slits is 0.8 mm.POQ is the line perpendicular to the plane AB andpassing through O, the middle point of S1 and S2. Amonochromatic light source is kept at S, 40 cm belowP and 2 m from the vessel, to illuminate the slits asshown in the figure alongside. Calculate the positionof the central bright fringe on the other wall CD withrespect to the line OQ. Now, a liquid is poured into thevessel and filled upto OQ. The central bright fringe isfound to be at Q. Calculate the refractive index of theliquid. (2001; 5M)

197

P

S

40cm

2m

S2

S1

O

10cm

D

Q

C

A

B32. A thin biconvex lens of refractive index 3/2 is placed

on a horizontal plane mirror as shown in the figure.The space between the lens and the mirror is thenfilled with water of refractive index 4/3. It is found thatwhen a point object is placed 15 cm above the lens onits principal axis, the object concides with its ownimage. On repeating with another liquid, the object andthe image again coincide at a distance 25 cm from thelens. Calculate the refractive index of the liquid.

(2001; 5M)

33. A poin source S emitting light of wavelength 600 nmis placed at a very small height h above a flat reflectingsurface AB (see figure). The intensity of the reflectedlight is 36% of the incident intensity. Inteferencefringes are observed on a screen placed parallel to thereflecting surface at a very large distance D from it.

(2002; 5M)

S

B

P Screen

D

h

(a) What is the shape of the inteference fringes onthe screen?

(b) Calculate the ratio of the minimum to the maximumintensities in the interference fringes formed nearthe point P (shown in the figure).

(c) If the intensity at point P corresponds to amaximum, calculate the minimum distance throughwhich the reflecting surface AB should be shiftedso that the intensity at P again becomes maximum.

34. In the figure, light is incident on the thin lens asshown. The radius of curvature for both the surface isR. Determine the focal length of this system.

(2003; 2M)

µ1µ2 µ3

35. A prism of refracting angle 30° is coated witha thinfilm of transparent material of refracting indx 2.2 onface AC of the prism. A light of wavelength 6600Å isincident on face AB such that angle of incidence is 60°.Find (2003; 4M)

B C

A

60°

30°

µ = 2.2f

µ = 3p

(a) the angle of emergence and(b) the minimum value of thickness of the coated film

on the face AC for which the light emerging fromthe face has maximum intensity. [Given refractive

index of the material of the prism is 3 ]

36. Figures shows an irregular block of material of refractive

index 2 . A ray of light strikes the face AB as shown

in the figure. After refraction it is incident on a sphericalsurface CD of radius of curvature 0.4 m and enters amedium of refractive index 1.514 to meet PQ at E. Findthe distance OE upto two places of decimal.

(2004; 2M)

60°A

µ = 1

45° µ = 2

B C

O E Q

µ = 1.514

D

P

37. An object is approaching a thin convex lens of focallength 0.3 m with a speed of 0.01 m/s. Find themagnitudes of the rates of change of position andlateral magnification of image when the object is at adistance of 0.4 m from the lens. (2004; 4M)

38. In a Young's double slit experiment, two wavelengthsof 500 nm and 700 nm were used. What is the minimumdistance from the central maximum where their maximascoincide again? Take D/d = 103. Symbols have theirusual meanings. (2004; 4M)

39. AB and CD are two slabs. The medium between theslabs has refractive index 2. Find the minimum angle of

198

incidence of Q, so that the ray is totally reflected byboth the stabs. (2005; 2M)

µ = 2

µ = 3pCP

A Q B

D

µ=2

40. A ray of light is incident on a prism ABC of refractive

index 3 as shown in figure. (2005; 4M)

P0

P1

P2

S1

S2

AC

E

B D

60°

60°

60°

(a) Find the angle of incidence for which the deviationof light ray by the prism ABC is minimum.

(b) By what angle the second prism must be rotated.so that the final say suffer net minimum deviation.

MATCH THE COLUMN

1. Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II. Column I Column II

(A) Intensity of light received by lens (p) radius of aperture (R)(B) Angular magnification (q) dispersion of lens(C) Length of telescope (r) focal length f0, fe(D) Sharpness of image (s) spherical aberration

2. An optical component and an object S placed along its optic axis are given in Column I. The distanc between theobject and the component can be varied. The propertices of images are given in Column II. Match all the propertiesof images from Column II with the appropriate components given in Column I. Indicate your answer by darkeningthe appropriate bubbles of the 4 × 4 matrix given in the ORS Column I Column II

(A) (p) Real image

(B) (q) Virtual image

(C) (r) Magnified image

(D) (s) Image at infinity

3. Column I shows four situations of standard Young's double slit arrangment with the screen placed far away fromthe slits S1 and S2. In each of these cases S1P0 = S2P0, S1P1 - S2P1 = λ/4 and S1P2 - S2P2 = λ/3, where λ is the wavelengthof the light used. In the cases B, C and D, a transparent sheet of refractive index µ and thickness t is pasted onslit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light wavesreaching a point P on the screen from the two slits is denoted by δ(P) and the intensity by I(P). Match each situationgiven in Column I with the statement(s) in Column II valid for that situation.

Column - I Column - II

(A) (p) δ (P0) = 0

199


This section contains, statement I (assertion) andstatement II (reasons).

1. Statement-I : The formula connecting u, v and f for aspherical mirror is valid only for mirrors whose sizesare very small compared to their radii of curvature.

(2007; 3M)Because :Statement II : Laws of reflection are strictly valid forplane surfaces, but not for large spherical surfaces.(a) Statement-I is true, statement-II is true; statement-

II is a correct explanation for statement -I(b) Statement-I is true, statement-II is true; statement-

II is NOT a correct explanation for statement -I(c) Statement-I is true, statement-II is false(d) Statement-I is false, statement-II is true

COMPREHENSION

PassageThe figure shows a surface XY separating twotransparent media, medium-I and medium-2. Teh linesab and cd represent wavefronts of a light wave travellingin medium-I and incident on XY. The lines ef and ghrepresent wavefronts of the light wave in medium-2after refraction.

X Y

b d

a cf h

e g

Medium-1

Medium-2

1. Light travels as a : (2007; 4M)(a) parallel beam in each medium(b) convergent beam in each medium(c) divergent beam in each medium(d) divergent beam in one medium and covergent

beam in the other medium

2. The phases of the light wave at c, d, e, and f are φc,φd, φe and φf respectively. It is given that φe ≠ φf.

(2007; 4M)(a) φc cannot be equal to φd(b) φd can be equal to φe(c) (φd – φf) is equal to (φc – φe)(d) (φd – φc) is not equal to (φf – φe)

3. Speed of light is : (2007; 4M)(a) the same in medium-1 and medium-2(b) large in medium-1 than in medium-2(c) large in medium-2 than in medium-1(d) different at b and d

P0

P1

P2

S1

S2

(B) (µ - 1) t = λ/4 (q) δ (P1) = 0

(C) (µ - 1) t = λ/2 (r) Ι (P1) = 0

(D) (µ - 1) t = 3λ/4 (s) Ι (P0) > I (P1)

(t) Ι (P2) > I (P1)

P0

P1

P2

S 1

S 2

P0

P1

P2

S1

S2

200

ANSWERS

FILL IN THE BLANKS

1. 2 × 108m/s, 4 × 10–7m 2. 15

3. 4000Å, 5 ×1014Hz 4. 2 5. 60 6.259

7. 30 cm o the right of P. Image will be virtual 8. –1.5 9. zero

10. smaller 11. 0 0

µµ

εε 12. 2.945 × 104

13. 0.125, 0.5 14. 15°

TRUE/FALSE1. T 2. F 3. T 4. T 5. F 6. T

OBJECTIVE QUESTION (ONLY ONE OPTION)1. (a) 2. (d) 3. (a) 4. (a) 5. (a) 6. (c) 7. (d)8. (a) 9. (d) 10. (c) 11. (c) 12. (c) 13. (d) 14. (c)15. (a) 16. (c) 17. (c) 18. (b) 19. (b) 20. (d) 21. (d)22. (a) 23. (d) 24. (a) 25. (a) 26. (c) 27. (d) 28. (b)29. (a) 30. (d) 31. (a) 32. (b) 33. (b) 34. (d) 35. (c)36. (b) 37. (c) 37. (a) 38. (a) 39. (b) 40. (b) 41. (b)42. (b) 43. (a) 44. (b) 45. (c) 46. (d) 47. (c) 48. (c)49. (d) 50. (c) 51. (c) 52. (c) 53. (c) 54. (b) 55. (a)56. (b) 57. (c)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)1. (b, d) 2. (a, c) 3.(b, d) 4. (a, b, c, d) 5. (a, c) 6. (b, c) 7. (c, d)8. (a, b)


1. (i) 15 cm (ii) 13.84 cm 2. 13.97 W/m2 3.5892Å 4. 75 cm

5. (i) 1.17 mm (ii) 1.56 mm 7. (i) 2 (ii) No 8. (i) – 6mm, – 5mm 9. 7 ×10–6 W

10. 149 11. (a) b = 8 × 105 (Å)2 (b) δ4000Å = 37Å, δ3000Å = 27.13 (C) 9I 12. 2

13. (i) 1 mm (ii) increase 15. (a) Slope = cot i (b) 4y1/4 = x (c) (4m, 1m) 16. 4200Å, 1.43

17. (i) i1 = sin–1 2 21 1

1– –

2n n n

(ii) 73° 18. (i) 0.63 mm (ii) 1.579 µm

19. 0.4 m, 0.6 m 20. 7 × 10–6m, 16, – 5.71 × 10–5m 21. 0.9 m from the lens (rightwards) or 0.1 m behind themirror

22. 9.3 µ m 23. (a) 600 nm (b) sin–134

24. (a) + 0.26 m, + 1.13 m (b) 0.26 m, 1.13 m

25. (a) 4.33 mm (b) I = max34

I= (c) 650 nm; 433.33 nm 26.

1

5 2 $(3i+4j–5k)$ $ 27.

43

28. 15 cm, – 3/2 29. 2µt 1

n –2

= λ with µ = 1.8 and n = 1, 2, 3..., 99 nm, tmin = 90 nm

201

30. (i) 4° (ii) – 0.04° 31. 2 cm above point Q on side CD, µ = 1.0016 32. 1.6

33. (a) circular (b)1

16 (c) 300nm 34. 3

3 1–µ R

µ µ 35. (a) zero (b) 1500Å 36. 6.06 m

37. 0.09 m/s, 0.3/s 38. 3.5 mm 39. 60° 40. (a) 60° (b) 60°

MATCH THE COLUMN

1. A-p, B-r, C-r, D-p, q, s.2. A-p, q, B-q, C-p, q, r, s, D-p, q, r, s.3. A-p, s; B-q; C-t; D-r, s, t.


1. (c)

COMPREHENSION

1. (a) 2. (c) 3. (b)

SOLUTIONS

FILL IN THE BLANKS

1. v 83 10

1.5cµ

×= = = 2.0 × 108 m/s

λ =8

142.0 10

5.0 10

vf

×=

× = 4.0 × 10–7 m

2. At I1, second focus of convex lens should coincidewith first focus of concave lens.orPower of combination of lens is zero.

∴ 21210 PdPPPp −+==

1005

1201

0d

+−=

Solving this, we get 15=d cmd

ll

5cm

20cm

3. λmedium air?

=µ

6000=1.5 = 4000Å

f air medium

air medium

v v= =

? ?=

8

–73.0×10

6.0 10×= 5.0 × 1014 Hz

Frequency remains unchanged.

4. In case of YDSE, at mid-point intensity will be Imax =4I0. In the second case when sources are incoherent,the intensity will be I = I0 + I0 = 2I0

Therefore, the desired ratio is 0

0

42

2II

=

Here I0 is the intensity due to one slit.

5. fair = 1 2

1 1(1.5–1) –R R

fmedium 1 2

1.5 1 1–1 –4 / 3 R R

=

Dividing Eq. (2) by E. (1), we get

medium

air

ff = 4

∴ fmedium = 4fair = 4 × 15 = 60 cm

6. I ∝ 21

r (in case of point source)

and I ∝ A2

A ∝1r

or1

2

AA =

2

1

259

rr

=

202

7. Rays starting from O will suffer single refraction fromspherical surface APB. Therefore, applying

A

P

15cm

B

OE

C

D

µ2µ1

+Ve

Ru

uvu 1212 µ−=µ−

1.0 2.0––15v =

1.0–2.0–10

1v =

1 1–10 7.5

or v = – 30 cmTherefore, image of O will be formed at 30 cm to theright of P.Note that image will be virtual. There will be no effectof CED.

8. F = focusC = centre of the curvature

P' CP F O

f3

Object

Image

When the object lies between F and C, image is real,elongated and inverted. As one end of rod just touchesits image, this end should lie at C. Because image ofobject at C is at C itself.Let P' be the image of other end of rod P.For P :

u = – (2f – f/3) =5–3f

Applying the mirror formula : 1 1 1v u f

+ =

or1 3

–5v f =

1– f

1v =

3 1–

5 f f

or v = –52f

or OP' 52f

=

∴ Length of image of rod

CP' = OP' – OC =5

– 22 2f f

f =

∴ Magnification =/ 2

– –1.5/ 3

ff

=

Here, negative sign implies that image is inverted.

9. Let δm be the angle of minimum deviation. Then

µsin

2sin( /2)

mA

A

+ δ =

(A = 60° for an equilateral prism)

∴ 2 =

60sin

260

sin2

m° + δ

°

Solving this we get δm = 30°The given deviation is also 30° (i.e. δm)Under minimum deviation, the ray inside the prism isparallel to base for an equilateral prism.

10. The resolving power of a microscope is inverselyproportional to the wavelength of the wave used De-Brogie matter wave is used in case of an electronmicroscope whose wavelength is less than thewavelength of visible light used in optical microscope.

11. Speed of light in vacuum, c 0 0

1

µ=

ε

and speed of light in same medium, v1

µ=

ε

Therefore, refractive index of the medium is

µ cv

= = 0 01/ e µ

1/ eµ 0 0= µ

µε

ε

12. GivenMinima

y

Centralmaximaf

d

y

Minima

203

2y = 2 × 10–3my = 1 × 10–3m

First minima is obtained at

d sin θ = λ but sin θ = tan θ yf

=

∴y

df

= λ

d fy

λ= =

–7

–35.89 10 0.5

1 10

× ×

×= 2.945 × 10–4m

13. When the lenses are in contact, the power of thesystem is

P = P1 + P2 or P1 + P2 = 10...(1)When lenses are separated by a distance d = 0.25 m

= 14

m

The power is P = P1 + P2 – d P1 P2

or P1 + P2 – 1 24

P P = 6 ...(2)

Solving Eqs. (1) and (2), we can find that P1 = 8 dioptreand P2 = 2 dioptre

∴ f1 =18

m = 0.125 m

A

B C

i = 901 r2

60° i2

f2 =12

m = 0.5 m

14. Ray falls normally on the face AB. Therefore, it willpass undeviated through AB.

r2 = 90 – 60 = 30°

µ = 2 = 2

2

sinsin

ir

i2 = 45°Deviation = i2 – r2 = 45° – 30° = 15°

(Deviation at face AC only)

TRUE FALSE

1. At a distance r from a line source of power P andlength I, the intensity will be,

I =PS 2

Prl

=π or I ∝

1r

2. At I1, second focus of convex lens coincides with firstfocus of concave lens. Hence, rays will become parallelto the optic axis after refraction from both the lenses.

ll

0.25m

1.0m

0.75m

3. Through a thin glass slabray of light almost passesundeviated. A hollowprism can be assumed tobe made up of three thinglass slabs as shown infigure.

Air

4. To obtain interference, source must be coherent. Twodifferent light sources can never be coherent.

5. With white light we ge coloured fringes (not onlyblack and white) with centre as white.

6. Focal length of concave is less i.e., power of concavelens will be more. Hence the combination will behavelike a concave lens. Further µV is greater than all othercolours. Hence, fV will be least.

RY


1.vf

λ =

In moving from air to glass, f remains unchanged whilev decreases. Hence, λ should decreases.

2. β = Dd

λ

d is halved and D is double∴ Fringe width ω will become four times.∴ Correct option is (d).

3. Let θc be the critical angle at face AC, then

204

sinθc = w

g

µµ

4/3 8=3/2 9

=

Angle of incidence at face AC is i = θTotal internal reflection (TIR) will take place on thissurface if.

i > θcor θ > θcor sin θ > sin θc

B A

θθ

or sin θ >89

∴ Correct option is (a)

4. P = P1 + P2 =1 2

1 1( ) ( )f m f m

+

=1 1–

0.4 0.25 = – 1.5 D

5. r + r' + 90° = 180°

i r 90°

r'

DenserRarer

i = r

∴ r' = 90° – rFurther, i = rApplying Snell's law,

µD sin i = µR sin r'or µD sin r = µR sin (90° – r) = µR cos r

∴ R

D

µµ

= tan r

θC = sin–1

µµ

D

R = sin–1 (tan r)

∴ Correct option is (a).

6. Imax ( ) ( )2 21 2 4 9I I I I I= + = + =

Imin ( ) ( )2 21 2– 4 –I I I I I= = =

∴ Correct option is (c)

7. From the mirror formula :

1 1v u

+ =1f (f = constant) ....(1)

–v–2 dv – u–2du = 0

or |dv| =2

2| |

vdu

u

Here, |dv| = size of image|du| - size of object (short) lying along the axis = bfurther, from Eq. (1), we can find

2

2v

u=

2

–f

u f

Substituting these values in Eq. (2) we get

Size of image 2

–f

bu f

=

∴ Correct option is (d).

8. The colours for which i > θc, will get total internalreflection :

i > θc or sin i > sin θc

i

45°CB

A

i = 45°

or sin 45° >1µ or

1 1>

µ2

or for which µ > 2 or µ > 1.414Hence, the rays for which µ > 1.414 will get TIRFor green and blue µ > 1.414 so, they will suffer TIRon face AC only red comes out from this face.∴ Correct option is (a).

205

9.

fef0

35cm

P1

O e

P2l1

P1I2 = 36 cm∴ f0 + fe = 36 ....(1)Further angular magnification is given as 5. Therefore,

0

e

ff = 5 ...(2)

Solving Eqs. (1) and (2), we getf0 = 30 cm and fe = 6 cm

∴ Correct option is (d).

10. Deviation : d = (µ – 1) AGiven δnet = 0∴ (µ1 – 1)A1 = (µ2 – 1)A2

or A2 1

12

( –1)( –1µ

Aµ

= =(1.54–1)

(4 ) 3(1.72–1)

° = °

∴ Correct option is (c).

11. From the first lens parallel beam of light is focussed atits focus i.e.. at a distance f1 from it. This image I1 actsas virtual object for second lens L2. Therefore, for L2.

O

d

x∆I1

I2

y

L1

L2

f1

f1

f ( f – d)f + f – d

2 1

2 1y=

u = + (f1 – d), f = + f2

∴1v =

1f u1

+2 1

1 1–f f d

= +

Hence, v = 2 1

2 1

( – )–

f f df f d+

x = d + v = d + 2 1

2 1

( – )–

f f df f d+

=1 2 1

1 2

( – )–

f f d f df f d

++

Linear magnification for L2

2 1 2

2 2 1 2 1

( – ) 1.

– – –f f d fv

mu f f d f d f f d

= = =+ +

Therefore, second image will be formed at a distance

of m∆ or 2

2 1 –f

f f d

∆ + below its optic axis.

Therefore, y - coordinate of the focus of system willbe

y = ∆ – 2

2 1

.–

ff f d

∆ +

or y =1

2 1

( – ).–

f df f d

∆+

12. Spherical aberration is caused due to spherical natureof lens. Paraxial and marginal rays are focussed atdifferent place on the axis of the lens. Therefore, imageso formed is blurred. This aberration can be reducedby either stopping paraxial rays or marginal rays,which can be done by using a circular annular markover the lens.

e

Paraxial rays Marginal rays

13. For first dark fringe on either side d sin θ = λ

ordyD

= l ∴ y Dd

λ=

Therefore, distance between two dark fringes on either

side = 2y 2 D

dλ=


distance –6 32(600×10 mm)(2×10 mm)

=(1.0mm) = 24 mm

Dark

y

Dd

Dark

θ

sin = y/Dθ

206

14. The diagrammatic representation of the given problemis shown in figure.

ri

120°

i

θ

δ

From figure it follows that ∠i = ∠A = 30°From Snell's law, we written

n1 sin i = n2 sin ror sin r =Now ∠δ = ∠r – ∠i = sin–1 (0.72) – 30°∴ θ = 2(δ) = 2 sin–1 (0.72) – 30°

15. Since, the final image is formed at infinity, the imageformed by the objective wil;l be at the focal point ofthe eyepiece, which is 3.0 cm. The image formed by theobjective will be at a distance of 12.0 cm (= 15.0 cm –3.0 cm) from the objective.

EO

O

V=∝

12.0m 3m

15.0m

If u is the distance of the object from the objective, wehave

1 1u v

+ =1f

⇒1 1

12.0u+ =

12.0

⇒ u (1.20)(2.0)12.0–2.0

= =24.010.0 = 2.4 cm

16. Image can be formed on the screen if it is real. Realimage of reduced size can be formed by a concavemirror or a convex lens as shown in figure.

1m

2f fO

t1 2tO

C F

1m

C

A diminished real image is formed by a convex lenswhen the object is placed beyond 2 f and the image ofsuch object is formed beyond 2 f on other side.Thus, d > (df + 2f)or 4f < 0.1 mo4 f < 0.25 m

17. Path difference between the opposite edges is λ.For a phase difference of 2π, we gets a path differenceof λ.

A

O

B

O'

P

18. The focal length of conbination is given by

1F

=1 1–40 25

1 2

1 1 1F F F

= +

or F =200–

3 cm 2–3

= m

∴ Power of the combination in dioptres,

P =3

–2

1( )

PF m

=

= – 1.5

19. In general spherical aberration is minimum when thetotal deviation produced by the system is equallydivided on all refracting surfaces. A planoconvex lensis used for this purpose. In order that the total deviationbe equally divided on two surfaces, it is essential thatmore parallel beam (or the incident and refracted) beincident on the convex side. Thus, when the object isfar away from the lens, incident rays will be moreparallel than the refracted rays, therefore, the objectshould face the convex side, but if the object is nearthe lens, the object should face the plane side. Thishas been shown in figure.

θb

A

B D

P

O

Firstminimum

20. At first minima, b sin θ = λ

207

or bθ = λ

ory

bD

= λ

or y =Db

λ

orybD

= λ sin θ = 0 ...(1)

Now, at P (First minima) path difference between therays reaching from two edges (A and B) will be

∆x ybD

= (Compare with ∆x ybD

= in YDSE)

or ∆x = λ (From Eq. 1)Corresponding phase difference (f) will be

φ =2

. xπ ∆ λ

f =2 . 2π λ = πλ

21. The ray digram is shown in figure. Therefore, theiamge will be real and between C and O.

O

C

O

C

P

Image

Normal at P willpass through C

22. Let us say PO = OQ = x

O QP

+ve

µ = 1 1.0 µ = 2 1.5

Appling2 2µ µ–

v u = 2 1µ – µR

Substituting the values with sign

1.5 1.0–

–X X+=

1.5–1.0R+

(Distance are measured from O and are taken as positivein the direction of ray of light)

∴2.5X

=0.5R

∴ X = 5R

23. Diffraction is obtained when the slit width is of theorder of wavelength of light (or any electromagneticwave) used. Here, wavelength of X-rays(1 – 100Å) < < slit width (0.6 mm). Therefore, nodiffraction pattern will be observed.

24. Locus of equal path difference are the lines runningparallel to the axis of the cylinder. hence, straightfringes are obtained.

Note : Circular rings (also called Newton's rings) areobserved in intereference pattern when a plano-convexlens of large focal length is placed with its convexsurface in contact with a plane glass plate becauselocus of equal path difference in this case is a circle.

25. R1 = – R, R2 = + R, µg = 1.5 and µm = 1.75

∴1f =

1 2

1 1–1 –gµ

µm R R

R1 R2


RRRf 5.3111

175.15.11

=

−

−

−=

∴ f = + 3.5RTherefore, in the medium it will behave like a convergentlens of focal length 3.5 R. It can be understood as,un > µg, the lens will change its behaviour.

26. The ray diagram is a as follows :

B

A A''FO A'

B'

E

F0O

B''

From the figure it is clear that image formed by objective(or the intermediate image) is real, inverted andmagnified.

27. The len Makers' formula is :

1f =

1 2

1 1–1 –L

m

nn R R

208

where nL = Refractive index of lens and nm = Refractive index of medium.In case of double concave lens, R1 is negative and R2

is positive. Therefore, 1 2

1 1–R R

will be negative. For

the lens to be diverging in nature, focal length f should

be negative or –1L

m

nn

should be positive or nL > nm

but since n2 > n1 (given), therefore, the lens should befilled with L2 and immersed in L1.

28. The path of rays become parallel to initial direction asthey emerge.Applying Snell is law at P

,sin

2sin

2α

α

=′n at α

β

=sin

2sin

21

,nQ

Ai i

Drr

α

α

i iEB

r

CF

O

29. In intereference we know that

Imax = 21 2( )I I+ and Imin = 2

1 2( ~ )I I

Under normal conditions (when the width of both theslits are equal)

I1 = I2 = I (say)∴ Imax = 4I and Imin = 0When the width of one of the slits is increased.Intensity due to that slit would increase, while that ofthe other will remain same. So, let :

I1 = I and I2 = I η ( η > I)

Then, Imax = I (1 + η )2 > 4I

and Imin = I ( )2–1η > 0

∴ Intensity of both maxima and minima is increased.

30. The ray diagram will be as follows :

ACD

G

H

I

Jr

L

S

FB

d

φφ

E

HI = AB = d

DS = CD 2d

=

Since, AH = 2.AD

∴ GH = 2CD 22d

d= =

Similarly, IJ = d

∴ GJ = GH + HI + IJ

= d + d + d = 3d

31. Rays come out only from CD, means rays afterrefraction from AB get total internally reflected at AD.From the figure :

A D

CB n2

n1

r 2r1αmax

r1 + r2 = 90°∴ r1 = 90° – r2(r1)max = 90° – (r2)min and (r2)min = θC (for totalinternal reflection at AD)

where sinθC= 2

1

nn or θC = sin–1

2

1

nn

∴ (r1)max = 90° – θCNow applying Snell's law at face AB :

1

2

nn =

max

1 max

sinsin( )r

α=

maxsinsin(90 – )C

α° θ

maxsincos C

α=

θ

or sin αmax = 1

2

nn

cosθC

∴ αmax = sin–1 1

2cos C

nn

θ

= sin–1–11 2

2 1cossinn n

n n

209

32. I (φ) = I1 + I2 + 2 1 2I I cos φ ..(1)

Here, I1 = I and I2 = 4I

At point A, φ = 2π

∴ IA = I + 4I = 5I

At point B, φ = π ∴ IB = I + 4I – 4I = 5I∴ IA – IB = 4I

Note : Eq. (1) for resultant intensity can be appliedonly when the sources are coherent. In the questionit is given that the rays interfere. Interference takesplace only when the sources are coherent. That is whywe applied equation number (1). When the sources areincoherent, the resultant intensity is given by I = I1 +I2

33. Fringe width, ω Dd

λ= ∝ λ

When the wavelength is decreased from 600 nm to 400

nm, fringe width will also decrease by a factor of 46 or

23 or the number of fringes in the same segment will

increase by a factor of 3/2.Therefore, number of fringes observed in the same

segment 3

12 182

= × =

Note : Since ω ∝ λ, therefore, if YDSE apparatus isimmersed in a liquid of refractive index µ, thewavelength λ and thus the fringe width will decreaseµ times.

34. Applying Snell's law at B and C,µ sin i = constant or

µ1 sin iB = µ4 sin iCBut AB| |CD∴ iB = iCor µ1 = µ4

A

BC

Di

i

µ1 µ4

Note : The Snell's law, which the students read in their

plus two syllabus µ sinsin

ir

= comes from µ sin i =

constant

µ1 = 1, µ1 sin i1 = µ2 sin i2

r

i12

µ2 = µ, or sin i = µ sin r

or µ =sinsin

ir

35. Figure (a) is part of an equilateral prism of figure (b)as shown in figure which is a magnified image of figure(c). Therefore, the ray will suffer the same deviation infigure (a) and figure (c).

Q

P R

Q

P R

(a) (b)

(c)

⇒

⇒

Note : Question are often asked based on part of aprism For example, section shown in figure (a) is partof a prism shonw in figure (c).

36. PQ = QR = 2h ∴ ∠i = 45°∴ ST = RT = h = KM = MN

210

So, KS = 2 2(2 )h h+ 5h=

∴ sin r =5

h

h =

1

5

h

2h

P

S

ri

M

2h

Q

2h

R

K N

T

∴ µ =sin sin45sin 1/ 5

ir

°= 5

2=

37.1 2

1 1 1( –1) –µf R R

=

For no dispersion, 1

0df

=

or dµ 1 2

1 1–R R

= 0

or R1 = R238. Path difference due to slab should be integral multiple

of λ or ∆x = nλor (µ – 1)t = nλ n = 1, 2, 3, ...

or t =–1

nµ

λ

For minimum value of t, n = 1

∴ t = 2–1 1.5–1µλ λ

= = λ

39. d = 0.2 tan 30° 0.2

3=

B

A

0.2m

2 3µ

30°

l =

30°

Id =

2 30.2/ 3

= 30

Therefore maximum number of reflection are 30.

40. PR = d∴ PO = d sec θand CO = PO cot 2θ= d secθ cos 2θpath difference between the two rays is,

∆x = CO + PO = (d sec θ + d sec θ cos 2θ)phase difference between the two rays is∆φ = π (one is reflected, while another is direct)Therefore, condition for constructive interference shouldbe)

O R

P

B

A

Cθ θ

∆x =3

, .....2 2λ λ

or d cosθ (1 + cos 2θ) = 2λ

or cosd

θ (2cos2θ) =

2λ

or cosθ = 4dλ

41. Image formed by convex lens at I1 will acg as a virtualobject for concave lens. For concave lens

ll

4cm26cm

l2

1 1–v u

1f

=

or1 1–

4v1

–20=

or v = 5 cmMagnification for concave lens

211

m =5 1.254

vu

= =

As size of the image at I1 is 2 cm. Therefore, size ofimages at I2 will be 2 × 1.25 = 2.5 cm

42. ApplyingSnell's law (µ sin i = constant)at 1 and 2, we have

r

i

r

Air

WaterGlass

µ1 sin i1 = µ2 sin i2Here, µ1 = µglass, i1 = i µ2 = µglass = 1 and i2 = 90°

µg sin i = (1) (sin 90°)

or ug =1

sin i

43. Critical angle θC = sin–11µ

Wavelength increases in the sequence of VIBGYOR.According to Cauchy's formula refractive index (µ)decreases as the wavelength increases. Hence therefractive index will increases in the sequence ofROYGBIV. The critical angle θC will thus increase in thesame order VIBGYOR. For green light the incidenceangle is just equal to the critical angle. For yellow,orange and red the critical angle will be greater thanthe incidence angle. So, these colours will emerge fromthe glass air interface.Hence, the correct option is (a).

44. During minimum deviation the ray inside the prism isparallel to the base of the prism in case of an equilateralprism.Hence, the correct option is (b).

45. When the object is placed at the centre of the glasssphere, the rays fro the object fall normally on thesurface of the sphere and emerge undeviated.Henc, the correct opion is (c).

46. Let nth minima of 400 nm coincides with mth minimaof 560 nm, then

(2n – 1) 400

2

= (2m – 1) 560

2

or2 –12 –1

nm =

7 14 ....5 10

= =

i.e., 4th minima of 400 nm coincides with 3rd minima of560 nm.Locaion of this minima is,

Y1 =–6(2 4–1)(1000)(400 10 )

2 0.4× ×

×= 14 mm

Next 11th minima of 400 nm will coincide with 8thminima of 560 nm.Location of this minima is,

+ Y2 =–6(2 11–1)(1000)(400 10 )

2 0.1× ×

×= 42 mm

∴ Required distance of Y2 – Y1 = 28 mmHence, the correct option is (2).

47. I = Imax cos2 2φ

∴ max4

I= Imax cos2

2φ

cos 2φ

=12

or2φ

= 3π

∴23πφ = =

2x

π ∆ λ ...(1)

where ∆x = d sin θSubstituting in Eq. (1), we get

sin θ = 3dλ

or θ = sin–13dλ

∴ Correct answer is (c).

48. Distance of object from mirror

=33.24151.33

+ = 40 cm

Distance of image from mirror

=2515

1.33+ = 33. 8 cm

212

From the mirror,1 1v u

+ =1f

∴1 1

–33.8 –40+ =

1f

∴ f = – 18.3 cm∴ Most suitable answer is (c).

49. Let focal length of convex lens is + f, lens length of

concave lens would be 3

–2

f .

From the given condition,

fff 31

321

301 =−=

∴ f = 10 cmTherefore, focal length of convex lens = + 10 cm andthat of concave lens = – 15 cm∴ Correct answer is (d).

50. Combination MPP += 22

215

1521 −⇒=− f

f

Using mirror formula,

152

2511111 −=−⇒=+

vfuv

12−=v cm51. From the lens formula :

1 1 1–

f v u= we have

1f =

1 1–10 –10

or f = + 5Further, ∆u = 0.1and ∆v = 0.1 (from the graph)Now, differentiating the lens formula we have,

2f

f

∆= 2 2

v u

v u

∆ ∆+

or ∆f =2

2 2v u

fv u

∆ ∆ +

Substituting the values we have,

∆f =2

2 20.1 0.1

(5)10 10

+

= 0.05

∴ f + ∆f = 5 + 0.05

52. θ r

f

r = f tan θor r ∝ f∴ πr2 ∝ f 2

53. Since θ < θC , both reflection and refraction will takeplace. From the figure we can see that angle betweenreflected and refracted rays α is less than 180° – 2θ.

AirWaterq

θα

θ

∴ Option (c) is correct

54. Since object and image move in opposite directions,the positioning should be as shown in the figure.Object lies between focus and centre of curvaturef < x < 2f.

Object

Image

right

left∴ Correct option is (b).

55. At minimum deviation (δ = δm) :

302

º60221 ==== A

rr (For both colours)

∴ Correct option is (a).

56. Critical angle from region III to region IV

43

6/8/

?sin0

0C ==

nn

or 81

43

61

?sin61

?sin =

== C

∴

= −

81

sin? 1

∴ Correct option is (b).

213

57. Let speed of ball in air at height of 12.8 m from watersurface be u.

122.71022 =××== ghu = 12 m/s

Speed of ball as observed by fish is =u34

16 m/s

∴ (c)


1.2

1 2max2

1 2

( )min ( – )

I III I I

+=

21 2

2 2

/ 1

/ –1

I I

I I

+=

= 9 (Given)

Solving this, we have 21

24

II

=

But I ∝ A2

∴ 1

2

AA

= 2

∴ Correct options are (b) and (d).

2. At P (directly infront of S1)

y =2b

∴ Path difference.

S1

S2

P

d

b y= b2

∆X = S2P – S1P = .( )y bd

2( )2

2

b bb

d d

= =

Those wavelengths will be missing for which

∆X = 31 2 53, , .....

2 2 2λλ λ

∴ λ1 = 2∆X =2b

d

dbX33

2 2

2 =∆

=λ

dbX53

2 2

3 =∆

=λ

∴ Correct option are (a) and (c).

3. When upper half of the lens is covered, image isformed by the rays coming from lower half of the lens.Or image will be formed by less number of rays.Therefore, intensity of image will decrease. But completeimage will be formed.

4. Distance between objective and eyepieceL = f0 + fe = (16 + 0.02) m = 16.02 m

Angular magnificationM = f0 / fe= – 16/0.02 = – 800

Image is inverted and objective is large than theeyepiece.

5. The intensity of light is I (θ) = I0cos2 2δ

where ( )2 xπδ = ∆λ = ( )2

sindπ θ λ

(i) for θ = 30°

8

63 10

10

cv

×λ = = = 300 m and d = 150 m

δ =2

(150)300

π

12 2

π =

∴2δ

=4π

∴ I (θ) = I0cos2 0

4 2Iπ =

(option a)

(ii) For θ = 90°

δ =2

(150)300

π

(1) = π

orand I (θ) = 0

(iii) For θ = 0°, δ = 0 or 2δ

= 0

∴ I (θ) = I0 (Option c)

6. For a lens : 1 1

–v u f

1=

O I

214

i.e.1v =

1 1f u

+

For a concave lens f and u are negative i.e., v willalways be negative and image will always be virtual.For a mirror :

IO

1 1v u

+ =1f

i.e.1v =

1 1–

f u

Here, f is positive and u is negative for a convex mirror.Therefore, v is always positive and image is alwaysvirtual.

7. For total internal reflection to take place :Angle of incidence, i > critical angle, θCor sin i > sin θC

or sin 45° >1n

or1

2>

1n

or n > 2or n > 1.414Therefore, possible values of n can be 1.5 or 1.6 in thegiven options.

8. for d = λ, there will be only one, central maxima.For λ < d < 2λ, there will be three maximas on thescreen coresponding to path difference,∆x = 0 and ∆x = ±λ.∴ Correct options are (a) and (b),


1. (i) Image of object will coincide with it if ray of lightafter refraction from the concave surface fall normallyon concave mirror so formed by silvering the convexsurface. Or image after refraction from concave surfaceshold form at centre of curvature of concave mirror orat a distance of 20 cm on same side of the combination.Let x be the distance of pin from the given opticalsystem. Applying.

2 1µ µ–v u = 2 1µ – µ

RWith proper signs :

1.5 1––20 –x

1.5–1–60

= or 1 3 1 8–

40 120 120x= =

∴ x =120

8 = 15 cm

(ii) Now, before striking with the concave surface, the rayis frist refracted fro a plane surface. So, let x be thedistance of pin, then the plane surface will form its

image at a distance 43

x (happ. = µh) from it.

Now, using 2 1µ µ–

v u = 2 1µ – µR

with proper signs,

we have1.5 4 / 3

––20 (4/3 )x =

1 . 5 – 4 / 3–60

or1x =

–3 1 –2640 360 360

+ =

2. Resultant intensity at P :IP = IA + IB + IC

= 2 24 ( ) 4 ( )A BP P

PA PB+

π πcos 60° + IC cos 60°

2 290 180

4 (3) 4 (1.5)= +

π π cos 60° + 20 cos 60°

= 0.79 + 3.18 + 10= 13.97 W/m2

3. Shifting of fringes due to introduction of slab in thepath of one of the slits, comes out to be,

∆y =(µ–1)tD

d ...(1)

Now, the distance between the screen and slits isdoubled. Hence, new fringe width will become

ω' =(2 )Dd

λ....(2)

Given, ∆y = ω'

ordD

dtD λ

=−µ 2)1(

∴(µ–1)

2t

λ =–6(1.6–1)(1.964 10 )

2×

=

= 0.5892 × 10–6m= 5892Å

215

4. Refer figure (a) :

O(a)

4cm

O(b)

4cm

In the case refraction of the rays starting from O takesplace from a plane surface. So, we can use

dapp =actualµ

d

or 3 =4µ

or µ =43

Refer figure (b) : In this case refraction takes placefrom a spherical surface. Hence, applying

1 2

2 1µ µ–v u = 2 1µ – µ

Rwe have,

1 4 / 3–

(–25/8) –4=

1 – 4 / 3R

or1

3R =1 8 1–3 25 75

=

∴ R = 25 cmNow, to find the focal length we will use the lensMaker's formula :

1f = (µ – 1)

1 2

1 1–R R

=4 1 1

–1 –3 –25

∝

=175

∴ f = 75 cm

25. (i) The desired distance will be :

y1 = 3ω1 = 3 1Dd

λ

=–10

–3(3)(6500 10 )(1.2)

(2 10 )

×

×= 11.7 10–4m = 1.17 mm

(ii) Let n1 bright fringe of λ1 coincides with n2 brightfringe of λ2. Then,

1 1n Ddλ

=2 2n D

dλ

or1

2

nn =

2

1

5200 46500 5

λ= =

λ

Therefore, 4th bright fringe of λ1 coincide with 5thbright fringe of λ2. Similarly, 8th bright fringe of λ1 willcoincide with 10th bright fringe of λ2 and so on. Theleast distance from the central maximum will thereforecorresponding to 4th bright fringe of λ1(or 5th brightof λ2). Hence,

m106.15)102(

)2.1)(106500(44 43

101

min−

−

−

×=××

=λ

=d

DY

= 1.56 mm

6. Given θ is slightly greater than sin–1 1

2

nn

(i) When n3 < n1 :

i.e., n3 < n1 < n2 or 3 1

2 2

n nn n

< or sin–1 3

2

nn

< sin–1 2

1

nn

Hence, critical angle for III and II will be less than thecritical angle for II and I. So, if TIR is taking palcebetween I and II, then TIR will definitely take placebetween I and III.(ii) When n3 < n1 : Now two cases may arise :Case I : n1 < n3 < n2In this case there will be no TIR between I and III butTIR will take place between III and II. This is because:

θ i > θ

iiP

II

IIII

Ray of light first enters from II to III i.e., from denserto rarer∴ i > θApplying Snell's law at P :

n2 sinθ = n3 sin i

216

or sin i =2

3

nn

sin θ

Since, sin θ is slightly greater than 1

2

nn

∴ sin i is slightly greater than 2 1

3 2

n nn n

× or 1

3

nn

but 1

3

nn is nothing but sin (θc)I,III

or TIR will now take place on I and III and the ray willbe reflected back.Case 2 : n1 < n2 < n3This time while moving from II to III, ray of light willbend towards normal. Again applying Snell's law at P:

θ i > θ

ii

PII

IIII

B

n2 sinθ = n3 sin i

sini =2

3

nn sin θ

Since, sin θ slightly greater than 1

2

nn

sin i will be slightly greater than 2 1

3 2

n nn n

× or 1

3

nn

but 1

3

nn is sin (θc)I,III

i.e, sin i > sin (θc)I,IIIor i > (θc)I,IIITherefore, TIR will again take place between I and IIIand the ray will be reflected back.

Note : Case I and case III of n3 > n1 can be explainedby one equation only. But two cases are deliberatelyformed for better understanding of refraction, Snell'slaw and total internal reflection (TIR).

7. Let x is the incident angle for reflection at AC. For total

internal reflection cix > (critical angle).

Let y be the incident angle of the ray on face BC. For

Total internal reflection ciy >

∴ ,2 ciyx >+ But CAx = and CBy =

∴ ciyx 290,90 >°°=+

⇒ °< 45ci

A

P

C

Q

BA B

(ii) To the refractive index of the medium for thishappen

245sin1

sin1

=°

==µci

For ,53

3/5111sin,

35 ==

µ==µ ci

°=°=′ 30,37 yic (Given)

′>°= cixx ,60

but ′< ciy⇒ Total internal reflection will take place on face

AC but not on BC.

8.

P Q

Applying 2 1µ µ–

v u = 2 1µ – µR

, one by one on two

spherical surfaces.First on left surfaces :

1

1 4 /3–

v ∞ =1 – 4 / 3

2+

or1

1v =

1–6

or v1 = – 6mmi.e., first image will be formed at 6 mm towards left ofP.Second on right surface : Now, distance of first imageI1 from Q will be 10 mm (towards left).

2

4 / 3 1–

–10v =4 / 3 – 1

–2

or2

43v =

1 1– –6 10

4–15

= or v2 = – 5mm

217

(ii) The ray diagram is shown in figure.

P Q

CI2I1

6mm 2mm5mm

2mm

Note :

(i) At P and Q both normal will pass through C.

(ii) At P ray of light is travelling from a denser medium(water) to rare medium (air) therefore, ray of light willbend away from the normal and on extending meet atI1. Similarly at Q, ray of light bends towards thenormal.

(iii) Both the images I1 and I2 are virtual.

9. Power received by aperture A.

PA = ( )2AI rπ

10=π (π) (0.001)2 = 10–5W

Power received by aperture, B

PB = ( )2BI rπ

10=π (π) (0.002)2 = 4 × 10–5 W

Only 10% of PA and PB goes to the original direction.Hence,% of PA = 10–6 = P1 (say) and% of PB = 4 × 10–6 = P2 (say)Path difference created by slab :∆x = (µ – 1)t = (1.5 – 1) (2000) = 1000ÅCorresponding phase difference,

φ = 2πλ ∆x =

26000

π× 1000 3

π=

Now, resultant power at the focal point :

P = P1 + P2 + 2 1 2 cosP P φ

= 10–6 + 4 × 10–6 + –6 –62 (10 )(4 10 )× cos 3π

= 7 × 10–6W

10. Each plate reflects 25% and transmits 75%.Incident beam has an intensity I. This beam undergoesmultiple reflections and refractions. The correspondingintensity after each reflection and refraction(transmission) are shown in fiugre.

1

2

314

3116

3264

9164

14

9116

Interference pattern is to take place between rays 1and 2I1 = 1/4 and I2 = 9I/64

∴min

max

II =

21 2

1 2

–I I

I I

+

149

=

11. (a) Total internal reflecion (TIR) will take place first forthe wavelength for which critical angle is small or µ islarge.

A

BC

i = θ

From the given expression of µ, it is more for thewavelength for which value of λ is less.Thus, condition of TIR is just satisfied for 4000Åor i = θc for 4000Åor θ = θcor sinθ = sinθc

or 0.8 =1µ (for 4000Å)

or 0.8 =

2

1

1.20(4000)

b+

Solving this equation, we getb = 8.0 × 105 (Å)2

(b)

A

BC

i δ = 90° – i

For 4000Å

A

BC

i δ

For 5000Å

For, 4000Å condition of TIR is just satisfied. Hence, itwill emerge from AC, just grazingly.or, 84000Å = 90° – i = 90° – sin–1 (0.8) = 37°

218

For 5000Å : µ 5

2 28.0 101.2 1.2 1.232(5000)

b ×= + = + =λ

Applying µ air

medium

sinsin

ii

=

or 1.232 = air airsin sinsin 0.8

i i= =θ

∴ iair = 80.26°δ5000Å = iair – imedium = 80.26° – sin–1 (0.8) = 27.13°

(c)i

P r

iR

r

S

1

2

Path difference between rays 1 and 2 :∆x = µ (QS) – PR ....(1)

FurtherQSPS = sin i

PRPS = sin r

∴//

PR PSQS PS =

sin µsin

ri

=

∴ µ (QS) = PRSubstituting in Eq. (1), we get ∆x = 0∴ Phase difference between rays 1 and 2 will be zro.Or these two rays will interfere constructively. So,maximum intensity will be obtained from theirinterference.

or Imax 2

1 2( )I I= + 2( 4 )I I= + = 9I

12. sin θc = 1n (θc = critical angle)

n1

θc

n – 12

r' = 90 – r(r') = 90° – r

(r')min = 90° – (r)max

and n =max

max

sin( )sin( )

ir = max

max

sin90( 90 )

sin( )i

r°

= °

Then, sin (r)max =1 sin cn

= θ

(r)max = θc or (r')min = (90° – θc)Now, if minimum value of r' i.e., 90° – θc is greater thanQc, then obviously all value of r' will be greater thanQc i.e., total internal reflection will take place at face ABin all conditions. Therefore, the necessary conditions is

(r')min ≥ θc or (90° – θc) ≥ θcor sin (90° – θc) ≥ sin θcor cos θc ≥ sin θcor cot θc ≥ 1

or 2 –1n ≥ 1

or n2 ≥ 2

or n ≥ 2

Therefore, minimum value of n is 2

13. (i) For the lens, u = – 0.15m; f = + 0.10 m

Therefore, using 1 1 1

–v u f

= we have

1 1 1v u f

= + =1 1

(–0.15) (0.10)+

or v = 0.3 m

Linear magnification, m 0.3 –2

–0.15vu

= = =

Hence, two images S1 and S2 of S will be formed at 0.3m from the lens as shown in figure. Image S1 due topart 1 will be formed at 0.5 mm above its optic axis(m = –2). Similarly, S2 due to part 2 is formed 0.5 mmbelow the optic axis of this part as shown.Hence, d = distance between S1 and S2 = 1.5 mmD = 1.30 – 0.30 = 1.0 m = 103 mmλ = 500 nm = 5 × 10–4mmTherefore, fringe width,

mm)5.1(

)10)(105( 34−×=λ=ωdD

mm31

=

Now, as the point A is at the third maxima

mm31

33

=ω=OA

or mm1=OA

Note : The language of the question is slightlyconfusing. The third intensity maximum may beunderstood as second order maximum (zero order, firstorder and the second order). In that case

.mm67.0mm31

22 =

=ω=OA

219

(ii) If the gap between L1 and L2 is reduced, d willdecrease. Hence, the fringe width ω will increaseor the distance OA will increase.

1.5m

O

A

0.25mmS

0.25m

0.5mmS1

S2

1

2

0.15m 0.3m1.3m

D = 1.0m

14. Steps (i) In case of a lens

A

XL

F B

Y

O

(a) Join X and Y. The point O, where the line XY cutsthe optic axis AB, is the optical centre of the lens.

(b) Draw a line parallel to AB from point X. Let it cutsthe lens at L. Join L and Y. The point F where theline LY cuts the optic axis AB is the focus of thelens F.

Note : As the image is inverted, lens should be aconvex because lens always forms a virtual and erectimage.

(ii) In case of a concave mirror

O

XF C

A PB

Y'

Y

(a) Draw a line YY′ perpendicular to AB from pointY. Let it cuts the line AB at point P. Locate apoint Y′ Such that PY = PY′.

(b) Extend the line XY′. Let it cuts the line AB atpoint O. Then O is the pole of the mirror.

(c) Join X and Y. The point C, where the line XYcuts the optic axis AB, is the centre of curvatureof the mirror.

(d) The centre point F of OC is the focus of themirror.

15. (a) Refractive index is a function of y. It variesalong y-axis i.e., the boundary separating two

media is parallel to x-axis or normal at any pointwill be parallel to y-axis.

Axθ

B (x, y)

t = 0

y

P

Secondly, refractive index increases as y is increased.Therefore, ray of light is travelling from rarer to densermedium i.e., it will bend towards the normal and shapeof its trajectory will be as shown below.

⇒Normal

Now, refer to figure (1):Let i be the angle of incidence at any point B on itspath

θ = 90° – ior tan θ = tan (90° – i) = cot ior slope = cot i

(b) but dxdy

=θtan

∴ idxdy

cot= ...(1)

Applying Snell's law at A and B

BBAA inin sinsin =

1=An because 0=y

1sin =Ai because °= 90Ai (Grazing incidence)

12/3 += KynB

12/3 += y because 2/3)(0.1 −= mK

∴ iy sin)1()1)(1( 2/3 +=

or 1

1sin2/3 +

=y

i

1

i

y +

1

3/2

y3/2

∴ 2/3cot yi = or 4/3y ...(2)

Equating equations (1) and (2), we get

220

4/3ydxdy

= or dxdyy =− 4/3

or ∫ ∫=−y x

dxdyy0 0

4/3

or xy =4/14 ...(3)

The required equation of trajectory is .4 4/1 xy =(c) At point P, where the ray emerges from the slab

y = 1.0 m∴ x = 4.0 m [From Eq. 3)

Therefore, coordinates of point P areP = (4.0 m, 1.0 m)

(d) As PPAA inin sinsin = and 1== PA nn

Therefore, °== 90AP ii i.e., the ray will emerge parallel

to the boundary at P i.e., at grazing emergence.

16. (a) (i) Given λ = 6000 ÅLet be the width of slit and D the distance betweenscreen and slit.

First minima

f

θθ

First minima is obtained at λ=θsinbor θ≈θλ=θ sinb

orbλ

=θ

Angular width of first maxima λ∝λ

=θ=b2

2

Angular width will decrease by 30% when λ is alsodecreased by 30%.Therefore, new wavelength–

−=′λ 6000

10030)6000( Å

4200=′λ Å(ii) When the apparatus is immersed in a liquid ofrefractive index µ, the wavelength is decreased µ times.Therefore,

µ=

Å6000Å4200

∴42006000

=µ or 43.1429.1 ≈=µ

17. (i) Critical angle θc at face AC will be given by

=θ −

nn

c11sin or

nn

c1sin =θ

Now, it is given that cr θ=2

∴ )45(21 crAr θ−°=−=Applying Snell's law at face AB, we have

1

1

sinsin

ri

n = or 11 sinsin rni =

∴ )sin(sin 11

1 rni −=

n1

n

B C

A

45°

45°

r1

r2

I1

Substituting value of ,1r we get

)45sin(sin 11 cni θ−°= −

)sin45coscos45(sinsin 1ccn θ°−θ°= −

=θ

θ−θ−= −

nnn

ccc121 sinsinsin1(

2sin

−−= −

nn

nnn 12

211 1

2sin

−−= − )(2

1sin 121

211 nnni

Therefore, required angle of incidence (i1) at face ABfor which the ray strikes at AC at critical angle is

−−= − )(2

1sin 121

211 nnni

(ii) The ray will pass undeviated through face AC when

either nna =1)( or

221

°= 02r i.e., ray falls normally on face AC.

Here nn ≠1 (because nn <1 is given)

∴ °= 02r

or °=°−°=−= 4504521 rAr

Now applying Snell's law at face AB, we have 1

1

sinsin

ri

n =

or°

=45sin

sin352.1 1i

∴

=

21)352.1(sin 1i

956.0sin 1 =i

∴ °≈= − 73)956.0(sin 11i

tµ

Therefore, required angle of incidence isi = 76°

18. Given, m33.1,mm1,33.1 ===µ Ddλ = 6300 Å

(i) Wavelength of light in the given liquid :

m104737Å4737Å33.1

6300 10−×=≈=µλ=′λ

∴ Fringe width, ω = 'Dd

λ

ω = –10

–34737 10 )(1.33)

(1 10 )

m m

m

× ×

= 6.3 × 10–6mω = 0.63 mm

(ii) Let t be the thickness of the glass slab.Path difference due to glass slab at centre O.

∆x = glass

liquid–1

µt

µ

= 1.53

–11.53

t

or ∆x = 0.15tNow, for the intensity to be minimum at O, this path

difference should be equla to '

2λ

∴ ∆x = '

2λ

or 0.15t =4737

Å2

∴ t = 15790 Åor t = 1.579µm

19. For both the halves, position of object and image issame. Only difference is of magnification. Magnificationfor one of the halves is given as 2 (> 1). This can ebfor the first one, because for this, | v | > | u |. Therefore,mangification, | m | = | v/u | > 1.So, for the first half

| v/u | = 2or | v | = 2 | u |Let u = – x then v = + 2x and | u | + | v | = 1.8 mi..e 3x = 1.8m or x = 0.6 mHere, u = – 0.6 m and v = + 1.2 m

Using1f

1 1–v u

= =1 1 1–

1.2 –0.6 0.4=

∴ f = 0.4mFor the second half

1f =

1 1–

1.2– –(0.6 )d d+

ordd +

+−

=6.01

2.11

4.01

Solving this, we get d = 0.6 mMagnification for the second half will be

m2 =0.6 1

––(1.2) 2

vu

= =

and magnification for the first half is :

m1 =1.2

–2–(0.6)

vu

= =

The ray diagram is as follows :

A

B

B1

A,A1 2

0.6m 0.6m 0.6m

d

t = 0.4m t = 0.4m

20. (i) Path difference due to the glass slab, ∆x = (µ – 1)t= (1.5 –1)t = 0.5tDue to this slab, 5 fringes have been shifted upwards.Therefore,

∆x = 5λredor 0.5t = (5) (7 × 10–2m)∴ t = thickness of glass slab = 7 ×10–6m(ii) Let µ' be the refractive index for given light then

∆x = (µ' – 1)t

222

Now the shifting is of 6 fringes of red light. Therefore,∆x = 6λred

∴ (µ' – 1)t = 6λred

∴ (µ' – 1)= –7

–6(6)(7 10 ) 0.6

7 10

× =×

∴ µ' = 1.6(iii) In part (i), shifting of 5 bright fringes was equal to10–3. Which implies that5ωred = 10–3m [Here ω = Fringe width]

∴ ωred =–310

5m = 0.2 × 10–3m

Now since ω =Dd

λ or ω ∝ l

∴green

red

ω

ω=

green

red

λ

λ

∴ ωgreen = ωred green

red

λ

λ–3(0.2 10 )= ×

–7

–75 10

7 10

× ×

ωgreen = 0.143 × 10–3m∴∆ω = ∴ ωgreen – ωred = (0.143 – 0.2) × 10–3 m

∆ ω = –5.71 × 10–5m

21. From lens Maker's formula,

1f = (µ – 1)

1 2

1 1–R R

We have

−−

−=

RR111

23

3.01

(Here, R1 = R and R2 = – R)∴ R = 0.3

Now applying 2 1µ µ–

v u = R

12 µ−µat air glass surface,

we get

1

3 / 2v

–1

–(0.9) =

3 / 2 – 10.3

∴ v1 = 2.7mi.e., first image I1 will be formed at 27 m from the lens.This will act as the virtual object for glass water

surface. Therefore, applying 2 1µ µ–

v u =R

12 µ−µat glass

water surface, we have

2

/ 3 / 2–

2.7v4 3

=4 / 3 – 3 / 2

–0.3∴ v2 = 1.2m

i.e., second image I2 is formed at 1.2 m from the lensor 0.4 m from the plane mirror. This will act as a virtualobject for mirror. Therefore, third real image I3 will beformed at a distance of 0.4 m in front of the mirror afterreflection from it. Now this image will work as a realobject for water-glass interface. Hence, applying

∴2 1µ µ–

v u = 2 1µ –µR

We get 4

3 / 2 / 3 3 / 2 – 4 / 3–

–(0.8–0.4) 0.3v4

=

∴ v4 = – 0.54mi.e., fourth image is formed to the right of the lens ata distance of 0.54 m from it. Now finally applying thesame formula for glass-air surface.

3.02/31

54.02/31

5 −−

=−

−v

∴ v5 = – 0.9mi.e., position of final image is 0.9 m relative to the lens(rightwards) or the image is formed 0.1 m behind themirror.

22. µ1 = 1.4 and µ2 = 1.7 and let t be the thickness of eachglass plates.Path difference at O, due to insertion of glass plateswill be

S1

S2

1

2

O

6th Minima

5th Minima

∆x = (µ2 – µ1)t = (1.7 – 1.4) t = 0.3tNow, since 5th maxima (earlier) lies below O and 6thminima lies above O.

This path difference should lie between 5λ and 5λ2λ

+

So, let ∆x = 5λ + ∆ ...(2)

Where ∆ <2λ

Due to the path different ∆x, the phase dfference at Owill be

φ = 2 xπ ∆λ = ( )2π 5 λ + ∆

λ

=2(10 . )ππ + ∆λ ...(3)

Intensity at O is given by max34

I and since

223

I (f) = Imaxcos2 2φ

∴ max34

I = Imaxcos2 2φ

or34

= cos2 2φ

....(4)

From Eqs. (3) and (4), we find that

∆ = 6λ

i.e., )3.0(3.0631

65 tttx =∆=λ=

λ+λ=∆

∴ t 31

6(0.3)λ

= = –10(31)(5400 10 )

1.8×

m

or t = 9.3 × 10–6m = 9.3µm

23. n1 =4

20

10.8 101.20 ×+λ

and n2=4

20

1.80 101.45 ( )

×+ λ = λ

λ

Here, λ is in nm.(a) The incident ray will no deviate at BC if n1 = n2

⇒ 4

20

10.8 101.20 ×+λ

=4

020

1.80 101.45 ( )

×+ λ = λ

λ

⇒4

20

9 10×

λ= 0.25

or, λ0

23 100.5×= or λ0 = 600 nm

(b) The given system is a part of an equilaterla prismof prism angle 60° as shown in figure.

A60°

CD

B

20°

40°

70°

60°

i

At minimum deviation,

r1 = r2 =60

302

°= ° = r (say)

∴ n1 =sinsin

ir

∴ sin i = n1 sin 30°

sin i 4

210.8 10 1 1.5 3

1.202 2 4(600)

× = + = = (l = l0 = 600 nm)

or i = sin–1 (3/4)

24. Given λ = 0.5 mm, d = 1.0 mm, D = 1m(a) When the incident beam falls normally :

S1

S2

Oθ1

d sinx

θ = 1

2

y2

α

P1

x = d sin

1

α

Path difference between that two rays S2P and S1P is∆x = S2P – S1P = d sin θ

For minimum intensity,

d sin θ = (2 –1)2

nλ

, n = 1, 2, 3, ....

or sin θ =(2 –1)

2n

dλ

=(2 –1)0.5

2 1.0n

× = 2 –1

4n

As sin θ ≤ 1 therefore (2 –1)

14

n≤ or n ≤ 2.5

So, n can be either 1 to 2

When n = 1, sinθ1 14

= or tan θ1

1

15=

n = 2, sin θ2 34

=

or tanθ2 =3

7∴ y = D tan θ = tan θ (D = 1m)So, the position of minima will be :

y1 = tan θ1 =1

15 m = 0.26m

y2 = tan θ2 =3

7 m = 1.13m

[In this case, net path dfference ∆x = ∆x1 = ∆x2][In this case ∆x = ∆x1 = ∆x2]

224

S1

S2

Oθ1

d sinx

θ = 1

2

y2

α

P1

x = d sin

1

α

[In this case, net path difference ]21 xxx ∆−∆=∆

S2

Oθ2

d sin

xθ = ∆2

2

y2

P2

∆

α

x = d sin

1

And as minima can be on either side of centre O.Therefore, there will be four minimas at positions +0.26 m and + 1.13 m on screen.(b) When α = 30°, path difference between the raysbefore reaching S1 and S2 is∆x1 = d sin α = (1.0) sin 30° = 0.5 mm = λSo, there is already a path differnce of λ between therays.Position of central maximum : Central maximum isdefined as a point where net path difference is zero.So

∆x1 = ∆x2or d sin α = d sin θor θ = α = 30°

or tan θ =1

30y

D=

∴ y0 =1

3nm (D = 1m)

y0 = 0.58m

O

θ2

θ1

θ

30°

P2

P1

Py = 0.58m

Centralmaximum

x = 0∆

OP = yand

OP = y

1 1

2 2

At point P, ∆x1 = ∆x2Above point P ∆x2 > ∆x1 andBelow point P ∆x2 > ∆x2Now, let P1 and P2 be the minimas on either side ofcentral maxima. Then, for P2

∆x2 – ∆x1 = 2λ

or ∆x2 = ∆x1 + 2λ

3

2 2λ λ

= λ + =

or d sin θ2 =32λ

or sin θ2= 32dλ (3)(0.5) 3

(2)(1.0) 4= =

tan θ2 =3

72y

D=

or y2 =3

7 m = 1.13m

Similarly, by for P1

∆x1 – ∆x2 2λ

= or ∆x1 = ∆x2 – 2λ

= –2 2λ λ

= λ =

or d sin θ1 2λ

= or sin θ1 2dλ=

(0.5) 1(2)(1.0) 4

= =

∴ tan θ1 =1

151y

D=

or y1 = 1

15m = 0.26 m

Therefore, y-coordinates of the first minima on eitherside of the central maixmum are y1 = 0.26 m and y2 =1.13 m.

Note : In this problem sin θ = tan θ = θ is not valid asθ is large

25. Given λ = 600 m = 6 × 10–7m,d = 0.4 mm = 0.45 × 10–3mD = 1.5mThickness of glass sheet, t = 10.4 µm = 10.4 × 10–6mRefractive index of medium µm = 4/3and refractive index of glass sheet, ug = 1.(a) Let central maximum is obtained at a distance y

below point O. Then ∆x1 = S1P – S2P ydD

=

S1

S2

O

Py

Path difference due to glass sheet

∆x1 =µ

– 1µ

g

mt

Net path difference will be zero when∆x1 = ∆x2

225

orydD

=µ

– 1µ

g

mt

y =µ

– 1µ

g

mt

Dd


y =1.5

–14 /3

1.5–1

4 /3

y = 4.33 × 10–3mor we can say y = 4.33 mm

(b) At O, ∆x1 = 0 and txm

−

µµ=∆ 12

2

∴ Net path difference, ∆x = ∆x2Corresponding phase difference, ∆φ or simple

φ =2 . xπ ∆λ


φ =–7

2

6 10

π

×

1.5–1

4 /3

(0.04 × 10–6)

φ =133

π

Now, I (φ) = Imax cos2 2φ

∴ I = Imax cos2 13

6π

I = 34

Imax

(c) At O : Path difference is txxm

g

−

µ

µ=∆=∆ 12

For maximum intensity at O∆x = nλ (Here n = 1, 2, 3,....)

∴ λ = , ,1 2 3x x x∆ ∆ ∆

..... and so on

∆x =–61.5

–1 (10.4 10 m)4 /3

×

=–61.5

–1 (10.4 10 nm)4 /3

× ∆x = 1300 nm

∴ Maximum intensity will be corresponding o

λ = 1300nm, 1300

2 nm,

13003 nm,

13004

nm .....

= 1300nm, 650 nm, 433.33 nm, 325 nm..The wavelength in the range 400 to 600 nm are 650 nmand 433.33 nm.

26. Incident ray $A = 6 3i+8 3j–10kur $ $

( ) $( )= 6 3i+8 3j –10k+$ $

z

x

y

z t

Q'

xO

P

Q

A

6 3i + 8 3j–10 k

→

QOuuur

+ PQuuur

(As shown in figure)

Note that QOuuur

is lying x-y plane.

Now, QQ' and Z-axis are mutually perpendicular. hence,we can shown them in two dimensionla figure asbelow:

z

Q'O

P

Q

A

k

→

q

r = 1

µ = 21

µ = 31

Vector Aur

makes an angle i with z-axis given by

i = cos–1

( ) ( )2 22

10

(10) 6 3 8 3

+ +

= cos–112

i = 60°Unit vecotr in the direction QOQ' will be

$q =

( )22

6 3i 8 3 j

(6 3) 8 3

+

+

$ $ ( )1 3i 4j

5= +$ $

226

Snell's law gives

32

=sinsin

ir

sin60sin r

°=

∴ sin r =3 / 2 1

3 / 2 2=

∴ r = 45°Now, we have to find a unit vector in refracted ray's

direction OR. Say it is r$ whose magnitude is 1. Thus,

r$ = (1 sin r) $q – 1 cos r) $k

=1

2 [ $q – $k ]

=1

2$1

(3i 4 j ) – k5

+ $ $

r$ = 1

5 2$(3i 4j––5k)+$ $

27. Applying 2 1µ µ–

v u = 2 1µ –µR

First on plane surface

1

1.5 1–

–AI mR =1.5–1

0=∞

(R – ∞ )

∴ AI1 = – (1.5 mR)Then on curved survace

1 1.5–

–(1.5 )mR R∞ +=

1–1.5–R

[v – ∞ because final image is at infinity]

⇒1.5

(1.5 1)m R+=

0.5R

⇒ 3 = 1.5m + 1

⇒32

m = 2

or m43

=

28. (a) Rays coming from object AB first refract from thelens and then reflect from the mirror.Refraction from the lens :

u = – 20 cm, f = + 15 cm

Using lens formula 1 1 1

–v u f

=

1 1––20v =

115

∴ v = +60 cm

and linear magnification, m1 60 –3

–20vu

+= = =

i.e., first image formed by the lens will be 60 cm fromit (or 30 cm form mirror) towards left and 3 timesmagnified but inverted. Length of first image A1, B1would be 1.2 × 3 = 3.6 cm (inverted)

B

A

B1

A1

0.6cm

3.0cm

3.0cm 60cm 20cm

Optics axis of lens

Optics axis of mirror

Reflection from mirror : Image formed by lens (A1B1)will behave like a virtual object for mirror at a distanceof 30 cm from it as shown. Therefore, u = + 30 cm, f= – 30 cm

Using mirror formula, 1 1 1v u f

+ =

or1 1

30v+ =

1–30

=

∴ v = – 15 cm

and linear magnification, m2 –15 1– –

30 2vu

= = = ++

i.e., final image A' B' will be located at a distance of 15cm from the mirror (towards right) and since

magnification is + 12

, length of final image would be

13.6 1.8

2× = cm

∴ A'B' = 1.8 cm,Point B1 is 0.6 cm above the optic axis of mirror,

therefore, its image B' would be (0.6) 12

= 0.3 cm

above optic axis. Similarly, point A1 is 3 cm below the

optic axis, therefore, its image A' will be 13 1.5

2× = cm

below the optic axis as shown below :

15cm

B'

A'A'B' = 1.8cm

227

Total magnificaion of the image,

m = m1 × m2 =1 3

(–3) –2 2

+ =

∴ A'B' = (m) (AB)3

– (1.2)2

= = –1.8cm

Note that, there is no need fo drawing the ray diagramif not asked in the question.

Note : With reference to the role of an optic instrument(whether it is a lens or a mirror) the coordinates of theobject (X0, Y0)

Xi is obtained using 1 1 1v u f

± =

Here, v is actual X1 and u is X0 i.e., the above formula

can be written as 0

1 1 1

iX X f± =

Similarly, Yi is obtained from 1mO

=

Here, I is Yi and O is Y0 .e., the above formula can bewritten as m = Yi/Y0 or Yi = mY0.

29. Incident ray AB is partly reflected as ray 1 from theupper surface and partly reflected as ray 2 from thelower surface of the layer of thickness t and refractiveindex µ1 = 1.8 as shown in figure. Path differencebetween the two rays would be

t

B

A 1 2

µ = 1.81

µ = 1.52

∆x = 2µ1t= 2 (1.8)t = 3.6t

Ray 1 is reflected from a denser medium, therefore, itundergoes a phase change of π, whereas the ray 2gets reflected from a rarer medium, therefore, there isnot change in phase of ray 2.Hence, phase difference between rays 1 and 2 wouldbe ∆f = π. Therefore, condition of constructiveintereference will be

∆x =1

–2

n λ where n = 1,2 3, ...

or 3.6t =1

–2

n λ Least values of t is corresponding to n = 1 or

tmin = 2 3.6λ

×

or tmin =6487.2 nm

or tmin = 90 nm

30.(i) When angle of prism is small and angle of incidenceis also small, the deviation is given by δ = (µ – 1)ANet deviation by the two prism is zero. So,

Flint

Crown

A1

A2

δ1 + δ2 = 0or (µ1 –1)A1 + (µ2 –1) A2 = 0 ...(1)Here, µ1 and µ2 are the refractive indices for crown andflint glasses respectively.

Hence,µ1= 1.51 1.492+ = 1.5 and µ2= 1.77 1.73

2+ =1.75

A1 = Angle of prism for crown glass = 6°Substituting the value in Eq. (1), we ge

(1.5 – 1)(6°) + (1.75 – 1)A2 = 0This gives A2 = –4°Hence, angle of flint glass prism is 4° (Negative signshown that flint glass prism is inverted with respect tothe crown glass prism.)(ii) Net dispersion due to the two prisms is

1 21 21 2(µ – µ ) (µ –µ )b r b rA A= +

= (1.51 – 1.49) (6°) + (1.77 – 1.73)(–4°)= – 0.04°∴ Net dispersion is – 0.04°

31. 2∆X at R will be zero if ∆X1= ∆X2or d sin α = d sin θor α = θor tan α = tan θ

1

1

yD =

2

2

yD

or y2 = 1

2

DD

, y1 =10200

(40) cm

or y2 = 2 cm

228

(ii) The central bright fringe will be observed at pointQ. If the path difference created by the liquid slab ofthickness t = 10 cm or 100 mm is equal to ∆X1, so thatthe net path difference at Q becomes zero.

S1

S2

S

t=100mm

Q

So, (µ – 1)t = ∆X1or (µ – 1) (100) = 0.16or µ – 1 = 0.0016or µ = 1.0016

32. Let R the radius of curvature of both the surfaces ofthe equi-convex. lens. In the first case :Let f1 be the focal length of equi-convex lens ofrefractive index µ1 and f2 and focal length ofplanoconcave lens of refractive index µ2. The focall;ength of the combined lens system will be given by:

1F

=1 2

1 1f f

+

µ =2

µ =132

43

= (µ1 – 1) 1 1

––R R

+ (µ2 – 1) 1 1

––R

∞

3 2 4 1–1 –1 –

2 3R R = +

1 1 2–3 3R R R

= = or 32R

F =

Now, image coincides with the object when ray of lightretraces its path or it falls normally on the plane mirror.This is possible only when object is at centre ofcurvature of the lens system.Hence, F = 15 cm (Distance of object = 15 cm)

or 32R

= 15 cm or R = 10 cm

In the second case, let µ be the refractive index of theliquid filled between lens and mirror and let F' be thefocal length of new lens system. Then,

µ1

µ

'F1

= (µ1 – 1) 1 1

––R R

+ (µ2 – 1) 1 1

––R

∞

or 'F

1 3 2 (µ–1)–1 –

2 R R =

or = 1 µ – 1 (2–µ)

–R R R

=

∴ F' =10

2 – µ 2 – µR

= (R = 10 cm)

Now, the image coincides with object when it is placedat 25 cm distance.Hence F' = 25

or 10

252–µ

= or 50 – 25 µ = 10 or 25 µ = 40

∴ µ =40 1.625

= or µ = 1.6

33 (a) Shape of the inference fringes will be circular.(b) Intensity of light reaching on the screen directlyfrom the source I1 = I0 (say) and intensity of light onmirror is I2 = 36 % of I0 = 0.36 I0

∴01

2 00.36II

I I= =

10.36 or

1

2

10.6

II

=

∴ max

min

II

=

21

22

1

2

–1

1

II

II

+

=

2

2

1 –110.6

1611

0.6

= +

(c) Initially path difference at P between two wavesreaching from S and S' is

h

h

S

P P

S

h+x

h+xS' S'

Initial Final

Therefore, for maximum intensity at P :

2h =1

–2

n λ ...(1)

229

Now, let the source S is displaced by x (away ortowards mirror) then new path difference will be 2h +2x or 2h – 2x. So, for maximum intensity at P

2h + 2x =1

1–2

n + λ ...(2)

or λ

−−=−

21

122 nxh ...(3)

Solving Eqs. (1) and (2) or Eqs. (1) and (3), we get

x = 6002 2λ

= = 300 nm

Note : Here, we have taken the condition of maximumintensity at P as :

Path difference λ

−=∆

21

nx

Because the reflected beam suffers a phase differenceof π.

34. For refraction at first surface,

2 1

1

µ µ–

–v ∞ = 2 1µ – µR+

...(1)

For refraction at second surface,

3 2

2 1

µ µ–

v v = 3 2µ –µR+

...(2)

Adding Eqs. (1) and (2), we get

µ1 µ2

µ3

v1v2

3

2 1

µv v = 3 1µ – µ

Rorv2 = 3

3 1

µµ – µ

R

Therefore, focal length of the given lens system is

3

3 1

µµ – µ

R

35. (a) sin i1 = µ sin r1

B C

A

60°

30°

30°90°

or sin 60° = 3 sin r1

∴ sin r1 =12

or r1 = 30°Now, r1 + r2 = A∴ r2 = A – r1 = 30° – 30° = 0°Therefore, ray of light falls normally on the face ACand angle of emergence i2 = 0°.

(b) Multiple reflection occurs between surfaces of film.Intensity will be maximum if intereference takes placein the transmitted wave.For maximum thickness

∆x = 2µt = λ (t = thickness)

∴ t = 2µλ 600 1500

2 2.2Å6= =

×

36. Applying Snell's law on face AB

(I) sin 45° = ( )2 sin r

∴ sin r =12

or r = 30°i.e., ray becomes parallel to AD inside the block. Nowapplying

2 1µ µ–v u = 2 1µ – µ

R on face CD,

1.514 2–OE ∞

=1.514– 2

0.4Solving this equaion, we get

OE = 6.06 cm

37. Differentiating the lens formula 1 1 1

–v u f

= with respect

to time, we get

dtdv

v.1

2− +

21

. 0µ

dvdt

= (as f = constant)

∴dvdt

=2

2.

v dudtu

...(1)

Further, substituting proper values in lens formula.We have

1 10.4v

+ =1

0.3(u = – 0.4m, f = 0.3 m)

or v = 1.2 mPuting the values in Eq. (1)

230

Magnitude of rate of change of position of image= 0.09 m/s

Laterel magnification, m vu

=

∴dmdt

= 2

– .dv duu vdt dt

u

= 2(–0.4)(0.09)–(1.2)(0.01)

(0.4)= 0.3/s

∴ Magnitude of rate of change of lateral magnification= 0.3/s

38. Let n1 bright fringe corresponding to wavelengthλ1 = 500 nm coincides with n2 bright fringecorresponding to wavelength λ2 = 700 nm.

∴1

1Dn

dλ

=2

2Dn

dλ

or1

2

nn =

2

1

λλ

75

=

This imples that 7th maxima of λ1 coincides with 5thmaxima of λ2. Similarly 14th maxima of l1 will coincidewith 10th maxima of λ2 and so on.

∴ Minimum distance 1 1n D

dλ=

= 7 × 5 ×10–7 × 103

= 3.5 × 10–3 m = 3.5 mm

39. Critical angles at 1 and 2

1Cθ–1 1

2

µ=sinµ

=–1 1

sin 452

= °

2Cθ–1 3

2

µ=sinµ

= –1 3sin 45

2

= °

Therefore, minimum angle of incidencefor total internal reflection to take place on both slabsshould be 60°.

imin = 60°'

40. (a) At minimum deviation r1 = r2 = 30°∴ From Snell's law

µ = 1

1

sinsin

ir or 3 =

1sinsin30

i°

∴ sin i1 =3

2 or i1 = 60°

(b) In the position shown net deviation suffered by

the ray of light should be minimum. Therefore, thesecond prism should be rotated by 60° (anticlockwise).

B,D

A C

E

MATCH THE COLUMNS

2. (A), (C) and (D) : In case of concave mirror or convexlens image can be real, virtual, diminished magnified orof same size.(B) : In case of convex image is always virtual (for realobject).


1. Law of reflection can be applied to any type of surface.∴ Correct option is (c).

COMPREHENSION

1. Wavefronts are parallel in both media. Therefore, lightwhich is prependicular to wavefront travels as a parallelbeam in each medium.Hence, the correct option is (a).

2. All points on a wavefront are at the same phase.∴ φd = φc and φf = φe∴ φd – φf = φc – φeHence, the correct option is (c).

3. In medium-2 wavefront bends away from the normalafter refraction. Therefore, ray of light which isperpendicular to wavefront bends towards the normalin medium-2 during refraction. So, medium-2 is denseror its speed in medium-1 is more.

X Y

b d

a cf

h

eg

Medium-1

Medium-2

Hence, option (b) is correct.

248

CHAPTER-14ELECTROSTATICS

FILL IN THE BLANKS1. Five identical capacitor, each of area A, are arranged such that adjacent plates are at a distance d apart, the plates

are connected to a source of emf V as shown in the figure. (1984; 2M)The charge on plate 1 is .... and on plate 4 is ....

2. Figure shows line of constant potential in a region in which an electric field is present. The values of the potentialare written in brackets of the points A, B and C, the magnitude of the electric field is greatest at the point .....

(1984; 2M)

A B

C(50V)

(40V)(30V)

(20V)(10V)

3. Two small balls having equal positive charge Q (coulomb) on each are suspended by two insulating strings of equallength L (metre) from a hook fixed to a stand. The whole set-up is taken in a satellite into space where there is nogravity (state of weightlessness). The angle between the strings is ..... and the tension in each string is .... newtons.

(1986; 2M)

4. Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential differenceV. The battery is then disconnected and the region between the plates of capacitor C is completely filled with amaterial of dielectric constant K. The potential difference across the capacitors now become ...... (1988; 2M)

5. A point charge q moves from point P to point S along the path PQRS (Fig.) in a uniform electric field E pointingparallel to the positive direction of the X-axis. The co-ordinates of points, P, Q, R and S are (a, b, 0), (2a, 0, 0), (a,– b, 0), (0, 0, 0) respectively. The work done by the field in the above process is given by the expression.

(1989; 2M)

Q

ER

S

Y

x

P

249

6. The electric potential V at any point x, y, z (all in metres) in space is given V = 4x2 volt. The electric field at thepoint (1m, 0.2 m) is .... V/m (1992; 1M)

7. Five point charges, each of value + q coulomb, are placed on five vertices of a regular hexagon of side L metre.The magnitude of the force on the point charge of value – q coulomb placed at the centre of the hexagon is ...newton. (1982; 1M)

q q

q

qq

–q

TRUE/FALSE1. The work done in carrying a point charge from one point to another in an electrostatic field depends on the path

along which the point charge is carried. (1981; 2M)

2. Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge Q coulomb andthe other an equal negative charge. Their masses after charging are different. (1983; 2M)

3. A small metal ball is suspended in a uniform electric field with the help of an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the direction of the field. (1983; 2M)

4. Two protons A and B are placed in between the two plates of a parallel plate capacitor charged to a potentialdifference V as shown in the figure. The forces on the two protons are identical. (1986; 3M)

Å

++++++++++++++++

–––––––––––––––––––––––––V

B

5. A ring of radius R carries a uniformly distributed charge + Q. A point charge – q is placed in the axis of the ringat a distance 2R from the centre of the ring and released from rest. The particle executes a simple harmonic motionalong the axis of the ring. (1988; 2M)

6. An electric line of forces in the x-y plane is given by the equation x2 + y2 = 1. A particle with unit positive charge,initially at rest at the point x = 1, y = 0 in the x-y plane, will move along the circular line of force. (1988; 2M)

OBJECTIVE QUESTIONSOnly One option is correct :1. An alpha particle of energy 5 MeV is scattered through

180° by a fixed uranium nucleus. The distance ofclosest approach is of the order of : (1981; 2M)(a) 1Å (b) 10–10 cm(c) 10–12cm (d) 10–15cm

2. A hollow metal sphere of radius 5 cm is charged suchthat the potential on its surface is 10 V. The potentialat the centre of the sphere is : (1983; 1M)(a) zero(b) 10 V

(c) same as at a point 5 cm away from the surface(d) same as at a point 25 cm away from the surface

3. Two equal negative charges – q are fixed at points(0, – a) and (0, a) on y-axis. A positive charge Q isreleased from rest at the point (2a, 0) on the x-axis.The charge Q will :(1984; 2M)(a) execute simple harmonic motion about the ori-

gin(b) move to the origin and remain at rest(c) move to infinity(d) execute oscillatory but not simple harmonic

motion

250

4. A charge q is placed at the centre of the line joiningtwo equal charge Q. The system of the three chargeswill be in equilibrium if q is equal to : (1987; 2M)

(a) –2Q

(b) –4Q

(c)4Q

+ (d)2Q

+

5. A solid conducting sphere having a charge Q issurrounded by an uncharged concentric conductinghollow spherical shell. Let the potential differencebetween the surface of the solid sphere and that of theouter surface of the hollow shell be V. If the shell isnow given a charge of – 3Q, the new potential differ-ence between the same two surfaces is :(1999; 2M)(a) V (b) 2V(c) 4V (d) 2V

6. Seven capacitors each of capacitance 2µF are con-nected in a configuration to obtain an effective capaci-

tance 10

µF11

. Which of the following combination will

achieve the desired result be ? (1990; 2M)

(a)

(b)

(c)

(d)

7. Two identical thin rings, each of radius R are coaxiallyplaced a distance R apart. If Q1 and Q2 respectively arethe charges uniformly spread on the two rings, thenwork done in moving a charge q from the centre of onering to that of the other is : (1992; 2M)(a) zero

(b) q (Q1 – Q2) ( ) ( )R042/12 ∈π−

(c) 1 2 02( / )/(4 )q Q Q Rπε

(d) q (Q1 – Q2) ( ) ( )R042/12 ∈π+

8. Two point charges + q and – q are held fixed at (–d,

0) and (d, 0) respectively of x-y co-ordinate system.Then: (1995; 2M)(a) the electric field E at all points on the x-axis has

the same direction(b) work has to be done in bringing at a test charge

from ∞ to the origin(c) electric field at all point on y-axis is along x-axis(d) the dipole moment is 2qd along the x-axis

9. A parallel plate capacitor of capacitance C is con-nected to a battery and is charged to a potentialdifference V. Another capacitor of capacitance 2C issimilarly charged to a potential difference 2V. Thecharging battery is now disconnected and the capaci-tors are connected in parallel to each other in such away that the positive terminal of one is connected tothe negative terminal of the other. The final energy ofthe configuration is : (1995; 2M)

(a) zero (b) 232

CV

(c)225

6CV (d) 29

2CV

10. The magnitude of electric field Eur

in the annular regionof a charged cylindrical capacitor : (1996; 2M)(a) is same through out(b) is higher, near the outer cylinder than the inner

cylinder(c) varies as 1/r, where r is the distance from the axis(d) varies as 1/r2, where r is the distance from the axis

11. A metallic solid sphere is placed in a uniform electricfield. The lines of force follow the path (s) shown infigure as : (1996; 2M)

1 12 23 34 4

(a) 1 (b) 2(c) 3 (d) 4

12. An electron of mass me, initially at rest moves througha certain distance in a uniform electric field in time t1.A proton of mass mp, also, initially at rest, takes timet2 to move through an equal distance in this uniformelectric field. Neglecting the effect of gravity, the ratiot2/t1 is nearly equal to (1997 1M)(a) 1 (b) (mp/me)

1/2

(c) (me/mp)1/2 (d) 1836

13. A non-conducting ring of radius 0.5 m carries a totalcharge of 1.11 × 10–10 C distributed non-uniformly on

251

it circumference producing an electric field E every-

where in space. The value of the integral ∫=

∞=

−0

.l

l

dlE

(l = 0 being centre of the ring) in volt is :(1997; 2M)

(a) +2 (b) – 1(c) – 2 (d) zero

14. A parallel combination of 0.1MΩ resistor and a 10µFcapacitor is connected across a 1.5 V source of neg-ligible resistance. The time required for the capacitor toget charged upto 0.75V is approximately (in second)

(1997; 2M)(a) infinite (b) loge2(c) log102 (d) zero

15. A charge + q is fixed at each of the points x = x0,x = 3x0, x = 5x0 .... ∞ . on the x-axis and a charge – qis fixed at each of the points x = 2x0, x = 4x0, x = 6x0..... ∞ . Here x0 is a positive constant. Take the electricpotential at a point due to a charge Q at a distance rfrom it to be Q/4πε0r. Then the potential at the origindue to the above system of charges is : (1997; 2M)

(a) zero (b) 2ln8 00xq

πε

(c) infinite (d)004

2lnx

qπε

16. Two identical metal plates are given positive chargesQ1 and Q2 (< Q1) respectively. If they are now broughtclose together to form a parallel plate capacitor withcapacitance C, the potential difference between themis: (1999; 2M)(a) (Q1 + Q2)/2C (b) (Q1 + Q2)/C(c) (Q1 – Q2)/C (d) (Q1 – Q2)/2C

17. For the circuit shown, which of the following state-ments is corret (1999; 2M)

= 30 V+

= 2pF

–V1

C1

S1

= 20 V+

= 3pF

–V2

C2

S3

S2

(a) With S1 closed, V1 = 15V, V2 = 20V(b) With S3 closed, V1 = V2 = 25V(c) With S1 and S2 closed, V1 = V2 = 0(d) With S1 and S3 . closed, V1 = 30V, V2 = 20V

18. Three charges Q, + q and + q, are placed at the verticesof a right angle triangle (isosceles triangle) as shown.The net electrostatic energy of the configuration iszero, if Q is equal to : (2000; 2M)

Q

(a)–

1 2

q

+(b)

–2

2 2

q

+(c) – 2q (d) + q

19. A parallel plate capacitor of area A, plate separation dand capacitance C is filled with three difference dielectricmaterials having dielectric constant K1, K2 and K3 asshown. If a single dielectric material is to be used toachieve the same capacitance C in this capacitor then itsdielectric constant K is given by : (2000; 2M)

d

A/2 A/2

k1 k2

k3

d2

A = Area of plates

(a)1 2 3

1 1 1 12K K K K

= + +

(b)1 2 3

1 1 12K K K K

= ++

(c)1 2

31 2

12

K KK

K K K= +

+

(d)2 31 2

1 2 2 3

K KK KK

K K K K= +

+ +

20. Three positive charges of equal value q are placed atthe vertices of an equilateral triangle. The resultantlines of forces should be sketched as in :

(2001; 1M)

(a) (b)

(c) (d)

252

21. Consider the situation shown in the figure. The capaci-tor A has a charge q on it whereas B is uncharged. Thecharge appearing on the capacitor B a long time afterthe switch is closed is : (2001; 1M)

q

S

+++++

–––––

A B

(a) zero (b) q/2(c) q (d) 2q

22. A uniform electric field pointing in positive x-directionexists in a region. Let A be the origin B be the pointon the x-axis at x = + 1 cm and C be the point on they-axis at y = + 1cm. Then the potential at the points A,B and C satisfy : (2001; 1M)(a) VA < VB (b) VA > VB(c) VA < VC (d) VA < VC

23. Two equal point charges are fixed at x = – a and x =+ a on the x-axis. Another point charge Q is placed atthe origin. The change in the electrical potential en-ergy of Q, when it is displaced by a small distance xalong the x-axis is approximately proportional to :

(2002; 1M)(a) x (b) x2

(c) x3 (d) 1/x

24. Two identical capacitors, have the same capacitanceC. One of them is charged to potential V1 and the otherto V2. The negative ends are also connected, thedecrease in energy of the combined system is :

(2002; 1M)

(a) 2 21 2

1( – )

4C V V (b) 2 2

1 21

( )4

C V V+

(c) 221 )(

41

VVC − (d) 221 )(

41

VVC +

25. A metallic shell has a point charge q kept inside itscavity. Which one of the following diagram correctlyrepresents the electric lines of forces? (2003; 1M)

(a) (b)

(c) (d)

26. Six charges, three positive and three negative of equalmagnitude are to be placed at the vertices of a regular

hexagon such that the electric field at O is double theelectric field when only one positive charge of samemagnitude is placed at R. Which of the followingarrangements of charge is possible for P, Q, R, S, T andU respectively? (2004; 1M)

P Q

U

T

O R

S(a) +, –, +, –, –, + (b) +, –, +, –, +, –(c) +, +, –, +, –, – (d) –, +, +, –, +, –

27. Consider the charge configuration and a sphericalGaussian surface as shown in the figure. When calcu-lating the flux of the electric field over the sphericalsurface, the electric field will be due to: (2004; 1M)

+q1

– q1

q2

(a) q2 (b) only the positive charge(c) all the charges (d) + q1 and – q1

28. Three infinitely long charge sheets are placed as shownin figure. The electric field at point P is :

(2005; 1M)z

z = 3a

z = 0xz = – a

– 2σ

–σ

σ.P

(a)$

0

2σε

k (b)$

0

2–

σε

k

(c)$

0

4σε

k (d)$

0

4–

σε

k

29. A long, hollow conducting cylinder is kept coaxiallyinside another long, hollow conducting cylinder oflarger radius. Both the cylinders are initially electricallyneutral. (2007; 3M)(a) a potential difference appears between the two

cylinders when a charge density is given to theinner cylinder

(b) a potential difference appears between the twocylinders when a charge density is given to theouter cylinder

(c) no potential difference appears between the twocylinders when a uniform line charge is kept alongthe axis of the cylinders

253

(d) no potential difference appears between the twocylinders when same charge density is given toboth the cylinders

30. Consider a neutral conducting sphere. A positive pointcharge is placed of outside the sphere. The net chargeon the sphere is then : (2007; 3M)(a) negative and distributed uniformly over the sur-

face of the sphere(b) negative and appears only at the point on the

sphere closest to the point charge(c) negative and distributed non-uniformly over the

entire surface of the sphere(d) zero

31. A spherical portion has been removed from a solidsphere having a charge distributed uniformly in itsvolume as shown in the figure. The electric field insidethe emptied space is : (2007; 3M)

(a) zero everywhere (b) non-zero and uniform(c) non-uniform (d) zero only at its centre

32. Positive and negative point charges of equal magni-

tude are kept at 0,0,2a

and

–0,0,

2a

, respectively.

The work done by the electric field when anotherpositive point charge is moved from (–a, 0, 0) to(0, a, 0) is : (2007; 3M)(a) positive(b) negative(c) zero(d) depends on the path connecting the initial and

final positions

33. Consider a system of three charges 3

2and

3,

3qqq

placed at point A, B and C respectively, as shown inthe fitgure. Take O to be the centre of the circle ofradius R and angle ∠ CAB = 60º (2008; 3M)

(a) The electric field at point O is 20pe8 Rq

directed

along the negative x-axis(b) The potential energy of the system is zero(c) The magnitude of the force between the charges

at C and B is 20

2

pe54 R

q

(d) The potential at point O is Rq

0pe12

34. A parallel plate capacitor C with plates of unit area andseparation d is filled with a liquid of direlectric con-

stant K = 2. The level of liquid is 3d

initially. Suppose

the liquid level decreases at a constant speed v, thetime constant as a function of time t is (2008; 3M)

(a)vtd

R35

e6 0

+(b) 222

0

932

e)915(

tvdvtd

Rvtd

−−

+

(c)vtd

R35

e6 0

−(d) 222

0

932

e)915(

tvdvtd

Rvtd

−+

−

35. Three concentric metallic spherical shells of radii R, 2R,3R, are given charges Q1, Q2, Q3, respectively. It isfound that the surface charge densities on the outersurfaces of the shells are equal. Then, the ratio of thecharges given to the shells, Q1 : Q2 : Q3, is

(2009; M)(a) 1 : 2 : 3 (b) 1 : 3 : 5(c) 1 : 4 : 9 (d) 1 : 8 : 18

36. A disc of radius a/4 having a uniformly distributedcharge 6C is placed in the x-y plane with its centre at(–a/2, 0, 0). A rod of length, 'a' carrying a uniformlydistributed charge 8C is placed on the x-axis fromx = a/4 to x = 5a/4. Two point charges –7C and 3C areplaced at (a/4, –a/4, 0) and (–3a/4, 3a/4, 0), respectively.Consider a cubical surface formed by six surfacesx =± a/2, y = ± a/2, z = ± a/2. The electric flux throughthis cubical surface is (2009; M)

254

x

y

(a)0

2ε

−C

(b)0

2εC

(c)0

10ε

C(d)

0

12ε

C

OBJECTIVE QUESTIONSMore than one options are correct?1. A parallel plate capacitor is connected to a battery.

The quantitites charge, voltage, electric field and en-ergy associated with this capacitor are given by Q0,V0, E0 and U0 respectively. A dielectric slab is nowintroduced to fill the space between the plates with thebattery still in connection. The corresponding quanti-ties now given as Q, V, E and U are related to theprevious one as : (1989; 2M)(a) Q > Q0 (b) V > V0(c) E > E0 (d) U > U0

2. A parallel plate capacitor is charged and the chargingbattery is then disconnected. If the plates of thecapacitor are moved further apart by means of insulat-ing handles. (1987; 2M)(a) the charge on the capacitor increases(b) the voltage across the plates increases(c) the capacitance increasees(d) the electrostatic energy stored in the capacitor

increases

3. A parallel plate capacitor of plate area A and plateseparation d is charged to a potential difference V andthen the battery is disconnected. A slab of dielectricconstant K is then inserted between the plates of thecapacitor so as to fill the space between the plates. IfQ, E and W denote respectively, the magnitude ofcharge on each plate, the electric field between theplates (after the slab is inserted), and work done on thesystem, in the process of inserting the slab, then :

(1999; 3M)

(a)0 AVQd

ε= (b)0 KAVQ

dε=

(c)VEKd

= (d)2

0 11–

2AV

Wd K

ε =

4. A dielectric slab of thickness d is inserted in a parallelplate capacitor whose negative plate is at x = 9 andpositive plate is at x = 3d. The slab is equidistant fromthe plates. The capactior is given some charge. As xgoes from 0 to 3d : (1998; 2M)(a) the magnitude of the electric field remains the

same(b) the direction of the electric field remains the same(c) the electric potential increases continuously(d) the electric potential increases at first, then de-

crease and again increases

5. A positively charged thin metal ring of radius R is fixedin the x-y plane with its centre at the origin O. Anegatively charged particle P is released from rest atthe point (0, 0, z0) where z0 > 0. Then the motion of Pis(1998; 2M)(a) periodic for all values of z0 satisfying 0 < z0 < ∞(b) simple harmonic for all values of z0 satisfying 0 <

z0 < R(c) approximately simple harmonic provided z0 < < R(d) such that P crosses O and continues to move

along the negative z-axis towards z = –∞

6. A non-conducting solid sphere of radius R is uni-formly charged. The magnitude of the electric field dueto the sphere at a distance r from its centre :

(1998; 2M)(a) `increases as r increases, for r < R(b) decreases as r increases, for 0 < r < ∞(c) decreases as r increases, for R < r < ∞(d) is discontinuous at r = R

7. An elliptical cavity is carved within a perfect conduc-tor. A positive charge q is placed at the centre of thecavity. The points A and B are on the cavity surfaceas shown in the figure. Then : (1999; 3M)(a) electric field near A in the cavity = elecrtic field

near B in the cavity(b) charge density at A = charge density at B(c) potential at A = potential at B(d) total electric field flux through the surface of the

cavity is q/ε0.

Aq B

8. For spherical symmetrical charge distribution, variationof electric potential with distance from centre is givenin diagram. Given that :

0 04q

VR

=πε for r ≤ R0

255

and 04

qV

r=

πε for r ≥ R0

Then which option (s) are correct : (2006; 5M)

V

r = R0

r

(a) Total charge within 2R0 is q(b) Total electrostatic energy for r ≤ R0 is zero(c) At r =R0 electric field is discontinuous(d) There will be no charge anywhere except at r =

R09. Under the influence of the Coulomb field of charge +Q,

a charge - q is moving around it in an elliptical orbit.Find out the correct statement(s). (2009; M)(a) The angular momentum of the charge - q is

constant(b) The linear momentum of the charge - q is constant(c) The angular velocity of the charge - q is constant(d) The linear speed of the charge - q is constant

SUBJECTIVE QUESTIONS1. A charge Q is distributed over two concentric hollow

spheres of radii r and R (> r) such that the surfacedensities are equal. Find the potential at the commoncentre. (1981; 3M)

2. The figure shows two identical parallel plate capacitorsconnected to a battery with the switch S closed. Theswitch is now opened and the free space between theplates of the capacitors is filled with a dielectric ofdielectric constant (or relative permittivity) 3. Find theratio of the total electrostatic energy stored in bothcapacitors before and after the introduction of thedielectric (1983; 6M)

V A C

S

B C

3. Two fixed, equal, positive charge, each of magnitudeq = 5 × 10–5C are located at points A and B separatedby a distance of 6 m. An equal and opposite chargemoves towards them along the line COD, the perpen-dicular bisector of the line AB. The moving charge,when it reaches the pointC at a distance of 4 m fromO as a kinetic energy of 4 J. Calculate the distance ofthe farthest point D which the negative charge willreach before returning towards C. (1985; 6M)

D

A +q

+q

C – q

B

O

4. Three point charge q, 2q and 8q are to be placed ona 9 cm long straight line. Find the positions where thecharges shold be placed such that the potential energyof this system is minimum. In this situaion, what is theelectric field at the position of the charge q due to theother two charges? (1987; 1M)

5. Three particles, each of mass 1 g and carrying a chargeq, are suspended from a common point by insulatedmassless strings, each 100 cm long. If the particles arein equilibrium and are located at the corners of anequilateral triangle of side length 3 cm calculate thecharge q on each particle. (Take g = 10 m/s2)

(1988; 5M)

6. Three concentric spherical metallic shells, A, B and Cof radii a, b and c (a < b < c) have surface chargedensities σ, –σ and σ respectively. (1990; 7M)(i) Find the potential of the three shells A, B and C.(ii) If the shells A and C are at the same potential,

obtain the relation between the radii a, b and c.

7. Two fixed charges – 2Q and Q are located at the pointswith coordinates (– 3a, 0) and (+ 3a, 0) respectively inthe x-y plane. (1991; 8M)(a) Show that all points in the x-y plane where the

electric potential due to the two charges is zero, lieon a circle. Find its radius and the location of itscentre.

(b) Give the expression V (x) at a general point on thex-axis and sketch the function V (x) on the wholex-axis.

(c) If a particle of charge + q starts form rest at thecentre of the circle, show by a short quantitativeargument that the particle eventually crosses thecircle. Find its speed when it does so.

8. (a) A charge of Q is uniformly distributed over aspherical volume of radius R. Obtain an expressionfor the energy of the system.

(b) What will be the corresponding expression for theenergy needed to completely disassemble theplanet earth against the gravitational pull amongstits constituent particles?Assume the earth to be a sphere of uniform massdensity. Calculate this energy, given the productof the mass and the radius of the earth to be 2.5× 1031 kg-m.

256

(c) If the same cahrge of Q as in part (a) above isgiven to a spherical conductor of the same radiusR, what will be the energy of the system?

(1992; 10M)

9. Two parallel plate capacitors A and B have the samespearation d = 8.85 × 10–4 m between the plates. Theplate areas of A and B are 0.04 m2 and 0.02m2 respec-tively. A slab of dielectric constant (relative permittiv-ity) K = 9 has dimensions such that it can exactly fillthe space between the plates of capacitor B.

(1993; 7M)

110 V(a)

A B

(b) (c)

B

(i) The dielectric slab is placed inside A as shown infigure (a). A is then charged to a potential differ-ence of 110 V. Calculate the capacitance of A andthe energy stored in it.

(ii) The battery is disconnected and then the dielec-tric slab is removed from A. Find the work done bythe external agency in removing the slab from A.

(iii) The same dielectric slab is now placed inside B,filling it completely. The two capacitor A and B arethen connected as shown in figure (c). Calculatethe energy stored in the system.

10. A circular ring of radius R with uniform positve chargedensity λ per unit length is located in the y-z planewith its centre at the origin O. A particle of mass m andpositive charge q is projected from the point

( )3,0,0P R on the positive x-axis directly towards O,

with an initial speed v. Find the smallest (non-zero)value of the speed v such that the particle does notreturn to P. (1993; 4M)

11. Two square metal plates of side 1 m are kept 0.1 mapart like a parallel plate capacitor in air in such a waythat one of their edges is perpendicular to an oilsurface in a tank filled with an insulating oil. The platesare connected to a battery of emf 500 V. The plates arethen lowered vertically into the oil at a speed of0.001ms –1. Calculate the current drawn from the batteryduring the process. (Dielectric constant of oil = 11,ε0 = 8.85 × 10–12C2N–1m–1). (1994; 6M)

12. The capacitance of a parallel plate capacitor with platearea A and separation d, is C. The space between theplates is filled with two wedges of dielectric constantK1 and K2 respectively (figure). Find the capacitance of

the resulting capacitor. (1996; 2M)

A

K1

K2d

13. Two isolated metallic solid spheres of radii R and 2Rare charged such that both of these have chargedensity σ. The spheres are located far away from achother and connected by a thin conducting wire. Findthe new charge density on the bigger sphere.

(1996; 3M)

14. Two capacitors A and B with capacities 3µ F and 2µFare charged to a potential difference of 100 V and 180V respectively. The plates of the capacitors are con-nected as shown in the figure with one wire of eachcapacitor free. The upper plate of A is positive and thatof B is negative. An unchanged 2 µF capacitor C withlead wires falls on the free ends to complete the circuit.Calculate : (1997; 5M)

A

C

B

+ –3 Fµ 2 Fµ

2 Fµ

100 V 180 V

(i) the final charge on the three capacitors and(ii) the amount of electrostatic energy stored in the

system before and after completion of the circuit.

15. A conducting sphere S1 and of radius r is attached toan insulating handle. Another conducting sphere S2 ofradius R is mounted on an insulating stand S2 isinitially uncharged.S1 is given a charge Q brought into contact with S2 andremoved. S1 is recharged such that the charge on it isagain Q and it is again brought into contact with S2and removed. This procedure is repeated n times.

(1998; 8M)(a) Find the electrostatic energy of S2 after n such

contacts with S1.(b) What is the limiting value of this energy as n → ∞ ?

16. A non-conducting disc of radius a and uniform posi-tive surface charge density σ is placed on the groundwith its axis vertical. A particle of mass m and positivecharge q is dropped, along the axis of the disc from aheight H with, zero initial velcoity. the particle has q/m = 4ε0g/σ. (1999; 10M)(a) Find the value of H if the particle just reaches the

disc.

257

(b) Sketch the potential energy of the particle as afunction of its height and find its equilibriumposition.

17. Four point charges + 8 µC, – 1µC, –1µC and + 8 µC

are fixed at the points – 27 /2 m, – 3 / 2 m,

3 / 2+ m and 27 /2+ m respectively on the y-axis.

A particle of mass 6 × 10–4kg and charge + 0.1µCmoves along the x-direction. Its speed at x = + ∞ is V0.Find the least value of V0 for which the particle willcross the origin. Find also the kinetic energy of theparticle at the origin. Assume that space is gravityfree. (1/4πε0 = 9 × 109 Nm2/C2). (2000; 10M)

18. A small ball of mass 2 × 10–3 kg having a charge of 1µCis suspended by a string of length 0.8 m. Anotheridentical ball having the same charge is kept at thepoint of suspension. Determine the minimum horizon-tal velocity which should be imparted to the lower ball,so that it can make complete revoltuion.

(2001; 5M)19. Eight point charges are placed at the corners of a cube

of edge a as shown in figure. Find the work done indisassembling this system of charges. (2003; 2M)

+ q

+ q

+ q

+ q

– q

– q

– q

– q

20. A positive point charge q is fixed at origin. A dipole

with a dipole moment pr

is placed along the x-axis

away from the origin with pr

pointing along positive

x-axis Find : (a) the kinetic energy of the dipole whenit reaches a distance d from the origin, and (b) the

force experienced by the charge q at this moment.(2003; 4M)

21. There are two large parallel metallic plates S1 and S2carrying surface charge densities S1 and S2 respec-tively (σ1 > σ2) placed at a distance d apart in vacuum.Find the work done by the electric field in moving apoint charge q a distance a (a < d) from S1 towards S2along a line making an angle π/4 with the normal to theplates. (2004; 2M)

22. A conducting bubble of radius a, thickness t (t < < a)has potential V. Now the bubble collapse into a drop-let. Find the potential of the droplet. (2005; 2M)

23. A solid sphere of radius R has a charge Q distributed

in its volume with a charge density ,? akr= where

k and a are constants and r is the distance from its

centre. If the electric field at 2R

r = is 81

times that at

r = R, find the value of a.


This question contains, statement-I (assertion) andstatement-II (reason).

1. Statement-I : For practical purposes, the earth is usedas a reference at zero potential in electrical circuits.

(2008; 3M)Because :Statement-II : The electrical potential of a sphere ofradius R with charge Q uniformly distributed on the

surface is given by RQ

04πε .

(a) Statement-I is true, statement -II is true, statement-II is a correct explanation for statmeent-I

(b) statemen-I is true, statement-II is true; statement-II is NOT a correct explanaion for statmeent-I

(c) statement-I is true, statement-II is false(d) statement-I is false, statement-II is true

ANSWERS

FILL IN THE BLANKS

1.02, –AV AV

d dε ε0

2. B 3. 180°, 2

2016

Q

Lπε 4. 3

2V

K +

5. – qEa 6. – 8 i$ 7. 2

21

Lkq

(Attraction)

TRUE/FALSE1. F 2. T 3. T 4. T 5. F 6. F

258

OBJECTIVE QUESTION (ONLY ONE OPTION)1. (c) 2. (b) 3. (d) 4. (a) 5. (a) 6. (a) 7. (b) 8. (c)9. (b) 10. (c) 11. (d) 12. (b) 13. (a) 14. (d) 15. (d) 16. (d)17. (d) 18. (b) 19. (d) 20. (c) 21. (a) 22. (b) 23. (b) 24. (c)25. (c) 26. (d) 27. (c) 28. (b) 29. (a) 30. (d) 31. (b) 32. (c)33. (c) 34. (a) 35. (b) 36. (a)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)1. (a, d) 2. (b, d) 3. (a, c, d) 4. (b, c) 5. (a, c) 6. (a, c) 7. (c, d)8. (a, b, c, d) 9. (a)


1.2 2

0

( )

4 ( )

Q R r

R r

+

πε +2.

35 3. Maximum distance from O = 8.48 m

4. (i) Charge q should be at a distance of 3 cm from 2q (ii) Eelctric field = 0 5. 3.17 × 10–9C

6. (i) 2

0 0( – ), –A B

aV a b c V b c

b

σ σ= + = + ε ε

, 2 2

0–C

a bV c

c c

σ= + ε

(iii) a + b = c

7. (a) Radius = 4a, Centre = (5a, 0) (b) Vs 04Q

=πε

1 2–

3 – 3a x a x +

for x ≤ 3a, Vx 04Q

=πε

1 2–

– 3 3x a a x +

for x > 3a (c) v 08

Qqma

=πε 8. (a)

3

0

320

QU

R=

πε (b)

23–5

GMUR

= E = 1.5 × 1022J (c) 2

08Q

UR

=πε

9. (i) CA = 2 × 10– 9F, UA = 1.21 × 10–5 J (ii) W = 4.84 × 10–5J (iii) U = 1.1 × 10–5J

10. vmin 02q

mλ=

ε 11. i = 4.43 × 10–9A 12. CR 1 2

2 1–CK KK K

= in 2

1

KK where

0 ACd

ε=

13. 56

σ 14. (i) q1 = 90 µC, q2 = 210µC, q3 = 150µC (ii) (a) Ui = 47.4mJ (b) Uf = 18 mJ

15. (a) Un 2

08nq

R=

πε (b)2

208

Q RUr

∞ =πε

Here qn 1–nQR R

r R r

= + 16. (a)

3

aH =

17. (v0)min = 3m/s, K = 3 × 10–4J 18. 5.86 m/s 19. W = 5.824 2

0

14

qa

πε

20. (a) 2

04

qpKE

d=

πε (b)

302

pq

d=

πεF iur $ 21.

1 2

0

( – )2

qaW

σ σ=

ε 22. V' = 1/3

3q

Vt

23. 2=a

ASSERTION AND REASION1. (b)

259

SOLUTIONS

6.$–

V V Vx y z

∂ ∂ ∂= + + ∂ ∂ ∂

E i j kur $ $

V = 4x2

Therefore, 8V xx

∂ =∂ and 0

V Vy z

∂ ∂= =

∂ ∂

Eur

= – 8 x i$

or Eur

at (1m, 0.2m) is – 8i$ V/m

7. Force on – q due to charges at 1 and 4 are equal andopposite. Similarly, forces on – q due to charges at 2and 5 are also equal and opposite. Therefore, net forceon – q due to charge at 1, 2, 4 and 5 is zero. Onlyunbalanced force is between – q and + q at 3 which

is equal to 2

2

041

Lq

πε and 9.0 × 109 2

2q

L (attraction)

TRUE FALSE1. Electrostatic force is conservative in nature and in

conservative force field work done is path indepen-dent.

2. Mass of negatively charged sphere will be slightlymore than the mass of positively charged spherebecause some electrons will be given to the negativelycharged sphere while some electrons will be taken outfrom the positively charged sphere.

3. When X-rays fall on the metalball, some electrons are emitedfrom it due to photoelectric effect.The ball thus gets positivelycharged and on a positivelycharged ball an electrostatic forcein the direction of electric fieldacts. due to this force ball getsdeflected in the direction of elec-tric field.

4. Electric field between the plates of capacitor is almostuniform. Therefore, force on both the protons will beidentical. It hardly matters whether they are placednear positive plate or negative plate.

5. Motion is simple harmonic only when charge – q is notvery far from the centre of ring on its axis. Otherwisemotion is periodic but not simple harmonic in nature.

6. Electric lines force does not represent the path of thecharged particle but tangent to the path at any pointon the line shows the direction of electric force on it

FILL IN THE BLANKS

1. dAC 0∈=

dAVVcQ 0

1 )0( ∈=−=

)0()0(4 VCVCQ −+−=

dAVCV 022 ∈−=−=

2. drdV

is greatest for B.

3. Due to electrostatic repulsion the charges will move asfarthest as possible and the angle between the twostring will be 180°. Tension in each string will be equalto electrostatic repulsion between the two charges.Thus,

T = Fe =2

2 20 0

14 (2 ) 16

Q Q Q

L L

× =πε πε

4. Total charge will remain uncharged.Hence Q = Q'or 3CV = (KC + 2C)'

∴ V' =3

2V

K +

5. WFe = .F dur r

( dr

= displacement)

= ). . –qE s p( i r rur uur$

= ). (– –qE a b ( i i j$ $ $

= – qEa

260

and it is not always necessary that motion of theparticle is in the direction of force acting on it.

OBJECTIVE QUESTIONS (ONLY ONE OPTION)1. From conservation of mechanical energy

decrease in kinetic energy = increase in potential energy

or0 min

1 ( )(2 )4

Ze erπ ∈ = 5 MeV

= 5 × 1.6 × 10–13J

+ Ze + 2e (Z = 92)

rmin

∴ rmin =2

–130

1 24 5 1.6 10

Zeπ∈ × ×

=9 –19 2

–13(9 10 )(2)(92)(1.6 10 )

5 1.6 10

× ×

× ×= 5.3 × 10–14m = 5.3 × 10 –12cm

i.e., rmin is of the order of 10–12 cm.∴ correct option is (c).

2. Electric potential at any point inside a hollow metallicsphere is constant. Therefore, if potential at surface is10 V, potential at centre will also be 10 V.

3. Motion is simple harmonic only if Q is released froma point very far from the origin on x-axis. Otherwisemotion is periodic but not simple harmonic.

4. Since, q is at the centre of two charges Q and Q, netforce on it is zero, whatever the magnitude and sign ofcharge on it. For the equilibrium of Q, q should benegative because other charge q will repel it, so qshould attract it. Simultaneously these attractions andrepulsions should be equal.

Q Q q

20

20 )2/(4

14

1rQq

rQQ

πε=

πε

or q =4Q

or with sign q = –4Q

∴ correct option is (b).

5. In such situation potential difference depends only onthe charge on inner sphere. Since, charge on innersphere is unchanged. Therefore, potential difference Vwill remain unchanged.

6. In series, C 1 2

1 2

C CC C

=+

∴ Cnet =(10)(1)10 1+

1011

= µF

10 Fµ

1 Fµ

7. 1CV = 1 2Q QV V+

=1 1

0 0

1 14 4 2

Q QR R

+πε πε

= 21

0

14 2

QQ

R + πε

Q1Q2

C2C1R

RR 2

Similarly, 2CV = 12

0

14 2

QQ

R + πε

∴ ∆V = 1 2–C CV V

= 0

14 Rπε

1 2 1 21

( – ) – ( – )2

Q Q Q Q

= ( )1 2

0

( – )2 – 1

2(4 )Q Q

Rπε

W = q∆V = q (Q1 – Q2) ( )2 – 1 / 02(4 )Rπε

8. The diagramatic representation of the given questionis shown in figure.

Eq

EE–q

q – q(– d, 0) (d, 0)

y

x

The electrical field Eur

at all points on the x-axis will nothave the same direcion.For – d ≤ x < d, electric field is along positive x-axiswhile for all other points it is along negative x-axis.

The electric field Eur

at all points on the y-axis will be

261

parallel to the x-axis (i.e., i$ )(option c)The electrical potential at the origin due to both thecharges is zero, hence, no work is done in bringing atest charge from infinity to the oriign.Dipole moment is directed from the –q charge to the

+ q charge (i.e., – i$ direction).

9. CVQQ 321 =′+′

Q2′–

+ Q1′

Q=CV1

+ –Q=4CV2

+ –

′=′⇒′=′12

21 22

QQC

QC

Q

⇒ CVQCVQCVQ 233 211 =′⇒=′⇒=′

CQ

CQE f 42

22

21 ′+′=

22

2

23

44

21

CVCV

CV =+=

10. The magnitude of electric field at a distance r from the

axis is given as r

E02πε

λ=

i.e., E ∝1r

Here, λ is the charge per unit length of the capacitor

11. Electric field lines never enter a metallic conductor (E= 0, inside a conductor) and they fall normally on thesurface of a metallic conductor (because whole surfaceis at same potential and lines are perpendicular toequipotential surface).

12. Electrostatic force, Fe = eE (for both the particle). Butacceleration of electron, ae = Fe/me and acceleration ofproton, ap = Fe/mp

22

21 2

121

tataS pe ==

∴2

1

tt =

e

p

aa

p

e

m

m=

13.0

– .l

l

=

= ∞∫ Edluruur

= 0

–l

ldV V

=

=∞=∫ (centre) – V (infinity)

but V (infinity) = 0

∴ 0

– .l

l

=

= ∞∫ Edluruur

corresponds to potential at centre of

ring. and V (centre) 0

1.

4qR

=πε

9 –10(9 10 )(1.11 10 )0.5

× ×= ≈ 2 volt

14. Since, the capacitor plates are directly connected tothe battery, it will take no time in charging.

C

R

V

15. Potential at origin will be given by

V = 04

qπε

00 0 0

1 1 1 1– – ....2 3 4

xx x x

+ +

= 04

qπε

.0

1 1 1 11– – ....

2 3 4x + +

= 04

qπε 0

2lnx

16. Electric field within the plates 1 2Q Q= +E E Eur ur ur

E

+Q1 +Q2 E1 E2

262

E = E1 – E2

=1 2

0 0–

2 2Q QA Aε ε

E =1 2

0

–2

Q QAε

Potential difference between the plates

VA – VB = E.d = 1 2

0

–2

Q Q dA

ε

= 1 2

0

–

2

Q QAdε

=1 2–2

Q QC

17. When S3 is closed, due to attraction with oppositecharge, no flow of charge takes place through S3.Therefore, potential difference across capacitor platesremains unchanged so, V1 = 30 V and V2 = 20 V.

18. Net electrostatic energy of the configuration will be

U. . .

2

q q Q q Q qa aa

= + +

Here, K

0

14

=πε

Q –2

2 2

q=

+

19.d

AKd

AKC 01

01

1

2

2 ∈=∈

=

dAKC 03

2∈=

A/2

K1 K2

K3 K4

C1

C3

C2

C4

d/2

d/2

d/2

d/2

dAKC 02

3∈=

dAKC 03

4∈=

Let Cd

A =∈0

43

43

21

21

CCCC

CCCC

Ceq ++

+=

32

32

21

21 ))(()(

))((KK

CKCKCKKCKCK

++

+=

CKCKK

KKKK

KKeq=

+

++

=32

32

21

21

⇒ 32

32

21

21

KKKK

KKKK

Keq ++

+=

20. Electric lines of force never form a closed loop. there-fore, options (b) and (d) are wrong. Electric lines offorce emanate from positive charge and terminate onnegative charge, therefore, option (a) is also wrong.

21. Due to attraction with positive charge, the negativecharge on capacitor A will not flow through the switchS.

22. Potential decreases in the direction of electric field.Dotted lines are equipotential lines.

∴ BACA VVVV >= and y

A B X

E

C

23.Q Q q q q q

x= – a x= – ax = 0 x = xx = + a x = aInitial position Final position

Ui =2KQq

a

and Uf =1 1

–KQq

a x a x + +

Here, K =0

14πε

∆U = Uf – Ui

or |∆U| =2

32KQqx

a for x < < a

∴ ∆U ∝ x2

24. ∆U = decrease in potential energy= Ui – Uf

( )2

2 2 1 21 2

1 1 (2 )2 2 2

V VC V V C

+ = + =

( )21 21

–4

C V V=

263

25. Electric field is zero everywhere inside a metal (con-ductor) i.e., field lines do not enter a metal.Simltaneously these are perpendicular to a metal sur-face (equipotential surfaces).

26. According to option (d) the electric field due to P andS and due to q and T add to zero. While due to U andR will be added up. Hence, the correct option is (d).

27. At any point over the spherical Gussian surface, netelectric field is the vector sum of electric fields due to+ q1, – q1 and q2,Hence, the correct option is (c).

28. All the three plates will produce electric field at Palong negative z-axis. Hence,

pEur

=$

0 0 0

2 (– )2 2 2

σ σ σ+ + ε ε ε k

= $0

2–

σε

k

∴ Correct answer is (b).

29. There will be an electric field between two cylinders(using Gauss theorem). This electric field will producea potential difference.∴ Answer is (a).

30. Charge will be induced on the conducting sphere, butnet charge on it will be zero.∴ Option (d) is correct.

31. Inside the cavity, field at any point is uniform and non-zero.Therefore, correct option is (b).

32. A = (–a, 0, 0)B = (0, a, 0)

y

–Q

xAQz

B

Point charge is moved from A to BVA = VB = 0 ∴ W = 0

or the correct option is (c).

33. Distance BC = AB sin 60º

RR 323)2( ==

∴ 20 )3(

)3/2)(3/(pe41

R

qqFeBC =

20

2

pe54 R

q=

34.21

2102

01 ,

3

2 ,

32 CC

CCC

vtd

AC

vtdA

C+

=−

∈=

+

∈=

vtd +3

2vtd +

32

K=2

−

+

∈=vtdvtd

ACeq

332

)(2 20

−

+

++−∈

−

+

∈

=

vtdvtd

vtdvtdA

vtdvtdA

Ceq

332

23

43

332

)(2

0

20

200 m1 35

6

352

=+

∈=

+

∈= A

vtdA

vtd

A

vtdR

RCeq 356 0

+∈

==τ

35. 2321

221

21

)3(4)2(44 RQQQ

RQQ

RQ

π++

=π

+=

π

Q1

-Q1

Q1+ Q2 Q1+ Q Q2 3+

3R2R

R

- Q( +1

Q2)

9432121

1QQQQQQ ++=+=

⇒ 5:3:1:: 321 =QQQ∴ (b)

36.000

. 2)723(

1ε

−=−+ε

=ε

=φC

CCCqencl

∴ (a)

OBJECTIVE QUESTIONS (MORE THAN ONEOPTION)1. When dielectric slab is introduced capacity gets in-

creased while potential difference remains uncharged.

264

∴ V = V0, C < C0Q = CV ∴ Q > Q0

U = 212

CV ∴ U > U0

E =VD

but E and d both are unchanged. Therefore, E

= E0 Therefore, correct options are (a) and (d).

2 Charging battery is removed. Therefore, q = constantDistance between the plates is increased. Therefore Cdecreases

Now, qVC

= , q is constant and C is decreasing.

Therefore, V should increase.

U 21

2qC

= again q is constant and C is decreasing.

Therefore U should increase.∴ Correct options are (b) and (d).

3. Battery is removed. Therefore, charge stored in theplates will remain constant.

Q = CV = 0 Ad

ε

V = ConstantNow, dielectric slab is inserted. Therefore, C will in-crease. New capacity will be,

C ' = KC =0 KAd

ε

V ' =Q VC K

=

And new electric field

E ='

.V Vd K d

=

Potential energy stored in the capacitor,

Initially, Ui = 212

CV

=2

02AVd

ε

Finally, Uf

22 01 1

' '2 2

K A VC V

d Kε = =

20 AVdKd

ε

Work done on the system will be

|∆U| =2

02AVd

ε 11–

K

∴ correct options are (a), (c) and (d).

4. The magnitude and direction of electric field at differ-

ent points are shown in figure. The direction of theelectric field remains the same. Hence, option (b) iscorrect. Simlarly, electric lines always flow from higherto lower potential, therefore, electric potential increasescontinuously as we move from x = 0 to x = 3d.

– –

–

–

– – – – – – – – –

+

+

+

+

+++++++++E /k0

qi qi q

x = 0 x = d x = 2d x = 3d E0 E0

Therefore, option (c) is also correct. The variation ofelectric field (E) and potential (V) with x will be asfollows :OA || BC and (Slope)OA > (Slope)ABBecause EO–d = E2d–3d and EO–d > Ed – 2d

E

d 2d 3d X

E

d 2d 3d X

V

V 0

A B

C

O

5. Let Q be the charge on the ring, the negative charge– q is released from point P (0, 0, z0). The electric fieldat P due to the charged ring will be along positive z-axis and its magnitude will be

E =0

2 2 3 / 20 0

14 ( )

Qz

R zπε +

E = 0 at centre of the ring because z0 = 0Therefore, force on charge P will be towards centre asshown, and its magnitude is

Fe = qE = 2/320

20

0 )(.

41

zRQz

qEFe +πε= ...(1)

Similarly, when it crosses the origin, the force is againtowards centre O.Thus, the motion of the particle is periodic for allvalues of z0 lying between 0 and

Secondly, if z0 < < R, ( )3 / 22 2 30R z R+ =

Fe = 030

1. .

4Qq

zRπε (from Eq. 1)

i.e., the restoring force Fe ∝ – z0. Hence, the motionof the particle will be simple harmonic. (Here negative

265

sign implies that the force is towards its mean position.

6. Inside the sphere

E =3

0

1.

4Q

rRπε

⇒ E ∝ r for r ≤ Ri.e., E at centre = 0 (r = 0)

and E at surface = )(4 2

0

RrR

Q=

πε

Outside the sphere

( )Rrr

QE ≥

πε= 2

04

or E ∝ 21

rThis variation of electric field (E) with distance (r) fromthe centre will be as follows :

21

rE ∝2

041

R

QE

πε=

E

O r=Rr

ER∝

7. Under electrostatic condition, all points lying on theconductor are at same potential. Therefore, potential atA = potential at B. Hence, option (c) is correct fromGauss theorem, total flux through the surface of thecavity will be q/ε0.

8. The given graph is of charged conducting sphere ofradius R0. The whole charge q distributes on thesurface of the sphere.

9. Force is always directed toward the centre of +Q,hence net torque on the charge - q is zero.

As 0≠Fr

so pr

will change.

Since moment of inertia is changing, ω will not beconstant.∴ (a)

SUBJECTIVE QUESTIONS1. Let q1 and q2 be the charges on them.

σ1 = σ2

∴1

24

q

rπ=

224

q

Rπ

∴1

2

qq =

2

2r

R

i.e., charge on them is distributed in above ratio.

or, q1 2

2 2r Q

r R=

+ and q2

2

2 2R Q

r R=

+Potential at centre V = potential due to q1 + potentialdue to q2.

or V =1 2

0 0

1 1. .

4 4q qr R

+πε πε

= 2 20

( )

4 ( )

Q R r

r R

+

πε +

2. Before opening the switch potential difference acrossboth the capacitors is V, as they are in parallel. Hence,energy stored in them is,

UA - UB 212

CV= ∴ Utotal = CV2 = U1 ...(1)

After opening the switch, potential difference across itis V and its capacity is 3C

∴ UA = 2 21 3(3 )

2 2C V CV=

In case of capacitor B, charge stored in it is q = CVand its capacity is also 3C. Therefore,

UB =2 2

2(3 ) 6q CV

C=

∴ Utotal 2 2

23 102 6 6

CV CV CV= + =25

3 fCV U= = ...(2)

From Eqs. (1) and (2) 35

i

f

UU

=

3. Equating the energy of (–q) at C and D :KC + UC = KD + UD

Here, KC = 4J

UC =0

1 ( )(– )24

q qAC

πε

=9 –5 2–2 9 10 (5 10 )

5× × × ×

= – 9JKD = 0 and

UD =0

1 ( )(– )24

q qAD

πε

=9 –5 2–2 9 10 (5 10 )AD

× × ×

= –45AD

266

Substituting these values in Eq. (1)

4 – 9 = 0– 45AD

∴ AD = 9 m

∴ OD 2 2–AD OA= 2 2(9) –(3)= 8 1 – 9== 8.48m

4. For potential energy to be minimum the bigger chargesshold be farthest. Let x be the distance of q from 2q.Then potential energy of the system shown in figurewould be :

0

(2 )( ) (8 )( ) (2 )(8 ) 1(9– ) 9 4

q q q q q qU K kx x

= + + = πε

For U to be minimum 2 8

9 –x x+ should be minimum.

2 89 –

ddx x x

+ = 0

∴ 2 2–2 8

(9– )x x+ = 0

∴ 9 –x

x =12

or x = 3cmi.e., distance of charge q from 2q should be 3 cm.Electric field at q :

0)106(

)8()103(

)2(2222 =

×−

×= −−

qkqkE

5.

F

mg

T

θ

F is the resultant of electrostatic force between twocharges.

30°30°

q

q q

aa

aF

Fe

Fe

F = 2Fe cos 30°

2

20

1 32 .

4 2q

a

=

πε

2

20

1 32 .

4 2q

a

=

πε

13 23 10 q= × ...(1)

θ is the angle of string with horizontal in equilibrium

lθ

r

a

a30°

r

a/2

cosθ =( /2)sec30r a

l l°=

=3

3 100 3

a

l=

∴ θ = 89°Now, the particle is in equilibrium under three concur-rent force, F, T and mg. Therefore, applying Lami'stheorem :

sin(90 )F

+ θ = sin(180 )mg

+ θ

or 3 ×1013q2 = (1 × 10–3)(10) cot 89°

Solving this equation, we getq = 0.317 × 10–8C

or q = 3.17 × 10–9C

6. (i) Potential at any shell will be due to all three charges.

VA0

14

CA B qq qa b c

= + + πε

σπ+

σ−π+

σππε

=c

cb

ba

a ))(4())(4())(4(4

1 222

0

0( – )a b c

σ= +

ε

VB0

14

CA B qq qb b c

= + + πε

σπ+

σ−π+

σππε

=c

cb

ba

a ))(4())(4())(4(4

1 222

0

2

0–

ab c

b

σ= +

ε Similarly,

267

VC0

14

CA B qq qc c c

= + + πε

σπ+

σ−π+

σππε

=c

cb

ba

a ))(4())(4())(4(4

1 222

0

2 2

0–

a bc

c c

σ= +

ε (ii) Given VA = VC

∴0

( – )a b cσ

= +ε =

2 2

0–

a bc

c c

σ= +

ε

∴ a – b + c =2 2

–a b cc c

+

or a + b = c

7. (a) Let P (x, y) be a general point on x-y plane. Electricpotential at point P would be,V = (potential due to Q) + (potential due to – 2Q)or

2 2 2 20 0

1 1 –24 4(3 – ) (3 )

Q QV

a x y a x y

= + πε πε+ + +

...(1)Given V = 0∴ 4 [(3a – x)2 + y2] = (3a + x2) + y2

On, simplifying we get(x – 5a)2 + y2 = (4a)2

This is the equation of a circle of radius 4a and centreat (5a, 0).

(b) on x-axis potential will be undefined (or say + ∞) at x= 3a and x = – 3a, because charge Q and – 2Q areplaced at these two points. So, between – 3a < x < 3awe can find potential by putting y = 0 in Eq. (1).Therefore,

0

1 2–

4 3 – 3Q

Va x a x

= πε + for – 3a < x < 3a

V = 0 at x = aV → – ∞ at x → – 3aand V → + ∞ at x → 3aFor x > 3a, there is again a point where potential willbecome zero so for x > 3a, we can write :

0

1 2–

4 – 3 3Q

Vx a a x

= πε + for x > 3a

V = 0 at x = 9aFor x < – 3a, we can write

0

1 2–

4 3 – 3 –Q

Va x a x

= πε for x < – 3a

In this region potential will be zero only at x → +∞ Thus, we can summarise it is under.

(i) At x = 3a, V = + ∞

(ii) at x = – 3a, V = – ∞

(iii) For x < – 3a, 0

1 2–

4 3 – 3Q

Va x a x

= πε +

(iv) For – 3a < x < 3a, expression of V is same i.e.,

0

1 2–

4 3 – 3Q

Va x a x

= πε +

(v) For x > 3a, 0

1 2–

4 – 3 3Q

Vx a a x

= πε + Potential on x-axis is zero at two places at x = a andx = 9a. The V-x graph is shown below

–3a a

V

+3a9a x

(c) Potential at centre i.e., at x = 5a will be,

0 0

1 2–

4 2 8 16Q Q

Va a a

= = πε πε = positive

Potential on the circle will be zero.Since, potential at centre > potential on circumferenceon it, the particle will cross the circle because positivecharge moves from higher potential to lower potential.Speed of particle, while crossing the circle would be,

v =2 ( )q V

m∆

08Qq

maπε

Here, ∆V is the potential difference between the centreand circumference of the circle.

8. (a) In this case the electric field exists from centre ofthe sphere to infinity. Potential energy is stored inelectric field with energy density

u = 20

12

Eε (Energy/Volume)

(i) Energy stored within the sphere (U1)Electric field at a distance r is

268

E = 30

1. .

4Q

rRπε

u 20

12

E= ε2

03

0

1 .2 4

Q rR

ε= πε Volume of element, dV = (4πr2) dVEnergy stored in this volume, dU = U (dV)

dU =2

2 03

0

1(4 ) .2 4

Qr dr rR

επ πε

dU =2

46

0

1. .

8Q

r drRπε

∴ U1= 0

RdU∫ =

24

60 0

1. .

8

RQr dr

Rπε ∫

[ ]Rr

RQ

05

60

2

40πε=

[ ]Rr

RQ

05

60

2

40πε=

U1 =2

0

1.

40QRπε

(ii) Energy stored outside the sphere (U2)Electric field at a distance r is

E = 20

1.

4Q

rπε

2

20

020 .

41

221

πεε=ε=

rQEu

dV = (4πr2) dV

dU = 4 dV =2

2 02

0

1(4 ) .

2 4Q

r drr

ε π πε

dU =2

20

1.

8Q

rπε

∴ U2= 0

RdU∫ ∫

∞

πε=

Rr

drQ2

0

2

.4

U2 =2

08Q

Rπε ...(2)

Therefore, total energy of the system is

U = U1 + U2 =2 2

0 040 8Q Q

R R+

πε πε

or U =2

0

320

QRπε

(b) Comparing this with gravitational forces, the gravita-tional potential energy of earth will be

U =23–

5GM

R

by replacing Q2 by M2 and

0

14πε

by G.

g = 2GM

R

∴ G =2gR

M

U =–35

MgR

Therefore, energy needed to completely disassemblethe earth against gravitational pull amongst itsconstitutent particle will be given by

E = |U| 35

MgR=

Substituting the values, we get

35

E = (10m/s2) (2.5 × 1031 kg–m)

E = 1.5 × 1032 J(c) This is the case of a charged spherical conductor of

radius R, energy of which is given by

=21

2QC

or U =2

0

12 4

QRπε

or U =2

08Q

Rπε

9. (i) Capacitor A is a combination of two capacitors CKand CO in parallel. Hence

CA = CK + CO 0 0K A A

d dε ε= + 0( 1) AK

dε= +

Here, A = 0.02m2. Substituting the values, we have

CA =–12

–48.85 10 (0.02)(9 1)

(8.85 10 )F×+

×

269

CA = 2.0 × 10–9 FEnergy stored in capacitor A, when connected with a110 V battery is

UA = 212 AC V –9 21

(2 10 )(110)2

= ×

UA = 1.21 × 10–5J(ii) Charge stored in the capacitor

qA = CAV = (2.0 × 10–9) (110)qA = 2.2 × 10–7C

Now, this charge remains constant even after batteryis disconnected. But when the slab is removed, capaci-tance of A will get reduced. Let it be C′A.

C'A =–12

0–4

(2 ) (8.85 10 )(0.04)

8.85 10

AF

dε ×

=×

C'A = 0.4 × 10–9FEnergy stored in this case would be

U 'A =2( )1

2 'Aq

C A

=–7 2

–91(2.2 10 )2 (0.4 10 )

×

×J

U 'A = 6.05 × 10–5 J > UATherefore, work done to remove the slab would beW = U'A – UA = (6.05 – 1.21) × 10–5Jor W = 4.84 × 10–5J(iii) Capacity of B when filled with dielectric is

CB –12

0–4

9)(8.85 10 )(0.02)

(8.85 10 )

K Adε ( ×= =

× F

CB= 1.8 × 10–9 FThese two capacitors are in parallel. Therefore, netcapacitance of the system is

C = C'A + CB = (0.4 + 1.8) × 10–9FC = 2.2 × 10–9F

Charge stored in the system is q = qA = 2.2 × 10–7C

Therefore, energy stored, U 21

2qC

=

U =–7 2

–91(2.2 10 )2 (2.2 10 )

×

×

or U = 1.1 × 10–5J

10. Total charge on the ring is Q = (2πR) λPotential due to a ring at a distance of x from its centreon its axis is given by

V (x) =2 20

1.

4Q

R xπε +

and at the centre is Vcentre = 0

1.

4QRπε

Using the above formula

VP = 2 20 0

1 2.

4 43

R

R R

π λ λ=

πε ε+

VO =0 0

1 2.

4 2RR

π λ λ=

πε ε

VO > VPPotential difference between points O and P is

V = VO – VP =0 0 0

–2 4 4λ λ λ

= =ε ε ε

∴ 212

mv ≥ qV

or v ≥2qV

m

or v ≥0

24

qmλ

ε

or v ≥02

qm

λε

Therefore, minimum value of speed v should be

vmin =02

qm

λε

11. Let a be the side of the square plate.As shown in figure C1 and C2 are in parallel. Therefore,total capacity of capacitors in the position shown is

C = C1 + C2

C = 0 0( – )a a x axd d

ε Κε+

∴ q = CV =0 ( – )aV a x Kxd

ε +

As plates are lowered in the oil, C increases or chargestored will increase.s

Therefore, i 0 ( – ).aVdq dxK tdt d dt

ε= =

Substituting the valuesε0 = 8.85 × 10–12C2 N-m3

a = 1m, V = 500 volt, d = 0.01 m, K = 1

and dxdt = speed of plate = 0.001 m/s

We get current

i = –12(8.85 10 )1(500)(11–1)(0.001)

(0.01)A×

i = 4.43 × 10–9 A

270

12 Let length and breadth of the capacitor be l and brespectively and d be the distance between the platesas shown in figure. Then, consider a strip at a distancex of width dx.Now, QR = x tan θand PQ = d – x tan θ where tan θ = d/l]Capacitance of PQ

C1 =1 0 1 0( ) ( )– tan –

K bdx K bdxxdd x dl

ε ε=

θ

C1 =1 0 1 0 ( )( – ) ( – )

K bldx K A dxd l x d l x

ε ε=

and C2 = capacitance of QR

1 0 ( )tan

K bdxxε=

θ

C2 1 0 ( )K Adx

xdε= (tan θ

dl

= )

Now, C1 and C2 are in series. Therefore, their resultantcapacity C0 will be given by

0

1C =

1 2

1 1C C

+

Then,0

1C =

1 2

1 1C C

+

=1 0 2 0

(1– ) .( ) ( )

d x x dK A dx K A dx

+ε ε

0

1C

=0 1 2

–( )d l x x

A d x K K

+ ε

=1

2 1

0 2

(1– ) ( )

d K x K xAK K dxε

∴ C0 =0 1 2

2 1 (1– ) AK K

dxd K x K x

ε+

C0 =0 1 2

2 1 2 ( – ) AK K

dxd K l K K x

ε+

Now, q the net capacitance of the given parallel platecapacitor is obtained by adding such infinitesimalcapacitors placed parallel from x = 0 to x = l

i..e, CR ∫=

==

lx

xC

00

0 1 20 2 1 2 ( – ) l AK K

dxd K l K K x

ε=

+∫

Finally we get CR 1 2 0

2 1( – )K K AK K d

ε= ln

2

1

KK

= 1 2

2 1–CK KK K in

2

1

KK where C

0 Ad

ε=

13. Let q1 and q2 be the charges on the two spheresbefore connecting them.Then q1 = σ (4πR2), and

q2 = σ (4π) (2R)2 = 16σπR2

Therefore, total charge (q) on both the spheres is —q = q1 + q2 = 20σπR2

Now, after connecting, the charge is distributed in theratio of their capacities, which in turn depends on theratio of their radii (C = 4πε0R)

∴1

2

''

qq =

12 2RR

=

∴ q1' = 3q

=203 σπR2

and q1' =23q

=403 σπR2

Therefore, surface charge densities on the spheres are:

σ1 = 1

2'

4

q

Rπ

2

2(20/3) 5

34

R

R

σπ= = σπ

and σ2 = 22

'

4 (2 )

q

Rπ

2

2(40/3) 5

616

R

R

σπ= = σπ

Hence, surface charge density on the bigger sphere isσ2

i.e., (5/6)σ.

14. (i) Charge on capacitor A, before joining with anuncharged capacitor

qA = CV = (100) (3)µC = 300µCSimilarly, charge on capacitor B

qB = (180) (2) µ c= 360 µC

Let q1, q2 and q3 be the charges on the three cpacitorsafter joining them as shown in figure alongside (q1, q2and q3 are in microcoulombs)Form conservation of chargeNet charge on plates 2 and 3 before joining= net charge after joining∴ 300 = q1 + q2 ...(1)Similarly, net charge on plates 4 and 5 before joining= net charge after joining

– 360 = – q2 – q3or 360 = q2 + q3 ...(2)Applying Kirchhoff's second law in closed loop ABCDA

31 2–3 3 2

qq q + = 0

or 2q1 – 3q2 + 3q2 = 0 ...(3)Solving Eqs. (1), (2) and (3), we get

q1 = 90 µCq2 = 210 µC

and q3 = 150 µCTherefore, final charges on the three capacitors are asshown below

271

(ii) (a) Electrostatic energy stored before, completingthe circuit

U1 –6 2 –6 21 1(3 10 )(100) (2 10 )(180)

2 2= × + ×

212

U CV = = 4.74 × 10–2Jor Ui = 47.4 mJ(b) Electrostatic energy stored after, completing thecircuit

Ui –6 2 –6 2

–6 –61(90 10 ) 1(210 10 )2 2(3 10 ) (2 10 )

× ×= +× ×

+ –6 2

–61 (150 10 )2 (2 10 )

×

×

212q

UC

=

= 1.8 × 10–2Jor Uf = 18 mJ

15. Capacities of conducting spheres are in the ratio oftheir radii. Let C1 and C2 be the capacities of S1 andS2, then

2

1

CC =

Rr

(a) Charges are distributed in the ratio of their capaci-ties. Let in the first contact, charge acquired by S2, isq1. Therefore, charge on S1 will be Q – q1. Say it is q'1

∴1 1

1 1' –q qq Q q

= =2

1

C RC r

=

It impleis that Q charge is to be distributed in S2 andS1 in the ratio of R/r.

∴ q1 =R

QR r

+

...(1)

In the second contact S1 again acquires the samecharge Q.Therefore, total charge in S1 and S2 will be

Q + q1 = 1R

QR r

+ + This charge is again distributed in the same ratio.Therefore, cahrge on S2 in second contact.

q2 = 1R R

QR r R r

+ + + =

2R RQ

R r R r

+ + + Similarly,

q3 = 2 3R R R

QR r R r R r

+ + + + +

qn = 2

....nR R R

QR r R r R r

+ + + + + +

or qn = 1–nR R

Qr R r

+ ..(1)

(1– )(1– )

n

na r

Sr

=

Therfore, electrostatic energy of S2 after n such con-tacts

Un

2

2nqC

=

2

02(4 )nq

R=

πε

or Un =2

08nq

Rπε

where qn can be written from Eq. (2).

(b) qn –1

1 .... ... 1nQR R R

R r R r R r

= + + + + + + + as n → ∞

1

1 –

QRq

RR rR r

∞

= + +

QR R r RQ

R r r r+ = = +

1 –a

Sr∞

=

∴2

2qUC∞

∞ =2 2 2

0

/8

Q R rRπε

or U∞ =2

208

Q R

rπε

16. Potential at a height H on the axis of the disc V (P): The charge dq contained in the ring shown in figure.

P (q, m)

Hx

O

rdr

a

dr

dq = (2πr dr)σ

272

Potential of p due to this ring

dV 0

14

=πε

.dqx where x 2 2H r= +

dV 0

14

=πε

.2 2 2 20

(2 )2

rdr rdr

H r H r

π σ σ=

ε+ +∴ Potential due to the complete disc

VP =0

r a

rdV

=

=∫

∫=

= +εσ

=ar

r rH

rdr0 2202

VP = 2 2

0–

2a H H

σ + ε

Potential at centre, (O) will be

VO =02aσ

ε H = 0

(a) Particle is released from P and it just reaches point O.Therefore, from conservation of mechancial energy—Decrease in gravitational potential energy = increase inelectrostatic potential energy

(∆KE = 0 becauseKi = Kf = 0)∴ mgH = q [VO – VP]

or gH 2 2

0–

2q a a H Hm

σ = + + ε ...(1)

qm =

04 gεσ

∴02

qm

σε = 2g

Substituting Eq. (1), we get

gH = 2g [a + H – 2 2a H+ ]

or2H

= ( ) –a H+ 2 2a H+

or 2 2a H+ =2H

a +

or a2 + H2 = 22H

a + + aH

or 234

H = aH

or H =43

a and H = 0

∴ H = (4/3)a(b) Potential energy of the particle at height H = Electro-

static potential energy + gravitational potential energy∴ U = qV + mgHHere V = potential at height H

U 2 2

0[ – ]

2q

a H H mgHσ

= + +ε

...(2)

At equilibrium position

F =– 0dUdH

=

differentiating Eq. (2) w.r.t. H

or 2 20

1 1(2 ) –1 0

2 2q

mg Ha H

+ σ = ε +

∴ 2 22 –1

Hmg mg

a H

+

+ = 0

or 2 2

21 – 2

H

a H+

+= 0

or 2 2

2H

a H+= 1

or2

2 2H

a H+=

14

or 3H2 = a2

or H =3

a

From Eq. (2), we can writeU – H equation as

U = mg (2 2 2a H+ – H)(Parabolic variation)

U = 2mga at H = 0

3

aH =

mga2

mga3

U

HO

and U = Umin = 3 mga at H 3

a=

Therefore, U-H graph will be as shown

Note that at H 3

a= , U is minimum

273

Therefore, H 3

a= is stable equilibrium position

17. In the figure q = 1µC = 10–6, q0 = + 0.1µC = 10–7C andm = 6 × 10–4kg and Q = 8 µC = 8 × 10–6C

V0 m

q0PO x

B

A

C

D

y

–q

+Q

–q

+Q

µ− 2/27

µ− 2/3

µ+ 2/3

µ+ 2/27

Let P be any point at a distance x from origin O. Then

AP = CP =23

2x+

BP = DP =227

2x+

Electric potential at point P will be

2 2–

KQ KqV

BP AP= where K

9

0

19 10

4= = ×

πε Nm2/C2

∴ V –6 –6

9

2 2

8 10 102 9 10 –

27 32 2

x x

× = × ×

+ +

V 4

2 2

8 11.8 10 –

27 32 2

x x

= ×

+ +

...(1)

∴ Electric field at P is

–3 /24 2–1 27

– 1.8 10 (8)2 2

dVE x

dX

= = × +

–3/221 3

– (2 )2 2

x x +

E = 0 on x-axis where – x = 0 or

3 / 22

8

272

x +

= 3/22

1

32

x +

⇒3 / 2

3 / 22

(4)

272

x +

= 3/22

1

32

x +

⇒227

2x +

=23

42

x +

This equation gives x 52

= ± m

The least value of kinetic energy of the particle atinfinity should be enough to take the particle upto x

52

= + m because at x 52

= + m

E = 0⇒ Electrostatic force on cahrge q is zero or Fe = 0

For at x > 52

m, E is repulsive (towards positive x-

axis) and for x < 52

m, E is attractive (towards

negative x-axis)

Now, from Eq. (1) potential at x = 52

m

4 8 11.8 10 –

27 5 3 52 2 2 2

V

= ×

+ +

V = 2.7 × 104 volt

Applying energy conservation at x = ∞ and 52

x = m

20

12

mv = q0V ...(2)

∴ v0 = 02q Vm


v0 =–7 4

–42 10 2.7 10

6 10

× × ×

×v0 = 3m/s

∴ Minimum value of v0 is 3m/sFrom Eq. (1), potential at origin (x = 0) is

V0 = 4 8 11.8 10 –

27 32 2

×

= 2.4 × 104 VLet K be the kinetic energy of the particle at origin.Aplying energy conservation at x = 0 and at x = ∞

274

K + q0V0 = 20

12

mv

But 20

12

mv = q0V [from Eq. (2)]

K = q0 (V – V0)K = (10–7) (2.7 × 104 – 2.4 × 104)K = 3 × 10–4J

Note : E = 0 or Fe on q0 is zeor at x = 0 and x 52

x = ± m

of these x = 0 is stable equilibrium position and

52

x = ± is Unstable equilibrium postion.

18. Given : q = 1µC = 106Cm = 8 × 10–3 kg and l = 0.8 mLet u be the speed of the particle at its lowest pointand v its speed at high point.

u

q

mqv

Fe

T = 0

l

At highest point three forces are acting on the particle.(i) Electrostatic repulsion

Fe =2

20

14

q

l=

πε (outwards)

(ii) weight W = mg (inwards)(iii) Tension T' (inwards)T = 0, if the particle has just to complete the circle andthe necessary centripetal force provided by W – Fe i.e.,

2mvl

= W– Fe

or v2 =2

20

1 1–

4q

mgm l

πε

v2 = 30.8

2 10×

9 6 2–3

29.0 10 (10 )

2 10 1 0 –(18)

× ×× ×

m2/s2

or v2 = 2.4m2/s2 ...(1)Now, the electrostatic potential energy at the lowestand highest points are equal. Hence, from conserva-tion of mechanical energyIncrease in grvaitational potential energy = Decreasein kinetic energy,

or mg (2l) =12

m (u2 – v2)

or u2 = v2 + 4glSubstituting the value of v2 from Eq. (1) we get

u2 = 2.4 + 4 (10) (0.8) = 34.4m2/s2

∴ u = 5.86 m/sTherefore, minimum horizontal velocity imparted to thelower ball, so that it can make complete revolution, is5.86 m/s

19. For potential energy of the system of charges, totalnumber of charge pairs will be 8C2 or 28 of these 28pairs 12 unlike charges are at a separation a, 12 like

charges are at separation 2 a and 4 unlike charges

are at separation 3 a. Therefore, the potential energy

of the system

0

1 (12)( )(– ) (12)( )( ) (4)( ) – ( )4 2 3

q q q q q qU

a a a

= + + πε

2

0

1–5.824 .

4qa

= πε

The binding energy of this system is therefore,

|U| =2

0

15.824

4qa

πε

So, work done by external force in disassembling, thissystem of charges is

W =2

0

15.824 .

4qa

πε

20. Applying energy conservation principle, increase iskinetic energy of the dipole = decrease in electrostaticpotential energy of the dipole.∴ Kinetic energy of dipole at distance d from origin

= Ui – Uf

or KE = 0 – (– pr

. Eur

) = pr

. Eur

2 20 0

1( ).4 4

q qppd d

= = πε πε

i i$ $

(b) Electric field at origin due to the dipole,

Eur

= 3

0

1 24

p

dπεi$ ( E

uraxis ↑↑ p

r)

∴ Force on charge q

Fur

= q Eur

=3

02

pq

dπεi$

21. Electric field near a large metallic plate is given by E= σ/ε0. In between the plates the two fields will be in

275

opposite direction. Hence,

Enet = 1 20

0

–E

σ σ=

ε (say)

Now, W = (q) (potential difference)= q (E0 a cos 45°)

1 2

0

–( )2

aq σ σ = ε

1 2

0

( – )2

qaσ σ=

ε

22. Let q be the charge on the bubble, then

V =Kqa

(Here K 0

14

=πε

)

∴ q =VaK

Let after collapsing the radius of droplet becomes R,then equating the volume, we have

(4πa2)t = 343

Rπ

∴ R = (3a2t)1/3

Now, potential of droplet will be V' KqR

=

V' =( )

2 1 / 3(3 )

VaKK

a t

or V' =1/3

3a

Vt

23. At ∫==2/

0

2p4;2

Ra

encl drkrrqR

r

32/

0

3

23p4

3p4

++

+=

+=

aRa Ra

kar

k

At r = R

;pe48

1pe4

;)(3

p42

02

0

3

Rq

rqR

akq enclencla

encl

′+′ =

+=

2

3

2

3

814

2 RR

RR aa ++

=

322 3 =+a

a + 3 = 5a = 2

ASSERTION AND REASION

1. No solution is required.

276

CHAPTER-15MAGNETICS

FILL IN THE BLANKS1. A neutron, a proton and an electron and alpha particle enter a region of constant magnetic field with equal velocities.

The magnetic field is along the inward normal to the plane of the paper. The tracks of the particles are labelled infigure. The electron follows track .... and the alpha particle follows track..... (1984; 2M)

A

BC

D

2. A wire of length L metres carrying a current i amperes is bent in the form of circle. The magnitude of its magneticmoment is .... in MKS units. (1987; 2M)

3. In a hydrogen atom, the electron moves in an orbit of radius 0.5Å making 1016 revolutions per second. the magneticmoment associated with the orbital motion of the electron is ..... (1988; 2M)

4. The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current I as shown. Themagnitude of the magnetic induction at the centre C is ..... (1988; 2M)

R2

R Q PS

R1

5. A wire ABCDEF (with each side of length L) bent as shown in figure and carrying I is placed in a uniform magneticinduction B parallel to the positive y-direction. The force experienced by the wire is ..... in the ..... direction.(1990; 2M)

O

Bx

A

y

FE

IC

z

6. A metallic block carrying current I is subjected to a uniform magnetic

induction Bur

as shown in figure. The moving charges experience a force Fur

given by .... which results in the lowering of the potential of the face ....Assume the speed of the charge carriers to be V. (1996; 2M)

B YE

FX

GH

I

D

BA

C

Z

277

OBJECTIVE QUESTIONS

Only One option is correct :

1. A magnetic needle is kept in a non-uniform magneticfield. It experiences : (1981; 2M)(a) a force and a torque(b) a force but not a torque(c) a torque but not a force(d) neither a force nor a torque

2. A conducting circular loop of radius r carries a constant

current i. It is placed in a uniform magnetic field 0Bur

such that 0Bur

is perpendicular to the plane of the loop.

The magnetic force acting on the loop is : (1983; 1M)

(a) ir 0Bur

(b) 2πir 0Bur

(c) zero (d) πir 0Bur

3. A rectangular loop carrying a current i is situated neara long straight wire such that the wire is parallel to oneof the sides of the loop and is in the plane of the loop.If steady current I is established in the wire as shownin the figure, the loop will : (1985; 2M)

i

i

(a) rotate about an axis parallel to the wire(b) move away from the wire(c) move towards the wire(d) remain stationary

4. Two thin long parallel wires separated by a distance bare carrying a current i ampere each. the magnitude ofthe force per unit length exerted by one wire on theother is (1986; 2M)

(a)2

02

µ i

b(b)

20µ

2ibπ

(c)0µ

2ibπ (d)

02

µ

2

i

bπ

5. Two particles X and Y having equal charges afterbeing accelerated through the same potential difference,enter a region of uniform magnetic field and describecircular paths of radii R1 and R2 respectively. The ratioof the mass of X to that of Y is : (1988; 2M)(a) (R1/R2)1/2 (b) R2/R1(c) (R1/R2)2 (d) R1/R2

6. A current I flows along the length of an infinitelylong, straight, thin-walled pipe. Then : (1993; 2M)(a) the magnetic field at all points inside the pipe is

the same, but not zero.(b) the magnetic field at any point inside the pipe is

zero.(c) the magnetic field is zero only on the axis of the

pipe.(d) the magnetic field is different at different points

inside the pipe.

7. A uniform magnetic field with a slit system as shown in figure is to be used as a momentum filter for high energycharged particles with a field B Tesla, it is found that the filter transmits α-particles each of energy 5.3 MeV. Themagnetic field is increased to 2.3 B Tesla and deutrons are passed into the filter. The energy of each deuterontransmitted by the filter is .... MeV. (1997C; 1M)

Source Detector

B

TRUE/FALSE1. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field.

(1981; 2M)2. There is no change in the energy of a charged particle moving in magnetic field although a magnetic force is acting

on it. (1983; 2M)3. A charged particle enters a region of uniform magnetic field at an angle of 85° to the magnetic line of force. The

path of the particle is a circle. (1983; 2M)4. An electron and proton are moving with the same kinetic energy along the same direction. When they pass through

a uniform magnetic field perpendicular to the direction of their motion, they describe circular path of the same radius.(1985; 3M)

278

7. A battery is connected between two points A and Bon the circumference of a uniform conducting ring ofradius r and resistance R. One of the arcs AB of thering subtends an angle θ at the centre. the value of themagnetic induction at the centre due to the current inthe ring is : (1995; 2M)(a) proportional to (180° – θ)(b) inversely proportional to r(c) zero, only if (θ = 180°)(d) zero for all values of θ

8. A proton, a deutron and an α-particle having the samekinetic energy are moving in circular trajectories in aconstant magnetic field. If rp , rd and rα denoterespectively the radii of the trajectories of theseparticles, then : (1997; 1M)(a) rα = rp < rd (b) rα > rd > rp(c) rα = rd > rp (d) rp = rd = rα

9. Two particles, each of mass m and charge q, areattached to the two ends of a light rigid rod of length2R. The rod is rotated at constant angular speed abouta perpendicular axis passing through its centre. Theratio of the magnitude of the magnetic moment of thesystem and its angular momentum about the centre ofthe rod is : (1995; 2M)(a) q/2m (b) q/m(c) 2q/m (d) q/πm

10. Two very long straight parallel wires carry steadycurrents I and – I respectively. The distance betweenthe wires is d. At a certain instant of time, a pointcharge q is at a point equidistant from the two wires

in the plane of the wires. Its instantaneous velocity vr

is perpendicular to this plane. This magnitude of theforce due to the magnetic field acting on the charge atthis instant is : (1998; 2M)

(a)0µ Iqv2 dπ (b)

0µ Iqvdπ

(c)02µ Iqvdπ (d) zero

11. A charged particle is released from rest in a region ofsteady and uniform electric and magnetic fields whichare parallel to each other. The particle will move in a:

(1999; 2M)(a) straight line (b) circle(c) helix (d) cycloid

12. a circular loop of radius R, carrying current I, lies in x-y plane with its centre at origin. The total magnetic fluxthrough x-y plane is : (1999; 2M)(a) directly proportional to I(b) directly proportional to R

(c) inversely proportional to R(d) zero

13. An infinitely long conductor PQR is bent to form aright angle as shown in figure A current I flowsthrough PQR, the magnetic field due to this current atthe point M is H1. Now, another infinitely long sraightconductor QS is connected at Q, so that current is I/2 in QR as well as in QS, the current in PQ remainingunchanged. The magnetic field at M is now H2. Theratio H1/H2 is given by : (2000; 2M)

–∞

–∞

∞90°

90°Q

M

P

R

S

(a) 1/2 (b) 1(c) 2/3 (d) 2

14. An ionized gas contains both positive and negativeions. If it is subjected simultaneously to an electricfield along the +x-direction and a magnetic field alongthe +z-direction, then : (2000; 2M)(a) positive ions deflect towards + y-direction and

negative ions towards –y direction(b) all ions deflect towards + y-direction(c) all ions deflect towards –y-direction(d) positive ions deflect towads – y direction and

negative ions towards + y directions

15. A particle of charge q and mass m moves in a circularorbit of radius r with angular speedh ω. The ratio ofthe magnitude of its magnetic moment to that of itsangular momentum depends on : (1997; 2M)(a) ω and q (b) ω, q and m(c) q and m (d) ω and m

16. Two long parallel wires are at a distance 2d apart. Theycarry steady equal currents flowing out of the plane ofthe paper as shown. The variation of the magnetic fieldB along the line XX' is given by : (2000; 2M)

d d

x xx' x''

d d

(a) (b)

279

d

d

x' x'x xd

d

(c) (d)

17. A non-planar loop of conducting wire carrying a currentI is placed as shown in the figure. Each of the straightsections of the loop is of length 2a. The magnetic fielddue to this loop at the point P (a, 0, a) points in thedirection : (2001; 2M)

x

y

z

(a) $1(– )

2+j k$ (b)

$1(– )

3+ +j k i$ $

(c)$1

( )3

+ +i j k$ $(d)

$1( )

2+i k$

18. A coil having N turns is wound tightly in the form ofa spiral with inner and outer radii a and b respectively.When a current I passes through the coil, the magneticfield at the centre is : (2001; 2M)

(a) 0µ NIb

(b) 02µ NIa

(c)0µ

2( – )NI

a b in ba (d)

0µ2( – )

NIb a In

ba

19. Two particles A and B of masses mA and mB respectivelyand having the same charge are moving in a plane. Auniform magnetic field exists perpendicular to thisplane. The speeds of the particles are vA and vBrespectively and the trajectories are as shown in thefigure. Then : (2001; 2M)

B

A

(a) mAvA < mBvB(b) mAvA > mBvB(c) mA < mB and vA < vB(d) ma = mb and vA = vB

20. a long straight wire along the z-axis carries a currenti in the negative z-direction. the magnetic vector field

Bur

at a point having coordinate (x, y) on the z = 0 planeis : (2001; 1M)

(a)0

2 2( – )

2 ( )

µ I y x

x yπ +

i j$ $(b)

02 2

( )

2 ( )

µ I x y

x y

+

π +

i j$ $

(c)0

2 2( – )

2 ( )

µ I x y

x yπ +

j i$ $(d)

02 2

( – )

2 ( )

µ I x y

x yπ +

i j$ $

21. The magnetic field lines due to a bar magnet arecorrect shown in : (2002; 2M)

(a)

N

(b)

N

S

(c)

N

S

(d)

N

S

22. A particle of mass m and charge q moves with aconstant velocity v along the positive x-direction. Itenters a region containing a uniform magnetic field Bdirected along the negative z-direction, extending fromx = a to x = b. The minimum value of v required so thatthe particle can just enter the region x > b is :

(2002; 2M)

(a)qbBm (b)

( – )q b a Bm

(c)qaBm (d)

( )2

q b a Bm

+

23. For a positively charged particle moving in a x-y planeinitially along the x-axis, there is a sudden change inits path due to the presence of electric and/or magneticfields beyond P. The curved path is shown in the x-y plane and is found to be non-circular.

P xO

y

Which one of the following combinations is possible?(2003; 2M)

280

(a) 0;=Eur $b c= +B j k

ur $ (b) ;a=E iur $ $c a= +B k i

ur$

(c) 0;=Eur $c b= +B j k

ur $ (d) ;a=E iur $ $c b= +B k j

ur $

24. A conducting loop carrying a current I is placed in auniform magnetic field point into the plane of the paperas shown. The loop will have a tendency to :

(2003; 2M)

YB

X

I

+

(a) contract(b) expand(c) move towards + ve x-axis(d) move towards -ve x-axis

25. A current carrying loop is placed in a uniform magneticfield in four different orientations, I, II, III and IV,arrange them in the decreasing order of potentialenergy : (2003; 2M)

n B

(I)

B

n(II)

n

B

(III)

n

B

(IV)

(a) I > III > II > IV (b) I > II > III > IV(c) I > IV > II > III (d) III > IV > I > II

26. An electrons moving with a speed u along the positivex-axis at y = 0 enters a region of uniform magnetic field

$0–B=B k

urwhich exists to the right of y-axis. The

electrons exits from the region after sometimes with thespeed v at co-ordinate y, then : (2004; 2M)

e– u

y

x

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

27. A magnetic field $0B=B J

ur exists in the region a < x

< 2a and $0B=B J

ur in the region 2a < x < 3a, where

B0 is a positive constant. A positive point charge

moving with a velocity 0v=v ir $ where v0 is a positive

constant, enters the magnetic field at x = a. Thetrajectory of the charge in this region can be like :

(2007; 3M)

B0

–B0

0 a 2a 3ax

(a) a 2a 3ax

z

(b)x

a 2a 3a

z

(c)x

a 2a 3a

z

(d)a 2a 3a x

z

OBJECTIVE QUESTIONS

More than one options are correct?

1. A proton moving with a constant velocity passesthrough a region of space without any change in itsvelocity. If E and B represent the electric and magneticfields respectively, this region of space may have :

(1985; 2M)(a) E = 0, B = 0 (b) E = 0, B ≠ 0(c) E ≠ 0, B = 0 (d) E ≠ 0, B ≠ 0

2. A particle of charge + q and mass m moving under the

influence of a uniform electric field E i$ and uniform

magnetic field B $k follows a trajectory from P to Q as

shown in figure. The velocities at P and Q are v i$ and

– 2j$ . Which of the following statement (s) is/arecorrect? (1991; 2M)

281

2a 2VQ

x

a

Py

v E

B

(a)23

4mv

Eqa

=

(b) Rate of work done by the electric field at P is

334

mva

(c) Rate of work done by the electric field at P is zero(d) Rate of work done by both the fields at Q is zero

3. H+, He+ and O2+ all having the same kinetic energypass through a region in which there is a uniformmagnetic field perpendicular to their velocity. Themasses of H+, He+ and O2+ are 1 amu, 4amu and 16 amurespectively. Then : (1999; 3M)(a) H+ will be deflected most(b) O2+ will be deflected most(c) He+ and O2+ will be deflected equally(d) all will be deflected equally

4. Which of the following statement is correct in thegiven figure. (2006; 5M)

B C

D

OO

A

l2

l1

infinitely long wire kept perpendicularto the paper carrying current inwards

(a) net force on the loop is zero(b) net torque on the loop is zero(c) loop will rotate clockwise about axis OO' when

seen from O(d) loop will rotate anticlockwise about OO' when

seen from O

5. A particle of mass m and charge q, moving withveloicty v enters Region II normal to the boundary asshown in the figure. Region II has a uniform magneticfield B perpendicular to the plane of the paper. Thelength of the Region II is l. Choose the correct choice(s). (2008; 4M)

(a) The particle enters Region III only if its velocity

mqlB

v >

(b) The particle enters Region III only if its velocity

mqlB

v <

(c) Path length of the particle in Region II is maximum

when velocity m

qlBv =

(d) Time spent in Region II is same for any velocityv as long as the particle returns to Region I


1. A potential difference of 600 V is applied across theplates of a parallel plate condenser. The separationbetween the plates is 3 mm. An electron projectedvertically, parallel to the plates, with a velocity of2 × 106 m/s moves undeflected between the plates.Find the magnitude and direction of the magnetic fieldin the region between the condenser plates. (Neglectthe edge effects). (Charge of the electron = 1.6 × 10–19C)

(1981; 3M)

+

+

–

–

600 V

2. A particle of mass 1 × 10–26 kg and charge + 1.6 × 10–

19C travelling with a velocity 1.28 × 106 m/s in the +Xdirection enters a region in which a uniform electricfield E and a uniform magnetic field of induction B arepresent such that Ex = Ey = 0, Ez = – 102.4 kV/m andBx = Bz = 0, By = 8 × 10–2 weber/m2. The particle entersthis region at the origin at time t = 0. Determine thelocation (x, y and z coordinates) of the paricle at t =5 × 10–6s. If the electric field is switched off at thisinstant (with the magnetic field still present), what willbe the position of the particle at t = 7.45 × 10–6s?

(1983; 6M)

282

are parallel to the y-axis such that OS = OR = 0.02 m.Find the magnitude and direction of the magneticinduction at the origin O. (1989; 6M)

y

OR

Pi

iL

iQ

∞

∞

∞

∞

S iM x

7. Two long parallel wires carrying currents 2.5 A and I(amperes) in the same direction (directed into the planeof the paper) are held at P and Q respectively suchthat they are perpendicular to the plane of paper. Thepoints P and Q are located at a distance of 5 m and2 m respectively from a collinear point R (see figure).

(1990; 8M)

P Q RX

2.5A 1A5m

2m

(a) An electron moving with a velocity of 4 × 105 m/s along the positive x-direction experiences aforce of magnitude 3.2 × 10–20 N at the point R.Find the value of I.

(b) Find all the positions at which a third long parallelwire carrying a current of magnitude 2.5 A may beplaced, so that the magnetic induction at R is zero.

8. A wire loop carrying a current I is placed at the x-yplane as shown in figure. (1991; 4 + 4M)

I

N

aP

+Q

v y

O x

M

120°

(a) If a particle with charge + Q and mass m is placed

at the centre P and given a velocity vr

along NP

(see figure), find its instantaneous acceleration.(b) If an external uniform magnetic induction field

B=B iur $ is applied, find the force and the torque

acting on the loop due to this field.

3. A particle of mass m = 1.6 × 10–27 kg and charge q =1.6 × 10–19C enters a region of uniform magnetic fieldof strength 1 T along the direction shown in figure.The speed of the particle is 107m/s. (1984; 8M)

E

45°

F

θ

(a) The magnetic field is directed along the inwardnormal to the plane of the paper. The particleleaves the region of the field at the point F. Findthe distance EF and the angle θ.

(b) If the direction of the field is along the outwardnormal to the plane of the paper, find the timespent by the particle in the region of the magneticfield after entering it at E.

4. A beam of protons with a velocity 4 × 105 m/s entersa uniform magnetic field of 0.3 T at an angle of 60° tothe magnetic field. Find the radius of the helical pathtaken by the proton beam. Also find the pitch of thehelix (which is the distance travelled by a proton inthe beam parallel to the magnetic field during oneperiod of rotation). (1986; 6M)

5. Two long straight parallel wiresare 2 m apart, perpendicular tothe plane of the paper.

10/11

1.2 m

1.6 m

2 m

A

B

P

SThe wire A carries a current of9.6 A, directed into the plane ofthe paper. The wire B carries acurrent such that the magneticfield of induction at the point P,

at a distance of 1011

m from the

wire B, is zero. (1987; 7M)Find :(a) The magnitude and direction of the current in B.(b) The magnitude of the magnetic field of induction

at the point S.(c) The force per unit length on the wire B.

6. A pair of stationary and infinitely long bent wires areplaced in the xy plane as shown in figure. The wirescarry currents of i = 10 A each as shown. The segmentsL and M are along the x-axis. The segments P and Q

283

9. A straight segment OC (of length L) of a circuitcarrying a current I is placed along the x-axis. Twoinfinitely long straight wires A and B, each extendingfrom z = – ∞ to + ∞, are fixed at y = – a and y = + arespectively, as shown in the figure. If the wires A andB each carry a current I into the plane of the paper,obtain the expression for the force acting on thesegment OC. What will be the force in OC if thecurrent in the wire B is reversed? (1992; 10M)

B

y

O

Az

I C x

10. An electron gun G emits electrons of energy 2 keVtravelling in the positive x-direction. The electrons arerequired to hit the spot S where GS = 0.1 m, and theline GS make an angle of 60° with the x-axis as shown

in figure. A uniform magnetic field Br

parallel to GSexists in the region outside the electron gun. Find theminimum value of B needed to make the electrons hitS. (1993; 4M)

90° vG X

S

B

11. A long horizontal wire AB, which is free to move in avertical plane and carries a steady current of 20 A, isin equilibrium at a height of 0.01 m over anotherparallel long wire CD which is fixed in a horizontalplane and carries a steady current of 30 A, as shownin figure. Show that when AB is slightly depresssed,it executes simple harmonic motion. Find the period ofoscillations. (1994; 6M)

A BDC

12. An electron in the ground state of hydrogen atom isrevolving in anticlockwise direction in a circular orbitof radius R (see figure). (1996; 5M)

30°

nB

(i) Obtain an expression for the orbital magneticmoment of the electron.

(ii) The atom is placed in a uniform magnetic induction

Bur

such that the plane normal of the electron orbitmakes an angle of 30° with the magnetic induction.Find the torque experienced by the orbitingelectron.

13. Three infinitely long thin wires each carrying currenti in the same direction, are in the x-y plane of a gravityfree space. The central wire is along the y-axis whilethe other two are along x = + d. (1997; 5M)(a) find the locus of the points for which the magnetic

field B is zero.(b) If the central wire is displaced along the z-direction

by a small amount and released, show that it willexecute simple harmonic motion. If the lineardensity of the wire is λ, find the frequence ofoscillation.

14. A particle of mass m and charge q is moving in aregion where uniform, constant electric and magnetic

fields Eur

and Bur

are present. Eur

and Bur

are parallel to

each other. At time t = 0, the velocity 0vuur

of the

particle is perpendicular to Eur

(Assume that its speedis always < < c, the speed of light in vacumm). Findthe velocity of the particle at time t. You must express

your answer in terms of t, q, m, the vector 0vuur

, Eur

and

Bur

and their magnitudes v0, E and B. (1998; 8M)

15. A uniform constant magnetic field Bur

is directed at anangle of 45° to the x-axis in xy plane. PQRS is rigidsquare wire frame carrying a steady current I0, with itscentre at the origin O. At time t = 0, the frame is at restin the position shown in the figure with its sidesparallel to x and y-axes. Each side of the frame is ofmass M and length L : (1998; 8M)

S

y

x

R

QPO

I0

(a) What is the magnitude of torque τr

about Oacting on the frame due to the magnetic field?

(b) find the angle by which the frame rotates underthe action of this torque in a short interval of time∆t and the axis about which this rotation occurs(∆t is so short that any variation in the torque

284

during this interval may be neglected). Given : themoment of inertia of the frame about an axis

through its centre perpendicular to its plance is 43

ML2 :

16. The region between x = 0 and x = L is filled with

uniform seteady magnetic field B0$k . A particle of

mass m, positive charge q and velocity 0v i$ travels

along x-axis and centers the region of the magneticfield. (1999; 10M)Neglect the gravity throughout the question.(a) Find the value of L if the particle emerges from

the region of magnetic field with its final velocityat an angle 30° to its initial velocity.

(b) Find the final velocity of the particle and the timespent by it in magnetic field, if the magnetic fieldnow expands upto 2.1 L.

17. A circular loop of radius R is bent along a diameter andgiven a shape as shown in figure. One of the semicircles(KNM) lies in the x-z plane and the other one (KLM)in the y-z plane with their centres at origin. Current Iis flowing through each of the semicircles as shown infigure. (2000; 10M)

KI

N

I

LM

z

y

x

(a) A particle of charge q is released at the origin with

a velocity 0–=r r

vv i . Find the instantaneous forceurF on the particle. Assume that space is gravityfree.

(b) If an external uniform magnetic field B0$j is applied

determine the force uur

1F and uur

2F on the semicircles

KLM and KNM due to the field and the net forceurF on the loop.

18. A current of 10 A flow around a closed path in acircular loop which is in the horizontal plane as shownin the figure. The circuit consists of eight alternatingarcs of radii r1 = 0.08 m and r2 = 0.12 m. Each subtendsthe same angle at the centre. (2001; 10M)

C

A

D

i

r1

r2

(a) Find the magnetic field produced by this circuit atthe centre.

(b) An infinitely long straight wire carrying a currentof 10 A is passing through the centre of the abovecircuit vertically with the direction of the currentbeing into the plane of the circuit. what is theforce acting on the wire at the centre due to thecurrent in the circuit? What is the force acting onthe arc AC and the straight segment CD due to thecurrent at the centre?

19. A rectangular loop PQRS made from a uniform wirehas length a, width b and mass m. It is free to rotateabout the arm PQ, which remains hinged along ahorizontal line taken as the y-axis (see figure). Take thevertically upwards direction as the z-axis. A uniform

magnetic field $0(3 4 )= +

ur $ BB i k exists in the region.

The loop is held in the x-y plane and a current I ispassed through it. The loop is now released and isfound to stay in the horizontal position in equilibrium.

(2002; 5M)

x Sb

a

Q xP

z

R

(a) What is the direction of the current I in PQ?(b) Find the magnetic force on the arms RS.(c) Find the expression for I in terms of B0, a, b and m.

20. A ring of radius R having uniformly distributed chargeQ is mounted on a rod suspended by two identicalstrings. The tension in strings in equilibrium is T0.Now a vertical magnetic field is switched on and ringis rotated at constant angular velocity ω. Find themaximum ω with which the ring can be rotated if the

strings can withstand a maximum tension of 032T

.

(2003; 4M)

285

MATCH THE COLUMN

1. Some laws/processes are given in Column I. Match these with the physical phenomena givne in Column II. Column I Column II(A) Dielectric ring uniformly charged (p) Time independent electrostatic field out of system(B) Dielectric ring uniformly charged rotating with (q) Magnetic field angular velocity ω(C) Constant current in ring i0 (r) Induced electric field(D) i = i0 cos ωt (s) Magnetic moment

2. Column I gives certain situations in which a straight metallic wire of resistance R is used and Column II gives someresulting effects. Match the statements in Column I with the statements in Column II. (2007; 6M) Column I Column II(A) A charged capacitor is connected to the ends (p) A constant current flows through the wire

of the wire(B) The wire is moved perpendicular to its length (q) Thermal energy is generated in the wire

with a constant velocity in a uniform magneticfield perpendicular to the plane of motion

(C) The wire is placed in a constant electric field (r) A constant potential difference develops betweenthat has a direction along the length of the wire the ends of the wire

(D) A battery of constant emf is connected to the (s) Charges of constant magnitude appear at the endswire of the wire

3. Two wires each carrying a steady current I are shown in four configuration in Column I. Some of the resulting effectare described in Column II. Match the statements in Column I with the statements in Column II. (2007; 6M) Column I Column II

(A) Point P is situated midway between the wires P (p) The magnetic field (B) at P due to

the currents in the wires are in thesame direction

(B) Point P is situated at the mid-point of the (q) The magnetic field (B) at P due to line joining the centers of the circular wires the currents in the wires are in

which have same radii. opposite directions.

P

21. A proton and an alpha particle, after being acceleratedthrough same potential difference, enter uniformmagnetic field the direction of which is perpendicularto their velocities. Find the ratio of radii of the circularpaths of the two particles. (2004; 2M)

ω0

T0 T0

B

d

22. A moving coil galvanometer experiences torque = k i

where i is current. If N coils of area A each and momentof inertia I is kept in magnetic field B.

(2005; 6M)(a) find k in terms of given parameters,

(b) if for current i deflection is 2π

, find out torsional

constant of spring,(c) if a charge Q is passed suddenly through the

galvanometer. Find out maximum angle ofdeflection.

23. A steady current I goes through a wire loop PQRhaving shape of a right angle triangle with PQ = 3x, PR= 4x and QR = 5x. If the magnitude of the magnetic

field at P due to this loop is

xI

kp48

µ0, find the value

of k .

286

(C) Point P is situated at the mid point of the (r) There is no magnetic field at P.line joining the centers of the circular wireswhich have same radii.

P

(D) Point P is situated at the common centre of (s) The wires repel each other.the wires.

P

4. Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column - II. Ineach case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electricpotential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the wholesystem is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at Mand µ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent toa steady current.

Column - I Column - II

(A) E = 0 (p) Charges are at the corners of a regular hexagon. M is atthe centre of the hexagon. PQ is perpendicular to theplane of the hexagon.

(B) V ≠ 0 (q) Charges are on a line perpendicular to PQ at equalintervals. M is the mid-point between the two innermostcharges.

(C) B = 0 (r) Charges are placed on the two coplanar insulating ringsat equal intervals. M is the common centre of the rings.PQ is perpendicular to the plane of the rings.

(D) µ ≠ 0 (s) Charges are placed at the corners of a rectangle of sidesa and 2a and at the mid points of the longer sides. M isat the centre of the rectangle. PQ is parallel to the longersides.

(t) Charges are placed on two coplanar, identical insulatingrings at equal intervals. M is the mid-point between thecentres of the rings. PQ is perpendicular to the linejoining the centres and coplanar to the rings.

ANSWERS

FILL IN THE BLANKS

1. D, B 2. 2

4L i

π3. 1.26 × 10–23Am2 4.

0

1 2

µ 1 1–4

IR R

(perpendicular to paper outwards)

5. IIB, positive Z 6. eVB $k , ABCD 7. 14.0185 Mev

TRUE/FALSE1. T 2. T 3. F 4. F

+

Q+

+ –

–

–

P

M

– ++ – + –

Q

P

M

M+Q

P

–+

––

P QM

+– –

+– –

++

Q

P

M

+

– –

287

OBJECTIVE QUESTION (ONLY ONE OPTION)1. (a) 2. (c) 3. (c) 4. (b) 5. (c) 6. (b) 7. (d)8. (a) 9. (a) 10. (d) 11. (a) 12. (d) 13. (c) 14. (c)15. (c) 16. (b) 17. (d) 18. (c) 19. (b) 20. (a) 21. (d)22. (b) 23. (b) 24. (b) 25. (c) 26. (d) 27. (a)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)1. (a, b, d) 2. (a, b, d) 3. (a, c) 4. (a, c) 5. (a, c, d)

SUBJECTIVE QUESTIONS1. 0.1 T (perpendicular to paper inwards) 2. (6.4m, 0,0), (6.4m, 0, 2m)3. (a) 0.14 m, 45° (b) 4.712 × 10–8s 4. 12 × 10–2m, 4.37 × 10–2m5. (a) 3 A, perpendicular to paper outwards (b) 13 × 10–7T (c) 2.88 × 10–6N/m6. 10–4 T, perpendicular to paper outwards7. (a) 4 A (b) At distance 1m from R to the left or right of it, current is outwards if placed to the left and inwardsif placed to the right of R.

8. (a) 00.11 ( – 3 )

2µ IQvam

j i$ $ (b) zero, (0.61 Ia2 B) j$ 9. 2

0µ2

I=π

Fur

In $2 2

2L a

a

+

k , zero

10. 4.73 × 10–3T 11. 0.2 s

12. (i) M 4eh

m=

π (ii) 8ehB

mτ =

π , perpendicular to both Muur

and Bur

. 13. (a) 3

dx = ± (b) 0

2µdf

d=

π πλ

14. ( ) ( ) 00cos sinqB q qBt t t

m m m B

× = + +

v Bv v Er urr r ur

15. 2

0I L Bτ =r

(b) θ 203( )

4I B

tM

= ∆

16. (a) 0

02mv

LB q

= (b) 00

– ,f ABm

v tB qπ

= =v ir $

17.(a) 0 04

µ qV IR

=Fur

(b) 1 2 2 , 4BIR BIR= = =F F i F iur ur ur$ $

18. (a) 6.54 × 10–5T (Vertically upwards or outward normal to the paper) (b) Zero, Zero, 8.1 × 10–6N (inwards)

19. (a) P to Q (b) IbB0 (3 $k – 4 i$ ) (c) 06

mgbB 20.

0max 2

DT

BQRω = 21.

1

2

22. (a) k = BNA (b) K 2BiNA=

π (c) 2

BNAQI

π

MATCH THE COLUMN

1. (A) → p, (B) → p, q, s (C) → q, s (D) → q, r, s 2. (A) → q, (B) → r, s (C) → s (D) → p, q, r3. (A) → q, r (B) → p (C) → q, r (D) → q, r 4. (A) → p, r, s; (B) → r, s; (C) → p, q, t; (D) - r, s.

SOLUTIONS

FILL IN THE BLANKS

1. Path C is undeviated, path. Therefore, it is of neutron'spath. From Fleming's left hand rule magnetic force onpositive charge will be leftwards and on negativecharge is rightwards. Therefore, track D is of electron.Among A and B one is of proton and other of α-particle.

Further, r =mvBq

or r ∝ mq

Since,mq α

>P

mq

∴ rα > rpor track B is of α - particle.

2. Let R be the radius of circle. Then,

2πR = L or R = 2Lπ

288

3. Equivalent current i = qfand magnetic moment m = (iπr2) = πqfr2

Substituting the values, we haveM = (π) (0.5 × 10–10)2 (1016) (1.6 × 10–19)= 1.26 × 10–23 A-m2

4. At C magneic field due to wires PQ and RS will bezero.Due to wire QR,

B1 0 0

1 1

12 2 4

µ I µ IR R

= =


And due to wire SP,

B20 0

2 2

12 2 4

µ I µ IR R

= =

(perpendicular to paper inwards)

∴ Net magnetic field would be,

B 0

1 2

µ 1 1–4

IR R

=


5. We can complete the loop EDCBE by assuming equaland opposite current I in wire BE.

x

z

yC

D

E

B AF

BNet force on loop EDCBE will be zero. Similarly, forceon wires FE and BA is also zero, because these areparallel to magnetic field.So, net force is only on wire EB.

Fur

= I [(L i$ ) × (B j$ )]

= ILB $k∴ Magnitude of force is ILB and direction of force ispositive z.

6.

x

z

yE

A

C D

BF H

I

G

In metal charge carriers are free electrons. Current is inpositive direction of x-axis. Therefore, charge carrierswill be moving in negative direction of x-axis.

mFur

= q ( Vur

× Bur

)

= (–e) (–V i$ ) × (B j$ )

nFur

= eVB $kDue to this magnetic force, electrons will be collectingat face ABCD, therefore, lowering its potential.

7. Radius of circular path is given by

R =2mv P Km

Bq Bq Bq= =

Radius in both the cases are equal. Therefore,

2K m

Bqα α

α=

22.3

d d

d

K mBq ,

dqqα

12 2ee

= = and d

mm

α 42

2= =

∴ Kd =2

2.3 .d

d

q m Kq m

αα

α

Kd =21

2.3 (2)(5.3)2

× MeV

Kd = 14.0185 MeV

TRUE/FALSE

1. A current carrying coil is a magnetic dipole. Netmagnetic force on a magnetic dipole in uniform field iszero.

2. Magnetic force acting on a charged particle is alwaysperpendicular to its velocity or work done by a magneticforce on a charged particle is always zero. Hence, amagnetic force cannot change the energy of chargedparticle.

3. The path will be a helix. Path is circle when it entersnormal to the magnetic field.

4.2km

rBq

=

or r ∝ m (k , q and B are same)

mp > me ∴ rp > re


1. In non-uniform magnetic field the needle will experienceboth a force and a torque.

2. Magnetic force on a current carrying loop in uniformmagnetic field is zero.

3. Straight wire will produce a non-uniform field to the

right of it. bcFur

and dcFur

will be calculated by integrationbut these two forces will cancel each other. Further

M = iA = iπR2 2

4L i=

π

289

force on wire ab will be towards the long wire and onwire cd will be away from the long wire. But since thewire ab is nearer to the long wire, force of attractiontowards the long wire will be more. Hence, the loop willmove towards the wire.

b c

da

I

∴ (c)

4. Force per unit lenght between two wires carryingcurrents i1 and i2 at distance r is given by :

Fl =

0µ2π

1 2i ir

Here, i1 = i2 = i and r = b

∴Fl =

20µ

2ibπ

∴ (b)

5. R =2qVmBq

or R ∝ m

or1

2

RR =

X

Y

mm

orX

Y

mm =

21

2

RR

∴ (c)

6. Using Ampere's circuital law over a circular loop ofany radius less than the radius of the pipe, we can seethat net current inside the loop is zero. Hence, magneticfield at every point inside the loop will be zero.

7. For a current flowing into a circular arc, the magneticinduction at the centre is

θ

l2

l1

A O

B

B = 0µ4

ir

π

θ

or B ∝ iθIn the given problem, the total current is divided intotwo arcs

i ∝1

resistanceof arc

∝1

lengthofarc

∝1

anglesubstendedatcentre( )θ

or iθ = constanti.e., magnetic field at centre due to arc AD is equal andopposite to the magnetic field at centre due to arcACD. Or the net magnetic field at centre is zero.

8. Radius of the circular path is given by

r =mVBq Bq

km2=

Here, K is the kinetic energy of the particle.

Therefore, r ∝ mq

if K and B are same.

∴ rp : rd : rα 1 2 4

: : 1: 2 :11 1 2

= =

Hence rα = rp < rd9. Current i = (frequency) (charge)

= (2 )2

qω

π

=qωπ

ω

(q,m) (q,m)RR

Magnetic moment, M = (i) (A)

=2( )

qR

ω π π = (qωR2)

Angular momentum, L = Iω = 2 (mR2)ω

∴ML

=2

22( )

q R

mR

ω

ω 2qm

=

10. Net magnetic field due to both the wires will bedownward as shown in the figure.

290

x

zy

dB

v

–II

Since, angle between vr

and Bur

is 180°.Therefore, magnetic force

mFur ( ) 0q v= × =B

r ur

11. The charged particle will be accelerated parallel (if it isa positive charge) or antiparallel (if it is a negativecharge) to the electric field, i.e., the charged particlewill move parallel or antiparallel to electric and magneticfield. Therefore, net magnetic force on it will be zeroand its path will be a straight line.

12. Total magnetic flux passing through whole of the X-Yplane will be zero, because magnetic lines from aclosed loop. So, as many lines will move in -Z directionsame will return to + Z direction from the X-Y plane.

13. H1 = Magnetic flux at M due to PQ + Magnetic fieldat M due to QRBut magnetic field at M due to QR = O∴ Magnetic field at M due to PQ (or due to currentI in PQ) = H1Now H2 = Magnetic field at M due to PQ (current I)+ magnetic field at M due to QS (current I/2) +(magnetic field at M due to QR

= 11 0

2H

H + +

= 132

H

1

2

HH =

23

Note : Magnetic field at any point lying on the currentcarrying straight conductor is zero.

B = 0

i

14. We can write iEE ˆ.=r

and $B=B kur

Velocity of the particle will be along the direction of

Er

.Therefore, we can write

Vur

= AqEi$

In Eur

, Bur

and Vur

, A, E and B are positive constantwhile q can be positive or negative.Now, magnetic force on the particle will be

mFuuur

= q ( )×V Bur ur

= q AqE i$ × B $k

= q2 AEB ( i$ × $k )

mFuuur

= q2 AEB (– j$ )

Since, mFuuur

is along negative y-axis, no matter what is

the sign of charge q. Therefore, all ions will deflecttowards negative y-direction.

15. Ratio of magnetic moment and angular momentum isgiven by

ML

= 2qm

which is a function of q and m only. This can bederived as follows :

M = i A= (q f). (πr2)

= (q) ( )22

rω π π

=2

2q rω

and L = Iω= (mr2ω)

∴ML

=

2

22r

q

mr

ω

ω

= 2qm

16. If the current flows out of the paper, the magnetic fieldat points to the right of the wire will be upwards andto the left will be downwards as shown in figure.

B

B

B

291

Now, let us come to the problem,Magnetic field at C = 0Magnetic field in region BX' will be upwards (+ve)because all points lying in this region are to the rightof both the wires.

X′XA C B

Magnetic field in region AC will be upwards (+ve),because points are closer to A, compared to B. Similarlymagnetic field in region BC will be downwards (–ve).Graph (b) satisfies all these conditions. Therefore,correct answer is (b).

17. The magnetic field at P (a, 0, a) due to the loop isequal to the vector sum of the magnetic field producedby loops ABCDA and AFEBA as shown in the figure.

kj

i

^^

^

P (a, 0, a)

C

B

FA

D

E

Magnetic field due to loop ABCDA will be along i$

and due to loop AFEBA, along $k . Magnitude ofmagnetic field due to both the loops will be equal.Therefore, direction of resultant magnetic field at P will

be $( )12

+i k$

18. Consider an element of thickness dr at a distance rfrom the centre. The number of turns in this element.

dN = –N

drb a

Magnetic field due to this element at the centre of thecoil will be

drb

ra

dB = 0µ ( )2dN Ir

= 0µ .2 –

I N drb a r

∴ B =r b

r adB

=

=∫ 0µ2( – )

NIb a

= ln ba

19. Radius of the circle mvBq

=

or radius ∝ mv if B and q are same.(Radius)A > (Radius)B

∴ mAvA > mBvB

20. Magnetic field at P is Bur

, perpendicular to OP in thedirection shown in figure.

O

P (x, y)r

y

θθ

i x

B

So, B = B sin θ i$ – Bcosθ j$

Here, B =µ2

Irπ

0

sin θyr

= and cos θ xr

=

∴ Bur 0

21

.2µ I

r=

π)ˆˆ( jxiy −

)(2)ˆˆ(

220

yxjxiyI

+π−µ=

(as r2 = x2 + y2)

21. Magnetic lines form closed loop. Inside magnet theseare directed from south to north pole.

22. If (b – a) ≥ r(r = radius of circular path of particle)The particle cannot enter the region x > bSo, to enter in the region x > b

r > (b – a)

or ( – )mv

b aBq

> or( – )q b a Bv

m>

23. Electric field can deviate the path of the particle in theshown direction only when it is along negative y-

direction. In the given options Eur

is either zero oralong x-direction. Hence, it is the magnetic field whichis really responsible for its curved path. Options (a)and (c) cannot be accepted as the path will be helix inthat case (when the velocity vector makes an angle

292

other than 0°, 180° or 90° with the magnetic field, pathis a helix) option (d) is wrong because in that casecomponent of net force on the particle also comes in

$k direction which is not acceptable as the particle ismoving in x-y plane. Only in option (b) the particle canmove in x-y plane.

24. Net force on a current carrying loop in uniformmagnetic field is zero. Hence, the loop cannot translate.So, options (c) and (d) are wrong. From Fleming's lefthand rule we can see that if magnetic field isperpendicular to paper inwards and current in the loop

is clockwise (as shown) the magnetic force mFur

oneach element of the loop is radially outwards. or theloops will have a tendency to expand.

Fm

y

x

25. U = – BMrr

. = – MB cos θ

Here, = magnetic moment of the loop

θ = angle between Mr

and Br

U is maximum when θ = 180° and minimum when θ = 0°So, as θ decrease from 180° to 0° its PE also decrease.

26. Magnetic force does not change the speed of chargedparticle. Hence v = u. Further magnetic field on theelectron in the given condition is along negative y-axisin the starting. Or it describes a circular path inclockwise direction. Hence, when it exists from thefield, y < 0.Therefore, the correct option is (d).

27. ( )m q= ×F v Bur r ur

Correct option is (a).


1. If both E and B are zero, then eFur

and mFur

both are

zero. Hence, velocity may remain constant. Therefore,option (a) is correct.If E = 0, B ≠ 0 but velocity is parallel or antiparallel to

magnetic field then also eFur

and mFur

both are zero.

Hence, option (b) is also correct.

If E ≠ 0, B ≠ 0 but eFur

+ mFur

= 0, then also velocity

may remain constant or option (d) is also correct.

2. Magnetic force does not do work. From work-energytheorem :

eFW = ∆KE

or (qE) (2a) =12

m [4v2 – v2]

or E =23

4mvqa

∴ Correct option is (a).At P, rate of work done by electric field

= eFur

. vr

= (qE) (v) cos 0°

=2 33 3

4 4mv mv

q vqa a

=

Therefore, option (b) is also correct.Rate of work done at Q :

of electric field = eFur

. vr

= (qE) (2v) cos 90° = 0and of magnetic field is always zero. Therefore, option(d) is also correct.

Note that eFur

= qE i$

3. r =2mv p Km

Bq Bq Bq= =

i.e., r ∝ mq

if K and B are same.

i.e., +Hr : +He

r : 2+Or

1 4 16: :

1 1 2=

= 1 : 2 : 2Therefore, He+ and O2+ will be deflected equally butH+ having the least radius will be deflected most.

4. BAFur

= 0, because magnetic lines are parallel to thiswire.

CDFur

= 0 because magnetic lines are antiparallel to this

wire.

CBFur

is perpendicular to paper inwards and ADFur

is

perpendicular to paper outwards. These two forces

293

(although calculated by integration) cancel each otherbut produce a torque which tend to rotate the loop inclockwise direction about an axis OO'.

5. Bvrr

⊥ in region II. Therefore, path of particle is circle

in region II. Particle enters in region III if, radius of

circular path, r > l or lBqmv > or m

Bqlv >

If ,m

Bqlv = ,lBqmv

r == particle will turn back and path

length will be maximum. If particle returns to region I,

,2 Bq

mTt

π== which is independent of v.

v = Bqlm

∴ Correct option is (a), (c) and (d).


1. Electron pass undeviated. Therefore,

eFuur

= mFuuur

or eE = eBv

or /E V dB

v v= = (V = potential difference between

the plates)

or B =Vdv

Substituing the values, we have

B = –3 6600

3 10 2 10× × × = 0.1T

Further, direction of eFuur

should be opposite of mFuuur

.

or e Eur

↑ ↓ e ( vr

× Bur

)

∴ Eur

↑ ↓ vr

× Bur

Here, Eur

is in positive x-direction.

There vr

× Bur

should be in negative x-direction or Bur

should be in negative z-direction (or perpendicular) topaper inwards, because velocity of electron is inpositive y-direction.

2. eFuur

= q Eur

= (1.6 × 10–19) (–102.4 × 103) $k

= – (1.6384 × 10–16) $k N

mFuuur

= q ( vr

× Bur

) = 1.6 × 10–19 (1.28 × 106 i$ × 8 × 10–

2 j$ )

= 1.6384 × 10–16 $k N

Since eFuur

+ mFuuur

= 0

∴ Net force on the charged particle is zero, particle willmove undeviated.In time t = 5 × 10–6s, the x-coordinate of particle willbecome,x = vxt = (1.28 × 106) (5 × 10–6) = 6.4 mwhile y and z-coordinates will be zero.At t = 5 × 10–6s electric field is switched off. Onlymagnetic field is left which is perpendicular to itsvelocity. Hence, path of the particle will now becomecircular. Plane of circle will be perpendicular to magneticfield i.e., x-z. Radius and angular velocity of circularpath will become :

mvr

Bq=

–26 6

–2 –19(10 )(1.28 10 )

(8 10 )(1.6 10 )

×=× ×

= 1m

Bqm

ω = –2 –19

–26(8 10 )(1.6 10 )

(10 )

× ×= = 1.28 × 106 rad/s

In the remaining time i.e., (7.45 – 5) × 10–6 = 2.45 × 10–6sAngle rotated by particle,θ = ωt = (1.28 × 106) (2.45 × 10–6) = 3.14 rad = 180°∴ z -coordinates of particle will become :

z = 2r = 2m

r

P v x

v

zQ

P = (6.4m, 0, 0)Q = (6.4m,0, 2m)

while y-coordinate will be zero.∴ Position of particle at t = 5 × 10–5s is P = (6.4 m, 0,0)and at t = 7.45 × 10–6 s is Q = (6.4 m, 0.2m)

3. Inside a magnetic field speed of charged particle doesnot change. Further, velocity is perpendicular tomagnetic field in both the cases hence path of theparticle in the magnetic field will be circular. Centre ofcircle can be obtained by drawing perpendicular tovelocity (or tangent to the circular path at E and F.Radius and angular speed of circular path would be

r =mvBq

294

and ω =Bqm

45°

45°

r

r

45°G

F

v

E

v

θ

C

45°

45° F

C

E

90°

(i) Refer figure (a) :∠CFG = 90° – θ and ∠CEG = 90° – 45° = 45°Since, CF = CE∴ ∠CFG = ∠CEGor 90° – θ = 45°or θ = 45°Further, FG = GE = r cos 45°

∴ EF = 2FG = 2r cos 45° 2 cos45mv

Bq°

=

=

–27 7

–19

12(1.6 10 )(10 )

2(1)(1.6 10 )

× ×

= 0.14m

Note : That in this case particle completes 14

th of circle

in the magnetic field.

(ii) Refer figure (b) : In this case particle will complete

34

th of circle in the magnetic field. Hence, the time

spent in the magnetic field :

t =34

(time period of circular motion)

3 2 34 2

m mBq Bq

π π= =

–27

–19(3 )(1.6 10 )

(2)(1)(1.6 10 )

π ×=×

= 4.712 × 10–8s

4. (i) sinmv

rBq

θ=

–27 5

–191.67 10 )(4 10 )(sin60 )

(0.3)(1.6 10 )

( × × °=×

= 1.2 × 10–2m

(ii) 2

( cos )m

p vBq

π= θ

–27 5

–19(2 ) 1.67 10 )(4 10 )(cos60 )

(0.3)(1.6 10 )

π ( × × °=×

= 4.37 × 10–2m

5. (a) Direction of current at B should be perpendicularto paper outwards. Let current in this wire be iB. Then,

0102 211

Aµ iπ +

= 0

2 (10/11)Bµ i

π

or B

A

ii

= 1032

90°

B2

B1

S

B

A

or iB = 1032 Ai× 10 9.6

32= × = 3A

(b) Since, AS2 + BS2 = AB2

∴ ∠ASB = 90°At S : B1 = Magnetic field due to iA

0µ2 1.6

Ai=π

–7(2 10 )(9.6)1.6

×= = 12 × 10–7T

B2 = Magnetic field due to iB

0µ2 1.2

Bi=π

–7(2 10 )(3)1.2

×= = 5 × 10–7T

Since, B1 and B2 are mutually perpendicular. Netmagnetic field at S would be :

B 2 21 2B B= + –7 2 –7 2(12 10 ) (5 10 )= × + ×

= 13 × 10–7 T(c) Force per unit length on wire B :

Fl =

0µ2

A Bi irπ (r = AB = 2m)

=–7(2 10 )(9.6 3)

2× ×

= 2.88 × 10–6 N/m

6. Magnetic field at O due to L and M is zero. Due to Pmagnetic field at O is :

–7–50

1µ1 (10 )(10)

T 5.0 10 T2 2 0.02

iB

OR = = = × π

295

(perpendicular to paper outwards)Similarly, field at O due to Q would be,

–7–50

2µ1 (10 )(10)

T 5.0 10 T2 2 0.02

iB

OS = = = × π

(perpendicular to paper outwards)Since, both the fields are in same direction, net fieldwill be sum of these two.∴ Bnet = B1 + B2 = 10–4TDirection of field is perpendicular to the paper outwards.

7. (a) Magnetic field at R due to both the wires P and Qwill be downwards as shown in figure. Therefore, netfield at R will be sum of these two.

P Q R

BP

BQ

v

B = BP + BQ

= 0 0µ µ2 5 2 2

QP II+

π π

=0µ 2.5

2 5 2I + π

= ( )0µ 14

I +π

= 10–7 (I + 1)Net force on the the electron will be,

Fm = Bqv sin 90°

M NR

1m 1mor (3.2 × 10–20) = (10–7) (I + 1) (1.6 × 10–19) (4 × 105)or I + 1 = 5∴ I = 4A

(b) Net field at R due to wires P and Q isB = 10–7 (I + 1)T

= 5 × 10–7 T (I = 4A)Magnetic field due to third wire carrying a current of2.5 A should be 5 × 10–7T in upwards direction so, thatnet field at R becomes zero. Let distance of this wirefrom R be r. Then,

0µ 2.52 rπ = 5 × 10–7

or–7(2 10 )(2.5)r

×= 5 × 10–7m

or r = 1 m

So, the third wire can be put at M or N as shown infigure.If it is placed at M, then current in it should beoutwards and if placed at N, then current be inwards.

8. (a) Magnetic field at P due to arc of circle.Subtending an angle of 120° at centre would be :

M

I

N

y

y v

x

x60°

60°

60°r

a

+QP

B1 =13 (field due to circle)

=0µ1

3 2I

a

aI

60µ

= (outwards)

00.16µ Ia

= (outwards)

or 1Bur

= $00.16µ Ia

k

Magnetic field due to straight wire NM at P :

B2 =0µ

4Irπ (sin 60° + sin60°)

Here, r = a cos 60°

∴ B2 =0

2µπ cos60

Ia ° (2 sin 60°)

or B2 =0µ

4πIa tan 60°

=00.27µ I

a (inwards)

or 2Bur

= $00.27µ– Ia

k

∴ netBur

= 1Bur

+ 2Bur $00.11µ I

a= k

Now, velocity of particle can be written as,

vr

= v cos 60° i$ + v sin 60° j$ 32

v v2

+= i j$ $

Magnetic force

296

mFuuur

Q= (v×B)r ur 00.11µ

2IQv

a= j$ 00.11 3µ–

2IQv

ai$

∴ Instantaneous acceleration

ar

mmF=

uuur

( )00.11 – 32µ IQvam

= j i$ $

(b) In uniform magnetic field, force on a current loop iszero. Further, magnetic dipole moment of the loop will

be, Muur

= (IA) $kHere, A is the area of the loop.

( )21 1–3 2

A a= π [2 × a sin 60°] [a sin60°]

2 2–

3 2a aπ= sin 120°

= 0.61a2

∴ Muur

= (0.61 Ia2) $k

Given, Bur

= B i$

∴rτ = M

uur × B

ur = (0.61 Ia2 B) j$

9. Let us assume a segment of wire OC at a point P, adistance x from the centre of length dx as shown infigure. Magnetic field at P due to current in wires Aand B will be in the directions perpendicular to AP andBP respectively as shown.

|Bur

| = πµ2

0 IAP

Therefore, net magnetic force at P will be along negativey-axis shown

B

y

O

A

90°

90°IC

x

BB

BA

Bnet = 2 |Bur

| cos θ

=0µ

22

π

IAP

xAP

Bnet =0µ

π 2.

( )

I x

AP

Bnet =0µπ 2 2( )

Ix

a x+Therefore, force on this element will be

(F = ilB)

dF = I 0

2 2µ Ix

dxa x

π +

(in negative z-direction)

∴ Total force on the wire will be

20

2 20 0

µx L L

x

I xdxF dF

x a

=

== =

π +∫ ∫

20µ2

I=π

In 2 2

2L a

a

+

(in negative z-axis)

Hence, Fur

π

µ−=

2

20I

In 2 2

2L a

a

+

$k

10. Kinetic energy of electron, K = 12

mv2 = 2 keV

∴ Speed of electron, v 2Km

=

B

v60°

S

G

v =–16

–312 2 1.6 10

9.1 10

× × ×

×= 2.65 × 107 m/s

Since, the velocity ( )vr

of the electron makes an angle

of θ = 60° with the magnetic field Bur

, the path will bea helix. So the particle will hit S if

GS = npHere n = 1, 2, 3 ......

p = pitch of helix 2 mqBπ

= v cos θ

But for B to be minimum, n = 1

Hence, GS = p = 2 mqBπ

v cos θ

B = Bmin 2 cos

( )mvq GS

π θ=


Bmin

–31 7

–19

1(2 )(9.1 10 (2.65 10 )2

(1.6 10 )(0.1)

π × × =×

T

or Bmin = 4.73 × 10–3T

297

11. Let m be the mass per unit length of wire AB. At aheight x above the wire CD, magnetic force per unitlength on wire AB will be given by

Fm

Fg

i1=20AA

C

B

D

x=d=0.01m

Fm =0 1 2µ

2i ixπ (upwards) ..(1)

Weight per unit length of wire AB isFg = mg (downwards)

Here, m = mass per unit length of wire ABAt x = d, wire is in equilibrium i.e.,

Fm = Fg or 0 1 2µ

2i idπ = mg

or0 1 2

2µ2

i i

dπ =mgd ...(2)

When AB is depressed, x decreased therefore, Fm willincrease, while Fg remains the same. Let AB isdisplaced by dx downwards. Differentiating Eq. (1)w.r.t. x we get

dFm = 0 1 22

µ2

i i

xπ.dx ...(3)

i.e. restoring force, F = dFm ∝ – dxHence, the motion of wire is simple harmonic.From Eqs. (2) and (3), we can write

dFm = –mgd

.dx (x = d)

∴ Acceleration of wire a – .g

dxd

= Hence, period of oscillation

T 2dxa

= π|displacement|

=2p|acceleration|

or 2d

Tg

= π 0.0129.8

= π or T = 0.2 s

12. (i) In ground state (n = 1) according to Bohr's theory:

mvR 2h=π

or v =2

hmRπ

Now, time period, T 2 R

vπ= 2

/ 2R

h mRπ=π

2 24 mRh

π=

Magnetic moment M = iA

where i charge

=timeperiod 2 2

e=

4 mRh

π 2 24

eh

mR=

π

and A = πR2

∴ M = (πR2) 2 24

eh

mR

π

or M = 4eh

mπ

Direction of magnetic moment Muur

is perpendicular tothe plane of orbit.

(ii) τr

= Muur

× Bur

Muur

= MB sin θ

where θ is the angle between Muur

and Bur

θ = 30°

∴ τ = 4eh

m π

(B) sin 30°

∴ τ = 8ehB

mπ

The direction of τr

is perpendicular to both Muur

and

Bur

.

13. (a) Magnetic field will be zero on the y-axis i.e.,x = 0 = z

z-axisi

x = – d O x = + dx

IVIIIIII

y

Magnetic field cannot be zero in region I and regionIV because in region I magnetic field will be alongpositive z-direction due to all the three wires, while inregion IV magnetic field will be along negative z-axis

298

due to all the three wires. It can zero only in region IIand III.

x d–x

d+x

1 2 3

i ix=x

Let magnetic field is zero on line (z = 0) and x = x. Thenmagnetic field on this line due to wires 1 and 2 will bealong negative z-axis and due to wire 3 along positivez-axis. Thus

B1 + B2 = B3

or0 0µ µ

2 2i i

d x x+

π + π =0µ

2 –i

d xπ

or1 1

d x x+

+ =1–d x

This equation gives x = + 3

d

x = – d x = 0 x = dx

32

zy-axis

1

where magnetic field is zero.

(b) In this part we change our coordinate axes system, justfor better understanding.There are three wires 1, 2 and 3 as shown in figure. Ifwe displace the wire 2 towards and z-axis.

F F2

θθr r

3z

d d

1

then force of attraction per unit length between wires(1 and 2) and (2 and 3) will be given by

F =2

0µ2

irπ

The components of F along x-axis will be cancelledout. Net resultant force will be towards negative z-axis(or mean position) and will be given by

Fnet =2

0µ2

irπ

)cos2( θ

=2

0µ2

2i zr r

π

Fnet =2

02 2

µ

( )

i zz dπ +

If z < < d, then

z2 + d2 = d2 and Fnet = – 2

02

µ.

2i

zd

π

Negative sign imples that Fnet is restoring in nature.Therefore, Fnet ∝ – zi..e, the wire will oscillate simple harmonically.Let a be the acceleration of wire in this position andλ is the mass per unit length of wire then

Fnet = λa = – 2

02

µ iz

d

π

or a = – 2

02

µ.

iz

d

πλ

∴ Frequency of oscillation

f =1

2πaccleration

displacement

1=2

azπ

=1

2idπ

0µπλ

or 0µ2

ifd

=π πλ

14. j$ =EEur

or 0

0:

B v=

B vi

ur r$

$k =00

0v B×v B

r ur

E and B

v0

x

z

299

Force due to electric field will be along y-axis. Magneticforce will not affect the motion of charged particle inthe direction of electric field (or y-axis). So,

ay eF qE

m m= = = constant. Therefore, vy = ayt .qE t

m=

The charged particle under the action of magnetic field

describes a circle in x-z plane (perpendicular to Bur

)with

T =2 mBqπ

or 2 qBT mπω = =

Initially (t = 0) velocity was along x-axis. Therefore,

magnetic force m( )Fur

will be along positive z-axis

( )m 0q = × F v Bur r ur

. Let it makes an angle θ with x-axis

at time t, then

zvz v0

v0

vx

x

∴ vx = v0 cos ωt = v0 cos qB

tm

...(2)

vz = v0 sin ωt = v0 sin qB

tm

...(3)

From Eq. (1), (2) and (3),

vr

= vx i$ + vy j$ + vz $k

∴ vr

= ( )00 cos qB qEv t tm m E

+

Evurr

+

××

BvBvt

mqBv

0

00 sin

rr

or vr

= ( ) ( )0cosqB q

t tm m

+ v Er ur

+

××

BBvt

mqB

rr0sin

Note : The path of the particle will be a helix of increasingpitch. The axis of the helix will be along y-axis.

15. Magnetic moment of the loop, $( )iA=M kuur

= $20( )I L k

Magnetic field field, Bur

= (B cos 45°) i$ + (B sin 45°) j$

= ( )2

B+i j$ $

(a) Torque acting on the loop, τ = ×M Br uur ur

= $20( )I L k × ( )

2

B +

i j$ $

∴ τr

=2

0 ( – )2

I L Bj i$ $

or | τr

| = I0L2 B

(b) Axis of rotation coincides with the torque and since

torque is in j$ – i$ direction or parallel to QS. Therefore,

the loop will rotate about an axis passing through Qand S as shown in the figure.

S

P Q

R

x

Angular acceleration, α I

τ=

r

where I = moment of inertia of loop about QS.IQS + IPR = IZZ

(From theorem of perpendicular axis)But IQS = IPR

∴ 2IQS = IZZ 24

3ML=

IQS =22

3ML=

∴ α I

τ=

r

=2

022 / 3

I L B

ML03

2I BM

=

∴ Angular by which the frame rotates in time ∆t is

21( )

2tθ = α ∆ or 203

.( )2

I Bt

Mθ = ∆

300

16. (a) θ = 30³

sin θ =LR

Here R =0

0

mvB q

∴ sin 30° =0

0

LmvB q

or12

=0

0

B qLmv

∴ L =0

02mvB q

(b) In part (a)

sin 30° =LR

or12

=LR

or L = R/2Now when L' = 2.1 L

or2.12

R

⇒ L' > RTherefore, deviation of the particle is θ = 180° is asshown.

∴ ivv fˆ

0−=r

and tAB = T/20

mB qπ

=

17. (a) Magnetic field ( )Bur

at the origin = magnetic field

due to semicircle KLM + Magnetic field due to othersemicircle KNM

∴ Bur

= )ˆ(4

)ˆ(4

00 jRI

iRI µ

+−µ

Bur

)ˆˆ(4

ˆ4

ˆ4

000 jiRI

jRI

iRI

+−µ

=µ

+µ

−=

∴ Magnetic force acting on the particle

Fur

= ( )q ×V Bur ur

RI

jiivq4

)ˆˆ()ˆ( 00

µ+−×−=

kR

IqvF ˆ

400µ

−=r

(b) = =KLM KNM KMF F Fur ur ur

and KMFur

= B I (2R) i$

= 2BIR i$

1Fur

= 2Fur

= 2 BIR i$

Total force on the loop, 1 2= +F F Fur ur ur

or Fur

= 4 BIR i$

Then ADC AC=F Fur ur

or ADC ( )AC B=F iur $

From this we can conclude that net force on a currentcarrying loop in uniform magnetic field is zero. In thequestion, segments KLM and KNM also form a loopand they are also placed in a uniform magnetic fieldbut in this case net force on the loop will not be zero.It would had been zero if the current in any of thesegments was in opposite direction.

18. (a) Given : i = 10 A, r1 = 0.08 m and r2 = 0.12 m. Straightportions i.e., CD etc., will produce zero magnetic fieldat the centre. Rest eight arcs will produce the magneticfield at the centre in the same direction i.e.,perpendicular to the paper outwards or verticallyupwards and its magnitude is

B = Binner arcs + Bouter arcs

=0

1

µ12 2

ir

0

2

µ12 2

ir

+

=0µ

4 π

(πi) 1 2

1 2

r rr r

+


B –7(10 )(3.14)(10)(0.08 0.12)

(0.08 0.12)+

=× T

B = 6.54 × 10–5T (Vertically upward or outward normalto the paper)

(b) Force an ACForce on circular portions of the circuit i.e., AC etc.,due to the wire at the centre will be zero becausemagnetic field due to the central wire at these arcs willbe tangential (θ = 180°) as shown.Force on CDCurrent in central wire is also i = 10A. Magnetic fieldat P due to central wire.

B = 0µ .2

ixπ

∴ Magnetic force on element dx due to this magneticfield

dF = (i) 0µ

.2

ixπ

. dx 20µ

2dx

ixπ

=

301

(f = ilB sin 90°)Therefore, net force on CD is

F =

πµ

=π

µ= ∫∫

−

= 23

ln22

2012.0

08.0

202

1

ix

dxidF

rx

rx

Substituting the value, F = (2 × 10–7) (10)2 ln (1.5) orF = 8.1 × 10–6N (inwards)Force on wire at the centreNet magnetic field at the centre due to the circuit is invertical direction and current in the wire in centre isalso in vertical direction. Therefore, net force on thewire at the centre will be zero. (θ = 180°). Hence,(i) Force acting on the wire at the centre is zero.(ii)Force on are AC = 0(iii) Foce on segment CD is 8.1 × 10–6 N (inwards)

19. Let the direction of current in wire PQ is from P to Qand its magnitude be I

x S

I

Q yP

z

R

The magnetic moment of the given loop is :

Muur

=– Iab $kTorque on the loop due to magnetic force is :

1τr

= Muur

× Bur

= (–Iab $k ) × (3 i$ + 4$k ) B0 i$

= – 3lab B0j$

Torque of weight of the loop about axis PQ is :

2τr

= rr

× Fur

=$( )–

2 ×

i k$amg

=2

jmga

We see that when the current in the wire PQ is from

P to Q, 1τr

and 2τr

are in opposite direction, so theycan cancel each other and the loop may remain inequilibrium. So, the direction of current I in wire PQ isfrom P to Q. Further for equilibrium of the loop :

| 1τr

| = | 2τr

|

or 3IabB0 =2

mga

I =06

mgbB

(b) Magnetic force on wire RS is :

Fur

= I ( lr

× Bur

)

= I [(–b j$ ) × (3 i$ + 4$k ) B0]

Fur

= IbB0 (3 $k – 4 i$ )

20. In equilibrium,2T0 = mg

or T0 =mg2

Magnetic moment, M = iA ωπ

= Q2 ( )2πR

τ = MB sin 90° 2

2ω

=BQR

Let T1 and T2 be the tensions in the two strings whenmagnetic field is switched on (T1 > T2).For translational equilibrium of ring is vertical direction,

T1 + T2 = mg ...(2)For rotational equilibrium,

(T1 – T2)2D 2

2ω

= τ =BQR

or T1 – T2

2

2ω

=BQR

..(3)

Solving Eqs. (2) and (3), we have

T1 =2

2 2ω

+mg BQR

D

As T1 > T2 and maximum values of T1 can be 032T

, we

have

032T

= T0 2

max2

ω+

BQRD

02 =

mgT

∴ ωmax =02

DT

BQR

21. BqqVm

r2

= or ∝m

rq

α

pr

rα

α= p

p

m qm q

1 2 14 1 2

= =

22. (a) τ = MB = ki (θ = 90°)

302

∴ k =( )=MB NiA B

i i= NBA

(b) τ = k .θ = NiAB (k = Torsional constant)

∴ π= NiABk 2

(as θ = π/2)

(c) τ = NiAB

or 0 0

τ =∫ ∫t t

dt BNA idt

Iω = BNAQ

or ω =BNAQ

I..(1)

At maximum deflection whole kinetic energy (rotational)will be converted into potential energy of spring.

Hence, 212

ωI = 2max

12

θk

Substituting the value, we get

θmax =2

πBN AQI

23. )53sin37(sinp4

µ0 °+°=rIBp

+

×π

µ=

54

53

53

44

0

x

I

R

Q

P

3xr

5x

4x

37°53°

==

xI

kxI

p48µ

p48µ7 00

∴ k = 7

303

CHAPTER-16ELECTROMAGNETIC INDUCTION

FILL IN THE BLANKS

1. A uniformly wound solenoidal coil of self-inductance 1.8 × 10–4 H and resistance 6Ω is broken up into two identicalcoils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The time constantfor the current in the circuit is ..... s and the steady state current through the battery is ..... A. (1989; 2M)

2. In a striaght conducting wire, a constant current is flowing from left to right due to a source of emf. When the sourceis switched off, the direction of the direction of the induced current in the wire will be .... (1993; 1M)

3. The network shown in figure is part of a complete circuit. If at a certain instant the current (I) is 5A and is decreasingat a rate of 103. As then VB – VA = ...V (1997C; 1M)

15VA 1Ω 5 mH B

TRUE/FALSE1. A coil of metal wire is kept stationary in a non-uniform magnetic field. An emf is induced in the coil. (1986; 3M)2. A conducting rod AB moves parallel to the x-axis (see fig.) in a uniform magnetic field pointing in the positive z-

direction. The end A of the rod gets positively charged. (1987; 2M)

B

y

o x

A

2. A thin semicircular conducting ring of radius R isfalling with its plane vertical in a horizontal magnetic

induction Bur

. At the position MNQ the speed of thering is v and the potential difference across the ringis : (1986; 2M)

B

N

v

QM(a) zero(b) BvπR2/2 and M is at higher potential(c) πBRv and Q is at higher potential(d) 2RBv and Q is at higher potential

OBJECTIVE QUESTIONSOnly One option is correct :1. A conducting square loop of side L and resistance R

moves in its plane with a uniform velocity vperpendicular to one of its sides. a magnetic inductionB, constant in time and space, pointing perpendicularto and into the plane of the loop exists everywhere.

++

++ + +

+ ++ +

+++

+

+

+++ +

+ ++

+

+

+

++

+

+

+

V

B

The current induced in the loop is : (1989; 2M)(a) BLv/R clockwise(b) BLv/R anticlockwise(c) 2BLv/R anticlockwise(d) zero

304

3. A metal rod moves at a constant velocity in a directionperpendicular to its length. A constant uniformmagnetic field exists in space in a direction perpendicularto the rod as well as its velocity. Select the correctstatement (s) from he following : (1988; 2M)(a) The entire rod is at the same electric potential(b) There is an electric field in the rod.(c) The electric potential is highest at the centre of

the rod and decrease towards its ends(d) The electric potential is lowest at the centre of the

rod and increases towards its ends

4. A small square loop of wire of side l is placed insidea large square loop of wire of side L ( L > > l). Theloops are coplanar and their centres coincide. Themutual inductance of the system is proportional to:

(1998; 2M)(a) l/L (b) l2/L(c) L/l (d) L2/l

5. Two identical circular loops of metal wire are lying ona table without touching each other. Loop A carries acurrent which increases with time. In response, theloop B : (1999; 2M)(a) remains stationary(b) is attracted by the loop A(c) is repelled by the loop A(d) rotates about its CM, with CM fixed

6. A coil of inductance 8.4 mH and resistance 6Ω isconnected to a 12 V battery. The current in the coil is1 A at approximately the time : (1999; 2M)(a) 500 s (b) 20 s(c) 35 ms (d) 1 ms

7. A uniform but time varying magnetic field B (t) existsin a circular region of radius a and is directed into theplane of the paper as shown. The magnitude of theinduced electric field at point P at a distance r from thecentre of the circular region : (2000; 2M)

a

B(t)

rP

(a) is zero (b) decreases as 1/r(c) increases as r (d) decreases as 1/r2

8. A coil of wire having finite inductance and resistancehas a conducting ring placed co-axially within it. Thencoil is connected to a battey at time t = 0, so that a timedependent current I1 (t) starts flowing through the coil.

If I2 (t) is the current induced in the ring and B (t) isthe magnetic field at the axis of the coil due to I1 (t)then as a function of time (t > 0), the product I2 (t) B(t) : (1997; 1M)(a) increases with time(b) decreases with time(c) does not vary with time(d) passes through a maximum

9. A metallic square loop ABCD is moving in its ownplane with velocity v in a uniform magnetic fieldperpendicular to its plane as shown in the figure.Electric field is induced : (2001; 2M)

V

CD

A B

(a) in AD, but not in BC(b) in BC, but not in AD(c) neither in AD nor in BC(d) in both AD and BC

10. Two circular coils can be arranged in any of the threesituations shown in the figure. Their mutual inductancewill be : (2001; S)

(a) (b) (c)(a) maximum in situation (a)(b) maximum in situation (b)(c) maximum in situation (c)(d) the same in all situation

11. As shown in the figure, P and Q are two coaxilconducting loops separated by some distance. Whenthe switch S is closed, a clockwise current Ip flows inP (as seen by E) and an induced current IQ, flows inQ. The switch remains closed for a long time. When S

is opened, a current 2QI flows in Q. Then the direction

1QI and 2QI (as seen by E) are : (2002; 2M)

S

EQP

Battery

305

(a) respectively clockwise and anticlock wise(b) both clockwise(c) both anticlockwise(d) respectively anticlockwisee and clockwise

12. A short-circuited coil is placed in a time varyingmagnetic field. Electrical power is dissipated due to thecurrent induced in the coil. If the number of turns wereto be quadrupled and the wire radius halved, theelectrical power dissipated would be : (2002; 2M)(a) halved (b) the same(c) doubled (d) quadrupled

13. When an AC source of emf e = E0 sin (100t) isconnected across a circuit, the phase differencebetween the emf e and the current i in the circuit is

observed to be 4π

, as shown in the diagram. If the

circuit consists possibly only of R-C or R-L or L-C inseries, find the relationship between the two elements:

(2003; 2M)

t

ei

(a) R = 1kΩ , C = 10 µF(b) R = 1kΩ , L = 1H(c) R = 1kΩ , C = 10 µF(d) R = 1kΩ , L = 1H

14. The variation of induced emf (e) with time (t) in a coilif a short bar magnet is moved along its axis with aconstant velocity is best represented as: (2004; 2M)

(a)

e

t(b)

e

t

(c)

e

t(d)

e

t

15. An infinitely long cylinder is kept parallel to an uniformmagnetic field B directed along positive z-axis. Thedirection of induced current as seen from the z-axis willbe : (2005; 2M)(a) clockwise of the +ve z-axis(b) anticlockwise of the +ve z-axis(c) zero(d) along the magnetic field

16. The figure shows certain wire segments joined togetherto form a coplanar loop. The loop is placed in aperpendicular magnetic field in the direction going intothe plane of the figure. The magnitude of the fieldincreases with time. I1 and I2 are the currents in thesegments ab and cd. Then, (2009; M)

c d

a b

(a) I1 > I2(b) I1 < I2(c) I1 is in the direction ba and I2 is in the direction

cd(d) I1 is in the direction ab and I2 is in the direction

dc

OBJECTIVE QUESTIONS

More than one options are correct?1. Two different coils have self-inductances L1 = 8 mH

are L2 = 2 mH. The current in one coil is increased ata constant rate. The current in the second coil is alsoincreased at the same constant rate. at a certain instantof time, the power given to the two coils is the same.At that time, the current, the induced voltage and theenergy stored in the first coil are i1, V1 and W1respectively. Corresponding values for the second coilat the same instant are i2, V2 and W2 respectively.Then : (1994; 2M)

(a)1

2

14

=ii (b)

1

24=

ii

(c)1

2

14

=WW (d)

1

24=

VV

2. A field line is shown in the figure. This field cannotrepresent. (2006; 2M)

306

(a) Magnetic field(b) Electrostatic field(c) Induced electric field(d) Gravitational field

3. Two metallic rings A andB, identical in shape andsize but having different

resistivities A? and

,? B are kept on top of two

identical solenoidsas shown in the figure.

A B

When current I is switched on in the both the solenoidsin identical manner, the rings A and B jump to heights

Ah and ,Bh respectively, with .BA hh > The possible

relation(s) between their resistivities and their masses

Am and Bm is (are) (2009; M)

(a) BA ?? > and BA mm =

(b) BA ?? < and BA mm =

(c) BA ?? > and BA mm >

(d) BA ?? < and BA mm <


1. The two rails of a railway track, insulated from eachother and the ground, are connected to a millivoltmeter.What is the reading of the millivoltmeter when a traintravels at a speed of 180 km/h along the track giventhat the vertical components of earth's magnetic fieldis 0.2 × 10–4 weber/m2 and the rails are separated by1 m? Track is south to north. (1981; 4M)

2. Three identical closed coils A, B and C are placed withtheir planes parallel to one another. Coils A and B carryequal currents as shown in figure. Coils B and C arefixed in position and coil A is moved towards B withuniform motion. Is there any induced current in B. Ifno, give reasons. If yes, mark the direction of theinduced current in the diagram. (1982; 2M)

P

OS

B

QV0

3Ω 3Ω

3Ω3Ω

3ΩA C

3. A square metal wire loop of side 10 cm and resistance1Ω is moved with a constant velocity v0 in a uniformmagnetic field of induction B = 2 weber/m2 as shownin the figure. The magnetic field lines are perpendicularto the plane of the loop (directed into the paper). Theloop is connected to a network of resistors each ofvalue 3Ω . The resistance of the lead wires OS and PQare negligible. what should be the speed of the loopso as to have a steady current of 1 mA in the loop?Give the direction of current in the loop. (1983; 6M)

A B C

4. Space is divided by the line AD into two regions.Region I is field free and the region II has a uniformmagnetic field B directed into the plane of the paper.ACD is a semicircular conducting loop of radius r withcentre at O, the plane of the loop being in the planeof the paper. The loop is now made to rotate with aconstant angular velocity ω about an axis passingthrough O and perpendicular to the plane of the paper.The effective resistance of the loop is R.(1985; 6M)

D

CO B

Region IIRegion I

r

Aω

(a) Obtain an expression for the magnitude of theinduced current in the loop.

(b) Show the direction of the current when the loopis entering into the region II.

(c) Plot of graph between the induced emf and thetime of rotation for two periods of rotation.

5. Two long parallel horizontal rails, a distance d apartand each having a resistance λ per unit length, arejoined at one end by a resistance R. A perfectlyconducting rod MN of mass m is free to slide along therails without friction (see figure). There is a uniformmagnetic field of induction B normal to the plane of thepaper and directed into the paper. A variable force Fis applied to the rod MN such that, as the rod moves,a constant current flows through R. (1988; 6M)

307

dF

B

R

M

N(i) Find the velocity of the rod and the applied force

F as functions of the distance x of the rod fromR.

(ii) What fraction of the work done per second by Fis converted into heat?

6. A circuit containing a two position switch S is shownin figure. (1991; 4+4M)

S2

1 A

3VL

10 mH

2Ω

2Ω

2Ω

3Ω

10Ω2µF

C

BR2

R4

R5E1

R3

12V

(a) The switch S is in position 1. Find the potentialdifference VA – VB and the rate of production ofjoule heat in R1.

(b) If now the switch S is put in position 2 at t = 0.Find :

(i) steady current in R4 and(ii) the time when current in R4 is half the steady

value. Also calculate the energy stored in theinductor L at that time.

7. A rectangular frame ABCD, made of a uniform metalwire, has a straight connection between E and F madeof the same wire, as shown in figure AEFD is a squareof side 1 m and EB = FC = 0.5 m. The entire circuit isplaced in a steadily increasing, uniform magnetic fielddirected into the plane of the paper and normal to it.the rate of change of the magnetic field is 1 T/s. Theresistance per unit length of the wire is 1Ω/m. Find themagnitudes and directions of the currents in thesegments AE, BE and EF. (1993; 5M)

B

A

D F C

BE

8. Two parallel vertical metallicrails AB and CD areseparated by 1 m. they areconnected at two ends byresistances R1 and R2 asshown in figure. Ahorizontal metallic bar ofmass 0.2 kg slides withoutfriction vertically down the

R1

R2

C

D

L

A

B

rails under the action of gravity. There is a uniformhorizontal magnetic field of 0.6T perpendicular to theplane of the rails. It is observed that when the terminalvelocity is attained, the powers dissipated in R1 and R2are 0.76 W and 1.2 W respectively. Find the terminalvelocity of the bar L and the values of R1 and R2.(1994; 6M)

9. A metal rod OA and mass m and length r kept rotatingwith a constant angular speed ω in a vertical planeabout 1 horizontal axis at the end O. The free end Ais arranged to slide without friction along a fixedconducting circular ring in the same plane as that ofrotation. A uniform and constant magnetic induction

Bur

is applied perpendicular and into the plane ofrotation as shown in figure. An inductor L and anexternal resistance R are connected through a switchS between the point O and a point C on the ring toform an electrical circuit. Neglect the resistance of thering and the rod. Initially, the switch is open.

(1995; 10M)

BA

OX

CR

S

Y

ωθ

(a) What is the induced emf across the terminals ofthe switch?

(b) The switch S is closed at time t = 0.(i) Obtain an expression for the current as a function

of time.(ii) In the steady state, obtain the time dependence of

the torque required to maintain the constant angularspeed. Given that the rod OA was along thepositive x-axis at t = 0.

10. A solenoid has an inductance of 10 H and a resistanceof 2Ω . It is connected to a 10 V battery. How long willit take for the magnetic energy to reach 1/4 of itsmaximum value? (1996; 3M)

308

11. A pair of parallel horizontal conducting rails ofnegligible resistance shorted at one end is fixed on atable. The distance between the rails is L. A conductingmassless rod of resistance R can slide on the railsfrictionlessly. The rod is tied to a massless stringwhich passes over a pulley fxied to the edge of thetable. A mass m tied to the other end of the stringhangs vertically. A constant magnetic field B existsperpendicular to the table. If the system is releasedfrom rest. calculate : (1997; 5M)

m

L R

(a) the terminal velocity achieved by the rod, and(b) the acceleration of the mass at the instant when

the velocity of the rod is half the terminal velocity.

12. An infinitesimally small bar magnet of dipole moment

Muur

is pointing and moving with the speed v in the Xur

direction. A small closed circular conducting loop ofradius a and negligible self inductance lies in the y-zplane with its centre at x = 0, and its axis coincidingwith the x-axis. Find the force opposing the motion ofthe magnet, if the resistance of the loop is R. Assumethat the distance x of the magnet from the centre of theloop is much greater than a. (1997C; 5M)

13. An inductor of inductance 2.0 mH is connected acrossa charged capacitor of capacitance 5.0 µF and theresulting L-C circuit is set oscillating at its naturalfrequency. Let Q denote the instantaneous charge onthe capacitor and I the current in the circuit. It is foundthat the maximum value of Q is 200 µC. (1998; 8M)(a) When Q = 100 µC, what is the value of |dI| dt|?(b) When Q = 200 µC, what is the value of I?(c) Find the maximum value of I.(d) When I is equal to one-half its maximum value,

what is the value of |Q|?

14. A magnetic field B = (B0y/a) $k is acting into the paper

in the +z direction. B0 and a are positive constants. asquare loop EFGH of side a, mass m and resistance Rin x-y plane starts falling under the influence of gravity.Note the directions of x and y in the figure. Find:

(1999; 10M)

E

O

y

F

g

x

H G

(a) the induced current in the loop and indicate itsdirection.

(b) the total Lorentz force acting on the loop andindicate its direction

(c) an expression for the speed of the loop v (t) andits terminal velocity.

15. A thermocole vessel contains 0.5 kg of distilled waterat 30°C. A metal coil of area 5 × 10–3m2, number ofturns 100, mass 0.06 kg and resistance 1.6 Ω is lyinghorizontally at the bottom of the vessel. A uniformtime varying magnetic field is setup to pass verticallythrough the coil at time t = 0. The field is first increasedfrom 0 to 0.8 T at a constant rate between 0 and0.2 s and then decreased to zero from 0.8 T at the samerate between 0.2 and 0.4s. The cycle is repeated 12000times. Make sketches fo the current rhough the coiland the power dissipated in the coil as a function oftime for the first two cycles. Clearly indicate themagnitudes of the quantities on the axes. Assume thatno heat is lost to the veseel or the surroundings.Determine the final temperature of the water underthermal equilibrium. Specific heat of metal= 500 J/Kg-K and the specific heat of water.= 4200 J/kg-K. Neglect he inductance of coil.

(2000; 10M)s

16. An inductor of inductance L = 400 mH and resistorsof resistance R1 = 2Ω and R2 = 2Ω are connected toa battery of emf E = 12 V as shown in the figure. Theinternal resistnace of the baterry is negligible. Theswitch S is closed at time t = 0. (2001; 5M)

E

S

R1

R2

L

What is the potenial drop across L as a function oftime? After the steady state is reached, the switch isopened. What is the direction and the magnitude ofcurrent through R1 as a function of time?

17. A metal bar AB can slideon two parallel thickmetallic rail separated bya distance I. Aresistance R and aninductance L areconnected to the railsas shown in the figure.

X0

l0

L

A

B

Rl

309

A long straight wire, carrying a constant current I0 isplaced in the plane of the rails and perpendicular tothem as shown. The bar AB is held at rest at a distancex0 from the long wire. At t = 0, it made to slide on therails away from the wire. Answer the followingquestions. (2002; 5M)

(a) Find a relation among i, didt

and ddtφ , where i is

the current in the circuit and φ is the flux of themagnetic field due to the long wire through thecircuit.

(b) It is observed that at time t = T, the metal bar ABis at a distance of 2 x0 from the long wire and theresistance R carries a current i 1. Obtain anexpression for the net charge that has flownthrough resistance R from t = 0 to t = T.

(c) The bar is suddenly stopped at time T. The currentthrough resistance R is found to be i1/4 at time 2T.Find the value of L/R in terms of the other givenquantities.

18. Two infinitely long parallel wires carrying currents I =I0 sin ωt in opposite directions are placed a distance3a part. A square loop of side a of negligible resistancewith a capacitor of capacitance C is placed in the planeof wires as shown. Find the maximum current in thesquare loop. Also sketch the graph showing thevariation of charge on the upper plate of the capacitoras a function of time for one complete cycle takinganticlockwise direction for the current in the loop aspositive. (2003; 4M)

T/4 T/2 3T/2 2Tt

a a3a

19. In the circuit shown A and B are two cells of same emfE but different internal resistance r1 and r2 (r1 > r2)respectively. Find the value of R such that the potentialdifference across the terminals of cell A is zero a longtime after the key K is closed. (2004; 4M)

R

R

RR

R

RS

C

L

A B

r2r1

20. In an L-R series circuit, a sinusoidal voltage V = V0 sinωt is applied. It is given that L = 35 mH, R = 11 Ω , Vrms= 220 V, ω/2π = 50 Hz and π = 22/7. Find the amplitudeof current in the steady state and obtain the phasedifference between the current and the voltage. Alsoplot the variation of current for the one cycle on thegiven graph. (2004; 4M)

T/4 T/2 3T/2 2Tt

V= V t0sin ω

21. A long solenoid of radius a and number of turns perunit lngth n is enclosed by cylindrical shell of radiusr thicness d (d << R) and length L. A variable currenti = i0 sin ωt flows through the coil. If the resistivityof the materal of cylindrical shell is ρ, find the inducedcurrent in the shell. (2005; 4M)

d

aR

L


This questions contains, Statement I (assertion) andStatement II (reasons).

1. Statement-I : A vertical iron rod has a coil of wirewound over it at the bottom end. An alternatingcurrnet flows in the coil. The rod goes through aconducting ring as shown in the figure. The ring canfloat at a certain height above the coil. (2007; 3M)Because :Statement II : In the above situation, a current isinduced in the ring which interacts with the horizontalcomponent of the magnetic field to produce ananverage force in the upward direction.

310

(a) Statement-I is true, statement-II is true; statement-II is acorrect explanation forstatement -I

(b) Statement-I is true, statement-II is true; statement-II is NOTa correct explanation forstatement -I

(c) Statement-I is true, statement-II is false

(d) Statement-I is false,statement-II is true

COMPREHENSION

Passage I :The capacitor of capacitance C can be charged (with

the help of a resistance R) by a voltage source V, byclosing switch S1 while keeping switch S2 open. Thecapacitor can be connected in series with an inductor L byclosing switch S2 and opening S1.

L

R

V

C S1

S2

1. Initially, the capacitor was uncharged. Now, switch S1is closed and S2 is kept open. If time constant of thiscircuit is τ , then : (2006; 6M)(a) after time interval τ , charge on the capacitor is

CV/2(b) after time inerval 2τ , charge on the capacitor is

CV (1 – e–2)(c) the work done by the voltage source will be half

of the heat dissipated when the capacitor is fullycharged

(d) after time interval 2τ , charge on the capacitor isCV (1– e–1)

2. After the capacitor gets fully charged S1 is opened andS2 is closed so that the inductor is connected in serieswith the capacitor. Then, (2006; 6M)(a) at t = 0, energy stored in the circuit is purely in the

form of magnetic energy.(b) at any time t > 0, current in the circuit is in the

same direction.(c) at t > 0, there is no exchange of energy between

the inductor and capacitor.(d) at any time t > 0, maximum instantaneous current

in the circuit may CVL

3. If the total charge stored in the LC circuit is Q0, thenfor t ≥ 0. (2006; 6M)(a) the charge on the capacitor is Q = Q0 cos

2t

LC

π +

(b) the charge on the capacitor is Q = Q0

cos –2

t

LC

π

(c) the charge on the capacitor is Q = – LC 2

2d Q

dt

(d) the charge on the capacitor is Q =2

21– d QLC dt

Passage II :Modern trains are based on Maglev technology in

which trains are magnetically leviated, whcih runs its EDSMaglev system.

There are coils on both sides of wheels. Due to motionof train, current induces in the coil of track which levitateit. This is in accordance with Lenz's law. If trains lowerdown then due to Lenz's law a repulsive force increasesdue to which train gets uplifted and if it gos much highthen there is a net downward force due to gravity. Theadvantage of Maglev train is that there is no frictionbetween the train and the track, thereby reducing powerconsumption and enabling the tain to attain very highspeeds.

Disadvantage of Maglev train is that as it slows downthe electromagnetic forces decreases and it becomesdifficult to keep it leviated and as it moves forwardaccording to Lenz law there is an electromagnetic dragforce.

4. What is the advantage of this system?(a) No friction hence no power consumption(b) No electric power is used(c) Gravitation force is zero(d) Electrostatic force draws the train

5. What is the disadvantage of this system?(2006; 6M)(a) Train experience upward force according to Lenz's law(b) Friction froce create a drag on the train(c) Retardation(d) By Lenz's law train experience a drag

6. Which force causes the train to elevate up?(2006; 6M)

(a) Electrostatic force(b) Time varying electric field(c) Magnetic force(d) Induced electric field

311

ANSWERS

FILL IN THE BLANKS

1. 3 × 10–3, 10 2. left to right 3. 15

TRUE/FALSE

1. F 2. T


1. (d) 2. (d) 3. (b) 4. (b) 5. (c) 6. (d) 7. (b)8. (d) 9. (d) 10. (a) 11. (d) 12. (b) 13. (a) 14. (b)15. (c) 16. (d)


1. (a, c, d) 2. (a, d) 3. (b, d)


1. 1 mV 2. yes, in the direction opposite to A.

3. 0.02 m/s, direction of induced current is clockwise. 4. (a) 21

2Br

Rω (b) anticlockwise (c) see the solution

5. (i) ( 2 )R x iv

Bd+ λ= (ii)

22

2 22 ( 2 )l mF R x idBB d

λ= + λ + 6. (a) –5V, 24.5 W (b) (i) 0.6 A (ii) 1.386×10–3s, 4.5×10–4J

7.722

A (E to A), 622

A (B to E), 122

A (F to E) 8. v = 1m/s, R1 = 0.47Ω , R2 = 0.3 Ω

9. (a) 2

2B r

eω

= (b) (i) 2

2B r

iR

ω=

–1–

Rt

Le

(ii) 2 4

cos4 2net

B r mgrt

Rω

τ = + ω

10. 3.465 s 11. (a) 2 2mgR

vB L

= (b) 2g

a =

12. 2 2 40

8214

µ M a VF

Rx= (Repulsion) 13. (a) 10–4 A/s (b) zero (c) 2.0 A (d) 1.732 × 10–4C

14. (a) 0B avi

R= , anticlockwise (b) F

ur

2 20–

B a vR

= j$ (c) –(1– )kTgv e

K= where K

2 20

1,B a gvmR K

= =2 20

gmR

B a=

15. 35.6°C 16. 12 e– 5t V, 6e–10t A (clockwise)

17. (a) d diiR Ldt dtφ = + (b) 0 0

11

(2)–2

µ I IIn Li

R π

(c) In(4)T

18. imax 2

0 0 (2)µ aCI Inω=π

19. 1 24 ( – )3

R r r= 20. 20A,4π

21. 2

0 0 cos2

µ Ldna I ti

Rω ω

=ρ

ASSERATION AND REASON1. (a)

COMPREHENSION1. (b) 2. (d) 3. (c) 4. (a) 5. (d) 6. (c)

312

1. Inductance of the circuit L –40.9 10

2×

= = 0.45 × 10–4H

(in parallel)

0.9×10–4 H

0.9×10–4 H

3Ω

3Ω

15V

Resistance of the circuit 3

1.52

R = = Ω (in parallel)

∴ Lτ (time constant) $L=

= 3.0 × 10–5sSteady state current in the circuit through the battery

i0 =15 10A1.5

VR

= =

2. When source is switched off left to right currentdecrease to zero. Therefore, from Lenz's law. inducedcurrent will oppose the change i.e., it will be from leftto right.

3.310di

dt= A/s

A 1Ω 15V

I

5 mH B

∴ Induced emf across inductance, |e| diLdt

=

|e| = (5 × 10–3) (103) V= 5V

Since, the current is decreasing, the polarity of this emfwould be so as to increase the existing current. Thecircuit can be redrawn as

A 1Ω 15V B

|e|=5V

I=5A

Now,VA – 5 + 15 + 5= VB∴ VA – VB = – 15Vor VB – VA = 15V

TRUE/FALSE

1. If the coil is kept stationary, emf will be induced whenmagnetic field is time varying. If the field is non-uniform emf will be induced when the coil is moved,because for emf to induce magnetic flux passingthrough the coil should change.

2. Magnetic force on freeelectrons will be towards B .Therefore, at B, there is excessof electrons (means negativecharge) and at A , there isdeficiency of electron (meanspositive charge). +

+++

++

Fm

v

BB

A


1. Net change in flux passing through the coil is zero.∴ current (or emf) induced in the loop is zero.

2. Induced motional emf in MNQ is equivalent to themotional emf in an imaginary wire MQ i.e.,eMNQ = eMQ = Bvl = Bv (2R) [l = MQ = 2R]Therefore, potential difference developed across thering is 2RBv with Q at higher potential.

3. A motional emf, e = Blv isinduced in hthe rod. Or wecan say a potential differenceis induced between the twoends of the rod AB, with A athigher potential and B a lower

V

BA

Bpotential. Due to this potential difference, there is anelectric field in the rod.

4. Magnetic field produced by a current i in a largesquare loop at its centre,

B ∝iL

say B =i

KL

∴ Magnetic flux linked with smaller loop,φ = B.S.

φ =2( )

iK l

L

Therefore, the mutual inductance

M =2lK

i Lφ = or

2lM

L∝

SOLUTIONS

FILL IN THE BLANKS

313

Note : Dimensions of self inductancde (L) or mutualinductance (M) are :

[Mutual inductance] = [self inductance]

= [µ0] [length]

Similarly, dimensions of capacitance are :

[capacitance] = [∈ 0] [length]

From the point of view options (b) and (d) may becorrect.

5. For understanding let us assume that the two loopsare lying in the plane of paper as shown. The currentin loop 1 will produce magnetic field in loop 2. Therefore,increase in current in loop 1 will produce an inducedcurrent in loop 2 which produces ⊗ magnetic fieldpassing through it i.e., induced current in loop 2 willalso be clockwise as shown in the figure.

1 2

F FF F

Perpendicular to paper outwards

Perpendicular to paper inwards

The loops will now repel each other as the currents atthe nearest and farthest points of the two loops flowin the opposite directions.

6. The current-time (i-t) equation on L-R circuit is givenby [Growth of current in L-R circuit]

i = i0 ( )– /1– Lte τ..(1)

where i0 =12 26

V AR

= =

and Lτ =–3

–38.4 10 1.4 106

LR

×= = × s

and i = 1 A (given)t = ?

Substituting these values in Eq. (1), we gett = 0.97 × 10– 3s

or t = 0.97 mst = 1 ms

7. . l∫E dur r

=ddtφ

=dB

Sdt

or E (2πr) =2 dB

adt

π for r ≥ a

∴ E =2

2a

rdBdt

∴ Induced electric field ∝ 1r

For r ≤ a

or E = 2r dB

dt or E ∝ r

At r = a, E 2a

= dBdt

Therefore, variation of E with r (distance from centre)will be as follows :

1 dB2 dt 1E

rµ

Er∝

r=a

E

r

8. The equations of I1 (i), I2 (t) and B (t) will take thefollowing forms :

I1 (t) = K1 2–(1– )k te → current growth in L-R circuit

B (t) = K3 2–(1– )k te B (t) ∝ I1 (t); B = µ0 Ni in case

of solenoid coil and 0µ2

NiR

in case of circular coil i.e.,

B ∝ i

I2 (t) = 2–4

k tK e

2 1 12 2 2( ) and : – = ∝ =

e dI dII t e e M

R dt dt

Therefore the product I2 (t) B (t) = 2 2– –5 (1– )k t k tK e e

The value of this product is zero at t = 0 and t = ∝.Therefore, the product will pass through a maximumvalue (K1 : K2 : K3 : K4 and K5 are positive constantand M is the mutual inductance between the coil andthe ring). The corresponding graphs will be as follows:

314

I ( t )2

t

B(t)

t

I(t)B(t)2

t

9. Electric field will be induced in both AD and BC.

10. When current flows in any of the coils, the flux linkedwith the other coil will be maximum in the first case.Therefore, mutual inductance will be maximum in case(a).

11. When switch S is closed magnetic field lines pasisngthrough Q increases in the direction from right to left.So, according to Lenz's law induced currnet in Q i.e.,

1QI , will flow in such a direction, so that the magnetic

field lines due to passes 1QI from left to right throuugh

Q. This is possible when 1QI flows in anticlockwise

direction as seen by E. Opposite is the case when

switch S is opened i.e., 2QI will be clockwise as seen

by E.

12. Power 2

=e

PR

Here, e = induced emf –φ =

ddt where φ = NBA

e = – NA

dBdt

Also R ∝ 2rl

where R = resistance, r = radius, l = length

∴ P ∝lrN 22

∴ 1

2

PP

= 1

13. As the current i leads the emf e by 4π

it is an R-C

circuit.

tan φ = CXR

or tan

1

4π ω= C

R

As ω = 100 rad/s

The product of C-R should be 1

100 s–1

∴ correct answer is (a).

14. Polarity of emf will be opposite in the two cases whileentering and while leaving the coil. Only in option (b)polarity is changing. Hence, the correct option is (b).

15. In uniform magnetic field, change in magnetic flux iszero.Therefore, induced current will be zero.∴ current answer is (c).

16. Using Lenz's law, the current is as shown in the figure.∴ (d)

c dI

a I b


1. From Faraday's law, the induced voltageV ∝ L if rate of change of current is constant

– =

diV L

dt

∴ 2 2

1 1

2 18 4

= = =V LV L

or 1

24=

VV

Power given to the two coils is same i.e.,

V1i1 = V2i2 or 1 2

2 1

14

= =i Vi V

Energy stored

W = 212

Li

∴

22 2 2

1 1 1

=

W L iW L i =

21(4)

4

or 1

2

14

=WW

2. Electrostatic and gravitational field do not make closedloops.

3. Due to induced currents in the rings they will be

315

repelled by the magnetic field of the solenoids. Nowsince change of flux is same in both so induced emfwill be same but induced current will be different asresistivity is not same. The ring with lesser resistivitywill get higher impulse.Given that hA > hB

This is possible if BA ?? < and BA mm = or BA mm <∴ (b), (d)


1. Potential difference between the two rails :

V = Bvl (when Bur

, vr

and lr

all are mutuallyperpendicular)

= (0.2 × 10–4) 5

180 (1)18

× = 10–3V= 1 mV

2. Due to the current in A a magnetic field is from rightto left. When A is moved towards B, magnetic linespassing through B (from right to left) will increase i.e.,magnetic flux passing through B will increase.Therefore, current will be induced in B. The inducedcurrent will have such a direction that it gives amagnetic field opposite to that, was passing throughB due to current in A. Therefore, induced current in Bwill be in opposite direction of current in A.

Note : The current C will decrease due to movement of coilA and this will give rise to an induced current in B insame direction as that of C, but since B is more closerto A, therefore net induced current will be opposite tocurrent in A and C.

3. Given network forms a balanced Wheatstone bridge.The net resistance of the circuit is therefore 3Ω + 1Ω= 4Ω . Emf of the circuit is Bv0l. Thereore, current in thecircuit would be

i = 0Bv lR

or v0 =iRBl

–3(1 10 )(4) 0.022 0.1

×= =×

m/s

Cross magnetic field passing through the loop isdecreasing. therefore, induced current will producedmagnetic field in cross direction. Or direction of inducedcurrent is clockwise.

4. (a) At time t : θ = ωt

ωθ

I II

∴ Flux passing through coil φ = BS cos 0°

or φ = 2( )2θ π π

B r

or φ =2 2

2 2

θ = ω

Br Brt

Magnitude of induced emf e 2

2φ ω= =d B r

dt R

∴ Magnitude of induced current i 2

2ω

= =e B rR R

(b) When the loop enters in region II, magnetic field incross direction passing through the loop is increasing.Hence, from the Lenz's law induced current will producemagnetic field in dot direction or the current will beanticlockwise.

(c) for half rotation 2

π = = ω

Tt , current in the loop will

be of constant magnitude 2

2ω

=B r

iR

and anticlockwise.

In next half rotation when loop comes out fo region IIcurrent will be clockwise, but again magnitude isconstant. So, taking anticlockwise current as thepositive i – t graphs for two rotations will be as under.

t

i

π/ω 2π/ω 3π/ω 4π/ω

Βωr2

2R

−Βωr2

2R

5. Total resistance of the circuit as function of distancex from resistance R is :

Rnet = R + 2λxLet v be velocity of rod at this instant, then motionalemf induced across the rod,

e = Bvd

316

∴ current i = 2=

+ λnet

e BvdR R x

∴ v =( 2 )+ λR x i

BdNet force on the rod, Fnet

2 ( 2 ).λ= = + λdv im dxm R xdt Bd dt

but =dx vdt =

( 2 )+ λR x iBd

Fnet =2

22 2

2 ( 2 )λ + λt m R xB d

This net force is equal to F – Fmwhere Fm = idB

∴ F = Fnet +Fm =2

22 2

2 ( 2 )λ + λ +i m R x idBB d

6. In steady state no current will flow through capacitor.Applying Kirchhoff's second law in loop 1 :

A 3V10 mH

2Ω

2Ω

2Ω

3Ω

1 Ω

2µF

B

i2

i1 i1i1– i2i1

12V1

2

–2i2 + 2 (i1 – i2) + 12 = 0∴ 2i1 – 4i2 = – 12or i1 – 2i2 = – 6 ...(1)Applying Kirchhoff's second law in loop 2 :

12 – 2 (i1 – i2) + 3 – 2i1 = 0or 4i1 – 2i2 = – 9 ...(2)Solving Eqs. (1) and (2), we get

i2 = 2.5 A and i1 = – 1ANow, VA + 3 – 2i2 = VBor VA – VB = 2i1 – 3 = 2 (–1) – 3 = 5V

12

1 2 1( – )=RP i i R =2(–1–2.5) (2) 24.5W=

(b) In position (2) : Circuit is as under

10 mH

3Ω

2Ω3V

Steady current in R4 :

i0 =3

3 2+ = 0.6A

Time when current in R4 is half the steady value:

t1/2 =In(2)1/

= ττ L

L (in 2) =

LR

in (2)

=–3(10 10 )

5×

In (2)

= 1.386 × 10–3s

Note : Compare with radioactive. Energy stored in inductor

at that time 0When 0.3A2

= =

ii

2 –3 2 –41 1(10 10 )(0.3) 4.5 10 J

2 2= = × = ×U Li

7. Induced emf in two loops AEFD and EBCF would be

e1 11

φ = =

d dBS

dt dt= (1 × 1) (1) V = 1V

Similarly,e2 2

2φ = =

d dBS

dt dt = (0.5 × 1) (1) V = 0.5V

Now, since the magnetic field is increasing the inducedcurrent will produce the magnetic field in • direction.Hence, e1 and e2 will be applied as shown in the figure.

1Ω 1Ω 1Ω

C0.5Ω1Ω

1Ω EA

0.5Ω

i2i1e2=0.5V

e1=1V

i2

i

i1FD

B

Kirchhoff's first law at junction F givesi1 = i + i2 ...(1)

Kirchhoff's second law in loop FEADF gives3i2 – i = 1 ...(2)

Kirchhoff's second law in loop FEBCF gives2i2 + i = 0.5 ....(3)

Solving Eqs. (1), (2) and (3), we geti1 = (7/22) A and i2 = (6/22)A and i = (1/22) ATherefore, current in segment AE is (7/22) A from E toA, current in segment BE is 6/22 A from B to E andcurrent in segment EF is (1/22) A from F to E.

317

8. Let the magnetic field be perpendicular to the plane ofrails and inwards ⊗ . If V be the terminal velocity ofthe rails, then potential difference across E and Fwould be BVL with E at lower potential and F at higherpotential. The equivalent circuit is shown in figure (2).In figure (2)

R1

R2

F

A

E

B D

(1)

F

C

B

⇒

i1

i2R2

(2)

R1

Ei e=BVL

F

i1 =1

eR

...(1)

i2 =2

eR

...(2)

Power dissipated in R1 is 0.76 WTherefore, ei1 = 0.76 W ...(3)Similarly, ei2 = 1.2 W ...(4)Now the total current in bar EF is

i = i1 + i2 (from E to F) .... (5)Under equilibiurm condition, magnetic force (Fm) onbar EF = weight (Fg) of bar EFi.e. Fm = Fgor iLB = mg ...(6)

iFE

Fm

Fg

From Eq. (6)

i (0.2)(9.8)(1.0)(0.6)

= =mg

ALB

or i = 3.27 AMultiplying Eq. (5) by e, we get

ei = ei1 + ei2= (0.76 + 1.2) W

(From Eqs. 3 and 4)= 1.96 W

e =1.96

i V

=1.963.27 V

or e = 0.6 VBut since e = BVL

V =(0.6)

(0.6)(1.0)=

eBL

m/s

= 1.0 m/sHence, terminals velocity of bar is 1.0 m/sPower in R1 is 0.76 W

∴ 0.76 =2

1

eR

∴ R1 =2

0.76e

=2(0.6)

0.76Ω = 0.47 Ω

R1 = 0.47 Ω

Similarly, R2 =2

1.2e

=2(0.6) 0.3

1.2Ω = Ω

R2 = 0.3Ω

9. (a) Consider a small element of length dx of the rod OAsituated at a distance x from O.

xdx

A

V=xω

B

Speed of this element, V = xωTherefore, induced emf developed across this elementin uniform magnetic field B

de = (B) (xω) dx (e = Bvl)Hence, total induced emf across OA,

e 0

=

== ∫

x r

xde =

2

0 2ω

ω =∫r B r

B xdx

e =2

2ωB r

(b) (i) A constant emf or P.D. 2

2ω

=B r

e is induced across

O and A.

318

The equivalent circuit can be drawn as shown in thefigure.

LR

S

iΒωr2

2

Switch S is closed at time t = 0. Therefore, it is caseof growth of current in an L-R circuit. Current at anytime t is given by

i = i0 ( )– /1– τLte

i0 =2

2ω

=e B rR R

τl = L/R

i =2

2ωB r

R

–1–

Rt

Le

The i- t graph will be as follows :

i0

i

t

(ii) At constant angular speed, net torque = 0

i=i0

t=t

t=0

0= tω

θ

B

Fm

r/2

Or/2cosθ

mg

θ

The steady state current will be i = i0 2

2ω

=B r

RFrom right hand rule we can see that this currentwould be inwards (from circumference to centre) andcorresponding magnetic force Fm will be in the directionshown in figure (4) and its magnitude is given by:

Fm = (i) (r) (B) R

rB2

32ω= [Fm = ilB]

Torque of this force about centre O is

.2

τ =mF m

rF =

2 4

4ωB rR

(clockwise)

Similarly, torque of weight (mg) about centre O is

( )2mgr

mgτ = cos θ cos2

mgrt= ω (clockwise)

Therefore, net torque at any time t (after steady statecondition is achieved) about centre O will be

τnet = mF mgτ + τ

=2 4

4 2B r mgr

Rω

+ cos ωt

(clockwise)Hence, the external torque applied to maintain a constant

angular speed is extτ2 4

4 2B r mgr

Rω

= + cos ωt (but in

anticlockwise direction)

Note that for 32 2π π

< θ < , torque of weight will be

anticlockwise the sign of which is automatically adjusted

because cos θ = negative for 32 2π π

< θ <

10. 212

U Li= i.e., U ∝ i2

U will reach 14

th of its maximum value when current

is reached half of its maximum value. In L-R circuit,equation of current growth is written as

i = i0 (1 – – / Lte τ )Here, i0 = Maximum value of current

Lτ = Time constant = L/R

Lτ =102

HΩ

= 5s–1

Therefore i = i0/2 = i0 ( )– / 51– te

or12

= 1 – e–t/5

or e–t/5 =12

or –t/5 = In 12

or t/5 = In (2) = 0.693∴ t = (5) (0.693)sor t = 3.465 s

319

11. (a) Let V be the velocity of the wire (as well as block)at any instant of time t.Motional emf, e = BVL

Motional current, e BVL

ir R

= =

and magnetic force on the wire

Fm = iLB 2 2VB LR

=

Net force in the system at this moment will be —

Fnet = mg – Fm = mg – 2 2VB LR

or ma = mg – 2 2VB LR

a =2 2

– VB LgmR

...(1)

Velocity will acquire its terminals value i.e., V = VTwhen Fnet or acceleration (a) of the particle becomeszero.

Thus 0 = g2 2

– TV B LmR

or VT =2 2

mgR

B L

(b) When V = 2TV

= 2 22

mgR

B LThen from Eq. (1) acceleration of the block,

a = g – 2 2

2 22

mgR B LmRB L

–2g

g=

or a = g/2

12. Given that x > > a.

zy

k j

ixa

x

V

^^

^

Magnetic field at the centre of the coil due to the barmagnet is.

B = 03

µ 24

M

xπ=

03

µ2

M

xπ

Due to this, magnetic flux linked with the coil will be,

φ = BS = 0

3µ2

M

xπ (πa2) = 2

03

µ

2

Ma

x∴ Induced emf in the coil, due to motion of hthemagnet is,

e = –ddt

φ= –

20µ

2Ma

31d

dt x

20

4µ 3

2Ma dx

dtx

=

= V

xMa

4

20

23 µ

dx

Vdt

=

Therefore, induced current in the coil is,

i =32

eR

=2

04

µ MaV

RxMagnetic moment of the coil due to this inducedcurrent will be,

M' = iS32

= 2

04

µ MaV

Rx(πa2)

M' =32

40

4µ Ma V

Rx

π

Potential energy of Muur

in Bur

will beU = – M' B cos 180°U = M'B

=23 4

04

µ Ma V

Rx

π 03

µ.

2M

x

π

i

M(of coil)

B(due to magnet)

S N

V

U =2 2 40

7µ3 1

4M a V

R x

∴ F = 21–4

dUdx

=2 2 40

8µ M a V

RxPositive sign of F implies that there will be a repulsionbetween the magnet and the coil.Note that here we cannot apply

F =0

4µ 6 '4

MM

xπ (directily)...(i)

because here M' is a function of x however Eq. (1) canbe applied where M and M' both are constants.

320

13. This is a problem of L-C oscillations.Charge stored in the capacitor oscillates simpleharmonically as

Q = Q0 sin (ωt + φ)Here, Q0 = maximum value of Q = 200 µC

= 2 × 10–4C

ω =1

LC

=–3 –6

1

(2 10 )(5.0 10 )H F× ×

= 104s–1

Let at t = 0, Q = Q0 thenQ (t) = Q0 cos ωt ...(1)

I (t) =dQdt

= – Q0ω sin ωt and ....(2)

( )dl tdt

= – Q0ω2 cos (ωt) ...(3)

(a) Q = 100µC or 02

Q

or cos ωt =12

or ωt 3π=

At cos (ωt) 12

= , from Eq. (3) :

dIdt = (2.0 × 10–4C) (104s–1)2

12

dIdt = 104 A/s

(b) Q = 200 µ C or Q0 when cos (ωt) = 1 i.e., ωt = 0.2π....At this time I (t) = – Q0 ω sin ωtor I (t) = 0 [sin 0° = sin 2π = 0]

(c) I (t) = – Q0ω sin ωt∴ Maximum value of I is Q0 ω

Imax = Q0 ω= (2.0 × 10–4C) (104s–1)

Imax = 2.0 A(d) From energy conservation

2max

12

LI =2

21 12 2

QLIC

+

or Q = 2 2max( – )LC I I

I = max 1.02

IA=

∴ Q = –3 –6 2 2(2.0 10 )(5.0 10 )(2 – 1 )× ×

Q = –43 10 C×or Q = 1.732 × 10–4C

14. When the side EF is at a distance y from the x-axis,magnetic flux passing through the loop is

Y=y

Y=y+a

z-direction

dY

y

x

ka

yBB ˆ0

=r

φ = dφ∫ =0 ( )

= +

=∫

Y y a

Y y

B YadY

a

φ = 2 20 [( ) – ]2

+B

y a y

(a) Induced emf is

e = – φddt =

0– [2( ) – 2 ]2

+B dy

y a ydt

e = B0adydt

e = B0va

where v =dydt = speed of loop

∴ Induced current i = 0=B ave

R R

Directions : Bur

∝ y i.e., as the loop comes down ⊗

magnetic field passing through the loop increase,therefore the induced current will producer • , magneticfield or the induced current in the loop will be counter-clockwise.Alternate Solution (of part a)

Motional emf in EH and FG = 0 as ||V Iur r

Motional emf in EF is e1 0

0( ) = =

B ya v B yv

a

(e = Blv)Similarly motional emf in GH will be

e2 =0 ( )

( )( )+

B y aa v

a

321

= B0 (a + y)vPolarities of e1 and e2 are shown in adjoining figure.

e1

e2

E

H

F

G

e2

e1

e2>e1 i

Net emf, e = e2 – e1e = B0av

∴ i = 0•=

B aveR R

and direction of current will be counter-clockwise.(b) Total lorentz force on the loop :

E

H GV

yx

y

B

F

We have seen in part (a) that induced current passingthrough the loop (when its speed is v) is

i = 0B avR

Now magnetic force on EH and FG are equal inmagnitude and in opposite directions, hence theycancel each other and produce no force on the loop.

FEF =0 0( )

B av B ya

R a (downwards)

(F = ilB)

=20B avyR

and FGH =0

B avR (a)

0 ( )+

B y aa (upwards)

=

20 ( )

+

B avy a

R

FGH > FEF∴ Net Lorentz force on the loop

= FGH – FEF =2 20=

B a vR

(upwards)

Fur

=2 20– j$

B a vR

(c) Net force on the loop will beF= weight – Lorentz force (downwards)

or F = mg 2 20–

B a vR

or

dvm

dt = mg 2 20–

B a vv

R

∴dvdt

= g2 20– –

=

B av g Kv

mR

where K =2 20B amR

= constant

or–dv

g Kv= dt

or 0 –∫v dv

g Kv =

1

0∫dt

This equation gives –(1– )= Ktgv e

K

vT=g/K

v

O t

Here K =

2 20

B amR

i.e., speed of the loop is increasing exponentially withtime t. Its terminal velocity will be

VT = 2 20

=

g mgRK B a at t → ∝

15. Magnetic field (B) varies with time (t) as shown infigure.

322

0 0.2 0.4

0.8

B(T)

0.6 0.8 t(s)Induced emf in the coil to change in magnetic fluxpassing through it,

e =φd

dt=

dBNA

dt

Here, A = Area of coil = 5 × 10–3m2

N = Number of turns = 100Substituting the values, we get

e = (100) (5 × 10–3)(4)V= 2V

Therefore, current passing through the coil

=e

iR

(R = Resistance of coil = 1.6Ω)

or i =2

1.6 = 1.25 A

Note that from 0 to 0.2 s and from 0.4 to 0.6s, magneticfield passing through the coil increase, while duringthe time 0.2s to 0.4s and from 0.6s to 0.8s magneticfield passing through the coil decrease. Therefore,direction of current through the coil in these two timeintervals will be opposite to each other. The variationof current (i) with time (t) will be as follows :

t(s)

i(A)

+ 1.25

0 0.2 0.4 0.6 0.8

– 1.25

Power dissipated in the coil isP = i2 R = (1.25)2 (1.6) W = 2.5 W

Power is independent of the direction of current throughthe coil. Therefore, power (P) versus time (t) graph forfirst two cycles will be as follows :

2.5

0 0.8t(s)

P(watt)

Total heat obtained in 12,000 cycles will beH = P. t = (2.5) (12000) (0.4) = 12000J

This heat is usd in raising the temperature of the coiland the water. Let θ be the final temperature. Then

H = mwsw (θ – 30) + mcSc (θ – 30)Here mw = mass of water = 0.5 kgSw = specific heat of water = 4200 J/kg– Kmc = mass of coil = 0.06 kgand Sc = specific heat of coil = 500 J/kg-KSubstituting the values, we get1200 = (0.5) (4200) (θ – 30) + (0.06) (500) (θ – 30)or θ = 35.6°C

16. (a) Given R1 = R2 = 2Ω , E = 12 Vand L = 400 mH = 0.4 H. Two parts of the circuit arein parallel with the applied battery. So, the uppercircuit can be broken as :

E

SR1

R2

L

E

S

R1

+

E

S R2

L

(a) (b)Now refer figure (b) :This is a simple L-R circuit, whose time constant

τL = L/R2 =0.4

0.22

= s

and steady state current i0 = E/R2 = 12/2 = 6ATherefore, if switch S is closed at time t = 0, thencurrent in the circuit at any time t will be given by

i (t) = i0 ( )– /1– τLte

i (t) = 6 ( )– / 0.21– te

= 6 ( )–51– te = i (say)

Therefore; potential drop across L at any time t is :

V = di

Ldt = L (30e–5t) = (0.4) (30)e–5t

or V = 12 e–5t volt(b) The steady state current in L or R2 is

i = 6ANow, as soon as the switch is opened, current in R1is reduced to zero immediately. But in L and R2 itdecreases exponentially. The situation is as follows :Refer figure (e) :

323

R1

i

R2

L

t = t

R1

R2

L

1

Ei 6A

R= =

i=6A0

Steady state condition

i=0

i0

t = 0S is open

(c) (d) (e)Refer figure (e)Time constant of this circuit would be

τL' =

1 2

0.4(2 2)

=+ +L

R R = 0.1s

∴ Current through R1 at any time t is

i = '– /0

τLti e = 6e–t/0.1 or i = 6e– 1 0t A

Direction of current in R1 is as shown in figure orclcokwise.

17. (a) Applying Kirchhoff's second law :

– – 0φ =d diiR Ldt dt or

φ = +d diiR Ldt dt

This is the desired relation between i, didt

and φddt

(b) Eq. (1) can be written asd φ = iRdt + Ldi

Integrating we get∆ φ = R.∆q + Li1

∆q = 1–∆φ LiR R

...(2)

Here ∆φ = φf – φi =

0

0

0 0

2 2

=

= π∫x x

x x

µ Ildx

x

=0 0µ2πI l

In (2)

So, from Eq. (2) charge flown through the resistanceupto time t = T, when current is i1 is

∆q =0 0

11

( 2 ) –2

π

µ I lIn Li

R

(c) This is the case of current decay in an L-R circuit.Thus,

i = – /0

τLti e ...(3)

Here, 10 1, , (2 – )

4= = = =

ii i i t T T T and

τL =LR

Substituting these values in eq. (3), we get :

τL =LR

In(4)=

T

18. (a) For an elemental strip of thickness dx at a distancex from left wire, net magnetic field (due to both wires)

a

aI i

3a

dxx

B = 0 0µ µ2 2 3 –

+π π

I Ix a x

(outwards)

=0µ 1 1

2 3 – + π

Ix a x

Magnetic flux in this strip,

dφ = BdS =0µ 1 1

2 3 – + π

Iadx

x a x

∴ Total flux φ 2

= φ∫a

ad

=2

0µ 1 12 3 –

+ π ∫a

a

Iadx

x a x or

φ =0µπIa

In (2)

φ =0µ In(2)

πIa

(I0 sin ωt) ...(1)

Magnitude of induced emf,

e = –φd

dt =0 0µ In(2)ω

πaI

cosωt = e0 cosωt

where e0 =0 0µ In(2)ω

πaI

Charge stored in teh capacitor,q = Ce = Ce0 cosωt ...(2)

and current in the loop

i =dqdt = Cωe0 sinωt ..(3)

imax = Cωe0 =2

0 0µ In(2)ωπ

aI

(b) Magnetic flux passing through the square loopφ ∝sin ωt [From eq. (1)]

i.e., • magnetic field passing through the loop isincreasing at t = 0. Hence, the induced current willproduce ⊗ magnetic field (from Lenz's law). Or the

324

current in the circuit at t = 0 will be clockwise (ornegative as per the given convention). Therefore,charge on upper plate could be written, as

q = q0 cos ωt [From Eq. (2)]

Here, q0 = Ce0 0 0µ In(2)ω

πaI

The corresponding q - t graph is shown in figure.

i+–

q

q0

q0

T4

T2

3T4

Tt

19. After a long time, resistance across an inductor be-comes zero while resistance across capacitor becomesinfinite. Hence, net external resistance.

Rnet =2

2

+R R

=34R

Current through the batteries,

i =

1 2

234

+ +

ER

r r

Given that potential across the terminals of cell A iszero.∴ E – ir1 = 0

or 11 2

2– 03 / 4

= + +

EE rR r r

Solving this equation, we get, 1 24 ( – )3

=R r r

20. Inductive reactanceXL = ωL = (50) (2π) (35 × 10–3) = 11Ω

Impedence Z 2 2= + LR X

= 2 2(11) (11)+ = 11 2Ω

Given Vrms = 220VHence, amplitude of voltage

v0 = rmsV2 220 2= V

or i0 = 20 A

Phase difference φ = tan–1

LXR

= tan–1 1111 4

π = In L-R circuit voltage leads the current. Hence, instan-

taneous current in the circuit is,i = (20A) sin(ωt – π/4)

Corresponding i - t graph is shown in figure.

5T/8T/2

i =20 sin ( t– /4) πω

T/4T/8

T 9T/8 t10 2-

O20

V,I

V= 220 2 sin tw

21. Out side the solenoid net magnetic field zero. It can beassumed only inside the solenoid and equal to µ0nl.

Induced )(BAdtd

dtde −=φ−=

)( 20 anI

dtd πµφ−=

or |e| = (µ0 nπa2) (I0 ω cos ωt)Resistance of the cylindrical vessel

R =(2 )ρ ρ π=l R

s Ld

∴ Induced current i 2

0 0µ cos| |2

ω ω= =

ρLdna I te

R R


1. The induced current in the ring will interact withhorizontal component of magnetic field and both willrepel each other.This repulsion will balance the weight of ring.Hence, option (a) is correct.

COMPREHENSION

1. Change on capacitor at time t is :q = q0 (1 – e–t/τ)

Here, q0 = CV and t = 2τ∴ q = CV (1 – e–2τ /τ)

= CV (1 – e–2)

2. From conservaion of energy,

2max

12

LI = 212

CV

∴LCVI =max

3. Comparing the LC oscillation with normal SHM we get,

2

2d Q

dt= – ω2Q

Here, ω2 =1

LC

∴ Q = –LC2

2d Q

dt

CHAPTER-5 CENTRE OF MASS€¦ · CENTRE OF MASS TRUE/FALSE 1. Two particles of mass 1 kg and 3 kg...

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Transcript of CHAPTER-5 CENTRE OF MASS€¦ · CENTRE OF MASS TRUE/FALSE 1. Two particles of mass 1 kg and 3 kg...