PREFACE - SMAROSA

PREFACE

This book contains the Daily Practice Problems (DPPs) designed for the aspirants

JEE(Main+Advanced). It is a collection of problems (Physics, Chemistry &

Mathematics in separate booklets) from multiple topics to understand the application

of concepts learned in theory. Each DPP is kind of a timed test with marking scheme

and prescribed time to be spent on each problem. It enables a student to practice time

management while solving a problem.

It covers all the pattern of problems asked in Target exam. Answer Key and Hints &

Solutions are also given for self evaluation. In all, it is a great tool for regular

practice of problems in a systematic manner.

Every effort has been taken to keep this book error free, however any suggestions to

improve are welcome [email protected]@resonance.ac.in.

© Copyright reserved.

All rights reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to enrolled

student of RESONANCE only. Any sale/resale of this material is punishable under law, subject to Kota Jurisdiction only. 1 1 3 3 R R D D L L P P

DPPs FileDPPs File(Class-XII)(Class-XII)

INDEXINDEXSS..NNoo. . TTooppiiccs s PPaagge e NNoo..

01 DPPs (01 to 82) 001 – 154

02 Answer Key 155 – 159

03 Hints & Solutions 160 – 250

DPPDPPSS F FILEILE # # 11

PHYSICSPHYSICS

Topics Topics : Kinetic : Kinetic Theory of Theory of Gases and Gases and ThermodynamicsThermodynamics, Motion in T, Motion in Two Dimensions, Newtonwo Dimensions, Newton’s Law of s Law of

Motion, Sound Wave, Projectile Motion,Simple Harmonic MotionMotion, Sound Wave, Projectile Motion,Simple Harmonic Motion

DPP No. 1DPP No. 1Total Marks : 32Total Marks : 32

Max. Time : 32 min.Max. Time : 32 min.

TTyyppe e oof f QQuueessttiioonnss MM..MM.., , MMiinn..

Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.51' negative marking) Q.1 to Q.5 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1155, , 1155]]

Multiple choice Objective ('Multiple choice Objective (' –1' negative marking) Q.6 to Q.71' negative marking) Q.6 to Q.7 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[88, , 88]]

Comprehension ('Comprehension (' –1' negative marking) Q.8 to Q.101' negative marking) Q.8 to Q.10 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. One mole of an ideal gas at a temperature T 1 expands slowly according to the lawp

V= constant. Its

final temperature is T 2. The work done by the gas is

(A) R(T2 ! T1) (B) 2R(T 2 ! T1) (C)2

R(T2 ! T1) (D)

3

R2(T2 ! T1)

2.2. A particle moves along the parabolic path y = ax 2 in such a way that the y-component of the velocity

remains constant, say c. The x and y coordinates are in meters. Then acceleration of the particle at

x = 1 m is

(A) ac k (B) 2ac 2 j (C) ia4

c2

2

! (D) ia2

c!

3.3. A bead of mass m is attached to one end of a spring of natural length R and spring constant

k =R

mg)13( ". The other end of t he spring is fixed at point A on a smooth vertical ring of radius R as

shown in figure. The normal reaction at B just after it is released to move is

(A)2

mg(B) mg3 (C) mg33 (D)

2

mg33

4.4. A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it

crosses a balloon moving upwards with a constant velocity of 2m/s one second after it started to fall. The

difference in the frequency observed by the man in balloon just before and just after crossing the body will be:

(given that -velocity of sound = 300m/s; g = 10m/s2)

(A) 12 (B) 6 (C) 8 (D) 4

5.5. A particle is projected at angle 60 º with speed 10 3 , from the point ' A

' as

shown in the fig. At the same time the wedge is made to move with speed

10 3 towards right as shown in the figure. T hen the time after which particle

will strike with wedge is (g = 10 m/sec 2) :

(A) 2 sec (B) 2 3 s ec (C)3

4 s ec (D) none of these


6.6. A particle performing S.H.M. undergoes displacement of2

A (where A = amplitude of S.H.M.) in one second.

At t = 0 the particle was located at either extreme position or mean position. The time period of S.H.M. can

be : (consider all possible cases)

(A) 12s (B) 2.4 (C) 6s (D) 1.2s

7.7. In the figure shown all the surface are smooth. All the blocks A, B and C are movable, x-axis is horizontal and

y-axis vertical as shown. Just after the system is released from the position as shown.

B

A

C

#

#

y

x

Horizontal Surface

(A) Acceleration of 'A' relative to ground is in negative y-direction

(B) Acceleration of 'A' relative to B is in positive x-direction

(C) The horizontal acceleration of 'B' relative to ground is in negative x-direction.

(D) The acceleration of 'B' relative to ground along the inclined surface of 'C' is greater than g sin #.

COMPREHENSIONCOMPREHENSION

A large tank of cross-section area A contains liquid of density$ . A cylinder of density$ / 4 and length !, and

cross- section area a (a <<A) is kept in equilibrium by applying an external vertically downward force as

shown. The cylinder is just submerged in liquid. At t = 0 the external force is removed instantaneously.

Assume that water level in the tank remains constant.

8.8. The acceleration of cylinder immediately after the external force is removed is

(A) g (B) 2g (C) 3g (D) zero

9.9. The speed of the cylinder when it reaches its equlibrium position is

(A) !g2

1(B) !g

2

3(C) !g2 (D) !g2

10.10. After its release at t = 0, the time taken by cylinder to reach its equilibrium position for the first time is

(A)g8

!&(B)

g3

!&(C)

g4

!&(D)

g2

!&


Topics : Kinetic Theory of Gases and Topics : Kinetic Theory of Gases and Thermodynamics, Simple Harmonic Motion, Circular Motion, Friction,Thermodynamics, Simple Harmonic Motion, Circular Motion, Friction,Work, Power and Energy, String WaveWork, Power and Energy, String Wave

PHYSICSPHYSICS






Comprehension ('Comprehension (' –1' negativ1' negativ e markie marki ng) ng) Q.10Q.10 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. In the figure shown the pressure of the gas in state B is:

(A)25

63 P0 (B)

25

73 P0 (C)

25

48 P0 (D) none of these

2.2. Figure shows the kinetic energy K of a simple pendulum versus its angle # from the vertical. The

pendulum bob has mass 0.2 kg. The length of the pendulum is equal to (g = 10 m/s 2).

-100 1000

5

10

15

K(mJ)

#(mrad)

(A) 2.0 m (B) 1.8 m (C) 1.5 m (D) 1.2 m

3.3. A particle is revolving in a circle increasing its speed uniformly. Which of the following is constant?

(A) centripetal acceleration (B) tangential acceleration

(C) angular acceleration (D) none of these

4.4. A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the srcin as shown in

figure and whose equation is x2 = 4ay. The wire frame is fixed in vertical plane and the bead can slide on it

without friction. The bead is released from the point y = 4a on the wire frame from rest. The tangential

acceleration of the bead when it reaches the position given by y = a is :

(A)2

g(B)

2

g3(C)

2

g(D)

5

g


5.5. In the shown arrangement if f 1, f2 and T be the frictional forces on 2 kg block, 3k g block and tension in

the string respectively , then their values are:

(A) 2 N, 6 N, 3.2 N (B) 2 N, 6 N, 0 N

(C) 1 N, 6 N, 2 N (D) data insufficient to calculate the required values.

6.6. A block is attached with a spring and is moving toward s a fixed wall with sp eed v as shown in figu re. As

the spring reaches the wall, it starts compressing. T he work done by the spring on the wall during the

process of compression is :

(A) 1/2 mv 2 (B) mv2 (C) Kmv (D) zero


Figure shows a clamped metal string of length 30 cm and linear mass density 0.1 kg/m. which is taut

at a tension of 40 N. A small rider (piece of paper) is placed on string at point P as sh own. An external

vibrating tuning fork is brought near this string and oscillations of rider are carefully observed.

7.7. At which of the following frequencies of turning fork, r ider will not vibrate at all :

(A) 3

100Hz (B) 50 Hz (C) 200 Hz (D) None of these

8.8. At which of the following frequencies the point P on string will have maximum oscillation amplitude

among all points on string :

(A)3


9.9. Now if the tension in the string is made 160 N, at which of the following frequencies of tur ning fork, rider

will not vibrate at all

(A)3


10.10. In each situation of column-I, the x-coordinate of a particle moving along x-axis is given in terms of time t. ('is a positive constant). Match the equation of motion given in column-I with the type of motion given in

column-II.

Column-Column- Column-Column-

(A) sin 't – cos 't (p) SHM

(B) sin3 't (q) Periodic

(C) sin 't + sin3

't + sin5

't (r) Periodic but not SHM

(D) exp ( – '2 t2) (s) Non periodic


Topics : Simple Harmonic Motion, Circular Motion, Work, Power and Energy, NewtonTopics : Simple Harmonic Motion, Circular Motion, Work, Power and Energy, Newton’s Law of Motion,s Law of Motion,

Kinetic Theory of Kinetic Theory of Gases and Thermodynamics, Sound Waves, Geometrical Gases and Thermodynamics, Sound Waves, Geometrical OpticsOptics

PHYSICSPHYSICS






Subjective Question ('Subjective Question (' –1' negativ1' negativ e markine markin g) g) Q.8Q.8 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]

MMaatthhcch h tthhe e ffoolllloowwiinng g ((nno o nneeggaattiivve e mmaarrkkiinngg) ) QQ..99 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal

direction with constant acceleration 3 g m/s 2. The period of small oscillations of the pendulum about

its equilibrium position is (g = &2 m/s2) :

(A) 1.0 sec (B) 2 s ec (C) 1.53 sec (D) 1.68 sec

2.2. A smooth wire is bent into a vertical circle of radius a a . A bead P can slide smoothly on the wire. The

circle is rotated about vertical diameter AB as axis with a constant speed ' as shown in figure. The

bead P is at rest w.r.t. the wire in the position shown. Then '2 is equal to :

(A)2 g

a(B)

2

3

g

a

(C)g

a

3(D)

2

3

a

g

3.3. The potential energy of a particle varies with x according to the relation U(x) = x 2 ! 4 x. The point

x = 2 is a point of :

(A) stable equilibrium (B) unstable equilibrium

(C) neutral equilibrium (D) none of above

4.4. Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in

the figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB and

CD of the two blocks are made reflecting. The acceleration of two images form ed in those two reflecting

surfaces w.r.t. each other is :

(A) 5g/6 (B) 5g/3 (C) g/3 (D) 17 g/6


5.5. A cylinder of mass M and radius R is resting on two corner edges A and B as

shown in figure. The normal reaction at the edges A and B are : (Neglect friction)

(A) NA = B N 2 (B) NB = A N 3

(C) NA =2

Mg(D) NB =

5

32 Mg

6.6. An ideal gas undergoes a cyclic process abcda which is shown by

pressuredensity curve.

(A) Work done by the gas in the process 'bc' is zero

(B) Work done by the gas in the process 'cd' is negative

P

$

d

a

b

c

$1 $2

(C) Internal energy of the gas at point 'a' is greater than at state 'c'

(D) Net work done by the gas in the cycle is negative.

7.7. A car moves towards a hill with speed vc. It blows a horn of frequency f which is heard by an observer following

the car with speed v0. The speed of sound in air is v.

(A) the wavelength of sound reaching the hill isv

f

(B) the wavelength of sound reaching the hill isv v

f

c!

(C) The wavelength of sound of horn directly reaching the observer isf

vv c"

(D) the beat frequency observed by the observer is( )

2c

2

oc

vv

fvvv2

!

"

8.8. Power delivered to a body varies as P = 3 t2. Find out the change in kinetic energy of the body from

t = 2 to t = 4 sec.

9.9. Four particles are m oving with different velocities in fron t of stationary plane mirror (lying in y-z plane).At t = 0, velocity of A is ivA *

", velocity of B is j3ivB "!*

", velocity of C is j6i5vC "*

", velocity of D

is ji3vD !*"

. Acceleration of particle A is ji2aA "*"

and acceleration of particle C is jt2aC *"

. TheThe

particle B and D move with uniform velocity (Assume no collision to take place till t = 2 seconds). All

quantities are in S.I. Units. Relative velocity of image of object A w ith respect to object A is denoted by

A,AV +

". Velocity of images relative to corr esponding objects are given in column I and their values are

given in column II at t = 2 second. Match column I with corresponding values in column II and indicate

your answer by darkening appropriate bubbles in the 4 × 4 matrix given in OMR.

ColumnColumn II ColumnColumn IIII

(A) A,AV +

"

(p) i2

(B) B,BV +

"

(q) i6!

(C) C,CV +

"

(r) j4i12 "!

(D) D,DV +

"

(s) i10!


Topics : Kinetic Theory of Topics : Kinetic Theory of Gases ,Thermodynamics, Projectile Motion, Friction, Geometrical Optics, StringGases ,Thermodynamics, Projectile Motion, Friction, Geometrical Optics, String

WavesWaves

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.31' negative marking) Q.1 to Q.3 ((3 3 mmaarrkks s 3 3 mmiinn..)) [[99, , 99]]

Multiple choice Objective ('Multiple choice Objective (' –1' negative marking) Q.41' negative marking) Q.4 ((4 4 mmaarrkks s 4 4 mmiinn..)) [[44, , 44]]

Subjective Question ('Subjective Question (' –1' negative marking) Q.51' negative marking) Q.5 ((4 4 mmaarrkks s 5 5 mmiinn..)) [[44, , 55]]

ComprehensioComprehensio n n ('(' –1' negative marking) Q.6 to Q.81' negative marking) Q.6 to Q.8 ((3 3 mmaarrkks s 3 3 mmiinn..)) [[99, , 99]]

1.1. A gas undergoes an adiabatic process and an isotherma l process. The two processes are plotted on a

P-V diagram. The resulting curves intersect at a point P. Tangents are drawn to the two curves at P.

These make angles of 135 º & 121º with the positive V-axi s. If tan 59º = 5/3, the gas is likely to be:

(A) monoatomic (B) diatomic

(C) triatomic (D) a mixture of monoatomic & diatomic gases

2.2. A particle is projected from a point P (2, 0, 0)m with a velocity 10 m/s mak ing an angle 45 º with the

horizontal. The plane of projectile motion passes throu gh a horizontal line PQ which makes an angle of

37º with positive x-axis, xy plane is horizontal. The coordinates of the point where the particle willstrike the line PQ is: (Take g = 10 m/s 2)

(A) (10, 6, 0)m (B) (8, 6, 0)m (C) (10, 8, 0)m (D) (6, 10, 0)m

3.3. A ray of light is incident at an , of 30º on a plane mirror M 1. Another plane mirr or M2 is inclined at angle

# to M1. What is the value of angle # so that light reflected from M 2 is parallel to M 1.

(A) 60 º (B) 75 º (C) 67.5 º (D) none of these

4.4. A variable force F = 10 t is applied to block B placed on a smooth surface. The coefficient of friction between

A & B is 0.5. (t is time in seconds. Initial velocities are zero, A is always on B)

(A) block A starts sliding on B at t = 5 seconds

(B) the heat produced due to friction in first 5 seconds is 312.5J

(C) the heat produced due to friction in first 5 seconds is (625/8) J

(D) acceleration of A at 10 seconds is 5 m/s2.


5.5. A point source S is centered in front of a 70 cm wide plane mirror. A man starts walking from the source along

a line parallel to the mirror in a single direction. Maximum distance that can be walked by man without losing

sight of the image of the source is _____.


A sinusoidal wave is propagating in negative x –direction in a string stretched along x-axis. A particle of string

at x = 2m is found at its mean position and it is moving in positive y direction at t = 1 sec. If the amplitude

of the wave, the wavelength and the angular frequency of the wave are 0.1meter, & /4 meter and 4& rad/sec

respectively.

6.6. The equation of the wave is(A) y = 0.1 sin (4&(t –1)+ 8(x – 2)) (B) y = 0.1 sin ((t –1) – (x – 2))

(C) y = 0.1 sin (4&(t –1) –8(x – 2)) (D) none of these

7.7. The speed of particle at x = 2 m and t = 1sec is

(A) 0.2& m/s (B) 0.6 & m/s

(C) 0.4& m /s (D) 0

8.8. The instantaneous power transfer through x=2 m and t= 1.125 sec, is

(A) 10 J/s (B)3

4& J /s (C)

3

2& J /s (D) zero


Topics : Rigid Body Dynamics, Geometical OpTopics : Rigid Body Dynamics, Geometical Optics., Calorimetry, Work, Power and Energy , Fluid Mechanics,tics., Calorimetry, Work, Power and Energy , Fluid Mechanics,Kinetic Theory Kinetic Theory of of Gases and Gases and thermodynamicsthermodynamics

PHYSICSPHYSICS





Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.31' negative marking) Q.3 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44, , 44]]

Subjective Questions ('Subjective Questions (' –1' negative marking) Q.4 to Q.61' negative marking) Q.4 to Q.6 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[1122, , 1155]]

Comprehension ('Comprehension (' –

1' negative marking) Q.7 to Q.91' negative marking) Q.7 to Q.9 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.10× 4) Q.10 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its centre of

mass and perpendicular to its plane is :

(A)2

RM 2

! M

2

3

R4--

.

/001

2

&(B)

2

RM 2

! M

2

3

R42 --

.

/001

2

&

(C)2

RM 2

+ M

2

3

R4--

.

/001

2

&(D)

2

RM 2

+ M

2

3

R42 --

.

/001

2

&

2.2. Angle of incidence of the incident ray for which reflected ray intersect perpendiculaly the principal axis.

i

C

(A) 0° (B) 30° (C) 45° (D) 60°

3.3. Heat is supplied to a certain homogeneous sample of matter at a uniform rate. Its temperature is

plotted against time as shown in the figure. Which of the following conclusions can be drawn? kk

(A) its specific heat capacity is greater in the solid state than in the liquid state.

(B) its specific heat capacity is greater in the liquid state than in the solid state.

(C) its latent heat of vaporization is greater than its latent heat of fusion.

(D) its latent heat of vaporization is smaller than its latent heat of fusion.

4.4. A block of weight W is dragged across the horizontal floor from A to B by the constant vertical force P acting

at the end of the rope. Calculate the work done on the block by the force P = ( 3 + 1)N.Assume that block

does not lift off the floor. (g = 10 m/s2)


5.5. A Plane mirror revolves as shown at constant angular velocity making 2 rps about its normal. With what

velocity will the light spot move along a spherical screen of radius of 10 m if the mirror is at the centre of

curvature of the screen and the light is incident from a fixed direction.

6.6. A water tank stands on the roof of a building as shown. Find the value of h (in m) for which the horizontal

distance 'x' covered by the water is greatest.

1m h

3m

x


A quantity of an ideal monoatomic gas consists of n moles initially at temperature T 1. The pressure and

volume are then slowly doubled in such a manner so as to trace out a straight line on a P-V diagram.

7.7. For this process, the ratio1nRT

W is equal to (where W is work done by the gas) :

(A) 1.5 (B) 3 (C) 4.5 (D) 6

8.8. For the same process, the ratio1nRT

Q is equal to (where Q is heat supplied to the gas) :

(A) 1.5 (B) 3 (C) 4.5 (D) 6

9.9. If C is defined as the average molar specific heat for the process thenR

C has value

(A) 1.5 (B) 2 (C) 3 (D) 6

10.10. Consider a system of particles (it may be rigid or non rigid). In the column- 3 some condition on force

and torque is given. Column- 33 contains the effects on the system. (Letters have usual meaning)


(A) 0Fres *"

(p) systemP"

will be constant

(B) 0res *4"

(q) systemL"

will be constant

(C) External force is absent (r) total work done by all forces will be zero

(D) No nonconservative force acts. (s) total mechanical energy will be constant.


Topics : Topics : Centre of Centre of Mass, Circular Motion, Geometrical Optics., Mass, Circular Motion, Geometrical Optics., Kinetic Throry of Gases Kinetic Throry of Gases and Thermodynamicsand Thermodynamics

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.51' negative marking) Q.1 to Q.5 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1155, , 1155]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.61' negative marking) Q.6 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]


Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.10× 4) Q.10 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. If a ball is dropped from rest, it bounces from the floor repeatedly. The coefficient of restitution is 0.5 and the

speed just before the first bounce is 5ms –1. The total time taken by the ball to come to rest finally is :

(A) 1.5s (B) 1s (C) 0.5s (D) 0.25s

2.2. A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is releasedfrom position A and leaves the track at B. The radius of curvature of its trajectory when it just leaves the trackat B is:

(A) R (B)4

R (C)

2

R(D) none of these

3.3. A bob is attached to one end of a string other end of which is fixed at peg A. The bob is taken to aposition where string makes an angle of 30 0 with the horizontal. On the circular path of the bob invertical plane there is a peg ‘B’ at a symmetrical position with respect to the position of release asshown in the figure. If V c and V a be the minimum tangential v elocity in clockwise an d anticlockwisedirections respectively , given to the bob in order to hit the peg ‘B’ then ratio V c : Va is equal to :

(A) 1 : 1 (B) 1 : 2 (C) 1 : 2 (D) 1 : 4

4.4. A virtual erect image in a concave mirror is represented, in the above figures, by

(A) (B) (C) (D)

5.5. A driving mirror on a car is never concave because :(A) its field of view is too sm all(B) the image would be inverted(C) the image would be virtual and therefore useless for the driver

(D) only a plane mirror forms true images.


6.6. A point object is placed at a distance 20 cm from the focus of a concave mirror of radius of curvature 20 cm.Find the distance (in cm) of the image from the focus.


A point object is placed on principal axis of a concave mirror ( of focal length 15 cm) at a distance u = 61 cm

from pole. A slab of thickness t = 3 cm and refractive index 5 = 1.5 is placed with two sides perpendicular to

principal axis, such that its nearest face is x0 cm from pole. The final image of object is to be considered after

refraction by slab, reflection by mirror and final refraction by slab.

x0

f=15cm

t=3cm 5=1.5

object

7.7. If x0 = 30 cm, then the distance of final image from pole is

(A) 19 (B) 21 (C) 23 (D) 24

8.8. If the slab is shifted parallel to itself by 3 cm then the final image(A) shifts towards left (B) shifts towards right(C) may shifts towards left or right (D) does not shift

9.9. If x0 = 30 cm and the object is given velocity 18 m/s towards left then the speed of image at that instant is

(A) 2 m/s (B) 6 m/s (C) 9 m/s (D) 162 m/s

10.10. A sample of gas goes fr om state A to state B in four different m anners, as shown by the graphs. Let Wbe the work done by the gas and 6U be change in internal energy along the path AB. Correctly matchthe graphs with the statements provided.

(A)

V

P

A B (p) Both W and 6U are positive

(B)

P

T

A

B

(q) Both W and 6U are negative

(C)

T

V

B

A

(r) W is positive w hereas 6U is negative

(D)

V

P

A

B (s) W is negative whereas 6U is positive


Topics : Simple Harmonic Motion, Friction, Fluid MTopics : Simple Harmonic Motion, Friction, Fluid Mechanics, Rigid Body Dynamics, Kinematics, Geometricalechanics, Rigid Body Dynamics, Kinematics, GeometricalOpticsOptics

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.41' negative marking) Q.1 to Q.4 ((3 3 mmaarrkks s 3 3 mmiinn..)) [[1122, , 1122]]

Subjective Questions ('Subjective Questions (' –1' negative marking) Q.5 to Q.61' negative marking) Q.5 to Q.6 ((4 4 mmaarrkks s 5 5 mmiinn..)) [[88, , 1100]]

Comprehension ('Comprehension (' –1' negative marking) Q.7 to Q.91' negative marking) Q.7 to Q.9 ((3 3 mmaarrkks s 3 3 mmiinn..)) [[99, , 99]]

1.1. A particle performs S.H.M. on x!axis with amplitude A and time period T. The time taken by the particle totravel a distance A/5 starting from rest is:

(A)20

T(B) &2

T cos!1 -

.

/01

2 5

4 (C) &2

T cos!1 -

.

/01

2 5

1(D) &2

T sin!1 -

.

/01

2 5

1

2.2. A body of mass 10 kg lies on a rough inclined plane of inclinatio n # = sin !1

5

3 with the horizontal. When

a force of 30 N is applied on the block par allel to & upward the plane, the total reaction by the plane on

the block is nearly along:

(A) OA (B) OB (C) OC (D) OD

3.3. A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side 'L'

at a depth '4y' from the top and the other is a circular hole of radius 'R' at a depth ‘y’ from the top. When the

tank is completely filled with water, the quantities of water flowing out per second from both holes are the

same. Then, 'R' is equal to :

4yv1

v2L

2Ry

(A)&2

L(B) 2&L (C)

&2

. L (D)&2

L

4.4. A ring of m ass m and r adius R rolls on a horizontal rough surface without slipping due to an applied

force ‘F’. The friction force acting on ring is : –

(A)3

F(B)

3

F2(C)

4

F(D) Zero


5.5. A particle is projected from the ground level. It just passes through upper ends of vertical poles A, B, C of

height 20 m, 30 m and 20 m respectively. The time taken by the particle to travel from B to C is double of the

time taken from A to B. Find the maximum height attained by the particle from the ground level.

6.6. A man is standing at the edge of a 1m deep swimming pool, completely filled with a liquid of refractive

index 3 2 / . The eyes of the man are 3 m above the ground. A coin located at the bottom of the pool

appears to be at an angle of depression of 300 with reference to the eye of man. Then horizontal distance

(represented by x in the figure) of the coin from the eye of the man is ____________ mm.


A block with a concave mirror of radius of curvature 1m attached to one of its sides floats, with exactly half of

its length immersed in water and the other half exposed to air.

Any ray srcinating from an object O 1 and O2 (as shown in figure) floating on the surface of water first gets

reflected by the mirror. This then gets refracted by the water surface if the image formed by reflection is real

(31). If the image is virtual (32), then the reflected ray never encounters the air-water interface and hence there

is no refraction. The image for the next three questions refers to the final image, formed after both the

reflection and refraction (if it occurs at all) has taken place.

7.7. The final image formed is unique (i.e. only one image is formed) if the point object floats on the surface of

water at a distance x in front of the mirror where

(A) x is less than 50 cm (B) x is less than 1m

(C) x is between 50 cm and 1m (D) for any value of x

8.8. There is no refraction of light rays reflected from the mirror if the point object floats on the surface of water at

a distance x in front of the mirror where

(A) x is less than 50 cm (B) x is less than 1m

(C) x is between 50 cm and 1m (D) for any value of x

9.9. The final image is not unique (i.e. more than one image for the same object) if the point object is placed a

distance ‘y’ above the surface of water and a distance x in front of the mirror (rays are not paraxial), where

y = 20 cm and x is :

(A) less than 50 cm (B) less than 1m

(C) between 50 cm and 1m (D) for any value of x


Tpoics : Simple Harmonic MotTpoics : Simple Harmonic Motion, Friction, Rigid Body Dynamics, String Wavesion, Friction, Rigid Body Dynamics, String Waves, Sound Waves, GeometricalSound Waves, GeometricalOpticsOptics

PHYSICSPHYSICS







1.1. A particle is executing SHM according to the equation x = A cos ' t. Average speed of the particle during the

interval 0 7 t 7'

&6

.

(A)2

A3 '(B)

4

A3 '(C)

&'A3

(D)&

'A3 ( )32 !

2.2. A particle of mass 5 kg is moving on rough fixed inclined plane (making an angle 30 ° with horizontal)

with constant velocity of 5 m/s as shown in the figure. Find the friction force acting on a body by the

inclinedplane. ( take g = 10m/s 2)

(A) 25 N (B) 20 N(C) 30 N (D) none of these

3.3. A sphere rolls without sliding on a rough inclined plane (only mg and contact forces are acting on the

body). The angular momentum of the body:

(A) about centre is conserved

(B) is conserved about the point of contact

(C) is conserved about a point whose distance from the inclined plane is g reater than the

radius of the sphere

(D) is not conserved about any point.

4.4. A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the

centre of the string is 4 m m. Distance between the two points having amplitude 2 mm is:

(A) 1 m (B) 75 cm (C) 60 cm (D) 50 cm

5.5. The source (S) of sou nd is moving constant velocity v0 as shown in diagram . An obserever O list ens to

the sound emmited by the source. T he observed frequency of the sound :

(A) continuously decreases (B) continuously increases

(C) first decreases then increases (D) first increases then decreases.


6.6. A composite slab consisting of different media is placed infront of a concave mirror of radius of curvature 150

cm. The whole arrangement is placed in water. An object O is placed at a distance 20 cm from the slab. The

R.I. of different media are given in the diagram. Find the position of the final image formed by the system.

7.7. Find out the moment of inertia of the following structure (written as ) about axis AB made of

thin uniform rods of mass per unit length 8.


All objects referred to the subsequent problems lie on the principle axis.

0000

8.8. If light is incident on surface 1 from left, the image formed after the first refraction is definitely :

(A) Real for a real object (B) Virtual for a real object

(C) Real for a virtual object (D) Virtual for a virtual object

9.9. In above question if the object is real, then the final image formed after two refractions :

(A) may be real (B) may be virtual (C) must be virtual (D) both A and B

10.10. If light is incident on surface 2 from right then which of the following is true for image formed after a single

refraction.

(A) Real object will result in a real image (B) Virtual object will result in a virtual image

(C) Real object will result in a virtual image (D) Virtual object will result in a Real image


Topics : Rigid Body Dynamics, Relative MotioTopics : Rigid Body Dynamics, Relative Motion, Sound Wavesn, Sound Waves,, Geometrical OpticsGeometrical Optics

PHYSICSPHYSICS








1.1. A uniform disk of mass 300kg is rotating freely about a vertical axis through its centre with constant angular

velocity '. A boy of mass 30kg starts from the centre and moves along a radius to the edge of the disk. The

angular velocity of the disk now is

(A)'0

6(B)

'0

5(C)

4

50'

(D)5

60'

2.2. A man is holding an umbrella at angle 30° with vertical with lower end towards himself, which is appropriate

angle to protect him from rain for his horizontal velocity 10 m/s. Then which of the following will be true-

(A) rain is falling at angle 30° with vertical, towards the man

(B) rain may be falling at angle 30° with vertical, away from the man

(C) rain is falling vertically

(D) none of these

3.3. In Resonance tube experim ent, if 400 Hz tuning fork is used, the first resonan ce occurs when length of

air column in the tube is 19 cm. If the 400 Hz. tuning fork is replaced by 1600 Hz tuning fork then to get

resonance, the water level in the tube should be further lowered by (take end correction = 1 cm )

(A) 5 cm (B) 10 cm (C) 15 cm (D) 20 cm

44 A parallel beam of light is incident on a lens of f ocal length 10 cm. Aparallel slab of refractive index 1.5 and thickne ss 3 cm is placed on the

other side of the lens. Find the distance of the final image from the

lens.

5.5. A small object stuck on the surface of a glass sphere (n = 1.5) is viewed from the diametrically opposite

position. Find transverse magnification.


Two bodies A and B of masse s 10 kg and 5 kg are placed very slightly

separated as shown in figure. The coeff icient of friction between the

floor and the blocks is 5 = 0.4. Block A is pushed by an external force

F. The value of F can be changed. When the welding between block A

and ground breaks, block A will start pressing block B and when welding

of B also breaks, block B will start pressing the vertical wall –

6.6. If F = 20 N, with how much f orce does block A presses the block B

(A) 10 N (B) 20 N (C) 30 N (D) Zero

7.7. What should be the minimum value of F, so that block B can press the vertical wall

(A) 20 N (B) 40 N (C) 60 N (D) 80 N

8.8. If F = 50 N, the friction force (shear force) acting between block B and ground will be :

(A) 10 N (B) 20 N (C) 30 N (D) None

9.9. The force of f riction acting on B varies with the applied force F according to curve :

(A) (B) (C) (D)


Topics : Sound Waves, Rigid Body Dynamics, Projectile MotionTopics : Sound Waves, Rigid Body Dynamics, Projectile Motion,, Geometrical Optics, Simple HarmonicGeometrical Optics, Simple HarmonicMotionMotion

PHYSICSPHYSICS





Subjective Questions ('Subjective Questions (' –1' negative marking) Q.71' negative marking) Q.7 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]

Comprehension ('Comprehension (' –1' negative marking) Q.8 to Q.101' negative marking) Q.8 to Q.10 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99,,99]]

1.1. A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the

other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the

speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s.

[ Use g = 10 m/s2 ]

(A) 1 m/s (B) 2 m/s (C) 0.5 m/s (D) 1.5 m/s

2.2. A plank P is placed on a solid cylinder S, which rolls on a horizontal surface. The two are of equal

mass. There is no slipping at any of the surfaces in contact. T he ratio of the k inetic energy of P to the

kinetic energy of S is:

(A) 1: 1 (B) 2: 1 (C) 8: 3 (D) 1: 4

3.3. A stone is projected from gro und and hits a smooth vertical wall after 1 sec. and again falls back on the

ground. The time taken by stone to reach the ground after the collision is 3 secs. The m aximum height

reached by the same stone if the vertical wal l were not to be present is. (g = 10 m/s 2)

(A) 10 m (B) 12.5 m (C) 15 m (D) 20 m

4.4. A cylinde rical optical fibre (quarter circular shape) of refractive index n = 2 and diameter d = 4mm is surrounded

by air. A light beam is sent into the fibre along its axis as shown in figure. Then the smallest outer radius R

(as shown in figure) for which m

(A) 2mm (B) 4mm (C) 8 mm (D) 6 mm


5.5. A light is incident on face AB of an equilateral glass prism ABC. After refraction at AB, the ray is incident

on face BC at the angle slightly greater than critical angle so that it gets reflected from face BC and

finally emerges out from face AC. Net deviation angle of the ray is 112 ° anticlockwise. The angle of

incidence 'i' has value :

A

B C

i

#c

(A) 22 ° (B) 24 ° (C) 26° (D) 28°

6.6. In a compound microscope

(A) the objective has a shorter focal length (B) the objective has a shorter aperture

(C) (A) and (B) (D) the aperture of objective and eyepiece are same.

7.7. A particle initially at rest experiences a force F = a sin( 't) where a and ' are positive constants. The

direction of force is always towards positive x-axis. State with reason or explanation or derivation

whether the particle will change its direction of velocity or not.


An extended object of size 2 cm is placed at a distance of d (cm) in medium (refractive index n = 3) from pole,

on the principal axis of a spherical curved surface.The medium on the other side of refracting surface is air

(refractive index n = 1).

2cm

d

n=1n=3

ROC =20cm

8.8. For d = 20 cm, the distance of the image from the pole is

(A) 2 cm (B) 3 cm (C) 4 cm (D) 5 cm

9.9. For d = 20 cm, The size of image is

(A)6

1 c m (B)

15

2 c m (C)

5

6 c m (D)

2

3cm

10.10. For all nonzero and finite values of d (the object is placed to the left of pole as shown), the nature of image

formed always is

(A) Image is real and erect (B) Image is real and inverted

(C) Image is virtual and erect (D) None of these


Topics : Geometrical Optics, Fluid, Work, Power and Energy, Center of Mass, Circular Motion, Heat andTopics : Geometrical Optics, Fluid, Work, Power and Energy, Center of Mass, Circular Motion, Heat and

Thermodynamics, Rigid Body DynamicsThermodynamics, Rigid Body Dynamics

PHYSICSPHYSICS





Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.6 to Q.71' negative marking) Q.6 to Q.7 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[88, , 88]]



1' negative marking) Q.9 to Q.101' negative marking) Q.9 to Q.10 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[66, , 66]]

1.1. An object moves in front of a fixed plane mirror. The velocity of the image of the object is

(A) Equal in the magnitude and in the direction to that of the object.

(B) Equal in the magnitude and opposite in direction to that of the object.

(C) Equal in the magnitude and the direction will be either same or opposite to that of the object.

(D) Equal in magnitude and makes any angle with that of the object depending on direction of motion of the

object.

2.2. A point object is m oving along principal axis of a concave m irror with uniform velocity towards pole.

Initially the object is at infinite distance from pole on right side of the mirror as shown. Before the

object collides with mirror, the number of times at which the distance between object and its image is

40 cm are.

(A) one time (B) two times (C) three times (D) Data insufficient

3.3. Three containers of same base are a, same height are filled with three different liquids of same mass as

shown in the figure. If F 1, F2, F3 are the force exerted by the liquid on the base of the container in case

I, II and III respect ively, then we have the relation:

(A) F 1 = F2 = F3 (B) F 1 > F2 > F3 (C) F 3 > F2 > F1 (D) F 2 < F3 < F1

4.4. A particle is projected with a velocity u making an angle # with the horizontal. The instantaneous

power of the gravitational force

(A) varies linearly with time (B) is constant throughout

(C) is negative for complete path (D) None of the above

5.5. A body of mass 'm' sliding down a movable inclined plane of mas s 2 m,

assuming friction is absent everywhere the k inetic energy of 2 m as a

function of time is: (m remains on 2m)

(A) (B) (C) (D) none of these


6.6. A car is moving along a circle with constant speed on an inclined plane as shown in diagram. Then frictionforce on car may be in horizontal direction at least at one point :

(A) in portion 'AB' including point A and B(B) in portion 'BC' including point B and C(C) in portion 'CD' including point C and D

(D) in portion 'DA' including point D and A

7.7. In a cyclic process, a gas is taken from state A to B viapath-3 as shown in the indicator diagram and taken backto state A fromstate B via path-33. In the complete cycle :(A) work is done on the gas.(B) heat is given to the gas(C) no work is done by the gas.(D) nothing can be said about work as data is insufficient

8.8. In the figure, an object is placed at distance 25 cm from the surface of a convex mirror, and a plane mirror isset so that the image formed by the two mirrors lie adjacent to each other in the same plane. The plane mirroris placed at 20 cm from the object. What is the radius of curvature of the convex mirror?


A uniform ring of radius 4 cm placed on a rough horizontal surface is given a sharp imp ulse as in figure.As a consequence it acquires a linear velocity of 2 m/s. If coefficient of f riction between the ring and thehorizontal surface be 0.4. Then answer the following questions based on given information.

9.9. The velocity of centre of mass after which the ring will start pure rolling is -(A) 2 m/s (B) 1 m/s (C) 4 m/s (D) 1/2 m/s

10.10. The time af ter which the ring will start pure rolling is -(A) 1/2 s (B) 1 s (C) 1/4 s (D) 2 s


TopTopics : Geometics : Geometrical Orical Optics., Rectilinear ptics., Rectilinear MotiMoti on, on, StriString Waveng Wave, Project, Project ile ile MotiMoti onon ,, Rigid Body Dynamics,Rigid Body Dynamics,

Heat and ThermodynamicsHeat and Thermodynamics

PHYSICSPHYSICS







1.1. Two plane mirrors are inclined to each other at 90 º. A ray of light is incident on one mirror and the

reflected light goes to the other mirror. The ray will undergo a total deviation of :

(A) 180º (B) 90 º

(C) 45º (D) cannot be found because angle of incidenc e is not

given.

2.2. AB is an incident beam of light and CD is a reflected beam (the num ber of reflections for this may

be 1 or more than 1) of light. AB & CD are separated by some distance (may be large). It is

possible by placing what ty pe of mirror on the right side.

(A) one plane mirror (B) one concave mirror

(C) one convex mirror (D) none of these

3.3. A particle A of mass 10

7 kg is moving in the positive direction of x. Its initial position is x = 0 &

initial veloci ty is 1 m/s. The velocity a t x = 10 is: (use the graph given)

(A) 4 m/s (B) 2 m/s

(C) 3 2 m /s (D) 100/3 m/s

4.4. Two wave functions in a medium along x direction are given by -

m)t3x2(2

1y

21!"

*22

)6t3x2(2

1y

!""!* m

where x is in metres and t is in seconds

(A) There is no position at which resultant displacement will be zero at all times.

(B) There is no time at which resultant displacement will be zero everywhere.

(C) Both waves travel along the same direction.

(D) Both waves travel in opposite directions.


5.5. A bullet is fired with speed 50 m/s at 45° angle find the height of the bullet when its direction of motion makes

angle 30° with the horizontal.

6.6. A thin uniform rod AB of mass ‘m’ translates with an acceleration ‘a’,

when two anti parallel forces F1 and F

2 act on it as shown in figure. If

the distance between F1 and F

2 is ‘b’, the length of the bar is _______.

7.7. In the figure shown M 1 and M2 are two spherical mirrors of focal length 20 cm each. AB and CD are theirprincipal axes respectively whi ch are separated by 1 cm . PQ is an object of height 2 cm and kept atdistance 30 cm from M 1. The separation between the mirrors is 50 cm. Consider two successivereflections first on M 1 then on M 2. Find the size of the 2 nd image. Also find distances of end points P 9and Q 9 of that image f rom the line AB.


One mole of an ideal monatomic gas undergoes a linear process from A to B, in which is pressure P

and its volume V change as shown in figure

8.8. The absolute temperature T versus volume V for the given process is

(A) (B) (C) (D)

9.9. The maximum tem perature of the gas during this process is

(A) R2

VP 00

(B) R4

VP 00

(C) R4

VP3 00

(D) R2

VP3 00

10.10. As the volume of the gas is increased, in some range of volume the gas expands with absorbing the heat

(the endothermic process) ; in the other range the gas emits the heat (the exothermic process). Then the

volume after which if the volume of gas is further increased the given process switches from endothermic to

exothermic is

(A)8

V2 0(B)

8

V3 0(C)

8

V5 0(D) none of these


Topics : Rigid Body Dynamics, Geometrical Optics, Simple HarTopics : Rigid Body Dynamics, Geometrical Optics, Simple Har monic Motion, Center of monic Motion, Center of MassMass

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.61' negative marking) Q.1 to Q.6 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1188, , 1188]]Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.71' negative marking) Q.7 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44, , 44]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.81' negative marking) Q.8 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]Comprehension ('Comprehension (' –1' negative marking) Q.1' negative marking) Q. 9 to Q.119 to Q.11 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. A car is initially at rest, 330 m away from a stationary observer. It begins to move towards the obser verwith an acceleration of 1.1 m/s 2, sounding its horn continuously. 20 second later, the driver stopssounding the horn. Th e velocity of sound in air is 330 m/s. The obs erver will hear the sound of the hornfor a duration of:(A) 20 sec (B) 21 sec (C) 62/3 sec (D) 58/3 sec

2.2. A point moves in a straight line under the retardation a v2 , where ‘a’ is a positive constant and v is

speed. If the initial speed is u , the distance covered in ' t ' seconds is :

(A) a u t (B)1

a l n (a u t) (C)

1

a l n (1 + a u t) (D) a ln (a u t)

3.3. A ring of radius R lies in vertical plane. A bead of mass‘m’ can move along the ring without friction. Initially the

bead is at rest at the bottom most point on ring. The minimum constant horizontal speed v with which the ringmust be pulled such that the bead completes the vertical circle

(A) gR3 (B) gR4 (C) gR5 (D) gR5.5

4.4. Minimum number of prisms required to m ake a combination which is 'Achromatic' as well as 'DirectVision' is:(A) 2 (B) 3 (C) 4 (D) 11

5.5. As shown in figure, S is a point on a uniform disc rolling with uniform angular velocity on a fixed roughhorizontal surface. The only forces acting on the disc are its weight and contact forces exerted byhorizontal surface. Which graph best represents the magnitude of the acceleration of point S as afunction of time

s

(A) (B) (C) (D)


6.6. Consider a uniform disc of mass ‘m’ performing pure rolling with velocity‘v’ on a fixed rough surface

(A) Kinetic energy of upper half will be8

3 mv2

(B) Kinetic energy of upper half will be less than8

3 mv2

(C) Kinetic energy of upper half will be more than8

3 mv2

(D) Kinetic energy of upper half will be more than4

3 mv2

7.7. A block of mass m = 1 kg is placed on a smooth surface and is connected w ith a spring of springconstant k = 100 N/m and another end of spring is connected to a fixed wall as shown. The block is

pulled by a distance A = 0.10 m from its natural length and released at t = 0.

(A) The maximum speed is after t =20

&s. (B) Time taken to cover f irst 0.10 m, t =

20

&s.

(C) Time taken to cover first 0.05 m, t =40

&s. (D) Time taken to cover fi rst 0.05 m, t =

30

&s.

8.8. A block of mass m1 = 2 kg slides on a frictionless table with speed of 10 m/s. In f ront of it, another

block of mass m2 = 5 kg is moving with speed 3 m/s in the sam e direction. A massless spring of spring

constant k = 1120 N/m is attached on the backside of m2 as shown. Find the maximum compression

of the spring in cm when the block of mass m1 in contect with spring. (Take g = 10 m/s 2)


Magnification (by a lens) of an object at distance 10 cm from it is – 2. Now a second lens is placed exactly

at the same position where first was kept and first lens is removed. The magnification by this lens is – 3.

9.9. Find position of image formed by combination of both in contact. (relative to combination) :

(A)9

60 c m (B)

11

60 c m (C)

13

60 c m (D)

17

60 cm

10.10. What is the focal length of the combination when both lenses are in contact :

(A)17

60 c m (B)

17

5 c m (C)

7

12(D)

9

13 cm

111.1. When both the lenses are kept in contact at the same place, what will be the new magnification :

(A)

5

13!(B)

7

12!(C)

11

6!(D)

7

5!


Topics : Rigid Body Dynamics, Geometrical Optics, Simple Harmonic Motion, Work, Power andTopics : Rigid Body Dynamics, Geometrical Optics, Simple Harmonic Motion, Work, Power and

Energy , Electrostatics, FluidEnergy , Electrostatics, Fluid

PHYSICSPHYSICS






Subjective Questions ('Subjective Questions (' –1' negativ1' negativ e marking) e marking) Q.7 Q.7 to Q.8to Q.8 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[88, , 1100]]


1.1. A circular platform is mounted on a frictionless vertical axle. Its radius R = 2m and its moment of inertia about

the axle is 200 kgm2. it is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk

along the edge at the speed of 1ms –1 relative to the ground. Time taken by the man to complete one

revolution with respect to disc is :

(A) &s (B) s2

3&(C) 2&s (D) s

2

&

2.2. For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of

a material whose refractive index :

(A) lies between 2 a nd 1 (B) lies between 2 and 2

(C) is less than 1 (D) is greater than 2

3.3. The potential energy of a particle executing SHM changes from maximum to minimum in 5 s. Then the time

period of SHM is :

(A) 5 s (B) 10 s (C) 15 s (D) 20 s

4.4. A force N) j4i3(F "*"

acts on a 2 kg movable object that moves fr om an initial position m) j2i3(di !!*"

to a final position m) j4i5(df "*"

in 6 s. The average power delivered by the force during the interval is

equal to :

(A) 8 watt (B)6

50 w att (C) 15 watt (D)

3

50watt.

5.5. Two blocks A (5kg) and B(2kg) attached to the ends of a spring constant 1 120N/m are placed on a

smooth horizontal plane with the spring undeformed. Simultaneously velocities of 3m/s and 10m/s

along the line of the spring in the same direction are imparted to A and B then

k = 1120 N/m

5kg 2kg

(A) when the extension of the spring is maximum the velocities of A and B are zero.

(B) the maximum extension of the spring is 25cm.

(C) the first maximum compression occurs 3& /56 seconds after start.(D) maximum extension and maximum compression occur alternately.

6.6. Two free point charges +q and +4q are placed a distance x apart. A third charge is so placed that all the three

charges are in equilibrium. Then

(A) unknown charge is -4q/9

(B) unknown charge is -9q/4

(C) It should be at (x/3) from smaller charge between them

(D) It should be placed at (2x/3) from smaller charge between them.


7.7. A fixed container of height ' H

' with large cross-sectional area '

A

' is completely filled with water. Two

small orifice of cross-sectional area ' a

' are made, one at the bottom and the other on the vertical side

of the container at a distance H/2 from the top of the container. Find the time taken by the water level

to reach a height of H/2 from the bottom of the container.

8.8. A standing wave pattern of maximum amplitude 2mm is obtained in a string whose shape at t = 0 is represented

in the graph. If the speed of the travelling wave in the string is 5 cm/s then find the component waves.


Consider the situation in figure. The bottom of the pot is a reflecting plane mirror, S is a small f ish and

T is a human eye. Refractive index of water is 4/3.

60 cm

60 cm30 cm

9.9. At what distance from itself will the fish see the image of the eye in upward direction?

(A) 35 cm (B) 45 cm (C) 55 cm (D) 110 cm

10.10. At what distance from itself will the fish see the im age of the eye in downward direction?

(A) 90 cm (B) 110 cm (C) 170 cm (D) 180 cm


Topics : ElectTopics : Elect rostaticsrostatics , Surface Tensio, Surface Tensio n, Frictionn, Friction, Simple Harmon, Simple Harmon ic Motic Mot ino, Thermal ino, Thermal ExpaExpansion, nsion, Work,Work,

Power and EnergyPower and Energy ,, Rigid Body Dynamics Rigid Body Dynamics ,, Geometrical Optics Geometrical Optics

PHYSICSPHYSICS








1.1. A point charge + Q is placed at the centroid of an equilateral triangle. When a second charge + Q is

placed at a vertex of the triangle, the magnitude of the electrostatic force on the central charge is 8 N.

The magnitude of the net force on the central charge when a third charge + Q is placed at another

vertex of the triangle is:

(A) zero (B) 4 N (C) 4 2 N (D) 8 N

2.2. The figure shows a soap film in which a closed elastic thread is lying. The

film inside the thread is pricked. Now the sliding wire is moved out so that

the surface area increases. The radius of the circle formed by elastic thread

will

(A) increase (B) decrease

(C) remain same (D) data insufficient

3.3. A block of mass M rests on a rough horizontal surface. The co-efficient of friction between the block

and the surface is 5. A force F = M g acting at angle # with the vertical side of the bloc k pulls it. In which

of the following cases, the block can be pulled along the surface?

(A) tan # : 5 (B) cot # : 5 (C) cot ( # /2) : 5 (D) none

4.4. A particle is executing SHM between points -X m and X m, as shown in figure-I. T he velocity V(t) of the

particle is partially graphed and shown in figure-I I. Two points A and B corresponding to time t 1 and time

t2 respectively are marked on the V(t) curve.

-Xm

O Xm

Figure-I

+x

A

B

V

t

Figure-II

t1

t2

(A) At time t 1 , it is going towards X m.

(B) At time t 1, its speed is decreasing.(C) At time t 2, its position lies in between –Xm and O.

(D) The phase difference 6; between points A and B must be expressed as 90 ° < 6; < 180 °.

5.5. When the temperature of a copper coin is raised by 80 oC, its diameter increases by 0.2%,

(A) percentage rise in the area of a face is 0.4%

(B) percentage rise in the thickness is 0.4%

(C) percentage rise in the volume is 0.6%

(D) coefficient of linear expansion of copper is 0.25x10-4 / oC.


6.6. Starting at rest, a 5 k g object is acted upon by only one force as indicated in figure. Find the total work

done by the force.

7.7. A 2 kg uniform cylinde r is placed on a plank of mass 4 kg which in turn

rests on a smooth ho rizontal plane. A constant horizontal force of 20 N

is applied on the plank. If no slipping occurs between cylinder andplank obtain the acceleration of the cylinder and the plank.


There is a slab of refractive index n2 placed as shown. Medium on two side are n1 and n3. Width of slab is d2.

An object is placed at distance d1 from surface AB and observer is at distance d3 from surface CD. Given

n1 = air [ref. index = 1] and n2 is glass

[ref. index = 3/2], d1 = 12 cm and d2 = 9cm, d3 = 4cm.

8.8. If object starts to move with speed of 12cm/sec towards the slab then find the speed of image as seen by

observer [given n3 = air = (ref. index = 1)]

(A) 8 cm/sec (B) 180 cm/sec (C) 12 cm/sec (D) none of these

9.9. If n3 = 3

4 then distance of image of object seen by observer is :

(A) 25 cm (B) 30 cm (C) 28 cm (D) none of these

10.10. If n3 = 3

4 and object start to move towards the slab with speed of 12 cm/sec then speed of image is :

(A) 16 cm/sec (B) 9 cm/sec (C) 12 cm/sec (D) none of these


Topics : Electrostatics, Fluid,Topics : Electrostatics, Fluid, Circular MotionCircular Motion,, Work, Power and EnergyWork, Power and Energy ,, Sound Wave, String Wave,Sound Wave, String Wave,

Geometrical OpticsGeometrical Optics

PHYSICSPHYSICS









1.1. Four positive charges (2 <2-1) Q are arranged at corner of a square. Another charge q is placed at the

centre of the square. Resultant force acting on each corner is zero If q is

(A) – 7Q/4 (B) – 4Q/7 (C) -Q (D) None

2.2. A vessel contains oil (density = 0.8 gm/cm3) over mercury (density = 13.6 gm/cm3). A uniform sphere floats

with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in gm/

cm3 is:

(A) 3.3 (B) 6.4 (C) 7.2 (D) 12.8

3.3. A small coin of mass 40 g is placed on the horizontal surface of a rotating

disc. The disc starts from rest and is given a constant angular acceleration == 2 rad/s2. The coefficient of static friction between the coin and the disc is µ

s

= 3/4 and coefficient of kinetic friction is µk = 0.5. The coin is placed at a

distance r = 1 m from the centre of the disc. The magnitude of the resultant

force on the coin exerted by the disc just before it starts slipping on the disc

is : (Take g = 10 m/s2)

(A) 0.2 N (B) 0.3 N (C) 0.4 N (D) 0.5 N

4.4. A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1m. Its speed

at the bottom is 2 m /s. The work done by frictional force is : (g = 10 m/s 2)

(A) 8 J (B) ! 8 J

(C) 4 J (D) ! 4 J

5.5. Which of the following is/ are correct.

(A) (B) (C) (D)


6.6. An electron is placed just in the middle between two long f ixed l ine

charges of charge density + each. The wires are in the xy plane (Do

not consider gravity)

(A) The equilibrium of the electron will be unstable along x-direction

(B) The equilibrium of t he electron will be stable along y-direction

(C) The equilibrium of the electron will be neutral along y-direction

(D) The equilibrium of the electron will be stable along z-direction

7.7. The ratio of the intensities of the mechanical waves propagating in the same medium Y1=10 sin(t – kx) and

Y2= 5 [sin (t – kx) + 3 cos (kx – t)] is


There is an insect inside a cabin eying towards a thick glass plate P1. Insect sees the images of light source

across the glass plate P1 ouside the cabin. Cabin is made of thick glass plates of refractive index =

2

3 and

thickness 3 cm. Insect is eying from the middle of the cabin as shown in figure. (glass plates are partially

reflective and consider only paraxial rays)

8.8. At what distance (from eye of insect) will the eye see first image?

(A) 5 cm (B) 7 cm (C) 9 cm (D) 14 cm

9.9. At what distance (from eye of insect) will the eye see second image?

(A) 11 cm (B) 13 cm (C) 16.5 cm (D) 18 cm

1010.. Number of image seen by insect ?

(A) 2 (B) 4 (C) 8 (D)


Topic : Rigid Body Dynamics, String WTopic : Rigid Body Dynamics, String Wave, ave, Kinetic Throry of Gases, Kinetic Throry of Gases, Thermal Expansion, Simple HarmonicThermal Expansion, Simple HarmonicMotion, ElectrostaticsMotion, Electrostatics

PHYSICSPHYSICS







Comprehension ('Comprehension (' –1' negative marking) Q.7 to Q.91' negative marking) Q.7 to Q.9 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4)Q.10× 4)Q.10 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. A uniform cylinder of mass M lies on a fixed plane inclined at an angle #with horizontal. A light string is tied to the cylinder’s right most point, and

a mass m hangs from the string, as shown. Assume that the coefficient of

friction between the cylinder and the plane is sufficiently large to prevent

slipping. For the cylinder to remain static, the value of mass m is-

(A)#"#

sin1

cosM(B) M

#"#

sin1

sin

(C) M#

#sin –1

cos(D) M

##

sin –1

sin

2.2. At t = 0, a transverse wave pulse travelling in the positive x direction with a speed of 2 m/s in a long

wire is described by the fu nction y =2

x

6, given that x ? 0. Transverse velocity of a particle at x = 2m and

t = 2 seconds is :

(A) 3 m/s (B) – 3 m/s (C) 8 m/s (D) – 8 m/s

3.3. Graph shows a hypothetical speed distribution for a sample of N gas particle (for V > V0; dV

dN = 0,

dV

dN is rate

of change of number of particles with change veloicty)

a

V0

speed V

dNdV

(A) The value of aV 0 is 2N.

(B) The ratio V avg /V0 is equal to 2/3.(C) The ratio V rms /V0 is equal to 1/ 2 .

(D) Three fourth of the total particle has a speed between 0.5 V 0 and V 0.

4.4. A vessel is partly filled with liquid. When the vessel is cooled to a lower temperature, the space in the vessel,

unoccupied by the liquid remains constant. Then the volume of the liquid (VL), volume of the vessel (V

v), the

coefficients of cubical expansion of the material of the vessel (@v) and of the liquid (@

L) are related as

(A) @L>@

v(B) @

L<@

v(C) @

v / @

L=V

v /V

L(D) @

v / @

L=V

L /V

v


5.5. Two particles perform SHM with the same amplitude and same frequency about the same mean position and

along the same line. If the maximum distance between them during the motion is A (A is the amplitude of

either) then the phase difference between their SHM is _ _ _ _ _ _ _.

6.6. Calculate the magnitude of electr ostatic force on a charge placed at a vertex of a triangular pyramid (4

vertices, 4 faces), if 4 equal point charges are placed at all four vertices of pyramid of side ‘a’.


A bar of mass m is held as shown between 4 uniform discs , each of mass m + and radius r = 75 mm. The

bar has been released from rest, knowing that the normal forces exerted on the discs by vertical walls

are sufficient to prevent any slipping. Answer the following questions assuming attachments are massless

and fixed to the bar. (Acceleration due to gravity is g)

7.7. If m = 5 kg and m + = 2 kg, then acceleration of the bar is

(A) g (B) 13g/7 (C) 13g/17 (D)17

g

8.8. If the mass m + of the discs is negligible, then acceleration of the bar is

(A) g (B) 2g/3 (C) 13g/17 (D)17

g

9.9. If the mass m of the bar is negligible, then acceleration of the bar is

(A) g (B) 2g/3 (C) 13g/17 (D)17

g

10.10. Assume only electrostatic interaction forces :

ColumnColumn – ColumnColumn –

(A) Three charges are kept along a (p) The system may be in equilibrium with

straight line proper choice of the value of charges.

(B) Three charges are kept at the vertices of an (q) The system will be in equilibrium for any

equilateral triangle value of the charges.

(C) Three charges are kept at the three vertices (r) The system will not be in equilibrium for

of a square, and a fourth charge is kept at any choice of the value of charges.

the point of intersection of the diagonals.

(D) Three charges are kept at the vertices of an (s) The equilibrium is unstable.

equilateral triangle with the fourth charge

at the centroid.


TopiTopics : Calorimetrcs : Calorimetry & Thermal Exy & Thermal Expansion, Electrostatpansion, Electrostatics, ics, Work, PWork, Power and Enerower and Energygy, Rigid , Rigid Body Body DynamiDynamics,cs,String Wave, Geometrical OpticsString Wave, Geometrical Optics

PHYSICSPHYSICS







Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4× 4 )) ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. Water of mass m2 = 1 kg is contained in a copper calorimeter of mass m

1 = 1 kg. Their common temperature

t = 10°C. Now a piece of ice of mass m3 = 2 kg and temperature is –11°C dropped into the calorimeter.

Neglecting any heat loss, the final temperature of system is. [specific heat of copper = 0.1 Kcal/ kg°C, specific

heat of water = 1 Kcal/kg°C, specific heat of ice = 0.5 Kcal/kg°C, latent heat of fusion of ice = 78.7 Kcal/kg]

(A) 0°C (B) 4°C (C) – 4°C (D) – 2°C

2.2. Two identical spheres of same mass and specific gravity (which is the ratio of density of a substance and

density of water) 2.4 have different charges of Q and – 3Q. They are suspended from two strings of same

length ! fixed to points at the sam e horizontal level, but distant! from each other. When the entire set up is

transferred inside a liquid of specific gravity 0.8, it is observed that the inclination of each string in equilibrium

remains unchanged. Then the dielectric constant of the liquid is

(A) 2 (B) 3 (C) 1.5 (D) None of these

3.3. A block of mass 10 kg is released on a fixed wedge inside a cart which is

moved with constant velocity 10 m/s towards right. Take initial velocity of

block with respect to cart zero. Then work done by normal reaction (with

respect to ground) on block in two seconds will be: (g = 10 m/s2).

(A) zero (B) 960 J

(C) 1200 J (D) none of these

4.4. If the fr equency of a wave is increased by 25 %, then the change in its wavelength will be:

(medium not changed)

(A) 20 % increase (B) 20 % decrease (C) 25 % increase (D) 25 % decrease

5.5. The moment of inertia of a thin sheet of mass M of the given shape

about the specified axis is

(A) (3/2)MR 2 (B) (3/4)MR 2

(C) MR 2 (1 + 1/ <2) (D) MR 2 /(2<2)

6.6. AB wire is vibrating in its fundamental mode. W ireAB is in resonance with resonance tube in which air

column is also vibrating with its fundamental mode.

Sound speed is 400 m/sec and linear mass density

of AB wire is 10 –4 kg/m and g = 10 m/sec 2, value of

mass m = [ A(10 –1)] kg, then find value of A.Neglect

the masses of wires in comparison to block's mass

'm'.



A glass prism with a refracting angle of 600 has a refractive index 1.52 for red and 1.6 for violet light.

A parallel beam of white light is incident on one face at an angle of incidence, which gives minimum deviation

for red light. Find :

[Use:Use: sin (50º) = 0.760; sin (31.6º) = 0.520 ; sin (28.4º) = 0.475; sin (56º) = 0.832 ; & = 22/7]

7.7. The angle of incidence at the prism is :

(A) 30º (B) 40º (C) 50º (D) 60º

8.8. The angular width of the spectrum is :

(A) 6º (B) 4.8º (C) 9.6º (D) 12º

9.9. The length of the spectrum if it is focussed on a screen by a lens of focal length 100 cm is :

(A) cm3

10&(B) m

3

10&(C) cm

3

5&(D) m

3

5&

10.10. The column 3 gives the two point charge system separated by 2a and the column 33 gives the variation of

magnitude of electric field intensity at point on the x-axis. Match the situat ion in Column 3 with the results

in Column 33 and indicate your answer by darkening approp riate bubbles in the 4 × 4 matrix given in the

OMR.

ColumnColumn – ColumnColumn –

(A) + +(0, 0) a

q q

(a, 0)(-a, 0)

xx' (p) Increases as x increases

in the interval 0 7 x < a

(B) + –(0, 0) a

q -q

(a, 0)(-a, 0)

xx' (q) Decreases as x increases

in the interval 0 7 x < a

(C)

+

+

(0, 0)

q

q

(0,+a)

(0, –a)

x

y

(r) Zero at x = 0

(D)

–

+

(0, 0)

–q

q

(0,+a)

(0, –a)

x

y

(s) Decreases as x increases

in the interval a < x < >


Topics : Topics : Rigid Rigid Body DynBody Dyn amics, amics, ElectrostaElectrosta tics, tics, KinemaKinematics, Stics, Sound Wound Wave, Friave, Friction, Fluidction, Fluid

PHYSICSPHYSICS








Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4)Q.10× 4)Q.10 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. A uniform rod of mass M and length L leans against a frictionle ss wall,

with quarter of its length hanging over a corner as shown. Friction at

corner is sufficient to keep the rod at rest. Then the r atio of magnitude

of normal reaction on r od by wall and the magnitude of normal reaction

on rod by corner is

(A)#sin2

1(B)

#sin

2(C)

#cos2

1(D)

#cos

2

2.2. A non – conducting semicircular disc (as shown in figure) has a uniform

surface charge density B. The ratio of electric field to electric potential at

the centre of the disc will be :

(A))ab(

a / bn1

!&

!(B)

&2

(C))ab(

)a / b(n1 2

!&!

(D) )a / b(n2

)ab(

!

!&

3.3. Two particles P and Q start their journey simultaneously from point A. P

moves along a smooth horizontal wire AB. Q moves along a curved smoothtrack. Q has sufficient velocity at A to reach B always remaining in contact

with the curved track. At A, the horizontal component of velocity of Q is

same as the velocity of P along the wire. The plane of motion is vertical. If

t1, t2, are times taken by P and Q respectively to reach B then (Assume

velocity of P is constant) A BPQ

(A) t1 = t2 (B) t1> t2 (C) t1 < t2 (D) none of these

4.4. A receiver and a source of sonic oscillations of frequency 200 Hz are located on the x !

axis. The receiver

is fixed and the source swings harmonically along that axis with a circular frequency ' and an amplitude 50

cm. At what value of ' (in rad/sec) will the frequency band width (fmax

! fmin

) registered by the stationary

receiver be equal to 20 Hz. [ The velocity of sound is equal to 340 m/s ]

(A) 17 (B) 34 (C) 68 (D) 8.5

5.5. Block A is kept on block B as shown in figure. It is known that acceleration of

block A is 2 m/s2 towards right and acceleration of b lock B is 3 m/s2 towards right

under the effect of unknown forces. Direction of friction force acting on A by B

(5AB = 0.3)(A) is necessarily towards right (B) may be towards right

(C) may be towards left (D) may be zero

6.6. An infinitely large nonconducting plane of uniform surface charge

density B has circular aperture of certain radius carved out from it. The

electric field at a point which is at a distance ‘a’ from the centre of the

aperture (perpendicular to the

plane) is022 C

BD Find the radius of aperture :



A tank of base area 4 m2 is initially filled with water up to height 2m. An object of uniform cros s-section 2m2

and height 1m is now suspended by wire into the tank, keeping distance between base of tank and that of

object 1m. Density of the object is 2000kg/m 3. Take atmospheric pressure 1 × 105N/m2 ; g = 10m/s 2.

A=2m2

1m

4m2

1m

7.7. The downwards force exerted by the water on the top surface of the object is :

(A) 2.0 × 105 N (B) 2.1 × 105 N (C) 2.2 × 105 N (D) 2.3 × 105 N

8.8. The tension in the wire is :

(A) 0.1 × 105 N (B) 0.2 × 105 N (C) 0.3 × 105 N (D) 0.4 × 105 N

9.9. The buoyant force on the object is :

(A) 0.1 × 105 N (B) 0.2 × 105 N (C) 0.3 × 105 N (D) 0.4 × 105 N

10.10. In each situation of column-I a mass distribution is given and information regarding x and y-coordinate of

centre of mass is given in column-II. Match the figures in column-I with corresponding information of centre

of mass in column-II.

Column-Column-II Column-Column-IIII

(A) An equilateral triangular wire (p) xcm

> 0

frame is made using three thinuniform rods of mass per unit

lengths 8, 28 and 38 as shown

(B) A square frame is made using (q) ycm

> 0

four thin uniform rods of mass

per unit length lengths 8, 28,

38 and 48 as shown

(C) A circular wire frame is made (r) xcm

< 0

of two uniform semicircular wires

of same radius and of mass per

unit length 8 and 28 as shown

(D) A circular wire frame is made (s) ycm

< 0

of four uniform quarter circular

wires of same radius and

mass per unit length 8, 28, 38and 48 as shown


Topics : GTopics : Geometeometrical Orical Opticptics , Kinetic s , Kinetic Theory Theory of of Gases, Rigid Gases, Rigid Body Body DynamicsDynamics, Str, String Wing Wave, Electrosave, Electrostatics,tatics,Center of MassCenter of Mass

PHYSICSPHYSICS









1.1. An infinitely lo ng rectangular strip is placed on principal axis of a concave mirror as shown in figure.

One end of the strip coincides with centre of curvature as sho wn. The height of rectangular strip is very

small in comparison to focal length of the m irror. Then the shape of image of strip formed by concave

mirror is

F C

(A) Rectangle (B) Trapezium (C) Triangle (D) Square

2.2. N(< 100) molecules of a gas have velocities 1, 2, 3........ N km/s respectively. Then

(A) rms speed and average speed of molecules is same.

(B) ratio of rms speed to average speed is <(2N + 1)(N + 1)/6N

(C) ratio of rms speed to average speed is <(2N + 1)(N + 1)/6

(D) ratio of rms speed to average speed of molecules is)1N(6

)1N2(2 ""

3.3. A horizontal plane supports a fixed ver tical cylinder of radius R and a particle is attach ed to the cylinder

by a horizontal thr ead AB as shown in Fig. A horizontal vel ocity vo is imparted to the particle, normal to

the thread, then during subsequent motion

(A) Angular momentum of particle about O remains constant

(B) Angular mom entum about B rem ains constant

(C) Momentum and kinetic energy both remain constant

(D) Kinetic energy remain constant.

4.4. The equation of a string wave is given by (all quantity expressed in S.I. units) Y = 5 sin10& (t – 0.01x) along

the x-axis. The magnitude of phase difference between the points separated by a distance of 10 m along x-

axis is

(A) & /2 (B) & (C) 2 & (D) & /4.


5.5. Two infinite plane sheets A and B are shown in the figure. The surface charge densities on A and B are (2/ &)

× 10-9 C/m2 and ( –1/&) × 10-9 C/m2 respectively. C, D, E are three points where electric fields (in N/C) are EC,

ED and E

E respectively.

(A) EC = 18, towards right (B) E

D = 54, towards right

(C) ED = 18, towards right (D) E

E = 18, towards right

6.6. A sinusoidal wave propagates along a string. In figure (a) and (b) ' y

' represents displacement of particle

from the mean position. ' x ' and ' t ' have usual meanings. Find:

y(in mm) at t = 0

x(in m)1 4 7

(a)(a) wavelength, frequency and speed of the wave.

(b)(b) maximum velocity and maximum acceleration of the particles

(c)(c) the magnitude of slope of the string at x = 2 at t = 4 sec.

7.7. A point charge Q is located at centre of a fixed thin ring of radius R with uniformly distribute d

charge-Q. The magnitude of the electric field strength at the point lying on the axis of the ring at

a distance x from the centre is (x >> R) _______________.


Figure shows an irregular wedge of mass m placed on a smooth horizontal surface. Horizontal part BC is

rough.The other part of the wedge is smooth.

8.8. What minimum velocity should be imparted to a small block of same mass m so that it may reach point B :

(A) gH2 (B) gH2 (C) )hH(g2 ! (D) gh

9.9. The magnitude of velocity of wedge when the block comes to rest (w.r.t. wedge) on part BC is :

(A) gH (B) hH(g ! (C) gH2 (D) none of these

10.10. If the coefficient of friction between the block and wedge is 5, and the block comes to rest with respect to

wedge at a point D on the rough surface then BD will be

(A) 5H

(B) 5! hH

(C) 5h

(D) none of these


Topics : Topics : Calorimetry & Thermal Calorimetry & Thermal Expansion, Kinetic Theory of Expansion, Kinetic Theory of Gases, Center of Mass, Geometrical Optics,Gases, Center of Mass, Geometrical Optics,Circular Motion.Circular Motion.

PHYSICSPHYSICS







Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4)Q.10× 4)Q.10 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. A cubical block of copper of side 10 cm is floating in a vessel containing mercury . Water is poured into

the vessel so that the copper block just gets submerged. The height of water column is

($Hg

= 13.6 g/cc , $Cu

= 7.3 g/cc, $water

=1 gm/cc)

(A) 1.25 cm (B) 2.5 cm (C) 5 cm (D) 7.5 cm

2.2. Maxwell ’s velocity distribution curve is given for the same quantity two

different temperatures. For the given curves.

(A) T 1 > T2 (B) T 1 < T2

(C) T1 7 T2 (D) T 1 = T2

3.3. A particle of mass m is given initial horizontal velocity of magnitude u as shown in the figure. It transfers to

the fixed inclined plane without a jump, that is, its trajectory changes sharply from the horizontal line to

the inclined line. All the surfaces are smooth and 90E : # F 0E. Then the height to which the particle shall

rise on the inclined plane (assume the length of the inclined plane to be very large)

u

horizontal surface

(A) increases with increase in # (B) decreases with increase in #(C) is independent of # (D) data insufficient

4.4. The angle of deviation (G) vs angle of incidence (i) is plotted for a prism.

Pick up the correct statements.

(A) The angle of prism is 60 °

(B) The refractive index of the prism is n = 3

(C) For deviation to be 65° the angle of incidence i1 = 55°

(D) The curve of 'G' vs 'i' is parabolic

5.5. A particle is describing circular motion in a horizontal plane in contact with the smooth inside surf ace

of a fixed right circular cone with its axis vertical and vertex down. The height of the plane of motion

above the vertex is h and the semivertical angle of the cone is =. The period of revolution of the particle:

(A) increases as h increases (B) decreases as h increases

(C) increases as = i ncreases (D) decreases as = increases


6.6. Two identical straight wires are stretched so as to produce 6 beats/ sec. when vibrating simu ltaneously.

On changing the tension slightly in one of them , the beat frequency remains unchanged. Denoting by

T1, T

2, the higher & the lower initial tensions in the strings, then it could be said that while mak ing the

above changes in tension:

(A) T2 w as decreased (B) T

2 w as increased (C) T

1 w as increased (D) T

1 was decreased


Two point charges are placed at point a and b. The field strength t o the

right of the charge Qb on the line that passes through the two charges

varies according to a law that is represented graphically in the figure.

The electric field is tak en positive if its direction is towards right and

negative if its direction is towards left.

7.7. Choose the correct statement regarding the signs of the charges.

(A) Charge at point a is positive and charge at point b is negative.

(B) Charge at point a is negative and charge at point b is positive.

(C) Both charges are positive

(D) Both charges are negative

8.8. Ratio of magnitudes of chargesb

a

Q

Q will be equal to

(A) --001

2 "

1x1

!(B)

2

x1 -

/01

2 " !

(C)

2

x1 -

/01

2 "

!(D)

4

x1 -

/01

2 " !

9.9. The distance x2 from point b where the field is maximum, will be

(A)

1x

x 32

1

1 !-- .

/001

2 "!

!(B)

1x

x 31

1

1 !-- .

/001

2 "!

!(C)

1x

x2 32

1

1 !-- .

/001

2 "!

!(D)

1x

x2 31

1

1 !-- .

/001

2 "!

!

10.10. A uniform disc rolls without slipping on a rough horizontal surf ace with uniform angular velocity. Point O

is the centre of disc and P is a point on disc as shown. In each situation of column I a statement is

given and the corresponding results are given in column- II. Match the statements in column- I with the

results in column- II.

Column ColumnColumn Column

(A) The velocity of point P on disc (p) Changes in magnitude with time

(B) The acceleration of point P on disc (q) Is a lways directed from that point (the

point on disc given in column- I)

towards centre of disc.

(C) The tangential acceleration of point P on di sc (r) is always zero

(D) The acceleration of po int on disc which i s in (s) is non-zero and remains constant

contact with rough horizontal surface in magnitude


Topics : STopics : String Wave, tring Wave, Circular Motion, ProjeCircular Motion, Projectictile Mle Motiotion, on, Center of Mass, Rectilinear Center of Mass, Rectilinear Motion, Sound Wave,Motion, Sound Wave,Geometrical Optics, Rigid Body DynamicsGeometrical Optics, Rigid Body Dynamics

PHYSICSPHYSICS






1.1. A diatomic ideal gas undergoes a thermodynamic change according to the P –V diagram shown in the

figure. The total heat given to the gas is nearly (use !n2 = 0.7) :

C

2P0

P0

B

A

P

V2V0V0

Isothermal

(A) 2.5 P0V

0(B) 1.4 P

0V

0

(C) 3.9 P0V

0(D) 1.1 P

0V

0

2.2. Figure shows three circular arcs, each of radius R and total charge as indicated.

The net elecric potential at the centre of curvature is :

•

R+3Q

30°

45°

+Q

–2Q(A) R2

Q

0&H (B) R4

Q

0&H

(C) R

Q2

0&H (D) R

Q

0&H

3.3. A heavy body of mass 25 kg is to be dragged along a horizontal plane ( 5 = 1/ 3). The least force

required is

(A) 25 kg f (B) 2.5 kg f (C) 12.5 kg f (D) 25/ 3 kg f

4.4. A particle is acted upon by a force whose component's variations with time are shown in diagrams. Then the

magnitude of change in momentum of the particle in 0.1 sec will be

t=0.1 sec

30 N

Fx

0

t=0.1 sec

80 N

Fy

0

t=0.1 sec

– 50 N

Fz

0

(A) 2 kgsec

m(B) 10 kg

sec

m(C) 12 kg

sec

m(D) 25 kg

sec

m


5.5. A small block of mass m is pushed towards a movable wedge of mass Im and height h with initial

velocity u. All surfaces are smooth. T he minimum value of u for which the block will reach

the top of the wedge

hu

m

(A) gh2 (B) gh2I (C) -- .

/001

2

I"

11gh2 (D) --

.

/001

2

I!

11gh2

6.6. A bird flies for 4 seconds with a velocity of |t – 2| m/sec. in a straight line, where t = time in seconds. It

covers a distance of

(A) 4 m (B) 6 m (C) 8m (D) none of these

7.7. A violin string oscillating in its fundamental mode, generates a sound wave with wavelength 8. To

generate a sound wave with wavelength 8 /2 by the string, stil l oscillat ing in its fundamental mode,

tension must be changed by the multiple :

(A) 2 (B) 1/2 (C) 4 (D) 1/4

8.8. In displacement method, the distance between object and screen is 96 cm. The ratio of length of two images

formed by a convex lens placed between them is 4.84.

(A) Ratio of the length of object to the length of shorter image is 11/5.

(B) Distance between the two positions of the lens is 36 cm.

(C) Focal length of the lens is 22.5 cm.

(D) Distance of the lens from the shorter image is 30 cm.

9.9. A source emit sound waves of frequency 1000 Hz. The source moves to the right with a speed of 32 m/s

relative to ground. On the right a reflecting surface moves towards left with a speed of 64 m/s relative to

ground. The speed of sound in air is 332 m/s :

(A) wavelength of sound infront of source is 0.3 m

(B) number of waves arriving per second which meets the reflected surface is 1320

(C) speed of reflected wave is 268 m/s

(D) wavelength of reflected waves is nearly 0.2 m

10.10. A wheel (to be considered as a ring) of mass m and radius R rolls without sliding on a horizontal

surface with constant velocity v . It encounters a step of height R/2 at which it ascends without sliding.

(A) the angular velocity of the ring just after it com es in contact with the step is 3v/4R

(B) the normal reaction due to the step on the wheel just after the impact ismg

2 +

9

16

2mv

R

(C) the normal reaction due to the step on the wheel increases as the wheel ascends

(D) the friction will be absent during the ascent.


TopTopics : Strics : String Waing Wave, Circve, Circular Motiular Motion, Projectile on, Projectile MotioMotio n, Geometrican, Geometrical Optics, Electrostatl Optics, Electrostat ics, ics, Center Center ofof

MassMass

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.51' negative marking) Q.1 to Q.5 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1155, , 1155]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.61' negative marking) Q.6 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]Comprehension ('Comprehension (' –1' negative marking) Q.7 to Q.91' negative marking) Q.7 to Q.9 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.10× 4) Q.10 ((8 m8 m aarrkkss, 1, 1 0 m0 m inin ..)) [8[8, , 1100]]

1.1. A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode. Amplitude at the

centre of the string is 4 m m. Minimum distance between the two points having amplitude 2 mm is:

(A) 1 m (B) 75 cm (C) 60 cm (D) 50 cm

2.2. A particle is projected horizontaly with speed 10m/s from a certain point above ground. Find the tangential

acceleration of particle at t = 2 sec. (Take g = 10 m/s2).

(A)5

10(B)

5

25(C) 4 5 (D) 10 5

3.3. A ball is thrown eastward across level ground. A w ind blows horizontally to the east, and assume that

the effect of wind is to provide a co nstant force to the eas t, equal in magnitude to the weight of the ball.

The angle # (with respect to horizontal) at whi ch the ball should be projected so that it travels maximum

horizontal distance is

(A) 45 ° (B) 37 ° (C) 53° (D) 67.5 °

4.4. Two equilate ral glass prisms of refractive index 2 are placed a s shown in figu re. A ray is inciden t on

side AB of left prism as shown in figure. This ray further suffers refraction at sides AC, PQ and PR in

succession. The prisms are adjusted such that for each refraction the deviation is clockwise. Then the

angle between sides A C and PQ of two prisms for net minimum deviation of incident ray is

(A) 30 ° (B) 60°

(C) 90° (D) 120°

5.5. Figure given below shows uniformly positively charged, thin rod of length L and four points A, B, C and D at the

same distance d from the rod, with position as marked. If VA, V

B, V

C and V

D are their respective potentials then:

A B C

dL4

L2

L

d

D

(A) VB > V

A > V

C > V

D(B) V

B > V

A > V

C = V

D

(C) VA = V

B > V

C = V

D(D) V

D > V

B > V

A > V

C


6.6. A rod AB is moving on a fixed circle of radius R with constant velocity ‘v’ as shown in figure. P is the point of

intersection of the rod and the circle. At an instant the rod is at a distance x =5

R3 from centre of the circle.

The velocity of the rod is perpendicular to the rod and the rod is always parallel to the diameter CD.

(a) Find the speed of point of intersection P.

(b) Find the angular speed of point of intersection P with respect to centre of the circle.


The nuclear charge (Ze) is non –uniformly distributed within a nucleus of radius R. The charge density $(r)

[charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown

in figure. The electric field is only along the radial direction. Figure [JEE-2008 ; 12/163][JEE-2008 ; 12/163]

7.7. The electric field at r = R is :

(A) independent of a (B) directly proportional to a

(C) directly proportional to a

2

(D) inversely proportional to a8.8. For a = 0, the value d (maximum value of $ as shown in the figure) is :

(A) 3

2

R4

Ze3

&(B) 3R

Ze3

&(C) 3R3

Ze4

&(D) 3R3

Ze

&

9.9. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies :

(A) a = 0 (B) a =2

R(C) a = R (D) a =

3

R2

10.10. Two blocks A and B of mass m and 2m respectively are connected by a massless spring of spring

constant K. This system lies over a smooth horizontal surface. At t = 0 the block A has velocity u towards

right as shown while the speed of block B is zero, and the length of spr ing is equal to its natural length at

that instant. In each situation of column I, certain statements are given and corresponding results are

given in column II. Match the statements in column 3 corresponding results in column 33 and indicate your

answer by darkening appropriate bubbles in the 4 × 4 matrix given in the OMR.

ColumnColumn II Column Column IIII

(A) The velocity of block A (p) can never be zero

(B) The velocity of block B (q) may be zero at certain instants of time

(C) Th e kine tic energy of sys tem of tw o blocks (r) is minimum at maximum compression of spring

(D) The potential energy of spring (s) is maximum at maximum extension of spring


Topics : Projectile Motion, Sound Wave, Relative Motion, Center of Mass, Geometrical Optics, SimpleTopics : Projectile Motion, Sound Wave, Relative Motion, Center of Mass, Geometrical Optics, Simple

Harmonic Motion, Circular Motion, FluidHarmonic Motion, Circular Motion, Fluid

PHYSICSPHYSICS







Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.10× 4) Q.10 ((8 8 mmaarrkks s ,,110 0 mmiinn..)) [[88, , 1100]]

1.1. Two stones are projected simultaneously from a tower at different angles of projection with same speed ‘u’.

The distance between two stones is increasing at constant rate ‘u’. Then the angle between the initial

velocity vectors of the two stones is :

(A) 30° (B) 60° (C) 45° (D) 90°

2.2. A curve is plotted to represent the dependence of the ratio of the received frequency J to the frequency J0

emitted by the source on the ratio of the speed of observer Vob

to the speed of sound Vsound

in a situation in

which an observer is moving towards a stationary sound source. The curve is best represented by :

(A) (B) 1

2

0.50

1V /Vob sound

f/f0

(C) (D)

3.3. Car A and car B m ove on a straight road and their velocity versus time graphs are as shown in f igure.

Comparing the motion of car A in between t = 0 to t = 8 sec. and m otion of car B in between t = 0 to t

= 7 sec., pick up the correct statement.

v (m/s)

10m/s

t=2s t=8st(s)

v (m/s)

10m/s

t=3s t=7st(s)

Car A Car B

(A) Distance travelled by car A is less than distance travelled by car B.

(B) Distance travelled by car A is greater than distance travelled by car B.

(C) Average speed of both cars ar e equal.

(D) Average speed of car A is less than average speed of car B.


4.4. In the figure shown a hole of radius 2 cm is m ade in a semicircular disc of radius 6 & cm at a distance

8 cm from the centre C of the disc. The distance of the centre of mass of this system from point C is:

(A) 4 cm (B) 8 cm (C) 6 cm (D) 12 cm

5.5. A parallel glass slab of refractive index 3 is placed in contact with an equilateral prism of refr active

index2

. A ray is incident on left surface of slab as shown. The slab and prism combination is

surrounded by air. The magnitude of minimu m possible deviation of this ray by slab-prism com bination

is

(A) 30 ° (B) 45 ° (C) 60° (D) 60 ° – sin –1

3

2

6.6. The amplitude of a particle due to superposition of following S.H.Ms. Along the same line is

X1 = 2 sin 50 & t ; X2 = 10 sin (50 & t + 37 º)X3 = ! 4 sin 50 &

t ; X

4 = ! 12 cos 50 & t

(A) 4 2 (B) 4 (C) 6 2 (D) none of these

7.7. A bob of mass 2 kg is suspended from point O of a cone with an inextensible string of length 3 m. It is

moving in horizontal circle over the surface of cone as shown in the figure. Then : (g = 10 m/s2)

(A) bob looses contact with cone if 5v F m/s (B) normal force on bob is 19 N when v = 2 m/s

(C) tension in string is3

38N when v = 2 m/s (D) normal force on bob is

3

17N when v = 2m/s


8.8. Inside water - .

/01

2 *5

3

4 there is an air bubble of radius 4 cm as shown an observer O is looking into the

diametrical ax is AB of bubble. Find the distance in cm of a point objec t from p oint A on the axis in water

which appears to be at point A as seen by observer.

O

AirBubble

A B

9.9. A wooden cube (density 0.5 gm/cc) of side 10 cm is floating in water kept in a cylindrical beakerof base area 1500 cm 2. When a m ass m is kept on the wooden block the level of water rises in the

beaker by 2mm. Find the mass m.

1100.. MMaattcch h tthhe e ccoolluummnn::

In all cases in column –I, the blocks are placed on the smooth horizontal surface.

ColumnColumn –II ColumnColumn –IIII

(A) When spring is relaxed. (p) Centre of mass of the complete system shown

the initial velocities given will not move horizontally

to the blocks are as shown

(friction is absent)

(B) A constant force is applied on 2 kg block. (q) Centre of mass of the complete system shown

Springs are initially relaxed and friction is absent will move horizontally

(C) There is no friction between plank and ground (r) Mechanical energy of the system will be

and initially system is at rest. Man starts moving conserved

on a large plank with constant velocity.

(D) Two trolleys are resting on a smooth horizontal (s) Mechanical energy of the system will increase

surface and a man standing on one of the trolleys

jumps to the other with relative velocity of 4 m/s

(t) Linear momentum of the complete system

will always remain constant


Topics : Electrostatics, Circular Motion, Geometrical Optics , Relative Motion, Work, Power and Energy,Topics : Electrostatics, Circular Motion, Geometrical Optics , Relative Motion, Work, Power and Energy,

FluidFluid

PHYSICSPHYSICS








1.1. At distance ' r ' from a point charge, the ratio 2V

U(where ' U ' is energy density and ' V ' is potential) is

best represented by :

(A) (B) (C) (D)

2.2. The figure shows several equipotential lines. Comparing between points A and B, pick up the best possible

statement

(A) the electric field has a greater magnitude at point A and is directed to left.

(B) the electric field has a greater magnitude at point A and is directed to right.

(C) the electric field has a greater magnitude at point B and is directed to left.

(D) the electric field has a greater magnitude at point B and is directed to right.

3.3. A small block slides with velocity 0.5 gr on the horizontal

frictionless surface as shown in the Figure. The block leaves

the surface at point C. The angle # in the Figure is:

(A) cos !1 ( 4/9) (B) cos!1(3/4)

(C) cos !1(1/2) (D) none of the above

4.4. In the figure the variation of com ponents of acceleration of a particle of mass 1 kg is shown w.r .t. time.

The initial velocity of the particle is ) j4i3(u "!*"

m/s. The total work don e by the resultant force on the

particle in time interval from t = 0 to t = 4 seconds is :

(A) 22.5 J (B) 10 J

(C) 0 (D) None of these


5.5. A ray of light falls on a transparent sphere as shown in figure. If the final ray emerges from the sphere parallel

to the horizontal diameter, then the refractive index of the sphere is (consider that sphere is kept in air) :

(A) 2 (B) 3

(C)2

3(D) 2

6.6. A boat moves relative to river with a velocity which is n times the river flow velocity.(A) If n < 1, boat cannot cross the river

(B) If n = 1, boat cannot cross the river without drifting

(C) If n > 1, boat can cross the river along shortest path

(D) Boat can cross the river whatever is the value of n (excluding zero)

7.7. A force F = 20 N is applied to a block (at rest) as shown in figure. After the block has moved a distance

of 8 m to the right, the direction of horizontal component of the force F is reversed. Find the velocity

with which block arrives at its starting point.


A glass tube has three different cross sectional areas with the values indicated in the figure. A piston at the

left end of the tube exerts pressure so that the mercury within the tube flows from the right end with a speed

of 8.0 m/s. Three points within the tube are labeled A, B and C. The atmospheric pressure is 1.01× 105 N/m2;

and the density of mercury is 1.36 × 104 kg/m3. (use g = 10 m/s2)

A B

Vacuum

h

6cm2

C

12cm2

Piston

5.6 cm2

8.8. At what speed is mercury flowing through the point A ?

(A) 2.0 m/s (B) 4.0 m/s (C) 8.0 m/s (D) 12 m/s

9.9. The pressure at point A is equal to:

(A) 2.02 × 105 P a (B) 2.25 × 105 P a (C) 3.26 × 105 P a (D) 4.27 × 105 Pa.

10.10. The height h of mercury in the manometer is

(A) 136 mm (B) 169 mm (C) 272 mm (D) 366 mm


Topics : Topics : Kinetic Theory of Kinetic Theory of Gases, Surface Gases, Surface Tension, Rigid Tension, Rigid Body Dynamics, Body Dynamics, Center of Mass, ElectCenter of Mass, Electrosrostatitatics,cs,Sound Wave, Work, Power and EnergySound Wave, Work, Power and Energy

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.61' negative marking) Q.1 to Q.6 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1188, , 1188]]Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.71' negative marking) Q.7 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44, , 44]]


1.1. V –T diagram for a process of a given mass of ideal gas is as shown in the figure.

During the process pressure of gas.

(A) first increases then decreases

(B) continuously decreases

(C) continuously increases

(D) first decreases then increases.

2.2. A long capillary tu be of mass ' &' gm, radius 2mm and negligi ble thickness, is partially immersed in a

liquid of surface tension 0.1 N/m. Take angle of contact zero and neglect buoyant force of liquid. The

force required to hold the tube vertically , will be -- (g = 10 m/s 2)

(A) 10.4 & mN (B) 10.8 & mN

(C) 0.8 & mN (D) 4.8 & mN

3.3. A small solid sphere of mass m is released from a point A at a height h above the bottom of a rough

track as shown in the figure. If the sphere rolls down the track without slipping, its rotational kinetic

energy when it comes to the bottom of tra ck is

(A) mgh (B)7

10mgh (C)

7

5 m gh (D)

7

2 mgh

4.4. In the figure shown a particle P strikes the inclined smooth plane horizontally and rebounds vertically. If the

angle # is 60º, then the coefficient of restitution is:

(A)1

3(B)

1

3(C)

1

2(D) 1


5.5. A point charge ' q

' is placed at the corner of an equilateral prism. Then the electric flux through the

surface of the prism is:

(A)08

q

H(B)

06

q

H(C)

012

q

H(D)

024

q

H

6.6. There is a set of four tuning forks, one with the lowest frequency vibratin g at 550 Hz. By using any two

tuning forks at a time, the following beat frequencies are heard: 1, 2, 3, 5, 7, 8. T he possible frequencies

of the other three forks are:

(A) 552, 553, 560 (B) 557, 558, 560 (C) 552, 553, 558 (D) 551, 553, 558

7.7. Which of the following statements is/are true

(A) work done by kinetic friction on an ob ject may be posi tive.

(B) A rigid body rolls up an inclined plane without sliding. The friction force on it will be up the incline.

(only contact force and gravitational force is acting)

(C) A rigid body rolls down an inclined plane without sliding. The friction force on it will be up the incline.


(D) A rigid body is release from rest and having no angular velocity from the top of a rough inclined plane. It

moves down the plane with slipping. The friction force on it will be up the incline.


A stone is projected from level ground with speed u and at an angle # with horizontal. Some how the

acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the

maxim um height and remains sam e thereafter. Assume directio n of acceleration due to gravity always

vertically downwards.

8.8. The total time of flight of particle is :

(A) gsinu

23 # (B) g

sinu # -- . /00

1 2 "

211 (C) g

sinu2 # (D) -- . /00

1 2 "#

212

gsinu

9.9. The horizontal range of particle is

(A)g

2sinu

4

3 2 #(B) --

.

/001

2 "

#

2

11

g2

2sinu2

(C)g

u2

sin2# (D) -- .

/001

2 "

#

2

12

g2

2sinu2

10.10. The angle ; which the velocity vector of stone makes with horizontal just before hitting the ground is

given by:

(A) tan ; = 2 tan # (B) tan ; = 2 cot # (C) tan ; = 2 tan # (D) tan ; = 2 cot #


TopicTopics : s : Kinetic Kinetic Theory Theory of of Gases,Gases, GeometrGeometrical Opticsical Optics, El, Electrostatectrostat ics, ics, Center of Mass, Relative Motion,Center of Mass, Relative Motion,Rigid Body Dynamics,Rigid Body Dynamics, String WaveString Wave

PHYSICSPHYSICS








1' negative marking) Q.7 to Q.91' negative marking) Q.7 to Q.9 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.10× 4) Q.10 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. At pressure P and absolute temperature T a mass M of an ideal gas fills a closed container of volume V. An

additional mass 2M of the same gas is added into the container and the volume is then reduced to3

V and the

temperature to3

T. The pressure of the gas will now be:

(A)3

P(B) P (C) 3 P (D) 9 P

2.2. A ray hits the y-axis making an angle # with y-axis as shown in the figure.

The variation of refractive index with x-coordinate is 5 = - .

/01

2 5

d

x –10 for 0 7

x 7 d -- . /00

1 2

50

1 –1 and 5 = 50 for x < 0, where d is a positive constant. The

maximum x-coordinate of the path traced by the ray is

(A) d(1 – sin #) (B) d (1 – cos #) (C) d sin # (D) d cos #

3.3. A sphere of radius R contains a total charge +Q which is uniformly distributed

throughout its volume. At a distance 2R from the centre of sphere, a particle

having charge +q is fixed. P is a point on surface of sphere and lying on

line joining the centre of sphere and point charge. For what value of q will

the electric field at P be zero.

(A)2

Q(B) Q (C)

2

3Q (D) 2Q

4.4. A particle ‘ A’ of mass m collides head on with another stationary particle ‘B’ of the same mass ‘m’.The

kinetic energy lost by the colliding particle 'A' will be maximum if the coefficient of the restituition is

(A) 1 (B) 0 (C) 0.5 (D) none

5.5. An open elevator is ascending with zero acceleration and speed 10 m/s. A ball is thrown vertically up by a

boy when he is at a height 10 m from the ground, the velocity of projection is 30m/s with respect to elevator.

Choose correct option, assuming height of the boy very small : (g = 10 m/s2)

(A) Maximum height attained by the ball from ground is 90 m.

(B) Maximum height attained by the ball with respect to lift from the point of projection is 45 m.

(C) Time taken by the ball to meet the elevator again is 6 sec

(D) The speed of the ball when it comes back to the boy is 20 m/s with respect to ground.


6.6. A circular ring of mass m and radius R rests flat on a frictionless surface. A bullet also of mass m and

moving with a velocity v, strikes the ring and gets embedded in it. The thickness of the ring is much smaller

than R. Find the angular velocity with which the system rotates after the bullet strikes the ring.


One end of massless inextensible string of length ! is fixed and other end is tied to a small ball of mass m.

The ball is performing a circular motion in vertical plane. At the lowest position, speed of ball is !g20 .

Neglect any other forces on the ball except tension and gravitational force. Acceleration due to gravity is g.

7.7. Motion of ball is in nature of

(A) circular motion with constant speed

(B) circular motion with variable speed

(C) circular motion with constant angular acceleration about centre of the circle.

(D) none of these

8.8. At the highest position of ball, tangential acceleration of ball is -

(A) 0 (B) g (C) 5 g (D) 16 g

9.9. During circular motion, minimum value of tension in the string -

(A) zero (B) mg (C) 10 mg (D) 15 mg

10.10. In each of the four situations of column - I , a stretched string or an organ pipe is given along with the

required data . In case of string s the tension in string is T = 102.4 N and the mass per unit length of

string is 1 g/m. Speed of sound in air is 320 m/s . Neglect end corrections. The frequencies of resonance

are given in column - II. Match each situation in column- I with the possible resonance frequencies

given in Column - II.


(A) String fixed at both endsfixed fixed

0.5m (p) 320 Hz

(B) String fixed at one end and (q) 480 Hz

free at other endfixed end free end

0.5m

(C) Open organ pipe 0.5m (r) 640 Hz

(D) Closed organ pipe 0.5m (s) 800 Hz


Topics Topics : Geometr: Geometrical Oical Optics, Sound ptics, Sound wave, Kinetic wave, Kinetic Theory Theory of of Gases, Gases, RectRectilinilinear Moear Motiontion, Proje, Projectilctile Mote Motion,ion,Electrostatics,Rigid Body Dynamics, Work, Power and EnergyElectrostatics,Rigid Body Dynamics, Work, Power and Energy

PHYSICSPHYSICS







ComprehenComprehension sion ('(' –

1' negative marking) Q.7 to Q.91' negative marking) Q.7 to Q.9 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Match Match the the FollowinFollowing g (no (no negative negative marking) marking) (2(2 × 4)Q.10× 4)Q.10 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. Two point objects are placed on principal axis of a thin converging lens. One is 20 cm from the lens and other

is on the other side of lens at a distance of 40 cm from the lens. The images of both objects coincide. The

magnitude of focal length of lens is

(A)3

80cm (B)

3

40 c m (C) 40 cm (D)

3

20 cm

2.2. A stationary source 's' is producing sound of frequency ' f ' and an observer 'o' is at rest at some distance from

source. A reflector is moving with constant speed ' v ' along perpendicular bisector of line joining source and

observer. The variation of beat frequency registered by observer will be: [ Reflector is moving

from ! > to + > along perpendicular bisector ]

(A) (B) (C) (D)

3.3. During an experiment, an ideal gas is found to obey a condition$

2P= constant [$ = density of the gas]. The

gas is initially at temperature T, pressure P and density$. The gas expands such that density changes to2

$

(A) The pressure of the gas changes to 2 P..

(B) The temperature of the gas changes to 2 T..

(C) The graph of the above process on the P-T diagram is parabola.(D) The graph of the above process on the P-T diagram is hyperbola.

4.4. Consider two cars moving perpendicular to each other as shown. Initially distance between them is 100

m. Velocity of A is 310 m/s and velocity of B is 10 m/s . Then:

(A) magnitude of velocity of A w.r.t. B is 20 m/s

(B) minimum distance between them is 50 m

(C) minimum distance betw een them is 350 m


(D) at t = 2 sec. they will be nearest to each other

5.5. Two graphs of the same projectile motion (in the xy plane) projected from srcin are shown. X axis is along

horizontal direction & Y axis is vertically upwards. Take g = 10 m/s2.

Find (i) Y component of initial velocity and (ii) X component of initial velocity

6.6. Electric dipole of moment P"

= p i is kept at a point (x,y) in an electric field E"

= 4 xy2 i + 4 x2y j . Find the

magnitude of force acting on the dipole.


A horizontal uniform rod of mass 'm' has its left end hinged to the fixed

incline plane, while its right end rests on the top of a uniform cylinder of

mass 'm' which in turn is at rest on the fixed inclined plane as shown. The

coefficient of friction between the cylinder and rod, and between the cylinder

and inclined plane, is sufficient to keep the cylinder at rest.

7.7. The magnitude of normal reaction exerted by the rod on the cylinder is

(A)4

mg(B)

3

mg(C)

2

mg(D)

3

mg2

8.8. The ratio of magnitude of frictional force on the cylinder due to the rod and the magnitude of frictional force on

the cylinder due to the inclined plane is:

(A) 1 : 1 (B) 3:2 (C) 2 : 1 (D) 1:2

9.9. The magnitude of normal reaction exerted by the inclined plane on the cylinder is:

(A) mg (B)2

mg3(C) 2mg (D)

4

mg5

10.10. Match the statements in Column 3 with the results in Column 33 and indicate your answer by darkening

appropriate bubbles in the 4 × 4 matrix given in the OMR.

ColumnColumn – I I ColumnColumn – II II

(A) Work done by ideal gas during free expansion (p) zero

(B) A wedge block system is as shown in the fig. (q) non zero

The wedge lying on horizontal surface is accelerated to

right by a horizontal force F. All surfaces are smooth.

Work done by normal reaction exerted by wedge on

block in any time interval is

(C) Two identical conducting spheres of radius 'a' are separated (r) negative

by a distance 'b' (b>>a). Both spheres carry equal and

opposite charge. Net electrostatic potential energy of

system of both spheres is(D) A uniform cylinder lies over a rough horizontal platform. The (s) positive

platform is accelerated horizontally as shown with acceleration

a. The cylinder does not slip over the platform.The work done

by the force of friction on the cylinder w.r.t ground in any time interval is


Topics : NewtonTopics : Newton ’s Law of Motions Law of Motion, Center of Mass, Geo, Center of Mass, Geo metrimetrical Opticcal Optics, Surfacs, Surfac e Tensioe Tensio n, Elen, Ele ctrostatctrostat icsics

PHYSICSPHYSICS







1.1. A cylinder rests in a supporting carriage as shown. The side AB of carriage makes an angle 30

o

with thehorizontal and side BC is vertical. The carriage lies on a fixed horizontal surface and is being pulled towards

left with an horizontal acceleration 'a' . The magnitude of normal reactions exerted by sides AB and BC of

carriage on the cylinder be NAB

and NBC

respectively. Neglect friction everywhere. Then as the magnitude of

acceleration 'a ' of the carriage is increased, pick up the correct statement:

(A) NAB

increases and NBC

decreases. (B) Both NAB

and NBC

increase.

(C) NAB

remains constant and NBC

increases. (D) NAB

increases and NBC

remains constant.

2.2. On a smooth carom board, a coin moving in negative y !direction with a speed of 3 m/s is being hit at

the point (4, 6) by a striker moving along negative x-axis. The line joining centres of the coin and the

striker just before the collision is parallel to x-axis. After collision the coin goes into the hole located at

the srcin. Masses of the striker and the coin are equal. Considering the collision to be elastic, the

initial and final speeds of the striker in m/s will be:

(A) (1.2, 0) (B) (2, 0)

(C) (3, 0) (D) none of these

3.3. A prism of angle A and refractive index 2 is surrounded by

medium of refractive index 3 . A ray is incid ent on side PQ at

an angle of incidence i (0 < i < 90 °) as shown. The refracted

ray is then incident on side PR of prism. The minimum angle A

of prism f or which ray incident on side PQ does not emerge

out of prism from side PR (for any value of i) is

(A) 30 ° (B) 45°

(C) 60° (D) 120°


4.4. Assuming the xylem tissues through which water rises from root to the branches in a tree to be of

uniform cross-section find the maximum radius of xylum tube in a 10 m high coconut tree so that water

can rise to the top. (surface tens ion of water = 0.1N/m, Angle of contact of water wi th xylem tube= 60 °)

(A) 1 cm (B) 1 mm (C) 10 µm (D) 1 µm

5.5. A particle is attached to an end of a rigid rod. The other end of the rod is hinged and the rod rotates

always rem aining horizontal without in contact with any thing else. It ’s angular speed is increasing at

constant rate. The mass of the particle is 'm'. The force exerted by the rod on the particle is F"

, then :

(A) F : mg

(B) F is constant

(C) The angle between F"

and horizontal plane decreases.

(D) The angle between F

"

and the rod decreases.


An uncharged ball of radius R is placed at a point in space and the region out side (from R to > measured

from centre of the ball) the ball is non uniformly charged with a charge density $ = 3r

C coul/m 3 where ‘C’

is a positive constant and r is the distance of a point measured from centre of the ball.

6.6. Electric potential at the centre of the ball is :

(A) Directly proportional to R (B) Directly proportional to R 2

(C) Inversely proportional to R (D) Inversely proportional to R 2

7.7. Electric field intensity at a distance x from centre of the ball (x > R) is :

(A) Rxn

RC

20

!C (B) -

. /0

1 2 !

C RRxn

R2C

20

! (C) 20 R2C

C (R2 – x2) (D) Rxn

xC

20

!C

8.8. As we move away from ball ’s surface, electric potential :

(A) decreases. (B) increases.

(C) decreases then increases. (D) increases then decreases.


Topics : Electrostatics, Calorimetry & Thermal Expansion, Rectilinear Motion, Fluid, Center of Mass,Topics : Electrostatics, Calorimetry & Thermal Expansion, Rectilinear Motion, Fluid, Center of Mass,

String WaveString Wave

PHYSICSPHYSICS








1.1. A ring of radius R having a linear charge density 8 moves towards a solid imaginary sphere of radius2

R, so

that the centre of ring passes through the centre of sphere. The axis of the ring is perpendicular to the line

joining the centres of the ring and the sphere. The maximum flux through the sphere in this process is :

(A)0

R

C8

(B)02

R

C8

(C)04

R

C8&

(D)03

R

C8&

2.2. A rod of length 1000 mm and co-efficient of linear expansion = = 10!4 per degree is placed symmetrically

between fixed walls separated by 1001 mm. The Young's modulus of the rod is 10 11 N/m2. If the

temperature is increased by 20 ºC, then the stress developed in the rod is ( in N/m 2):

(A) 10 (B) 10 8

(C) 2 K 108 (D) cannot be calculated

3.3. Two particl es are projected under gravity with speed 4m/s and 3m /s simultaneously from same point

and at angles 53 º and 37º with the horizontal surface respectively as shown in figure. Then :

(A) Their relative velocity is along vertical direction.

(B) Their relative acceleration is non-zero and it is along vertical direction.

(C) They will hit the surface simultaneously

(D) Their relative velocity (for time interval 0 < t < 0.36s) is constant and has magnitude 1.4 m/s.


4.4. A long cyclindrical drum is filled with water. Two small holes are made on the side of the drum as shown in the

fig. Find the depth of the liquid in the drum if the ranges of water from the holes are equal.

5.5. A 500 g block P rests on a frictionless horizontal table at a distance of 400 mm from a fixed pin

O . The block is attached to pin O by an elastic cord of constant k = 100N/m and of undeformed length

900 mm . If the block is set in motion perpendicularly as shown in figure, (assume there is no loss in

energy due to jerk) determine :

(a)(a) the speed in the begining for which the distance from O to the block

P will reach the maximum value of 1.2 m.

(b)(b) the speed when OP = 1.2 m

(c)(c) the radius of curvature of the path of the block when OP = 1.2 m .

COMPREHENSIONCOMPREHENSIONA sinusoidal wave travels along a taut string of linear mass density 0.1 g/cm. The particles oscillate

along y-direction and wave moves in the positive x-direction. The amplitude and frequency of oscillation

are 2mm and 50 Hz respectively . The minimum distance between two particles oscillating in the same

phase is 4m.

6.6. The tension in the string is (in newton)

(A) 4000 (B) 400 (C) 25 (D) 250

7.7. The amount of energy transferred (in Joules) through any point of the string in 5 seconds is

(A)10

2&(B)

50

2&(C)

5

2&

(D) Cannot be calculated because area of cross-section of string is not given.

8.8. If at x = 2m and t = 2s, the particle is at y = 1mm and its velocity is in positive y-direction, then the

equation of this travell ing wave is : (y is in mm, t is in se conds and x is in metres)

(A) y = 2 sin (2

x& – 100 &t + 30 °) (B) y = 2 sin (

2

x& – 100 &t + 120 °)

(C) y = 2 sin (2

x& – 100 &t + 150 °) (D) None of these


Topics : String Wave, NewtonTopics : String Wave, Newton’s Law of Motion,s Law of Motion, ElectroElectro statics, Prostatics, Pro jectile jectile MotiMoti on, Circular on, Circular Motion, Motion, CenterCenterof Massof Mass

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.11' negative marking) Q.1 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[33, , 33]]



MaMatctch h ththe e FoFollllowowiningg (n(no neo ne gatgat ive ive marmar kinkin g) (g) ( 22 × 4)Q.8× 4)Q.8 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. Two identi cal pulses move in opposite directions with same uniform speeds on a stretched string. The

width and kinetic energy of each pulse is L and k respectively . At the instant they completely overlap,

the kinetic energy of the width L of the string where they overlap is :

(A) k (B) 2k (C) 4k (D) 8 k

2.2. A solid sphere of mass 10 kg is placed over two smooth inclined planes as shown in figure. The normal

reactions at 2 is 10x N. Find x (g = 10 m/s2)

2 160° 30°

3.3. A conducting sphere of radius R has two spherical cavities of radius a and b. The cavities have charges qa

and qb respectively at their centres. ‘ A’ is the centre of the sphere and ‘B’ is the centre of the cavity of radius

‘b’. Find:

(i) electric field and electrical potential at

(a) r (distance from A)> R,

(b) r (distance from B)< b

(ii) surface charge densities on the surface of radius R, radius a and radius b.(iii) What is the force on q

a and q

b?

4.4. A particle moves along the plane trajectory y = f(x) with velocity v whose modulus is constant. Find the

acceleration of the particle at the point x = 0 and the curvature radius of the trajectory at that point if the

trajectory has the form

(a) of a parabola y = ax2.

(b) of an ellipse (x/a)2 + (y/b)2 = 1; a and b are constants here.



A ball is hanging vertically by a light inextensible string of length L from fixed point O. The ball of mass m is

given a speed u at the lowest position such that it completes a vertical circle with centre at O as

shown. Let AB be a diameter of circular path of ball making an angle # with vertical as shown. (g is

acceleration due to gravity)

A

B

u

O

m

5.5. Let TA and T

B be the magn itude of te nsion in str ing when ball is at A and B respectively, then T

A – T

B is

equal to

(A) 6 mg cos # (B) 6 mg (C) 12 mg cos # (D) None of these

6.6. Let Aa"

and Ba"

be acceleration of ball when it is at A and B respectively, then |aa| BA

""" is equal to

(A) 2 g sin# (B) 4cos12g 2 "# (C) 4

g cos # (D) None of these

7.7. Let KA and K

B be kinetic energy of ball when it is at A and B respectively, then K

A – K

B is equal to

(A) mgL cos # (B) 2mgL cos # (C) 4mgL cos # (D) None of these

8.8. Two identical uniform solid smooth spheres each of mass m each approach each other with constant velocitiessuch that net momentum of system of both spheres is zero. The speed of each sphere before collision is u.

Both the spheres then collide. The condition of collision is given for each situation of column- I. In each

situation of column-33 information regarding speed of sphere(s) is given after the collision is over. Match the

condition of collision in column-3 with statements in column-33.Column-Column- Column-Column-

(A) Collision is perfectly elastic and head on (p) speed of both spheres after collision is u

(B) Collision is perfectly elastic and oblique (q) velocity of both spheres after

collision is different

(C) Coefficient of restitution is e =2

1 a nd (r) speed of both spheres after collision

collision is head on is same but less than u.

(D) Coefficient of restitution is e =2

1 a nd (s) speed of one sphere may be more than u.

collision is oblique


Topics : NewtonTopics : Newton’s Law of Motion,s Law of Motion, Fluid, Surface Tension, Gravitation, Fluid, Surface Tension, Gravitation, ElectrostaticsElectrostatics

PHYSICSPHYSICS



TTyyppe e oof f QQuueessttiioonnss MM..MM.., , MMiinn..Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.31' negative marking) Q.1 to Q.3 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.4 to Q.61' negative marking) Q.4 to Q.6 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[1122, , 1155]]Comprehension ('Comprehension (' –1' negative marking) Q.7 to Q.91' negative marking) Q.7 to Q.9 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. A mass m is supported as shown in the figure by ideal strings connected to a rigid wall and to a m ass

3m at rest on a fixed horizo ntal surface. The string connected to la rger mass is horizontal, tha tconnected to smaller mass is vertical and the one connected to wall makes an angle 60 ° withhorizontal.Then the minimum coefficient of static f riction between the larger mass and the horizontalsurface that permits the system to remain in equilibrium in the situation shown is:

60°3m

m

(A)3

1(B)

33

1

(C)2

3(D)

2

3

2.2. Water (density $) is flowing through the uniform tu be of cross-sec tional area A with a constant speed vas shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube

is(neglect viscous forces)

P

60°v

A

(A) 2Av3 $ (B) 2Av2$

(C) 2Av2$ (D)2

Av2$

3.3. A container is partially filled with a liquid of density $2. A capil lary tube of radius r is vertically inserted

in this liquid. Now another liquid of density $1 ($

1 < $

2) is slowly poured in the container to a height h as

shown. There is only denser liquid in the capillar y tube. The rise of denser liquid in the capillary tube isalso h. Assuming zero contact angle, the surface tension of heavier liquid is

(A) r $

2g

h (B) 2

&

r $

2g

h

(C)2

r($

2 – $

1)

g

h (D) 2 &

r

($

2 – $

1)

g

h


4.4. A point P lies on the axis of a f ixed ring of mass M and radius a, at a distance a from its centre C. A smallparticle starts from P and reaches C under gravitational attraction only. Its speed at C will be _______.

5.5. Three conducting concentric spherical shells of radius R, 2R and 3R have charges Q,3

Q and

–2Q respectively. Q = 1.6 × 10 –6C. The intermediate shell is grounded. Find the number of electrons that willflow through the connecting wire. Also tell whether, the electrons flow into the earth or into the shell.

6.6. A non!uniform string of ma ss 45 kg and length 1.5 m has a variable linear mass density given by 5 =kx, where x is the distance from one end of the string and k is a constant. T ension in the string is 15N which is uniform. Find the time (in second) required for a pulse generated at one end of the string totravel to the other end.


Two blocks A and B of equal masses m k g each are connected by a light thread, which passes over amassless pulley as shown. Both the blocks lie on wedge of mass m kg. Assume friction to be absenteverywhere and both the blocks to be always in contact with the wedge. The wedge lying over smoothhorizontal surface is pulled towards right with constant acceleration a (m/s 2). (g is acceleration due togravity).

7.7. Normal reaction (in N) acting on block B is

(A)5

m(3g + 4a) (B)

5

m(3g – 4a) (C)

5

m(4g + 3a) (D)

5

m(4g – 3a)

8.8. Normal reaction (in N) acting on block A.

(A)5m (3g + 4a) (B)

5m (3g – 4a) (C)

5m (4g + 3a) (D)

5m (4g – 3a)

9.9. The maximum value of acceleration a (in m/s 2) for which norm al reactions acting on the block A andblock B are nonzero.

(A)4

3g (B)

3

4g (C)

5

3g (D)

3

5g


Topics :Electrostatics, Gravitation,Work, Power and Energy, Projectile Motion, Elasticity & Viscosity,Topics :Electrostatics, Gravitation,Work, Power and Energy, Projectile Motion, Elasticity & Viscosity,Geometrical Optics,Sound Wave, FrictionGeometrical Optics,Sound Wave, Friction

PHYSICSPHYSICS







1.1. The figure shows a charge q placed inside a cavity in an uncharged

conductor. Now if an external electric field is switched on :

(A) only induced charge on outer surface will redistribute.

(B) only induced charge on inner surface will redistribute.

(C) both induced charge on outer and inner surface will redistribute.

(D) force on charge q placed inside the cavity will change.

2.2. A certain quarternary star system consists of three stars, each of mass m, moving in same circular orbit

about a stationary central star of mass M. The three identical stars orbit in same sense and are symmetrically

located with respect to each other. Considering gravitational force of all remaining bodies on every star, the

time period of each of three stars is :

(A) 2& - .

/01

2 "

3

mMG

r3

(B) 2&--

.

/001

2 "

3

mMG

r3

(C) 2& ( )m3MG

r3

" (D) 2&)m3M(G

r3

"

3.3. Block A in the figure is released from rest when the extension in the spring is x0.

(x0 < Mg/k). The

maximum downwards displacement of the block is (ther is no friction) :

(A) 0x2K

Mg2! (B) 0x

K2

Mg"

(C) 0xK

Mg2! (D) 0x

K

Mg2"

4.4. Three stones A, B, C are projected from surface of very long inclined plane with equal speeds and

differen t angles of projection as shown in figure. The incline makes an angle # with horizontal.If H A, HB

and H C are m aximum height attained by A, B and C r espectively above incli ned plane then : (Neglect

air friction)

A

B

C

(A) HA + H

C = H

B(B) 2

B2C

2A HHH *" (C) H

A + H

C = 2H

B(D) 2

B2C

2A H2HH *"


5.5. When a ball is released from rest in a very long column of viscous liquid, its downward acceleration is ‘a’(just

after release). Then its acceleration when it has acquired two third of the maximum velocity :

(A)3

a(B)

3

a2(C)

6

a(D) none of these

6.6. An isosceles trapezium of reflecting material of refractive index 2 and dimension of sides being 5cm, 5cm,

10cm and 5cm. The angle of minimum deviation by this when light is incident from air and emerges in air is:

(A) E2

122 (B) 45° (C) 30° (D) 60°

7.7. A triangular medium has varying refracting index n = n0 + ax, where x is the distance (in cm) along x –axis

from srcin and n0 =

3

4. A ray is incident normally on face OA at the mid –point of OA. The range of a so that

light does not escape through face AB when it falls first time on the face AB (OA = 4 cm, OB = 3 cm and

AB = 5 cm) : (Surrounding medium is air)

(A) a >9

1(B) a >

9

2(C) a >

3

1(D) None of these

8.8. A string 25 cm long fixed at both ends and having a mass of 2.5 g is under tension. A pipe closed from one

end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental

frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases

the beat frequency. If the speed of sound in air is 320 m/s. Find tension in the string.

9.9. Two blocks of masses 20 kg and 10 kg are kept or a rough horizontal floor . The coefficient of friction

between both blocks and floor is 5 = 0.2. T he surface of contact of both blocks are smooth. Horizon tal

forces of magnitude 20 N and 60 N are applied on both the blocks as shown in figure. Match the

statement in column- I with the statements in column- II.

F =20N1

F =60N2

20kg 10kg

rightleft

rough horizontal floor

=0.2


(A) Frictional force acting on block of mass 10 kg (p) has magnitude 20 N

(B) Frictional force acting on block of mass 20 kg (q) has magnitude 40 N

(C) Normal reaction exerted by 20 kg block on 10 kg b lock (r) is zero

(D) Net force on system consisting of 10 kg block (s) is towards r ight (in horizontal

and 20 kg block direction).


Topics : GravitationTopics : Gravitation ,Work, Power and Energy, Elasticity & Viscosity, Sound Wave, Relative Motion,Work, Power and Energy, Elasticity & Viscosity, Sound Wave, Relative Motion,Electrostatics.Electrostatics.

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.11' negative marking) Q.1 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[33, , 33]]




1.1. The orbital velocity of an artificial satellite in a circular orbit just above the earth ’s surface is V0. The

value of orbital velocity for another satellite orbiting at an a ltitude of half of earth ’s radius is

(A) 0V2

3- .

/01

2 (B)

3

20V (C)

2

30V (D)

2

30

2 1 0

/ .-

V

2.2. The potential energy (in joules) of a particle of mass 1kg moving in a plane is given by V = 3x + 4y, the

position coordinates of the point being x and y, measured in metres. If the particle is at rest at (6, 4) ; then

(A) its acceleration is of magnitude 5m/s2

(B) its speed when it crosses the y –axis is 10m/s

(C) it crosses the y-axis (x = 0) at y = –4

(D) it moves in a striaght line passing through the srcin (0, 0)

3.3. A wire of uniform cross section is hanging vertically and due to its own weight its length changes. There is a

point‘C

’ on the wire such that change in length AC is equal to the change in length BC. Points A, B & C are

shown in figure. FindBC

AC.

4.4. A wall is moving with velocity u and a source of sound moves with velocity2

u in the same direction as shown

in the figure. Assuming that the sound travels with velocity 10u. Find the ratio of incident sound wavelength

on the wall to the reflected sound wavelength by the wall.

5.5. Two particl es P and Q are moving with velocities of ) ji( " and ( ) j2i "! respect ively. At time t = 0, P

is at srcin and Q is at a point with position vector ji2 " . Find the equation of the path of Q with

respect to P.


6.6. A cone of mass 'm' falls from a height 'h' and penetrates into sand. The resistance force R of the sand is given

by R = kx2. If the cone penetrates upto a distance x = d where d < ! , then find the value of 'k'.


A solid, conducting sphere of radius 6 cm carries a charge 3nC. This sphere is located centrall y inside

a thick, conducting sphere with an inner radius of 18 cm and an outer radius of 27 cm. The hollow

sphere is also given a charge 3nC.Three points A, B and C are mark ed on the surfaces as shown.

27cm

Conducting shell

18cm 6cm

A

B

C

7.7. Which one of the following figures shows a qualitatively accurate sketch of the electric field lines in

and around this system ?

(A) (B) (C) (D)

8.8. Suppose VA,V

B and V

C are potentials at points A,B and C respectively then values of potential differences

VC – V

B and V

B – V

A respectively are :

(A) 0 V and – 300 V (B) 0

V and 300 V

(C) 450 V and 150 V (D) 0 V and – 150 V

9.9. Suppose the shell is given additional charge 3nC. The potential difference VB – V

A will become :

(A) –100 V (B) –200 V (C) 300 V (D) –300 V


Topics : FrictioTopics : Friction, Elecn, Electrosttrostaticatics, s, Geometrical Optics, Relative Motion, Rigid Body DyGeometrical Optics, Relative Motion, Rigid Body Dynamics,namics, NewtonNewton’s Laws Law

of Motionof Motion

PHYSICSPHYSICS







Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4× 4 )) Q Q ..88 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. A plank is held at an angle = to the horizontal (Fig.) on two fixed supports A and B. The plank can slide

against the supports (without friction) because of its weight Mg. With what acceleration and in what direction,

a man of mass m should move so that the plank does not move.

(A) g sin = - .

/01

2 "

M

m1 d own the incline (B) g sin = -

.

/01

2 "

m

M1 down the incline

(C) g sin = - .

/01

2 "

M

m1 u p the incline (D) g sin = -

.

/01

2 "

m

M1 up the incline

2.2. Two small electric dipoles each of dipole moment p i are situated at (0, 0, 0) and (r, 0, 0). The electric

potential at a point --

001

0,2

r3,

2

r is :

(A) 20 r4

p

C& (B) 0 (C) 20 r2

p

C& (D) 20 r8

p

C&

3.3. A mango tree is at the bank of a river and one of the branch of tree extends over the river. A tortoise lives

in river. A mango falls just above the tortoise. The acceleration of the m ango falling from tr ee appearing

to the tortoise is ( Refractive index of water is 4/3 and the tortoise is stationary)

(A) g (B)4

g3(C)

3

g4(D) None of these

4.4. A balloon is ascending vertically with an acceleration of 0.4 m/s !2. Two stones are dropped from it a t an

interval of 2 sec. Find the distance between them 1.5 sec. after the second stone is released.(g = 10 m/sec 2)



In the given figure F=10N, R=1m mass of the body is 2kg and moment of inertia of the body about an axis

passing through O and perpendicular to plane of body is 4kgm2. O is the centre of mass of the body.

2R

R

O

F F

5.5. Find the frictional force acting on the body if it performs pure rolling.

(A)3

20(B)

3

10(C)

3

5(D) None of these

6.6. The kinetic energy of the body in the above question after 3 seconds will be.

(A) 75J (B) 80J (C) 82J (D) 85J

7.7. If ground is smooth, then the total kinetic energy after 3 seconds will be :

(A) 105.5J (B) 112.5J (C) 115.5J (D) None of these

8.8. In the column- I, a system is described in each option and corres ponding time period is given in the

column- II. Suitably match them.


(A) A simp le pendulum of leng th ' !' oscillating (p) T = 2 &g3

2!

with small amplitude in a lift m oving downwith retardation g/2.

(B) A block attached to an end of a vertical (q) T = 2 &g

!

spring, whose other end is fixed to the ceiling

of a lift, stretches the spring by length ' !' in

equilibr ium. It's tim e period when lift moves

up with an acceleration g/2 is

(C) The time period of small oscillation of a (r) T = 2 &g

2!

uniform rod of length ' !' smoothly hinged at

one end. The rod oscillates in vertical plane.

(D) A cubical block of edg e ' !' and specific (s) T = 2 &g2

!

gravity 1/2 is in equilibrium with some volume insidewater filled in a large fixed container. Negl ect viscous

forces and surface tension. The time period of smalloscillations of the block in vertical direction is


Topics : Circular Motion, Gravitation, Rigid Body Dynamics, Work, Power and Energy, Center of Mass,Topics : Circular Motion, Gravitation, Rigid Body Dynamics, Work, Power and Energy, Center of Mass,Electrostatics.Electrostatics.

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.41' negative marking) Q.1 to Q.4 ((112 m2 m aarrkkss, 1, 12 m2 m iinn..)) [[1122, 1, 1 22]]



1.1. A particle is projected along a horizontal field whose coefficient of friction varies as 5 = 2rA where r is the

distance from the srcin in meters and A is a positive constant. The initial distance of the particle is 1 m from

the srcin and its velocity is radially outwards. The minimum initial velocity at this point so that particle never

stops is :

(A) > (B) 2 gA (C) Ag2 (D) 4 gA

2.2. An automobile enters a turn of radius R. If the road is banked at an angle of 45 0 and the coefficient of

friction is 1, the minimum and maximum speed with which the automobile can negotiate the turn

without skidding is:

(A)2

rg and rg (B)

2

rg and rg

(C)2

rg and rg2 (D) 0 and infinite

3.3. A hollow cylinder has mass M, outside radius R2 and inside radius R

1. Its moment of inertia about an axis

parallel to its symmetry axis and tangential to the outer surface is equal to :

(A)2

M (R

22 + R

12) (B)

2

M (R

22 – R

12)

(C)4

M (R

2 + R

1)2 (D)

2

M(3R

22 + R

12)

4.4. In the Figure, the ball A is released from rest when the spring is at its natural length. For the block B,

of mass M to leave contact with the ground at some stage, the m inimum mass of A must be:

(A) 2 M (B) M

(C) M/2 (D) A function of M and the force constant of the spring.


5.5. Two blocks A and B each of mass m are conne cted to a massless spr ing of natural length L and spring

constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural

length as shown in the figure. A third identical block C also of mass m moves on the floor with speed v

along the line joining A and B and collides elastically with A, then :

K

(A) the K.E. of the A !B system at maximum compression of the spring is zero

(B) the K.E. of the A !B system at maximum com pression of the spring is mv 2 /4

(C) the maximum compression of the spring is v ( / )m K

(D) the maximum com pression of the spring is v ( / )m K2


The sketch below shows cross-sections of equipotential surfaces between two charged conductors

that are shown in solid black. Some points on the equipotential surfaces near the conductors are

marked as A,B,C,........ . The a rrangement lies in air. (Take C0 = 8.85 × 10 –12 C2 /N m2]

6.6. Surface charge density of the plate is equal to

(A) 8.85 × 10 –10 C/m2 (B) –8.85 × 10 –10 C/m2

(C) 17.7 × 10 –10 C/m2 (D) –17.7 × 10 –10 C/m2

7.7. A positive charge is placed at B. When it is released :

(A) no force will be exerted on it. (B) it will move towards A.

(C) it will move towards C. (D) it will move towards E.

8.8. How much work is required to slowly move a – 15C charge from E to D ?

(A) 2 × 10 –5 J (B) –2 × 10 –5 J

(C) 4 × 10 –5 J (D) –4 × 10 –5 J


Topics : Center of Mass, Rigid Body Dynamics, String Wave, Fluid, Electromagnetic Induction, RigidTopics : Center of Mass, Rigid Body Dynamics, String Wave, Fluid, Electromagnetic Induction, RigidBody DynamicsBody Dynamics

PHYSICSPHYSICS



TTyyppe e oof f QQuueessttiioonnss MM..MM.., , MMiinn..Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.31' negative marking) Q.1 to Q.3 ((3 3 mmaarrkks s 3 3 mmiinn..)) [[99, , 99]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.41' negative marking) Q.4 ((4 4 mmaarrkks s 5 5 mmiinn..)) [[44, , 55]]Comprehension ('Comprehension (' –1' negative marking) Q.5 to Q.71' negative marking) Q.5 to Q.7 ((3 3 mmaarrkks s 3 3 mmiinn..)) [[99, , 99]]

Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.8× 4) Q.8 ((8 8 mmaarrkks s 110 0 mmiinn..)) [[88, , 1100]]

1.1. Particle 'A' moves with speed 10 m/s in a frictionless circular fixed horizontal pipe of radius 5 m and

strikes with 'B' of double mass that of A. Coefficient of restitution is 1/2 and particle 'A' starts its

journey at t = 0. The time at which second coll ision occurs is :

(A)2

&s (B)

3

2&s (C)

2

5&s (D) &4 s

2.2. A solid cylinder is sliding on a smooth horizontal surface with velocity v 0 without rotation. It enters on

the rough surface. After that it has travelled some distance, select the correct statement:

v 0

////////////////////

(A) Friction force increases its translational kinetic energy.

(B) Friction force increases its rotational kinetic energy.

(C) Friction f orce increases its total mechanical energy .

(D) Friction force increases its angular m omentum about an axis passing through point of contact of

the cylinder and the surface.

3.3. Equation of a standing wave is generally expressed as y = 2A sin't coskx. In the equation, quantity ' /krepresents

(A) the transverse speed of the particles of the string.

(B) the speed of either of the component waves.

(C) the speed of the standing wave.

(D) a quantity that is independent of the properties of the string.

4.4. A block of density 2000 kg/m3 and mass 10 k g is suspended by a spring of stiffness 100 N/m. The other end

of the spring is attached to a fixed support. The block is completely submerged in a liquid of density 1000

kg/m3 . If the block is in equilibrium position.

(A) the elongation of the spring is 1 cm.

(B) the magnitude of buoyant force acting on the block is 50 N.

(C) the spring potential energy is 12.5 J.

(D) magnitude of spring force on the block is greater than the weight of the block.



In a certain region, there are non-uniform electrica l potential (V e) as well as gravitational potential (V g).

The electrical potential varies only with x as shown in figure (i), and t he gravitational potential varies

only with y as shown in figure (ii).

Consider a particle of mass 200 kg and charge 20 5C in this field.

5.5. If the particle is released from point (5cm, 15 cm ), it will try to move toward

(A) +x direction and +y direction (B) +x direction and – y direction

(C) – x direction, +y direction (D) – x direction, – y direction

6.6. What will be acceleration of the particle at point (25, 35)

(A) ) ji2( ! × 10 –2 m/sec 2 (B) ) ji2( " × 10 –2 m/sec 2

(C) ) ji2( "! × 10 –2 m/sec 2 (D) ) j2i3( ! × 10 –2 m/sec 2

7.7. Minimum work required to bring the particle from (5, 15) to (25, 35) is :

(A) 0.2 J (B) 0.1 J (C) – 0.2 J (D) – 0.1 J

8.8. Match the columns : (All the rigid bodies lie on a smooth horizontal plane. No object is hinged. J is impulse

of an impulsive force. Initially system is at rest.)

CCoolluummnn--II CCoolluummnn--IIII

(A)

M

J

M

l

rod ismass less

(impulse)

(p) Translation occurs

(B)

J

( uniform mass distribution) (q) Rotation occurs

(C)M

M

J

rod ismass less (r) Angular momentum (about centre of mass )increases

(D)

J

(uniform mass distribution) (s) Linear momentum increases

(t) Angular momentum will be conserved about more

than one points in space.


Topics : Topics : ElasticitElasticit y & y & Viscosity, ElecViscosity, Elec trostatitrostati cs, Geometricacs, Geometrical Optics, Strl Optics, String Waing Wave, Rigive, Rigid Body Dyd Body Dynamicsnamics

PHYSICSPHYSICS








1.1. If I represents the coefficient of viscosity and T the surface tension, then the dimension ofIT

is same as that of:

(A) length (B) mass (C) time (D) speed

2.2. The elongation in a metallic rod hinged at one end and rotating in a horizontal plane becomes four

times of the initial value. The angular velocity of rotation becomes :

(A) two times the initial value (B) half of initial value

(C) one third of initial value (D) four times the initial value.

3.3. In the figure shown, blocks P and Q are in contact but do not stick to each other . The lower face of P

behaves as a plane mirror. The springs are in their natural lengths. The system is released fr om rest.

Then the distance between P and Q when Q is at the lowest point first time will be

(A)K

mg2(B)

K

mg4(C)

K

mg3(D) 0

4.4. In the above question, the velocity of the image of Q in plane mirror of block P with respect to ground

when Q is at the lowest point first time is :

(A)K

mg2 2

(B)K

mg4 2

(C)K

mg3 2

(D) 0

5.5. The length, tension, diameter and density of a wire B are double than the corresponding quantities for

another stretched wire A. Then (both are fixed at the ends)

(A) Fundamental frequency of B is 12 2

times that of A.

(B) The velocity of wave in B is1

2 times that of velocity in A.

(C) The fundamental frequency of A is equal to the third overtone of B.

(D) The velocity of wave in B is half that of velocity in A.


6.6. A sample of a liquid has an initial volume of 1.5 L. The volume is reduced by 0.2 mL, when the pressure

increases by 140 kPa. What is the bulk modulus of the liquid.


A square frame of m ass m is made of four identical uniform rods of length L each. This frame is placed

on an inclined plane such that one of its diag onals is parallel to the inclined plane as shown in figur e, and

is released.

7.7. The moment of inertia of square frame about the axis of the frame is :

(A)3

mL2

(B)3

mL2 2

(C)3

mL4 2

(D)12

mL2

8.8. The frictional force acting on the frame just after the release of the frame assuming that it does not slide

is :

(A)3

sinmg #(B)

7

sinmg2 #(C)

5

sinmg3 #(D)

5

sinmg2 #

9.9. The acceleration of the center of square frame just after the release of the frame assum ing that it does

not slide is :

(A)3

sing #(B)

7

sing2 #(C)

5

sing3 #(D)

5

sing2 #


TopicTopics : Elass : Elasticticity & Viity & Viscosscosity, Geometrical Optics, Sity, Geometrical Optics, String Wtring Wave, Friction, Simple ave, Friction, Simple Harmonic Motion, RigidHarmonic Motion, RigidBody DynamicsBody Dynamics

PHYSICSPHYSICS







1.1. A spherical ball of mass 4m, density B and radius r is attached to a pulley-mass system as shown in figure.

The ball is released in a liquid of coefficient of viscosityI and density$ (<2

B). If the length of the liquid column

is sufficiently long, the terminal velocity attained by the ball is given by (assume all pulleys to be massless

and string as massless and inextensible) :

(A)I

$!B g)2(r

9

2 2

(B)I

$!B g)2(r

9

2 2

(C)I

$!B g)4(r

9

2 2

(D)I

$!B g)3(r

9

2 2

2.2. Which of the following relations is correct for a spherical mirror if a point object is kept on the principal axis.

[‘P’ is pole, ‘C’ is centre object is at point ‘O’, image is at point ‘3’]

(A)OC

OP=

C

P

33

(B)OC

P

C

OP 3*

3(C)

PC

P

PO

PC 3* (D)

CO

P

CP

O 3*

3

3.3. A travelling wave y = A sin (k x ! '

t + #) passes from a heavier string to a lighter string. The reflected wave

has amplitude 0.5 A. The junction of the strings is at x = 0. The equation of the reflected wave is:

(A) y + = 0.5 A sin (k

x + '

t + #) (B) y

+ = ! 0.5 A sin (k

x + '

t + #)

(C) y + = ! 0.5 A sin ('

t ! k

x ! #) (D) y

+ = – 0.5 A sin (k

x + '

t ! #)


4.4. Figure shows an ideal pulley block of mass m = 1 kg, resting on a rough ground with friction coefficient µ =

1.5. Another block of mass M = 11 kg is hanging as shown. When system is released it is found that the

magnitude of acceleration of point P on string is a. Find value of 4a in m/s2. (Use g = 10 m/s2)

5.5. A 900 kg elevator hangs by a steel cable for which the allowable stress is 1.15× 108 N/m2. What is the

minimum diameter required if the elevator accelerates upward at 1.5 m/s2. Take g = 10m/s2 and leave your

answer in terms of&.

6.6. A 40 kg mass, hanging at the end of a rope of length!, oscillates in a vertical plane with an angular amplitude

of #0. What is the tension in the rope, when it makes an angle # with the vertical ? If the breaking strength of

the rope is 80 kg f, what is the maximum angular amplitude # with which the mass can oscillate without the

rope breaking ?


A uniform bar of length 6 a & mass 8 m lies on a smooth horizontal table. Tw o point masses m & 2 m

moving in the same horizontal plane with speeds 2 v and v respectively strike the bar as sho wn & stick

to the bar after coll ision.

7.7. Velocity of the centre of mass of the system is

(A)2

v(B) v (C)

3

2v(D) Zero

8.8. Angular velocity of the rod about centre of mass of the system is

(A)5a

v(B)

15a

v(C)

3a

v(D)

10a

v

9.9. Total kinetic energy of the system, just after the collision is

(A)

5

3mv2 (B)

25

3mv2 (C)

15

3mv2 (D) 3 mv 2


Topics : Topics : Surface Surface Tension, ElastTension, Elasticity icity & Visco& Viscosity, Geomsity, Geometrietrical Optcal Optics, Circular ics, Circular MotionMotion, Rigid Body D, Rigid Body Dynamynamicsics

PHYSICSPHYSICS






1.1. A capillary tube with inner cross-section in the form of a square of side a is dipped vertically in a liquidof density $ and surface tension B which wet the surface of capillary tube wit h angle of contact #. Theapproximate he ight to which liquid will be raised in the tube is : (Neglect the effect of surf ace tension atthe corners capillary tube)

(A) ga

cos2

$#B

(B) ga

cos4

$#B

(C) ga

cos8

$#B

(D) None of these

2.2. Four uniform wires of the sam e material are stretched by the same force. T he dimensions of wire are as

given below . The one which has the m inimum elongation has :

(A) radius 3mm, length 3m (B) radius 0.5 mm, length 0.5 m(C) radius 2mm, length 2m (D) radius 3mm, length 2m

3.3. A space 2.5 cm wide between two large plane surfaces is filled with oil. Force required to drag a verythin plate of area 0.5 m 2 just midway the surfaces at a speed of 0.5 m/sec is 1N. The coefficient ofviscosity in kg –s/m 2 is :

(A) 5 × 10 –2 (B) 2.5 × 10 –2 (C) 1 × 10 –2 (D) 7.5 × 10 –2

4.4. A glass beaker has diameter 4cm wide at the bottom. An observer observes the edge of bottom when beaker

is empty as shown in figure. When the beaker is completely filled with liquid of refractive index n = 2 / 5 , he

can just see the centre of bottom, then the height of glass beaker is :

(A) 4 cm (B) 2 / 5 cm

(C) 16 cm (D) None of these


5.5. The given figure shows a small m ass connected to a string, which is attached to a vertical post. If themass is released f rom rest when the string is horizontal as shown, the magnitude of the total acceleration

of the mass as a function of the angle # is

(A) 2 g sin # (B) 2 g cos # (C) 1cos3g 2 "# (D) 1sin3g 2 "#

6.6. A 40 cm long wire having a mass 3.2 gm and area of cross section 1 mm2 is stretched between the support40.05 cm apart. In its fundamental mode, it vibrates with a frequency 1000/64 Hz. Find the young’s modulus

of the wire in the form X × 108 N/m2 and fill value of X.


A uniform wheel is released on a rough horizontal floor af ter imparting it an initial horizontal velocity v 0

and angular velocity '0 as shown in the figure below. P oint O is the centre of m ass of the wheel and pointP is its instantaneous point of contact with the ground. The radius of wheel is r and its radius of gyrationabout O is k. Coefficient of friction between wheel and ground is 5. A is a fixed point on the grou nd.

7.7. Which of the following is conserved ?(A) linear momentum of wheel(B) Angular m omentum of wheel about O(C) Angula r momentum of wheel about A

(D) none of these

8.8. If the wheel comes to permanent rest after som etime, then :

(A) v 0 = '0r (B) v 0 = r

k20'

(C) v 0 = R

r20'

(D) V 0 = --

001

2 "'

r

kr

2

0

9.9. In above question, distance travelled by centre of mass of the wheel before it stops is -

(A) -- .

/001

2 "

5 2

220

k

r1

g2

v(B)

g2

v2

0

5(C) -

- .

/001

2 "

5 2

220

r

k1

g2

v(D) None of these


Topics : Circular Motion, Rigid Body Dynamics, Surface Tension, Geometrical Optics, GravitationTopics : Circular Motion, Rigid Body Dynamics, Surface Tension, Geometrical Optics, Gravitation

PHYSICSPHYSICS







1.1. With keys k1 and k2 closed : if PS < QR then

R S

Dk2

k1

Q

B

P

A CG

(A) VB = V

D(B) V

B > V

D

(C) VB < V

D(D) V

B

LF

VD

2.2. A ring of mass m and radius R has three particle attached to the ring and it is rolling on the surface as

shown in the figure. The ratio of kinetic energy of the system to the kinetic of the particle of mass 2m

will be (speed of centre of ring is V 0 and slipping is absent) :

(A)5

2(B)

2

1(C)

1

3(D)

3

2

3.3. A soap film is created in a small wire frame as shown in the figure. The

sliding wire of mass m is given a velocity u to the right and assume that u

is small enough so that film does not break. Plane of the film is horizontal

and surface tension is T. Then time to regain the srcinal position of wire is

equal to :

(A)!T

um(B)

um

T!

(C)T

mu2

!(D) It will never regain srcinal position


4.4. A tube of length ! open at only one end is cut into two equal halves. The sixth overtone frequency of piece

closed at one end equals to sixth overtone frequency of piece open at both ends. The radius of cross –section

of tube is :

(A)12

!(B)

72

5!(C)

24

!(D)

24

5!

(E)48

5!

5.5. A ray of light is incident from air to a glass rod at point A the angle of incidence being 450. The minimum value

of refractive index of the material of the rod so that T.I.R. takes place at B is _____.


Consider a star and two planet system . The star has mas s M. The planet A and B have the same mas sm, radius a and they revolve around the star in circular orbits of radius r and 4r r espectively (M > > m,r > > a ). Planet A has intelligent life, and the people of this planet have achieved a very high degree oftechnological advance. They wish to shift a geostationary satellite of t heir own planet to a geostationaryorbit of planet B. T hey achieve this through a series of high precision maneuvers in which the satellitenever has to apply brakes and not a single joule of energy is wasted. S 1 is a geostationary satellite ofplanet A and S 2 is a geostationary satellite of plan et B. Neglect interaction between A and B, S 1 and S2,S1 and B & S 2 and A.

A

Bstar

S1

S2

6.6. If the time period of the satellite in geostationary orbit of planet A is T, then its time period in geostationary

orbit of planet B is :

(A) T (B) 4T (C) 8 T (D) Data insufficient

7.7. If the radius of the geostationary orbit in planet A is given by r G =

3 / 1

M

mr -

/01

2 , then the time in which the

geostationary satellite will complete 1 revolution isI. 1 planet year = time in which planet revolves around the starII. 1 planet day = time in which planet revolves about its axis.

(A) I (B) II (C) both I and II (D) neither I nor II.

8.8. If planet A and B, both complete one revolution about their own axis in the same time, then the energy

needed to transfer the satellite of mass m 0 from planet A to planet B is.

(A)r4

Gmm0(B)

r4

GMm0(C)

r8

GMm3 0(D) Zero


Topics : Gravitation, Elasticity & Viscosity, Geometrical Optics, Sound Wave, Rigid Body Dynamics,Topics : Gravitation, Elasticity & Viscosity, Geometrical Optics, Sound Wave, Rigid Body Dynamics,Current ElectricityCurrent Electricity

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.21' negative marking) Q.1 to Q.2 ((112 2 mmaarrkkss, , 112 2 mmiinn..)) [[66, , 66]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.3 to Q.51' negative marking) Q.3 to Q.5 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[1122, , 1155]]Comprehension ('Comprehension (' –1' negative marking) Q.6 to Q.81' negative marking) Q.6 to Q.8 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. In a binary star system one star has thrice the mass of other. The stars rotate about their common centre of mass

then:

(A) Both stars have same angular momentum about common centre of mass.(B) Both stars have angular momentum of same magnitude about common centre of mass.(C) Both stars have same angular speeds.(D) Both stars have same linear speeds.

2.2. A thin rod of negligible mass and area of cross-section 2 × 10 –6 m2, suspended vertically from one end, hasa length of 0.5 m at 2000C. The rod is cooled to 00C, but prevented from contracting by attaching a mass atthe lower end. The value of this mass is : (Young's modulus = 1011 N/m2, Coefficient of linear expansion 10 –

5 K –1 and g = 10 m/s2):(A) 20 kg (B) 30 kg (C) 40 kg (D) 50 kg

3.3. A glass sphere with 10 cm radius has a 5 cm radius concentric spherical hole. A narrow beam ofparallel light from outside is incident on the sphere . Find the final image position with respect to thelast surface.The refractive index of the glass i s 1.5 .

4.4. A sound wave propagating along x-axis, in medium 3 of density $1 = 1.5 kg/m 3 is transmitted to a

medium 33 of density $2 = 3kg/m 3 as shown. The equation of excess pressure developed by wave in

medium 3 and that in medium 33 respectively are

p1 = 4 × 10 –2

cos ' - .

/

01

2

! 400

x

t (in S 3 units)

p2 = 3 × 10 –2 cos ' - .

/01

2 !

1200

xt (in S 3 units)

If the intensity of transmitte d wave is 32 (wave in medium 33) and intensity of incident wave is 31 (wave in

medium 3), then find the value of2

16

33

.


5.5. A uniform thin rod of mass 'm' and length '3 !' is released from rest from horizontal position as shown.When it passes through the vertical line AD, it gets broken at point 'C' due to some reason. Find theangle (in radian) rotated by rod BC till its centre of mass passes thro ugh the horizontal line PQ from theinstant of breakage. (g = 10 m/s 2) (hinge is smooth)


Consider a battery with a large internal resistance (r M >) N 10 kO. If we

connect a light resistance R (R << r) across it, current will be

Rr

E

" N

r

E = constant (Independent from external resistance R)

So if we put any light resistance across it, current in it would not change,

and will be constant (N r

E)

So a battery with a large internal resistance is equivalent to a constantcurrent source (A source providing a constant current)

A BJ

This concept is used in potentiometer circuit for comparing two resistancesR and X, as shown in figure.

Here a battery with large internal resistance supply a cosntant current (i) to either R or X. I f it is connectedto R = 10 O potential drop across it will be iR, and the balance length is found to be 58.3 cm. If it isconnected to X (unknown), Potential drop across it will be iX and the balance length iscoming = 68.5 cm.

6.6. Estimate the unknown resistance X(A) 11.25 (B) 11.75 (C) 12.25 (D) 12.50

7.7. If you fail to f ind any balanced point, as we slide the jocky along A B. What may be the r eason.(A) Resistance of potentiometer wire may be large.(B) Emf of the primary cell (V) may be very high(C) potential drop across R or X may exceed V

(D) Emf of the source (E), may be very large.

8.8. In the previous question, what can we do to get the balance point(A) Increase V(B) Reduce E(C) Increase the internal resistance (r) connected with E(D) Use a longer potentio meter wire


Topics : Current Electricity, Elasticity & Viscosity, Geometrical Optics, Gravitation, Sound Wave, RigidTopics : Current Electricity, Elasticity & Viscosity, Geometrical Optics, Gravitation, Sound Wave, RigidBody DynamicsBody Dynamics

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.21' negative marking) Q.1 to Q.2 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[66, , 66]]Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.31' negative marking) Q.3 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44, , 44]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.4 to Q.51' negative marking) Q.4 to Q.5 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]Comprehension ('Comprehension (' –1' negative marking) Q.6 to Q.81' negative marking) Q.6 to Q.8 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. In the given circuit all resistors are of R ohm each. If a wire isconnected between C and B, then the ratio of equivalent resistancebetween AB of original circuit to new circuit formed is:(A) 1 (B) 3 (C) 1/3 (D) 1/2

2.2. A sphere of mass m and radius r is projected in a gravity free space with speed v. If coefficient of viscosity is

&6

1, the distance travelled by the body before it stops is :

(A) r2

mv(B)

r

mv2(C)

r

mv(D) none of these

3.3. An object AB is placed parallel and close to the optical axis between focus F and centre of curvature C of aconverging mirror of focal length f as shown in figure.(A) Image of A will be closer than that of B from the mirror.

C F

A B

(B) Image of AB will be parallel to the optical axis.(C) Image of AB will be straight line inclined to the optical axis.(D) Image of AB will not be straight line.

4.4. A satellite is moving around the earth in a circular orbit and in this orbit magnitude of its acceleration is ‘a1

’.Now a rocket is fired in the direction of motion of satellite from the satellite due to which its speed instantaneouslybecomes half of initial, just after the rocket is fired acceleration of satellite have magnitude ‘a

2’ . Then fill the

ratio2

1

a

a in answer sheet (Assume there is no external force other then the gravitational force of earth before

and after the firing of rocket from the satellite)

5.5. A bird is singing on a tree and a man is hearing at a distance ‘r ’ from the bird. Calculate the displacement of

the man towards the bird so that the loudness heard by man increases by 20 dB. [ Assume that the motionof man is along the line joining the bird and the man]


A uniform rod AB of length ' !' is thrown upwards s uch that initially AB ishorizontal, veloci ty of centre is 'u' upwards and angular velocity ' '' issuch that velocity of 'B' at this mom ent is zero. The values of ' '' and 'u'are also such that the rod becomes vertical first time at the momentwhen the centre of rod reaches the highest point of its motion.

6.6. The value of ' '' in terms of 'u' and ' !' is equal to

(A)!

u (B)!

u2 (C)!2u (D)

!4u

7.7. The value of 'u' is equal to

(A)4

g!&(B) !g (C)

2

g!&(D)

2

g!

8.8. The angular acceleration of the rod during the m otion is

(A)!

g(B)

!

g2(C) 0 (D)

!

2u


Topics : Circular Motion, Relative Motion, Rectilinear Motion,Topics : Circular Motion, Relative Motion, Rectilinear Motion, Gravitation, Electrostatics, GeometricalGravitation, Electrostatics, GeometricalOpticsOptics

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.21' negative marking) Q.1 to Q.2 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[66, , 66]]Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.31' negative marking) Q.3 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44, , 44]]


Comprehension ('Comprehension (' –1' negative marking) Q.5 to Q.71' negative marking) Q.5 to Q.7 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q. 8× 4) Q. 8 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. A section of fixed smooth circular track of radius R in vertical plane is

shown in the figure. A block is released from position A and leaves the

track at B. The radius of curvature of its trajectory when it just leaves the

track at B is:

(A) R (B)4

R

(C)2

R(D) none of these

2.2. A cyclist observes a passenger in a bus. He finds that the passenger

closed his glass window displacing 20 cm in forward direction with constant

speed in 1 sec. Bus overtakes the cyclist in 3 sec. Initially he was at the

middle of the bus as shown in the figure. Length of the bus is 18 m. Both

cyclist and bus are moving with constant speed in the same direction.

Then velocity of the glass window with respect to cyclist was:

(A) 0.2 m/s (B) 2.8 m/s (C) 3.2 m/s (D) 3 m/s

3.3. A particle is moving rectilinea rly so that its acceleration is give n as a = 3t 2 + 1 m/s 2.Its initial velocity is

zero.

(A) The velocity of the particle at t=1 sec will be 2m/s.

(B) The displacement of the particle in 1 sec will be 2m.

(C) The particle will continue to move in positive direction.

(D) The particle will come back to its starting point after som e time.

4.4. A particle is projected from point A, that is at a distance 4R from the c entre of the earth, with speed V 1

in a direction making 30 ° with the line joining the centre of the earth and point A, as shown. Find the

speed V 1 if particle passes grazing the surface of the earth. Consider gravitational interaction only

between these two. (useR

GM = 6.4 × 107 m2 /s2) Express you answer in the form

2

X m/s and f ill value

of X.

V2

V1

R

4R

A

30°



A point charge q1 = +6e fixed at the srcin of a coordinate system, and another point charge q

2 = – 10e

is fixed at x = 8 nm , y = 0. The locus of all points in the xy plane for which potential V= 0 (other than

infinity) is a circle centered on the x-axis , as shown.

xc q1 q2

y

x

CircleV=0

5.5. Radius R of the circle is

(A) 3nm (B) 6nm(C) 7.5nm (D) 9nm

6.6. x-coordinate of the centre of the circle is -(A) -2nm (B) -3nm (C) -4.5nm (D) -7.5nm

7.7. The potential at the centre of the c ircle is(A) 0.32V (B) 0.77V (C) 1.2V (D) -1.2V

8.8. An object O (real) is placed at focus of an equi-biconvex lens as shown infigure 1. The refractive index of lens is 5 = 1.5 and the radius of curvature of either surface of lens is R. Thelens is surrounded by air. In each statement of column I some changes are made to situation given aboveand information regarding final image formed as a result is given in column-II. The distance between lens andobject is unchanged in all statements of column-I. Match

the statements in column-I with resulting image in column-II.

Column-Column-II Column- Column-IIII(A) If the refractive index of the lens is (p) final image is real

doubled (that is, made 2 5) then

(B) If the radius of curvature is doubled (q) final image is virtual(that is, made 2R) then

(C) If a glass slab of refractive index 5 = 1.5 (r) final image becomes smaller inis introduced between the object size in comparison to size of imageand lens as shown, then before the change was made

R R

O

slab

(D) If the left side of lens is filled with a medium (s) final image is of same size of object.of refractive index 5 = 1.5 as shown, then

R R

O

air


Topics : Topics : RectilinRectilin ear Mear M otionotion , Relative Moti, Relative Motion, Ciron, Circular cular Motion, SoMotion, Sound Wund Wave, Electrostaticsave, Electrostatics

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.21' negative marking) Q.1 to Q.2 ((112 m2 m aarrkks s 112 m2 m iinn..)) [[66, , 66]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.3 to Q.41' negative marking) Q.3 to Q.4 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[88, , 1100]]Comprehension ('Comprehension (' –1' negative marking) Q.5 to Q.71' negative marking) Q.5 to Q.7 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.8× 4) Q.8 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]

1.1. A particle moves along x-axis with initial position x = 0. Its velocityvaries with x-coordinate as shown in graph. The acceleration 'a' of thisparticle varies with x as

(A) (B) (C) (D)

2.2. A coin is released inside a lift at a height of 2 m from the floor of the lift. The height of the lift is10 m. The lift is moving with an acceleration of 9 m/s2 downwards. The time after which the coin will strikewith the lift is : (g = 10 m/s2)

(A) 4 s (B) 2 s (C)21

4 s (D) s

11

2

3.3. A particle is projected from ground with an initial velocity 20 m/sec making an angle 60° with horizontal. If R1

and R2 are radius of curvatures of the particle at point of projection and highest point respectively, then find

the value of2

1

R

R.

4.4. A point source of sound emiting sound of frequency 700 Hz and observer starts moving from a point alongmutually perpendicular directions with velocity 20 m/s and 15 m/s respectively. If change in observed frequencyby observer is 10x Hz then calculate 'x'. [speed of sound in 334 m/sec]


A thin ring of radius R metres is placed in x-y plane such that its

centre lies on srcin. The half ring in region x< 0 carries uniform linear

charge density + 8 C/m and the remaining half r ing in region x> 0

carries uniform linear charge density –8 C/m.

x

y

+ + +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ + + + ++

+

x´

y´

– – – – –

– – – –

–

– – – –

–

– –

–

–

– –

+8 8

5.5. Then the electric potential (i n volts) at point P w hose coordinates are (0m, +R

2m) is

(A) 24

1

o

8&H (B) 0 (C) 44

1

o

8

&H (D) cannot be determined

6.6. Then the direction of electric field at point P whose coordinates are ( 0m, +R

2m) is

(A) Along positive x-direction (B) Along negative x-direction

(C) Along negative y-direction (D) None of these

7.7. Then the dipole moment of the ring in C –m is

(A) – (2&R28) i (B) (2&R28) i (C) – (4R28) i (D) (4R28) i


88.. MMaattcch h tthhe e ccoolluummn n ::


(A) (p) Optical power will be positive

If 52 > 5

1

(B) (q) Optical power will be negative

If 52 > 5

1

(C) (r) System will converge a parallel beam of light incident on it

If 52 < 5

1

(D) (s) Focal length will be positive

(t) Focal length will be negative


Topics : Topics : ElectrostatElectrostat ics, Reics, Re ctilinectiline ar Mar Motion, otion, RelatRelative Motionive Motion, Simp, Simple Harle Harmonic Motimonic Motionon

PHYSICSPHYSICS







1.1. Two infinitely large charged planes having uniform surface charge density + B and –B are placed along x-y plane and yz plane respectiv ely as shown in the figure. Then the nature of electric lines of forces in

x-z plane is given by :

z

x

–B

+B

(A)

z

x (B)

z

x (C)

z

x (D)

z

x

2.2. A point charge placed on the axis of a uniformly charged disc experiences a force f due to the disc. If the

charge density on the disc is ! , the electric flux through the disc, due to the point charge will be :

(A)B&f2

(B)&B2

f(C)

B

2f(D)

Bf

3.3. A particle moves along x-axis in positive direction. Its acceleration 'a' is given as a = cx + d, where x

denotes the x-coordinate of particle, c and d are positive constants. For velocity-position graph of

particle to be of type as shown in figure, the value of speed of particle at x = 0 should be.

(A)c

d4 2

(B)c

d2

(C)c

d2 2

(D)c

d8 2

4.4. P is a point moving with constant speed 10 m/s such that its velocity vector

always maintains an angle 60° with line OP as shown in figure (O is a f ixed

point in space). The initial distance between O and P is 100 m. After what time

shall P reach O.

(A) 10 sec. (B) 15 sec. (C) 20 sec. (D) 20 3 sec



A solid cylinder attached with a horizontal massless spring (see figure) can roll without slipping along a

horizontal surface. If the system is released from rest at a position in which spring is stretched by 0.3 m.

5.5. Period of oscillation of c.o.m of cylinder is

(A)k

M2& (B)

k2

M32& (C)

k3

M22& (D)

k2

M2&

6.6. If k = 20 N/m, Rotational K.E. of cylinder as it passes through mean position is

(A) 0.3 J (B) 0.45 J (C) 0.6 J (D) 0.9 J

7.7. Translation K.E. of cylinder as it passes through mean position is

(A) 0.3 J (B) 0.45 J (C) 0.6 J (D) 0.9 J

8.8. instruments are given. In column 33, there are results corresponding to a given situation. Match the optical

instrument of column 3 with the results in column 33.

CCoolluummn n II CCoolluummnn

(A) concave mirror (p) Parallel rays can be obtained by using point source.

(B) convex mirror (q) Real image can be obtained for real object.

(C) concave lens (r) Virtual and diminished image can be obtained for real

object.

(D) convex lens (s) Real and magnified image can be obtained for real

object.

(t) For a point object moving along optical axis, the image also

moves in the same direction of motion of the object.


Topics : Topics : RectilinRectilin ear Mear M otionotion , Friction, Friction, Cir, Circular cular Motion, Gravitation, Motion, Gravitation, RelatRelative Motionive Motion, Electrostati, Electrostati cscs

PHYSICSPHYSICS


Max. Max. Time Time : 24 : 24 min.min.




1.1. A particle moves along a path ABCD as shown in the figure. Then the magnitude ofnet displacement of the

particle from position A to D is :

(A) 10 m (B) 25 m

(C) 9 m (D) 27 m

2.2. A block of mass 20 kg is acted upon by a force F = 30 N at an angle 53° with the horizontal in downward

direction as shown. The coefficient of friction between the block and the horizontal surface is 0.2. The friction

force acting on the block by the ground is (g = 10 m/s

2

)

F53°

(A) 40.0 N (B) 30.0 N (C) 18.0 N (D) 44.8 N

3.3. The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f.

2 volts and internal resistance 1 ohm. If a cell of e.m.f. 1.2 volts is balanced by it, the balancing lengt h

will be:

(A) 90 cm (B) 60 cm (C) 50 cm (D) 75 cm

4.4. If the apparent weight of the bodies at the equator is to be zero, then the earth should rotate with

angular velocity

(A)g

R r ad/sec (B)

2 g

R rad/sec

(C)g

R2 r ad/sec (D)

3

2

g

R rad/sec


5.5. A particle ‘ A’ is rotating in a circle of radius R with centre at O. Another particle B in the same plane is

resting at point Q which is at a distance d from O . The path of B as seen from A is :

(A) (B) (C) (D)


Two uniformly charged fixed rings, each of radius R are separated by a distance of 3r (r > > R). Ring, 1 has

charge Q and ring 2 has net charge 4Q. A neutral point is a point in space at which the value of electric field

intensity is zero, while its value is non zero at any neighbouring point.

Ring 1 Ring 2

+Q +4Q

66 .. The position of a neutral point N is at a distance of

(A) r from the center of ring 1 towards right (B) 2r from the center of ring 1 towards right

(C) r from the center of ring 1 towards left (D) None of these

7.7. A negatively charged particle is projected from the center of ring 1 towards the neutral point N with just

enough energy to pass the neutral point. Its subsequent motion will be

(A) Oscillatory (B) SHM

(C) Unidirectional (D) Non oscillatory rectilinear but not unidirectional

8.8. A positively charged particle is released from the center of ring 2 by slightly pushing it towards left. Its

subsequent motion will be

(A) Oscillatory (B) SHM

(C) Unidirectional (D) Non oscillatory rectilinear but not unidirectional


Topics : Electrostatics, Friction, Circular Motion, Current ElectricityTopics : Electrostatics, Friction, Circular Motion, Current Electricity

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.21' negative marking) Q.1 to Q.2 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[66, , 66]]Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.31' negative marking) Q.3 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44, , 44]]Comprehension ('Comprehension (' –1' negative marking) Q.4 to Q.81' negative marking) Q.4 to Q.8 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1155, , 1155]]

1.1. The figure shows two large, closely placed, parallel, nonconducting

sheets with identical (positive) uniform surface cha rge densities, and asphere with a uniform (positive) volume charge density . Four pointsmarked as 1, 2, 3 and 4 are shown in the space in between. If E

1, E

2,

E3 and E

4 are magnitude of net electric fields at these points respectively

then :

(A) E1 > E

2 > E

3 > E

4(B) E

1 > E

2 > E

3 = E

4

(C) E3 = E

4 > E

2 > E

1(D) E

1 = E

2 = E

3 = E

4

2.2. A particle of mass 5 kg is moving on rough fixed inclined plane (making anangle 30 ° with horizontal) with constant velocity of 5 m/s as shown in the

figure. Find the fr iction force acting on a body by the inclined plane.( take g = 10m/s 2)

(A) 25 N (B) 20 N (C) 30 N (D) none of these

3.3. AB is a potentiometer wire of length L, which is connected to an accum ulator.

The potential gradient in AB can be decreased by:

(A) increasing the radius of AB (B) decreasing the radius of AB

(C) in creasing a resis tance in se ries with AB (D) all the above.


A high tension wire is at a high potential with respect to a wire which is well grounded called‘Earth wire’. Youmust have seen such wires stretched parallel to roads. There is a high tension wire between two points A andB, 1 km apart. The distance between HT wire and earth wire is 1 m. The resistance of the HT (and also theearth wire) is 1 O / m. This wire is at a potential of 11 KV at point A w.r. to earth wire., and its is carrying 1Acurrent which returns back to the generator by through the earth wire. This wire is quite a thick wire. There isa sign board at a pole over which this wire is stretched reading ‘DANGER, 11 KV’. You might think what

would happen if one touched this wire. Will one feel a shock or not. Well ! it depends on whether the currentthrough our body exceeds a particular value, which we may call CRITICAL CURRENT.

4.4. Why is H.T. wire thick wire. Select the most appropriate option(A) so that more current may flow(B) so that resistance may be less thereby reducing power loss in the transmission line.(C) so that it may bear high tension & therefore sag less.(D) so that when in future population increases, the existing wire may serve the purpose.

5.5. Consider a bird having effective resistance 10O between its feet sitting on this high tension wire. The dis-tance between its feet is 10 cm. Find the potential difference between the feet of the bird is approximately.(A) 0.1 V (B) 1 V (C) 10 V (D) 0.05 V

6.6. In the above question find the current through the bird.

(A) 10 A (B) 1 A (C) 0.01 A (D) 0.005 A

7.7. If the potenti al difference between H. T. and earth wire is 11 kV at point A, find p.d. between these wiresat point B.(A) 1 KV (B) 2 KV (C) 9 KV (D) 10 KV

8.8. If the ‘Critical Current ’ for the bird is 0.1 A, find the maximum power at 11 KV can be transmitted at

point A so that the bird m ay not get shock. Assume that the distance between the feet is 10 cm .

(A) 111 KW (B) 11 KW (C) 101 KW (D) 110011 KW


Topics : Electrostatics, Kinematics, Current ElectrisityTopics : Electrostatics, Kinematics, Current Electrisity , Center of , Center of Mass, NewtonMass, Newton ’s Law of Motions Law of Motion

PHYSICSPHYSICS







1.1. There exists a uniform electric field in the space as shown. Four points

A, B, C and D are marked which are equidistant from the srcin. If

VA,V

B, V

C and V

D a re their potentials respectively, then 30°A

D

C

B

E

y

x

(A) VB > V

A > V

C > V

D(B) V

A > V

B > V

D > V

C

(C) VA = V

B > V

C = V

D(D) V

B > V

C > V

A > V

D

2.2. The displacement time graphs of two bodies A and B are shown in figure. The ratio of velocity of A, vA to

velocity of B, vB is :

(A)3

1(B) 3

(C)3

1(D) 3

3.3. In the circuit shown, the galvanometer shows zero current.

The value of resistance R is :

(A) 1 O (B) 2 O(C) 4 O (D) 9 O

4.4. A disc of mass ‘m’ and radius R is f ree to rotate in horizontal plane about

a vertical smooth fixed axis passing through its centre. There is a smooth

groove along the diameter of the disc and two small balls of mass2

m each

are placed in it on either side of the centre of the disc as shown in fig. The

disc is given initial angular velocity '0and released. The angular speed of

the disc when the balls reach the end of disc is :

(A)20'

(B)3

0'(C)

3

2 0'(D)

40'

5.5. A small block A is placed on a smooth inclined wedge B which is placed on a horizontal smooth surface. B

is fixed and A is released from top of B. A slide down along the incline and reaches bottom in time t1. In

second case A is released from top of B, but B is also free to move on horizontal surface. The block A takes

t2 time to reach bottom. Without actually calculating the values of t

1 and t

2 find which is greater.


A car battery with a 12 V emf and an internal resistance of 0.04 O is being charged with a cur rent of 50 A.

6.6. The potential difference V across the terminals of the battery are

(A) 10 V (B) 12 V (C) 14 V (D) 16 V

7.7. The rate at which energy is being dissipated as heat inside the battery is :

(A) 100 W (B) 500 W (C) 600 W (D) 700 W

8.8. The rate of energy conversion from electrical form to chemical form is :

(A) 100 W (B) 500 W (C) 600 W (D) 700 W


TopicTopics : Current s : Current ElectricityElectricity, Capacit, Capacitance, Sound Wance, Sound Wave, Kinematicsave, Kinematics

PHYSICSPHYSICS







11.. An ammeter and a voltmeter are initially connected in series to a battery of

zero internal resistance. When switch S1 is closed the reading of the

voltmeter becomes half of the initial, whereas the reading of the ammeter

becomes double. If now switch S2 is also closed, then reading of amm eter

becomes:

(A) 3/2 times the initial value

(B) 3/2 times the value after closing S1

(C) 3/4 times the value after closing S1

(D) 3/4 times the initial value

2.2. A circuit has a section AB shown in the figure. The emf of the source

equals H = 10V, the capacitance as of the capacitors are equal to C1 = 1.0

5F and C2 = 2.0 5F, the potential difference ;

A ! ;

B = 5.0V. The voltage

across capacitor C1 & C

2 i s respectively :

(A) 10/3 V, 5/3 V (B) 10/3 V, 10/3 V (C) 5/3 V, 5/3 V (D) 0 V,0 V

3.3. Under similar conditions of temperature and pressure, In which of the following gases the

velocity of sound will be largest.(A) H

2(B) N

2(C) He (D) CO

2

4.4. A rigid body is made of three identical thin rods fastened together in

the form of letter H. The body is free to rotate about a fixed horizontal

axis AB that passes through one of the legs of the H. The body is

allowed to fall from rest from a position in which the plane of H is

horizontal. What is the angular speed of the body when the plane of H

is vertical. To p View of the figure in the initial position.

L

L

LA

B

5.5. A man holding a flag is running in North-East direction with speed 10 m/s. Wind is blowing in east direction

with speed 25 m/s. Find the direction in which flag will flutter..


AB is a uniform wire of meter bridge, across which an ideal 20 volt cell is

connected as shown. Two resistor of 1 O and X

O are inserted in slo ts of

metre bridge. A cell of emf E volts and internal resistance r O and a

galvanometer is connected to jockey J as shown. 20V

A BJ

E,r

1 X

P

6.6. If E = 16 volts, r = 4 O and distance of balance point P from end A is 90 cm, then the value of X is :

(A) 3 O (B) 6

O (C) 9

O (D) 12

O

7.7. If E = 16 volts, r = 8 O and X = 9

O, then the distance of balance point P from end A is :

(A) 10 cm (B) 30 cm (C) 60 cm (D) 90 cm

8.8. If E = 12 volts, X = 9 O, then distance of balance point P f rom end A is

(A) 20 cm (B) 50 cm (C) 70 cm (D) Data insufficient


TopicTopics : Currs : Current ent ElectricityElectricity, , Capacitance, Capacitance, Sound, ElectroSound, Electrostatstaticsics

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.31' negative marking) Q.1 to Q.3 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.41' negative marking) Q.4 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44, , 44]]Comprehension ('Comprehension (' –1' negative marking) Q.5 to Q.81' negative marking) Q.5 to Q.8 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1122, , 1122]]

1.1. In the circuit shown, cells are of equal emf E but of different internalresistances r

1 = 6W and r

2 = 4W . Reading of the ideal voltmeter connected

across cell 1 is zero. Value of the external resistance R in ohm is equal to-

Cell1

E E

Cell2

r1

R

r2

V

(A) 2 (B) 2.4 (C) 10 (D) 24

2.2. The circuit was in the shown state for a long time. Now if the switch S isclosed then the net charge that f lows through the switch S, will be

S

4 F5

50V

2 F5 4 F5

2 F5

(A)3

4005C (B) 100 5C

(C)3

1005C (D) 50 5C

3.3. The equation of displacement due to a sound wave is s = s0 sin2 ('

t ! k

x). If the bulk modulus of the medium

is B, then the equation of pressure variation due to that sound is(A) B k s

0 sin (2 '

t ! 2 k

x) (B) ! B k s

0 sin (2 '

t ! 2 k

x)

(C) B k s0 cos2 ('

t ! k

x) (D) ! B k s

0 cos2 ('

t ! k

x)

4.4. Four point charges are fixed on y-axis as shown in figure. When apositive test charge at srcin is given a finite impulse along positivex-axis,which of the following is/are correct option/options :(A) Test charge may pass through srcin with same momentum once again.

–Q

2Q

–Q

2Q

(0, –2a)

(0, –a)

(0,a)

(0,2a)

x

y

(B) Test charge must return at srcin at least once.(C) Test charge may perform SHM on x-axis.(D) Test charge may reach upto infinite distance on positive x-axis.


A wire AB (of length 1m , area of cross section &m2) is used inpotentiometer experiment to calculate emf and internal resistance (r) ofbattery. The emf and internal resistance of driving battery are 15 V and 3Orespectively. The resistivity of wire AB varies as$=$

0x (where x is distance

from A in meters and $0 = 24& O)

15V

E

A B

rG

S

The distance of null point from A is obtained at3

2m when switch 'S' is open and at

2

1m when switch is

closed.

5.5. The resistance of whole wire AB is -(A) 6O (B) 12O (C) 18O (D) 24O

6.6. The current through 15 V battery (with 3G = 0)

(A) is 1 A only when switch S is closed (B) is 1A only when switch S is open(C) is 1A in both cases (D) can not be calculated.

7.7. The emf (E) of battery is -(A) 4V (B) 6V (C) 8V (D) 10V

8.8. Internal resistance of battery E is -(A) 4O (B) 3O (C) 2O (D) 1O


TopicTopics : s : Current ElectricityCurrent Electricity, Sound, Fluids, Capacitor, Rotation, Sound, Fluids, Capacitor, Rotation

PHYSICSPHYSICS






1.1. An ammeter A of finite resistance and a resistor R are joined in series to an ideal cell C. A potentiometer

P is joined in parallel to R. The ammeter reading is I 0 & the potentiometer reading is V 0. P is now

replaced by a voltmeter of finite resistance. T he ammeter read ing now is I and the voltmeter reading is

V.

(A) I > I 0, V > V 0 (B) I > I 0, V < V 0

(C) I = I 0, V < V 0 (D) I < I 0, V = V0.

2.2. The two pipes are submerged in sea water, arranged as shown in figure. Pipe A w ith length LA = 1.5 m

and one open end, contains a small sound source that sets up the standing wave with the secondlowest resonant f requency of that pipe. Sound f rom pipe A sets up resonance in pipe B, which has both

ends open. The reso nance is at the second lowest resonant f requency of pipe B. The length of the pipeB is :

(A) 1 m (B) 1.5 m (C) 2 m (D) 3 m

3.3. A body of density $ is dropped from rest from a height 'h' (fr om the surface of water) into a lake of density ofwater B (B > $). Neglecting all dissipative effects, the acceleration of body while it is in the lake is:

(A) g 1!$B u pwards (B) g 1!$

B downwards

(C) g $B u pwards (D) g $

B downwards

4.4. In the above problem, the maximum depth the body sinks before returning is:

(A)$!B

$h(B)

$"B$h

(C) hB$

(D) h$

B


5.5. Find the potential difference between points A and B of the system shown in the figure, if the emf is equal to

" = 110V and the capacitance ratio1

2

C

C = I = 2.0.


A thin uniform rod having mass m and length 4 ! is free to rotate about horizontal axis passing through

a point distant ! from one of its end, as shown in f igure. It is released, from the horizontal position asshown :

x

6.6. What will be acceleration of centre of mass at this instant

(A)7

g3(B)

7

g2(C)

5

g3(D)

5

g2

7.7. What will be normal reaction due to hinge at the instant of release

(A) mg (B)2

mg(C)

7

mg4(D)

7

mg2

8.8. What will be angular velocity of rod when it becomes vertical

(A)!7

g6(B)

!7

g12(C)

!2

g3(D)

!7

g3


Topics : Rotation, CapaciTopics : Rotation, Capaci tor, Center of Mass, Gravittor, Center of Mass, Gravit ation, Current ation, Current ElectElectricitricityy, , ElectrosElectros taticstatics

PHYSICSPHYSICS








1.1. A uniform thin rod is bent in the form of a closed loop ABCDEFA as shown in the figure.

(A) > 1 (B) < 1

(C) = 1 (D) = 1/2

2.2. An uncharged parallel plate ca pacitor is connected t o a battery. The electr ic field between the plates is

10V/m. Now a dielectric of dielectric constant 2 is inserted between the plates filling the entire space.

The electric field between the plates now is

(A) 5 V/m (B) 20 V/m

(C) 10 V/m (D) none of these

3.3. In the figure shown, coefficient of restitution between A and B is e =2

1, then :

(A) velocity of B after collision is2

v

(B) impulse on one of the balls during collision is4

3mv

(C) loss of kinetic energy in the collision is 16

3mv2

(D) loss of kinetic energy in the collision is4

1mv2


4.4. A spherical planet has uniform density2

& × 104 kg/m 3. Find out the minimum period for a satellite in a

circular orbit around it in seconds. (Use G =3

20 × 10 –111 2

2

kg

mN !).

55.. Two potentiometer wires w 1 and w 2 of equal length ! connected to a battery of emf HP and internal

resistance ' r ' through two switches s 1 and s 2. A battery of emf H is balanced on these potentiometer

wires. If potentiometer wire w 1 is of resistance 2r and balancing length on w 1 is ! /2 when only s1 is

closed and s 2 is open. On closing s 2 and opening s 1 the balancing length on w 2 is found to be - .

/01

2 3

2!,

then find the resistance of potentiometer wire w 2.


Two conducting spheres of radius R and 3R carry charges Q and –2Q. Between these spheres a neutral

conducting sphere of radius 2R is connected. The separation between the sphere is considerably large.

Charge flows through conducting wire due to potential difference.

6.6. The final charge on initially neutral conducting sphere is :

(A)6

Q! (B)

3

Q! (C)

3

Q(D)

2

Q!

7.7. The decrease in electric potential energy of sphere of radius R is :

(A)R4

KQ2

(B)R72

KQ35 2

(C)R72

KQ2

(D) none

8.8. The final electric potential of sphere of radius 3R will be :

(A)R6

KQ! (B)

R2

KQ! (C)

R3

KQ2! (D)

R

KQ3!


Topics : Center oTopics : Center o f Mf M ass, Rotation, Current ass, Rotation, Current ElectricityElectricity , Capacitor, Gravitation, Electrostatics, Capacitor, Gravitation, Electrostatics

PHYSICSPHYSICS







1.1. A projectile of range R bursts at its highest point in two fragments. Both pieces at the time of bursts have the

velocity in the horizontal direction. The heavier is double the mass of the lighter. Lighter fragment falls at2

R

horizontal distance from the point of projection in the opposite side of projection. The distance, where other

part falls, from point of projection.

(A)4

R7(B)

4

R5(C)

4

R8(D)

4

R6

2.2. There are four arrangements of a cylinder and a plank as shown in the figures. Some surfaces are smooth

and some are rough as indicated. There is no slipping at each rough surface. String always remanse tight.

The plank and/or centre of cylinder are given a horizontal constant velocity as shown in each of the situations.

Using this information fill in the blanks.

(i)

rough

Vplank

smoothC

The speed of center of mass of the cylinder is ___________.

(ii)rough

V

C

V

plank

plank

The angular velocity of the cylinder is __________.

(iii) rough

V

C

plank

The speed of center of mass of the cylinder is _________ .

(iv) Smooth

V

C3V

plank

The angular velocity of the cylinder is ___________ .

(a) V (b) V/R (c) 2V/R (d) 4V/R

(e) cannot be determined from the given information

(f) Zero.(A) (i) d (ii) b (iii) f (iv) c (B*) (i) e (ii) b (iii) f (iv) c(C) (i) e (ii) d (iii) f (iv) c (D) (i) e (ii) b (iii) f (iv) a


33.. In a practical wheat stone bridge circuit as shown, when one more resistance of 100 O is connected in

parallel with unknown resistance ' x ', then ratio !1 / !2 become ' 2 '. !1 is balance length. AB is a uniform

wire. Then value of ' x ' must be:

(A) 50 O (B) 100 O (C) 200 O (D) 400 O

4.4. A parallel plate capacitor (without dielectric) is charged and disconnected from a battery. Now a dielectric

is inserted between the plates. The electric force on a plate of the capacitor will:

(A) decrease (B) increase

(C) remain same (D) depends on the width of the dielectric.

5.5. Assuming that the law of gravitation is of the form F = 3r

GMmand attractive. A body of mass m revolves in a

circular path of radius r around a fixed body of mass M. Find on what power of r will the square of time period

depend.


Gravitational force is conservative and medium independent force. Its nature is attractive. Gravitational field

and gravitational potential are related as.

Eg = –

r

V

PP

where Eg is component of gravitational field along 'r'. Since gravitational force is attractive so

gravitational potential will always be zero or negative assuming potential to be zero at large distance. With

help of above discussions answer the following questions.

6.6. A person brings mass of 1 kg from infinity to a point 'A'. Initially the mass was at rest but it moves with speed

2 m/s as it reaches A. If the work done by the person on the mass is – 3J. The potential of A is :

(A) – 3 J/kg (B) + 3 J/kg (C) – 5 J/kg (D) 5 J/kg

7.7. The gravitational potential inside a hollow sphere of mass M (uniformly distributed), radius R at a distance r

from the centre is-

(A) zero (B) – r

GM(C) –

R

GM(D) – 2r

GM

8.8. Gravitational potential V/s x graph is shown. The magnitude of field intensity is equal to

(A) 8 N/kg (B) 4 N/kg

(C) 6 N/kg (D) 2 N/kg


Topics : Electrostatics, Current ElectricityTopics : Electrostatics, Current Electricity , Capacitor, Gravitation, Rotation, Center , Capacitor, Gravitation, Rotation, Center of Mof M assass

PHYSICSPHYSICS







1.1. Figure shows a solid metal sphere of radius a surrounded by a concentric thin metal shell of radius 2a.Initially both are having charges Q each. W hen the two are connected by a conducting wire as shown

in the figure, then amount of heat produced in this process will be

2aa

Q

Q

(A)a2

KQ2

(B)a4

KQ2

(C)

a6

KQ2

(D)

a8

KQ2

22.. Circuit for the measurement of resistance by potentiometer is shown. The galvanometer is first connected

at point A and zero deflection is observed at length PJ = 10 cm. In second case it is connect at point

C and zero deflection is observed at a length 30 cm from P. Then the unknown resistance X is

G

AR X

C

P QJ

(A) 2R (B)2

R

(C)3

R(D) 3R


3.3. In the figure shown P1 and P2 are two conducting plates having charges of equal magnitude and opposite

sign. Two dielectrics of dielectric constant K1 and K2 fill the space between the plates as shown in the figure.

The ratio of electrical energy in 1st dielectric to that in the 2nd dielectric is

(A) 1 : 1 (B) K 1 : K2

(C) K2 : K1 (D) K22 : K1

2

4.4. The gravitational potential energy of a satellite revolving around the earth in circular orbit is – 4 MJ. Find the

additional energy (in MJ) that should be given to the satellite so that it escapes from the gravitational field of

earth. Assume earth's gravitational force to be the only gravitational force on the satellite and no atmospheric

resistance.

5.5. A uniform circular disc has radius R and mass m. A particle also of mass m is fixed at a point A on the

periphery of the disc as shown in figure. The disc can rotate freely about a fixed horizontal chord PQ that is

at a distance R/4 from the centre C of the disc. The line AC is perpendicular to PQ. Initially the disc is held

vertical with the point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. Find

the linear speed of the particle when it reaches its lowest position. Acceleration due to gravity is g.


Two friends A and B (each weighing 40 kg) are sittin g on a frictionless platfor m some distance d apar t.

A rolls a ball of mass 4 kg on t he platform towards B which B catches. Then B rolls the ball towards A

and A catches it. Th e ball keeps on m oving back and f orth between A and B. The ball has a fixed speed

of 5 m/s on the platform.

6.6. Find the speed of A after he rolls the ball for the first time

(A) 0.5 m/s (B) 5m/s (C) 1 m/s (D) None of these

7.7. Find the speed of A af ter he catches the ball for the f irst time.

(A)21

10 m /s (B)

11

50 m /s (C)

11

10 m /s (D) None of these

8.8. Find the speeds of A and B after the ball has m ade 5 round trips and is held by A :

(A)11

10 m/s ,

11

50m/s (B)

11

50m/s,

11

10m/s (C)

11

50m/s, 5 m/s (D) None of these

9.9. How many times can A roll the ball ?(A) 6 (B) 5 (C) 7 (D) None of these

10.10. Where is the centre of mass of the system “ A + B + ball ” at the end of the nth trip? (Give the distance

from the initail position of A)

(A)11

10 d (B)

21

10d (C)

11

50 d (D) None of these


Topics : Current ElectricityTopics : Current Electricity , Heat, , Heat, Kinematics, Simple Harmonic Motion, Kinematics, Simple Harmonic Motion, CapacitanceCapacitance

PHYSICSPHYSICS







1.1. In the, Ohm's law experiment to f ind resistance of unknown resistor R, the arr angement is as shown.

The resistance measured is given by R measured =i

V, V = voltage reading of voltmeter, i = current

reading of ammete r. The ammete rs and the voltmeter used are not ideal, and have resistances R A and

RV respectively . For a rrangement shown , the measured resistance is

(A) R + R V (B) R + R A (C)V

V

RR

RR

" (D)V

V

RR

RR

" + RA

2.2. Two identical plates with thermal conductivities K and 3K are joined together to form a single plate of double

thickness. The equivalent thermal conductivity of one composite plate so formed for the flow of heat through

its thickness is:

(A) K (B) 1.5 K (C) 2.5K (D) 3K

3.3. The acceleration time graph of a particle moving along a straight line is as shown in the figure. At what

time the particle acquires its velocity equal to initial velocity ?

(A) 12 sec (B) 5 sec

(C) 8 sec (D) none of these

4.4. As the distance between the plates of a parallel plate capacitor is decreased

(A) chances of electrical break down will increase if potential difference between the plates is keptconstant.

(B) chances of electrical break down will decrease if potential difference between the plates is kept

constant.

(C) chances of electrical break down will increase if charge on the plates is kept constant.

(D) chances of electrical break down will decrease if charge on the plates is kept constant.


5.5. All linear dimensions are doubled then the capacitance of the parallel plate capacitor will

(A) Remain unchanged

(B) become double

(C) increase by eight times

(D) increases by four times

6.6. The x-coordinate of a particle moving on x-axis is given by x = 3 sin 100t + 8 cos2 50t, where x is in cm and

t is time in seconds. Which of the following is/are correct about this motion.

(A) the motion of the particle is not S.H.M.

(B) the amplitude of the S.H.M. of the particle is 5 cm

(C) the amplitude of the resultant S.H. M. is 73 cm

(D) the maximum displacement of the particle from the srcin is 9 cm.


Charges are concentrated exclusively on the external surface of a conductor. Therefore neither the

material of the conductor nor its mass are of any importance for its capacitance. Capacitance depends

on the shape and surface area of th e conductor. Since a conductor liable to be electrified by induction,

its capacitance is influenced by other conductors in its vicinity and by the medium they are in.

To fulfil its function the capacitor must be able to store accumulated charges and energy for appreciable

time to obtain a definite capacitance. One can conveniently ta ke two conductors and arrange them as

close to each other as possible and place a dielectric between them. The dielectric between the

conductors plays two fold role. Firstly, it increases the capacitance and secondly prevents the

neutralizatio n of the charges, i.e. it prevents them from jumping from one conductor to the other. For

this reason electrical breakdown strength (The maximum electrical field which a dielectric can withstand,

also called dielectric strength of dielectric) should be high.

In order to keep the capacitance constant & independent of surrounding bodies the entire electric field

should be contained between the plates. For this reason the distance between the plates should be

small as compound to their linear dimensions. To protect the capacitor from external influences it is

housed in a shell. Mathem atically we can relate q = CV and for a para llel plate capacitor fully filled with

a dielectric C = d

A0rHH

.

7.7. Capacitance does not depend on

(A) Area of plates (B) Separation between plates

(C) Surrounding bodies (D) Potential difference across plates

8.8. Two plates of a capacitor kept on ins ulating stand are fully charged. Now the ebonite plate between the

capacitor plates is removed then the capacitance of capacitor will

(A) Increase (B) Decrease

(C) Remains same (D) May increase or decrease

9.9. A table for parallel plate capacitors along with the properties of dielectrics used in these is given.

Choose the most appropriate capacitor.

(Assuming same potential difference across each capacitor)

CapacitorCapacitorDielectricDielectric

ConstantConstant

DielectricDielectric

strengthstrength

(V/m)(V/m)

DistanceDistance

betweenbetween

plates (m)plates (m)

Area ofArea of

plates (mplates (m 22))

A 2.8 3 × 107 0.01 0.125

B 3.3 6 × 107 0.01 0.125

C 2.2 7 × 107 0.01 0.25

D 4.4 1 × 107 0.01 0.125

(A) A (B) B (C) C (D) D


Topics : Heat, Simple Harmonic Topics : Heat, Simple Harmonic Motion, Current Electricity, Kinematics, CapacitanceMotion, Current Electricity, Kinematics, Capacitance

PHYSICSPHYSICS







1.1. A wall has two layer A and B each made of different material, both the layers have the same thickness. The

thermal conductivity of the material A is twice that of B. Under thermal equilibrium the temperature differenceacross the wall B is 360C. The temperature difference across the wall A for the flow of heat through itsthickness is(A) 60C (B) 12 0C (C) 18 0C (D) 24 0C

2.2. If the length of a simple pendulum is doubled then the % change in the time period is :(A) 50 (B) 41.4 (C) 25 (D) 100

3.3. In the circuit shown, the reading of the ammeter (ideal) is the same with both switches open as withboth closed. Find the value of r esistance R in ohm.

E

R

100O

+ –

A

50O

300O 1.5V

4.4. A particle moves along X axis. At t = 0 it was at x = – 1. It’s velocity varies with time as shown in the figure.Find the number of times the particle passes through the srcin in t = 0 to t = 10 sec.

COMPREHENSIONCOMPREHENSIONThe circuit contains ideal battery E and other elements arranged as shown. The capacitor is initiallyuncharged and switch S is closed at t = 0. (use e 2 = 7.4)

5.5. Time constant of the circuit is(A) 48 5s (B) 28.8 5s (C) 72 5s (D) 120 5s

6.6. The potential difference across the capacitor in volts, after two time constants, is approximately :

(A) 2 (B) 7.6 (C) 10.4 (D) 12

7.7. The potential difference across resistor R1 after two time c onstants, is approximately :

(A) 1.6 V (B) 7.6 V (C) 10 V (D) 12 V

8.8. The potential difference across resistor R2 after two time constants, is :

(A) 2V (B) 7.6V (C) 10V (D) 12 V


Topics : Heat, Kinematics, Simple Harmonic MotTopics : Heat, Kinematics, Simple Harmonic Motion, Viscosity, Elasticityion, Viscosity, Elasticity, Capacitance, Current Electricity, Capacitance, Current Electricity

PHYSICSPHYSICS







1.1. All the rods have same conductance‘K

’ and same area of cross section

‘ A

’. If ends A and C aremaintained at temperature 2T

0 and T

0 respectively then which of the f ollowing is/are correct:

(A) Rate of heat flow through ABC, AOC and ADC is same

(B) Rate of heat flow through BO and OD is not same

(C) Total Rate of heat f low from A to C isa2

TAK3 0

(D) Temperature at junctions B, O and D are same

2.2. For which of the following graphs the average velocity of a particle moving along a straight line for time interval

(0, t) must be negative -

(A) (B) (C) (D)

3.3. In the figure shown a block of mass m is attached at ends of two

springs. The other ends of the spring are fixed. The mass m is released

in the vertical plane when the spring are relaxed. The velocity of the

block is maximum when:

(A) k 1 is compressed and k 2 i s elongated (B) k 1 is elongated and k 2 is compressed

(C) k1 and k

2 b oth are compressed (D) k

1 and k

2 both are elongated.

4.4. Rigidity modulus of steel is I and its young’s modulus is Y. A piece of steel of cross –sectional area ‘ A’ is

changed into a wire of length L and area A/10 then :

(A) Y increases and I d ecrease (B) Y and I remains the same

(C) both Y and I i ncrease (D) both Y and I decrease


5. A container filled with viscous liquid is moving vertically downwards with constant speed 3v 0. At the

instant shown, a sphere of radius r is moving vertically downwards (in liquid) has speed v0. The coefficient

of viscosity is I. There is no relative motion between the liquid and the container. Then at the shown

instant, the magnitude of viscous force acting on sphere is

(A) 6 & I r v0 (B) 12 & I r v0

(C) 18 & I r v0 (D) 24 & I r v0

6.6. Find the equivalent capacitance between terminals ‘ A’ and ‘B’. The letters have their usual meaning.


A galvanometer measures current which passes through it. A galvanometer can measure typically

current of order of mA. To be able to measure currents of the order of am peres of main current, a shunt

resistance 'S' is c onnected in parallel with the galvanometer.

7.7. The resistance of the shunt 'S' and resistance 'G' of the galvanometer should have the following relation.

(A) S = G (B) S >> G (C) S << G (D) S < G

8.8. If resistance of galvanometer is 10 O and maximum current i g is 10mA then the shunt resistance

required so that the main current ' 3 ' can be upto 1A is (in O)

(A)10

99(B)

99

10(C) 990 (D)

1000

99


Topics : ElasticityTopics : Elasticity , Work, Power , Work, Power and Energyand Energy , Rotation, Heat, , Rotation, Heat, Current ElectricityCurrent Electricity

PHYSICSPHYSICS







1.1. A steel rod (Young’s modulus = 2 x 10 –

11 Nm –

2) has an area of corss -section 3 x 10 –

4 m2 and length 1 m.A force of 6 x 10 N stretches it axially . The elongation of the rod is

(A) 10 -4 m (B) 5 x 10 -3 m (C) 10-3 m (D) 5 x 10 -2 m

2.2. Which of the following statement is not true?

(A) Work done by conservative force on an object depends only on the initial and final states and not on

the path taken.

(B) The change in the potential energy of a system corresponding to conservative internal forces is equal to

negative of the work done by these forces.

(C) If some of the internal forces within a system are non-conservative, then the mechanical energy of the

system is not constant.

(D) If the internal forces are conservative, the work done by the internal forces is equal to the change in

mechanical energy.

3.3. A uniform rod of mass m and length L lies radially on a disc rotating with angular speed ' in a horizontal

plane about its axis. The rod does not slip on the disc and the centre of the rod is at a distance R from the

centre of the disc. Then the kinetic energy of the rod is :

(A)2

1m'2 -

- .

/001

2 "

12

LR

22

(B)2

1m'2 R2 (C)

24

1 m'2 L2 (D) None of these

4.4. The upper end of the string of a simple pendulum is f ixed to a vertical z - axis and set in motion such

that the bob moves along a horizontal circular pa th of radius 2 m , parallel to the x y plane, 5 m above the

srcin. The bob has a speed of 3 m/s. T he string breaks when the bob is vertically above the x - axis , and

it lands on the xy plane at a point (x, y)

(A) x = 2 m (B) x > 2 m(C) y = 3 m (D) y = 5 m

5.5. The rate of heat energy emitted by a body at an instant depends upon

(A) area of the surface

(B) difference of temperatur e between the surface and its surroundings

(C) nature of the surface

(D) None of these



AB is a uniform wire of length 70 cm and resistance 7 O. Part AC is 20 cm long.Two resistors and two

ideal cells are connected as shown.

A B

1V

C

2V

6.6. Potential gradient of the part CB of the wire is :(A) 2.5V/m (B) 2V/m

(C)3

10V/m (D) 7.5V/m

77. Potential gradient of the part AC is :

(A)6

5V/m (B) 2V/m

(C) 5 V/m (D) 7.5V/m

8.8. Of the points A, B and C the poten tial is maximum at point:

(A) A (B) B

(C) C (D) same at all of these three points


Topics : Heat, Work, Topics : Heat, Work, Power and Energy, Rotation, ElasticityPower and Energy, Rotation, Elasticity , Current Electricity, Current Electricity

PHYSICSPHYSICS







1.1. The energy radiated per unit area per sec. by a spherical black body will be doubled if its(A) radius is increased by nearly 41.5% (B) radius is doubled

(C) temp. (T) is increased by nearly 41.5% (D) T is increased by nearly 19%.

2.2. A body of mass 6 kg is acted upon by a force which causes a displacement in it given by x =4

2t

metre where

t is the time in second. The work done by the force is 2 seconds is:

(A) 12 J (B) 9 J

(C) 6 J (D) 3 J

3.3. In the figure shown, a small ball of mass 'm' can move without sliding in a fixed semicircular track of

radius R in vertical plane. It is released from the top. The resultant force on the ball at the lowest point

of the track is

(A)7

mg10(B)

7

mg17

(C)7

mg3(D) zero

4.4. An elastic rod will change its length, if

(A) the rod is suspended at one end

(B) The rod is allowed to fall freely under gravity

(C) the rod is rotated about one end on a frictionless horizontal table

(D) the rod is given a horizontal acceleration by a force applied at one end

5.5. A charged particle X moves directly towards another charged particle Y . For th e 'X + Y' system, the

total momentum is p and the total energy is E.

(A) p & E are conserved if both X & Y are free to move

(B) (A) is true only if X and Y have similar charges

(C) If Y is f ixed, E is conserved but not p

(D) If Y is fixed, neither E nor p is conserved.



Resistance value of an unknown resistor is calculated using the f ormulaI

VR * where V and I be the

readings of the voltmeter and the ammeter respectively. Consider the circuits below. The internal

resistances of the voltmeter and the ammeter (Rv and R

G respectively ) are finite and non zero.

E

R

r

V

A

Figure (A)

E

R

r

V

A

Figure (B)

Let RA and R

B be the calculated values in the two cases A and B respectively.

6.6. The relation between RA and the actual valu e R is

(A) R > RA

(B) R < RA

(C) R = RA

(D) dependent up on E and r.

7.7. The relation between RB and the actual value R is :

(A) R < RB

(B) R > RB

(C) R = RB

(D) dependent up on E and r.

8.8. If the resistance of voltmeter is Rv = 1 k O and that of ammeter is R

G = 1 O, the magnitude of the

percentage error in the measurement of R (the value of R is nearly 10 O) is :

(A) zero in both cases (B) non zero but equal in both cases

(C) more in circuit A (D) more in circuit B


Topics : ElasticityTopics : Elasticity , Rotation, Heat, , Rotation, Heat, NewtonNewton ’s Law of Ms Law of M otion, Work, Power and Energy, Capacitotion, Work, Power and Energy, Capacit anceance

PHYSICSPHYSICS








1.1. Two steel wires, where one has twice diameter and three t imes the length of t he other, are stretched by

the same force. The ra tio of the elastic strain energy stored in them is

(A) 2 : 3 (B) 3 : 4 (C) 3 : 2 (D) 6 : 1

2.2. Two men of equal masses stand at opposite ends of the diameter of a turntable disc of a certain mass,

moving with constant angular velocity . The two men make their way to the m iddle of the turntable at

equal rates. In doing so

(A) kinetic energy of rotation has increased while angular momentum remains sam e.

(B) kinetic energy of rotation has decreased while angular momentum remains same.

(C) kinetic energy of rotation has decreased but angular momentum has increased.

(D) both, kinetic energy of rotation and angular mom entum have decreased.

3.3. A straight nicrome wire is initially at room temperature 20ºC. It is connected to an ideal battery of 500 volt.

Jusr after switching on, the current detected is 5 amp. Due to heating effect its temperature increases, and

is also loosing heat to the environment according to newton's cooling law asdt

dQloss= 45(T – 20ºC)J/sec. At

steady state, the current detected is 4.5 amp.

(A) staedy state temperature of the wire is 70 ºC

(B) steady state temperature of the wire is 75.5ºC

(C) temperature co –efficient of resistance of the wire is nearly 2.2 × 10 –3 / ºC

(D) temperature co –efficient of resistance of the wire is nearly 1.57 × 10 –3 / ºC

4.4. A painter is applying force himself to raise him and the box with an acceleration of 5 m/s 2 by a massless

rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If

g = 10 m/s 2, then:

(A) tension in the rope is 1125 N

(B) tension in the rope is 2250 N

(C) force of contact between the painter and the f loor is 375 N

(D) none of these


5.5. A block of mass 60 kg is released from rest when compression in the spring is 2m (natural length of spring

is 8m). Surface AB is smooth while surface BC is rough. Block travels x distance before coming to complete

rest. Value of x is : [g = 10 m/s2]


The circuit consists of two resistors (of resistance R 1 * 20 O and R2 = 10 O) , a capacitor ( of capacitance

C = 10 5F) and two ideal cells. In the circ uit shown the capacitor is in steady state and the switch S is

open

6.6. The current through the resistor R 2 just after the switch S is closed is:

(A) 1 ampere (B) 2 ampere (C) 3 ampere (D) 4 ampere

7.7. The charge on capacitor in steady state with switch S closed.

(A) 1005C (B) 200 5C (C) 300 5C (D) 400 5C

8.8. The circuit is in steady state with switch S closed. Now the switch S is opened. Just after the switch

S is opened, the current through resistance R 1 is

(A) 1ampere (B) 2 ampere (C) 3 ampere (D) 4 ampere


Topics : Heat, Current Electricity, Magnetic Effect of Current and Magnetic Force on Charge/current,Topics : Heat, Current Electricity, Magnetic Effect of Current and Magnetic Force on Charge/current,Rotation, Kinamatics, Center of MassRotation, Kinamatics, Center of Mass

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.51' negative marking) Q.1 to Q.5 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1155, , 1155]]Comprehension ('Comprehension (' –1' negative marking) Q.6 to Q.81' negative marking) Q.6 to Q.8 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. A rod of length ! and cross section area A has a variable thermal conductivity given by k = = T, where= is a

positive constant and T is temperature in kelvin. Two ends of the rod are maintained at temperatures T1 andT

2 (T

1 > T

2). Heat current flowing through the rod will be

(A)!

)TT(A 22

21 !=

(B)!

)TT(A 22

21 "=

(C)!3

)TT(A 22

21 "=

(D)!2

)TT(A 22

21 !=

2.2. A potentiometer wire of length 10 m and resistance 10 ohm is connected in series with an ideal cell ofE.M.F. 2 V.

If a rheostat having range 0 –10 ohm is used in series with the cell then maximu m potential

gradient of the wire will be :(A) 2 V/m (B) 0.2 V/m (C) 2 5V/m (D) 0.2 5V/m

3.3. A charge particle A of charge q = 2 C has velocity v = 100 m/s . When itpasses through point A & has velocity in the direction shown . The strengthof magnetic field at point B due to this moving charge is (r = 2 m):

(A) 2.5 5 T (B) 5.0 5 T (C) 2.0 5 T (D) none of these

4.4. Two like parallel forces P and 3 P are 40 cm apart. If the direction of P is reversed, then the point of

application of their resultant shifts through a distance of :(A) 30 cm (B) 40 cm (C) 50 cm (D) 60 cm

5.5. The engine of a futuristic nuclear powered car for which power and speed can have fantastic values (say600 kph) can produce a maximum acceleration of 5 m/s 2 and its brakes can produce a maximumretardation of 10 m/s 2. The minimum time in which a person can reach his workplace, located 1.5 kmaway from his home using this car is(A) 5 sec. (B) 10 sec. (C) 15 sec. (D) 30 sec.


Two small ball s of same size having masses m and M with a slight separation are released from restsimultaneously on a smooth fixed inclined plane of inclination # = 53° and length L. There is a fix ed wall

at the bottom of the incline such that wall is perpendicular to inclined plane as shown. T he collision ofboth balls takes place only after the collision of m ass M with wall is over. Initial ly M is at a distanceL0 = 9 m from the wall. If M comes to stop just after its collision with m answer the following three

questions. (All collisions are perfectly elastic and g = 10 m/s 2 and sin 53° =5

4)

L 0

L

M

m

6.6. Just after collisio n of both bal ls, velocity of ball havin g mass m up the incli ne is:(A) 20 m/s (B) 22 m/s (C) 24 m/s (D) 12 m/s

7.7. The ratio m : M is :(A) 1 : 1 (B) 3 : 1 (C) 1 : 3 (D) 1 : 2

8.8. The minimum value of L (in meter) so that m does not leave the incline.(A) 12 (B) 24 (C) 36 (D) 48


Topics : Heat, Current Electricity, Magnetic Effect of Current and Magnetic Force on Charge/current,Topics : Heat, Current Electricity, Magnetic Effect of Current and Magnetic Force on Charge/current,Rotation, Kinematics, Center of MassRotation, Kinematics, Center of Mass

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.41' negative marking) Q.1 to Q.4 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1122, , 1122]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.51' negative marking) Q.5 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]Comprehension ('Comprehension (' –1' negative marking) Q.6 to Q.81' negative marking) Q.6 to Q.8 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. Two identica l long, solid cylinders are used to conduct heat from temp T1

to temp T2

. Originally thecylinder are connected in series and the rate of heat tra nsfer is H. If the cylinders are connected inparallel then the rate of heat transfer would be :(A) H

/4 (B) 2H (C) 4H (D) 8H

2.2. In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance Xis balanced aga inst anothe r resistan ce Y. If X < Y, then where will be the new positio n of the null pointfrom the same end, if one decides to balance a resistance of 4X against Y ?

(A) 50 cm (B) 80 cm (C) 40 cm (D) 70 cm

3.3. An electron (charge ! e, mass ' m ' ) is revolving around a fixed proton in circular path of radius ' r '. Themagnetic field at the centre due to electron is:

(A) 0 (B)rmr8

e

02

20

C&&

5(C)

rmr8

e

0

0

C&&

5 (D)

rmr4

e

02

0

C&&

5

4.4. A weightless rod is acted on by upward parallel forces of 2 N and 4

N ends A and B respectively. The

total length of t he rod AB = 3 m. To keep the rod in equilibrium a force of 6 N should act in the following

manner: (A) Downwards at any point between A and B(B) Downwards at mid point of AB

(C) Downwards at a point C such that AC = 1 m(D) Downwards at a point D such that BD = 1

5.5. Two cars A and B are travelling towards each other on a single-lane road at 24 m/s and 21 m/s respectively.They notice each other when 180 m apart and apply brakes simultaneously. They just succeed in avoidingcollision, both stopping simultaneously at the same position. Assuming constant retardation for each car,

the distance travelled by car A while slowing down is


Two identical masses are as shown in figure. One is thrown upwards with velocity 20 m/s and another is just dropped simultaneously. [g = 10 m/s²]

6.6. The masses collide in air and stick together. After how much time the combined mass will fall to the ground(Calculate the time from the starting when the motion was started).

(A) (1+ 2 ) second (B) 2 2 second

(C) (2+ 2 ) second (D) none of these

7.7. In the above problem, to what maximum height (from ground) will the combined mass rise .(A) 25 m (B) 18 m (C) 15 m (D) 20 m

8.8. If collision between them is elastic, find the time interval between their striking with ground.(A) zero (B) 2 sec (C) 1 sec (D) 3 sec


Topics Topics : H: Heat, eat, Magnetic Effect of Current and Magnetic Force on Charge/currMagnetic Effect of Current and Magnetic Force on Charge/current, Rotation, Current Eent, Rotation, Current Electricitylectricity,,Center of MassCenter of Mass

PHYSICSPHYSICS







1.1. Two identical rectangular rods of metal are welded end to end in seriesbetween temperature 0°C and 100°C and 10 J of heat is conducted (in

steady state process) through the rod in 2.00 min. If 5 such rods are

taken and joined as shown in figure maintaining the same temperature

difference between A and B, then the time in which 20 J heat will flow

through the rods is :

(A) 30 sec. (B) 2 min. (C) 1 min. (D) 20 sec.

2.2. An = particle is moving along a circle of radius R with a constant angular velocity'. Point A lies in the same

plane at a distance 2R from the centre. Point A records magnetic field produced by= particle. If the minimum

time interval between two successive times at which A records zero magnetic field is‘t’, the angular speed',

in terms of t is -

(A)t

2&(B)

t3

2&(C)

t3

&(D)

t

&

3.3. When a person throws a meter stick it is found that the centre of the stick is moving with speed 10 m/

s and left end of stick with speed 20 m/s. Both points move vertically upwards at that moment. Thenangular speed of the stick is:

(A) 20 rad/ sec (B) 10 rad/sec (C) 30 rad/sec (D) none of these

4.4. AB and CD are two uniform resistance wires of lengths 100 cm and 80 cm respectively . The connections

are shown in the figure. The cell of emf 5 V is ideal while the other cell of emf E has internal resistance

2 O.A length of 20 cm of wire CD is balanced by 40 cm of wire AB. Find the emf E in volt, if the r eading

of the ideal ammeter is 2 A. The other connecting wires have negligible resistance.

A B40 cm

5V

D Ammeter(ideal)

E2O

5.5. Figure showsv2 v/s s curve for a particle of mass 2 kg moving in a straight line. If the time (in seconds)

taken by the particle to achive a displacement of 10 m is t. (v = velocity, s = displacement), then find thevalue of (t – 20).



There are two blocks A and B placed on a smooth surface. Block A has mass 10 kg and it is moving with

velocity 0.8 m/s towards stationary B of unknown mass. At the time of collision, their velocities are given by

the following graph :

0.8

1

6.6. Coefficient of restitution of the collision is

(A) 1.5 (B) 1 (C) 0.5 (D) 0.8

7.7. Impulse of deformation is :

(A) 1 Ns (B) 3 Ns (C) 6 Ns (D) 5 Ns

8.8. Maximum deformation potential energy is:

(A) 1.2 J (B) 3.2 J (C) 2.0 J (D) 1.6 J


Topics : Center of Mass, Magnetic Effect of Current and Magnetic Force on Charge/current, Gravitation,Topics : Center of Mass, Magnetic Effect of Current and Magnetic Force on Charge/current, Gravitation,Heat, RotationHeat, Rotation

PHYSICSPHYSICS




Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.31' negative marking) Q.1 to Q.3 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.41' negative marking) Q.4 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44, , 44]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.51' negative marking) Q.5 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]



1.1. A man of m ass M stands at one end of a plank of length L which lies at rest on a frictionless horizon talsurface . The m an walks to the other end of the plank. If the mass of the plank is M/3, the distance thatthe man moves relative to the ground is

(A) 3 L/4 (B) 4 L/5 (C) L/4 (D) none of these

2.2. A particle is moving with velocity j3iv "*"

and it produces an electric field at a point given by k2E *"

.

It will produce magnetic field at that point equal to (all quantities are in S. 3. units and speed of l ight isc)

(A) 2c

j2i6 !(B) 2c

j2i6 "

(C) zero (D) can not be determined from the given data

3.3. Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth'ssurface is:(A) R/2 (B) 16R/9 (C) R/3 (D) R/8

4.4. Figure shows three different arrangements of materials 1, 2 and 3 (identical in shape) to form a wall. Thethermal conductivities are K

1, K

2 and K

3 respectively and K

1 > K

2 > K

3. The left side of the wall is 20°C higher

than the right side.

1 2 3

( )3

1 3 2

( )33

3 2 1

( )333

(A) In steady state, rate of energy conduction through the wall III is greatest.(B) In steady state, rate of energy conduction through all the walls (I), (II) and (III) are same.(C) In steady state, temperature difference across material 1 is greatest in wall II.(D) In steady state, temperature difference across material 1 is same in all the walls.

5.5. Find the moment of inertia (in kg.m 2) of a thin uniform square sheet of mass M = 3kg and side a = 2mabout the axis AB which is in the plane of sheet :


COMPREHENSIONCOMPREHENSIONA small ball (uniform solid sph ere) of mass m is released from the top of a wedge of the same mass m .The wedge is free to move on a sm ooth horizontal surface. The ball rolls without slidi ng on the wedge.The required height of the wedge are mentioned in the figure.

mh

h

m

P

Q

6.6. The speed of the wedge when the ball is just going to leave the wedge at point 'P' of the wedge is

(A)9

gh5(B) gh (C)

6

gh5(D) None of these

7.7. The total kinetic energy of the ball just before it falls on the ground

(A) 2 mgh (B) mgh (C)18

13mgh (D) None of these

8.8. The horizontal separation between the ball and the edge 'PQ' of wedge just bef ore the ball falls on

the ground is

(A) h2

103(B) h

3

102(C) 2 h (D) None of these


Topics : Center of Mass, Magnetic Effect of Current and Magnetic Force on Charge/current, Gravitation,Topics : Center of Mass, Magnetic Effect of Current and Magnetic Force on Charge/current, Gravitation,Heat, RotationHeat, Rotation

PHYSICSPHYSICS









1.1. Two identical rods are joined at one of their ends by a pin. Joint is smooth and rods are free to rotate about

the joint. Rods are released in vertical plane on a smooth surface as shown in the figure. The displacement

of the joint from its initial position to the final position is (i.e. when the rods lie straight on the ground) :

(A)4

L(B)

4

17 L (C)

2

L5(D) none of these

2.2. A conducting cylindrical rod of uniform cross -sectional area is kept between two large chamber s which

are at temperatures 100 °C and 0°C respective ly. The conductivit y of the rod increases with x, where x

is distance from 100 °C end. The temperature pr ofile of the rod in steady-state will be as :

100°C 0°C

(A)100°C 0°C

(B)100°C 0°C

(C)100°C 0°C

(D)100°C 0°C

3.3. Two observers moving with different velocities see that a point charge produces same magnetic field at

the same p oint A . Their relati ve velocity must be paralle l to r"

, where r"

is the position vector of point AA

with respect to point charge. This statement is :

(A) true

(B) false

(C) nothing can be said

(D) true only if the charge is moving perpendicular to the r


4.4. Suppose the earth suddenly shrinks in size, still remaining spherical and mass unchanged (All

gravitationa l forces pass through the centre of the earth).

(A) The d ays will become shorter.

(B) The kinetic energy of rotation about its own axis will increase

(C) The duration of the year will increase.

(D) The magnitude of angular mom entum about its axis will increase.

5.5. A symmetric lamina of mas s M consists of a square shape with a semicircular section over each of the

edge of the square as in figure. The side of the square is 2 a. The moment of inertia of the lam ina about

an axis through its centre of mass and perpendicular to the plane is 1.6 Ma 2. The moment of inertia of

the lamina about the tangent AB in the plane of lamina is _______________.


Uniform rod AB is hinged at the end A in a horizontal position as shown in the figure (the hinge is frictionless,

that is, it does not exert any friction force on the rod). The other end of the rod is connected to a block through

a massless string as shown. The pulley is smooth and massless. Masses of the block and the rod are same

and are equal to ' m

'.

6.6. Then just after release of block from this position, the tension in the thread is

(A)8

mg(B)

8

mg5(C)

8

mg11(D)

8

mg3

7.7. Then just after release of block from this position, the angular acceleration of the rod is

(A)!8

g(B)

!8

g5(C)

!8

g11(D)

!8

g3

8.8. Then just after release of block from this position, the magnitude of reaction exerted by hinge on the rod is

(A)16

mg3(B)

16

mg5(C)

16

mg9(D)

16

mg7


Topics : Rotation, Center of Mass, Heat, Magnetic Effect of Current and Magnetic Force on Charge/Topics : Rotation, Center of Mass, Heat, Magnetic Effect of Current and Magnetic Force on Charge/current, Gravitation, Rotationcurrent, Gravitation, Rotation

PHYSICSPHYSICS






1.1. Moment of inertia of uniform triangular plate about axis passing through sides AB, AC, BC are IP, IB & IH

respectively & about an axis perpendicular to the plane and passing through point C is IC. Then :

(A) IC > I

P > I

B > I

H(B) I

H > I

B > I

C > I

P

(C) IP > I

H > I

B > I

C(D) none of these

2.2. Two men ‘ A’ and ‘B’ are standing on a plank. ‘B’ is at the m iddle of the plank and ‘ A’ is at the left end

of the plank. System is initially at rest and masses are as shown in figure. ‘ A’ and ‘B’ start moving such

that the position of ‘B’ remains fixed with respect to ground then ‘ A’ meets ‘B’. Then the point where A

meets B is located at :

(A) the middle of the plank (B) 30 cm from the left end of the plank

(C) the right end of the plank (D) None of these

3.3. The wall of a house is made of two different materials of same thickness. The temperature of the outer wall

is T2 and that of inner wall is T

1 < T

2. The temperature variation inside the wall as shown in the figure. Then :

(A) thermal conductivity of inner wall is greater than that of outer.

(B) thermal conductivity of outer wall is greater than that of inner

(C) thermal conductivities of the two are equal

(D) no conclusion can be drawn about thermal conductivities


4.4. A uniformly charged ring of radius R is rotated about its axis with constant linear speed v of each of its

particles. The ratio of electric field to magnetic field at a point P on the axis of the ring distant x = R from

centre of ring is (c is speed of light)

(A)v

c2

(B)c

v2

(C)v

c(D)

c

v

5.5. A satellite is revolving around earth in a circular orbit. At some instant the speed of the satellite is increased

2 times its orbital speed keeping its direction unchanged. Then, the new path of the satellite is :

(A) circular (B) straight line (C) elliptical (D) parabolic


A thin uniform rod of length ! and mass m lies on a frictionless fixed horizont al surface. It is struck with

a quick horizontal blow (directed perpendicular to the rod) at a distance x from the center of rod. A dot

is painted on the horizontal surface at a distance d from the initial position of the center of the rod.

s t r

i k e

dotdm

x

Top view

6.6. Let v be the speed of centre of mass and ' be the angular speed of t he rod just after t he blow. Then the

value ofv

' is :

(A) 2

x

!(B) 2

x3

!(C) 2

x4

!(D) 2

x12

!

7.7. For the rod to make one com plete revolution by th e time the center reaches the dot, the value of x (in

terms of ! and d) should be :

(A)d2

2!&

(B)d3

2!&

(C)d6

2!&

(D)d12

2!&

8.8. For x as asked in pre vious question to exist, the minimum value of d (in terms of !) should be :

(A)2

!&(B)

3

!&(C)

4

!&(D)

6

!&


Topics : HeaTopics : Hea t, Emf, Rt, Emf, R otationotation , Cente, Cente r of Mr of M ass, Visoass, Viso sity, Geomsity, Geometricetrical Optal Optics, ics, CurrCurrent ent ElectElectricitricityy

PHYSICSPHYSICS






Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.8× 4) Q.8 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88,,1100]]

AAsssseerrttiioon n aannd d RReeaassoon (n ( nno o nneeggaattiivve e mmaarrkkiinngg) ) QQ. . 99 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[33, , 33]]

1.1. A simple microscope has a focal length of 5 cm. The magnification at the least distance of distinct vision is-

(A) 1 (B) 5 (C) 4 (D) 6

2.2. Two identical solid spheres have the same temperature. One of the sphere is cut into two identical pieces.

The intact sphere radiates an energy Q during a given small time interval. During the same interval, the two

hemispheres radiate a total energy Q'. The ratio Q'/Q is equal to :

(A) 2.0 (B) 4.0 (C)3

2(D) 1.5

3.3. A ring of radius 5 m is lying in the x-y plane and is carrying current of 1 A i n anti-clockwise sense. If a

uniform magnetic field B"

= j4i3 " is switched on, then the co-ordina tes of point about which the loop

will lift up is:

(A) (3, 4) (B) (4, 3)

(C) (3, 0) (D) (0, 3)

4.4. A ring of radius R rolls without slipping on a rough hor izontal surface with a constant velocity . The radius

of curvature of the path f ollowed by any particle of the ring at the highest point of its path will be :

(A) (B) 2 R

(C) 4 R (D) none of these


5.5. A small bob of mass ‘m’ is suspended by a massless string from a cart of the same mass ‘m’ as shown in

the figure.The friction between the cart and horizontal ground is negligible. The bob is given a velocity V0 in

horizontal direction as shown. The maximum height attained by the bob is,

(A)g

V22

0(B)

g

V2

0(C)

g4

V2

0(D)

g2

V2

0

6. Two identical spherical drops of water are falling (vvertically downwards) through air with a steady velocity of

5 cm/sec. If both the drops coalesce ( combine) to form a new spherical drop, the terminal velocity of the new

drop will be- (neglect bouyant force on the drops.)

(A) 5 × 2 cm/sec (B) 5 × 2 cm/sec (C) 5 × (4) 1/3 c m/sec (D)2

5 cm/sec.

7.7. A steel wire of length l has a magnetic moment M. It is then bent into a semicircular arc. What is the new

magnetic moment ?

8.8. In each situation of column-I a statement regarding a point object and its image is given. In column-II four

optical instruments are given which form the image of that object. Match the statement in column-I with the

optical instruments in column-II.

Column-Column-II Column- Column-IIII

(A) Real image of a real point object may be formed by (p) concave mirror

(B) Virtual image of a real point object may be formed by (q) convex mirror

(C) Real image of a virtual point object may be formed by (r) convex lens (surrounded by air)

(D) Virtual image of a virtual point object may be formed by (s) concave lens (surrounded by air)

99.. SSTTAATTEEMMEENNTT--1 1 :: Two cells of unequal emf E1 and E

2 having internal resistances r

1 and r

2 are connected as

shown in figure. Then the potential difference across any cell cannot be zero.

E , r1 1 E , r2 2

STASTATEMTEMENT-2 :ENT-2 :If two cells having nonzero internal resistance and unequal emf are connected across each

other as shown, then the current in the circuit cannot be zero.

E , r1 1 E , r2 2

(A) Statement-1 is True, Statement-2 is True; Statement-2 isis a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOTis NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True.


Topics : Heat, Rotation, Magnetic Effect of Current and Magnetic Force on Charge/current, Center ofTopics : Heat, Rotation, Magnetic Effect of Current and Magnetic Force on Charge/current, Center ofMass, Geometrical Mass, Geometrical Optics, Current Optics, Current ElectricityElectricity

PHYSICSPHYSICS





Multiple choice objective ('Multiple choice objective (' –1' negative marking) Q.41' negative marking) Q.4 ((4 4 mmaarrkkss, , 4 4 mmiinn..)) [[44,,44]]

Subjective Questions ('Subjective Questions (' –1' negative marking) Q.5 to Q.Q.71' negative marking) Q.5 to Q.Q.7 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[1122,,1155]]

Match Match the the Following Following (no (no negative negative marking) marking) (2(2 × 4) Q.8× 4) Q.8 ((8 8 mmaarrkkss, , 110 0 mmiinn..)) [[88, , 1100]]AAsssseerrttiioon n aannd d RReeaassoon (n ( nno o nneeggaattiivve e mmaarrkkiinngg) ) QQ. . 99 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[33,,33]]

1.1. In a compound microscope, the intermediate image is -

(A) virtual, erect and magnified (B) real, erect and magnified

(C) real, inverted and magnified (D) virtual, erect a nd reduced

2.2. The resolving power of a telescope is more when its objective lens has

(A) greater focal length (B) smaller focal length

(C) greater diameter (D) smaller diameter

3.3. Three separate segments of equal area A1, A

2 and A

3 are shown in the

energy distribution curve of a blackbody radiation. If n1, n

2 and n

3 are

number of photons emitted per unit time corresponding to each area

segment respectively then :A1 A2 A3 8

E8

(A) n2 > n

1 > n

3(B) n

3 > n

1 > n

2(C) n

1 = n

2 = n

3(D) n

3 > n

2 > n

1

4.4. Which of the following statements is/are true(A) work done by kinetic friction on an ob ject may be posi tive.

(B) A rigid body rolls up an inclined plane without sliding. The friction force on it will be up the incline.


(C) A rigid body rolls down an inclined plane without sliding. The friction force on it will be up the incline.


(D) A rigid body is left from rest and having no angular velocity from the top of a rough inclined plane. It

moves down the plane with slipping. The friction force on it will be up the incline.

5.5. The given fig. shows a coil bent with AB=BC=CD=DE=EF=FG=GH=HA =

1 m and carrying current 1 A. There exists in space a vertical uniform

magnetic field of 2 T in the y-direction. Then find out the torque (in vector

form) on the loop.

6.6. A bar magnet or magnetic moment M is aligned parallel to the direction of a uniform magnetic field B. What

is the work done to turn the magnet, so as to align its magnetic moment (i) opposite to the field direction and

(ii) normal to the field direction ?

7.7. Two blocks of mass m1 and m

2 are connected with an ideal spring on a sm ooth horizontal surface as shown

in figure. At t = 0 m1 is at rest and m

2 is moving with a velocity v towards right. At this time spring is in its

natural length. Prove that if m1 < m

2 block of mass m

2 will never come to rest.

mm22mm11

v


8.8. Match the Column if deviation in the Column –II is the magnitude of total deviation (between incident ray and

finally refracted or reflected ray) to lie between 0º and 180º. Here n represents refractive index of medium.


(A) (p) deviation in the light ray is greater than 90 º

(B) (q) deviation in the light ray is less than 90 º

(C) (r) deviation in the light ray is equal to 90 °

(D) (s) Speed of finally reflected or refracted light is same

as speed of incident light.

99.. SSTTAATTEEMMEENNTT--1 1 :: For calculation of current in resistors of resistance R1, R

2 and R

3 in the circuit shown in

figure 1, the circuit can be redrawn as shown in figure 2 (this means that circuit shown in figure 2 is

equivalent to circuit shown in figure 1). All the cells shown are ideal and identical.

STASTATEMENTEMENT-2 :T-2 : Whenever potential difference across two resistors is same, both resistors can be

assumed as a combination of two resistors in parallel.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1


(D) Statement-1 is False, Statement-2 is True


Topics Topics : H: Heat, eat, Magnetic Effect of Current aMagnetic Effect of Current and Magnetic Force on Charge/curnd Magnetic Force on Charge/current, Rotation, rent, Rotation, Center of Mass,Center of Mass,Geometrical Optics, Current, ElectricityGeometrical Optics, Current, Electricity

PHYSICSPHYSICS







AAsssseerrttiioon n aannd d RReeaassoon (n ( nno o nneeggaattiivve e mmaarrkkiinngg) ) QQ. . 99 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[3 3 , , 33]]

1.1. A Galileo telescope has an objective of focal length 100 cm & magnifying power 50. The distance

between the two lenses in normal adjustment will be

(A) 150 cm (B) 100 cm (C) 98 cm (D) 200 cm

2.2. Let the wavelength at which the the spectral emissive power of a black body (at a temperature T ) is

maximum, be denoted by 8max .As the temperature of the body is increased by 1 K , 8max decreases by

1 percent .The temperature T of the black body is

(A) 100K (B) 200K (C) 400K (D) 288K

3.3. In the above question, the emissive power of the black body at temperature T is :

(A) 17 Watt/m 2 (B) 12 Watt/m 2 (C) 6 Watt/m 2 (D) 24 Watt/m 2

4.4. A current carrying ring, carrying current&2

Amp., radius 1m, mass3

2kg

and having 10 windings is free to rotate about its tangential vertical axis. A

uniform magnetic field of 1 tesla is applied perpendicular to its plane. How

much minimum angular velocity (in rad/sec.) should be given to the ring in

the direction shown, so that it can rotate 270º in that direction. Write your answer in nearest single digit in rad/sec.

5.5. A uniform disc of mass 2 kg and radius 50 mm rolls from rest on a fixed rough incline. If the length of

incline is 1 m the time taken by the disc to arrive at bottom equals _______.

6.6. Horizontal and vertical components of earth’s magnetic field at a place are 0.22 tesla and 0.38 tesla respec-

tively. Find the resultant intensity of earth’s magnetic field.

77 A special wedge of mass m is kept on a smooth horizontal surface as shown in figure. Another smooth block

of mass m is projected towards the wedge with a horizontal velocity u0 which then climbs up the wedge and

leaves the wedge at the top most point. After this when block of mass m falls on horizontal surface distancebetween block and wedge is ‘x’. If the maximum height attained by the mass m from ground is 125m then

find x.

u0

m m

A

37°

105m

SmoothSmooth


88.. MMaattcch h tthhe e CCoolluummn n ::

For light passing from different surfaces shown, graph between deviation (G) versus angle of incidence (i) is

drawn. Choose appropriate graph.


(A) Light goes from denser to rarear medium through plane surface (p)

(B) Light goes from rarear to denser medium through plane surface (q)

(C) Light goes through prism placed in air (r)

(D) Light pass through slab placed in air (s)

99.. SSttaatteemmeenntt –1:1: If potential difference between two points is non zero in an electric circuit, electric current

between those two points may be zero.

StatementStatement –2:2: Current always flows from high potential to low potential






Topics Topics : : Magnetic Effect of Current and Magnetic Force on Charge/currMagnetic Effect of Current and Magnetic Force on Charge/current, Electromaent, Electromagnet Induction, Rotation,gnet Induction, Rotation,Center of Mass, GeometricCenter of Mass, Geometrical Optics, Current al Optics, Current ElectricityElectricity

PHYSICSPHYSICS







1.1. The direction of the field B at P is :

The V shaped wire is in x -y plane.

(A) along + x-axis

(B) along + z-axis

(C) along ( –x) –axis

(D) along + y-axis

2.2. If the magnetic field at 'P' can be written as K tan - .

/01

2 =2

then K is :

[Refer to the figure of above question ]

(A)d4

0

&

35(B)

d20

&

35(C)

d0

&

35(D)

d

2 0

&

35

3.3. A circular loop of radius r is moved with a velocity v as shown in the diagram. The work needed to

maintain its velocity constant is :

(A)a2

rvi0

&

5(B)

)ra(2

rvi0

"&

5(C) -

.

/01

2 "&

5

a

ar2n

2

rvi0! (D) zero

4.4. The magnifying power of a simple microscope can be increased if an eyepiece of :

(A) shorter focal length is used (B) longer focal length is used

(C) shorter diameter is used (D) longer diameter is used5.5. A rod of negligible mass and le ngth ! is pivoted at its centre. A particle of mass m is fixed to its left end

& another particle of mass 2 m is f ixed to the right end. If the system is released from rest,

(a)(a) what is the speed v of the two masses when the rod is vertical.

(b)(b) what is the angular speed ' of the system at that instant.


6.6. A ball is given velocity !g3 as shown. If the ratio of centripital acceleration to tangential acceleration is

2y:1 at the point where the ball leaves circular path then write the value of y. [Neglect the size of ball]

ball

(string)

vertical plane


A fixed cylindrical tank having large cross-section area is filled with two liquids of densities $ and 2 $

and in equal volumes as shown in the figure. A small hole of area of cross !section a = 6cm2 is made

at height h/2 from the bottom.

h

h

area = 6

h/2

R

cm2

7.7. Velocity of efflux will be :

(A) gh2 (B) gh3 (C) gh (D) 2 gh

8.8. Distance (R) of the point at which the liquid will strike from container is :

(A) 2h (B) h (C)2

h(D) h2

9.9. Area of cross section of stream of liquid just before it hits the ground.

(A) 2 cm 2 (B) 3 cm2 (C) 1 cm 2 (D) 5 cm2


Topics Topics : : Magnetic Effect of Current and Magnetic Force on Charge/currMagnetic Effect of Current and Magnetic Force on Charge/current, Electromaent, Electromagnet Induction, Rotation,gnet Induction, Rotation,Current ElectricityCurrent Electricity, Fluid, Center of , Fluid, Center of MassMass

PHYSICSPHYSICS







1.1. The magnetic field at the srcin due to the current flowing in the wire is

(A) )ki(a8

0 "&

35! (B) )ki(

a20 "&

35

(C) )ki(a8

0 "!&

35(D) )ki(

2a4

0 !&

35

2.2. A U-shaped conducti ng frame is fixed in space. A conducting rod CD lies at rest on the smooth frame

as shown. The frame is in uniform magnetic field B0, which is perpendicular to the plane of fram e. At

time t = 0, the magnitude of magnetic field begins to change with time t as, B =

kt1

B0

"

, where k is a

positive constant. For no curr ent to be ever induced in fram e, the speed with which rod should be pulled

starting from time t = 0 is (the rod CD should be moved such that its velocity must lie in the p lane of

frame and perpendicular to rod CD)

(A) ak (B) bk

(C) a(1 + kt) (D) b(1 + kt)

3.3. A uniform smooth rod is placed on a sm ooth horizontal floor is hit by a particle moving on the floor, at

a distance4

! from one end. Then the distance tra velled by the centre of the rod after the collision when

it has completed three r evolution wil l be:

[ e ? 0 & ' !

' is the length of the rod ]

(A) 2&! (B) can't be determined

(C) &! (D) none of these


4.4. The convex lens is used in-

(A) Microscope (B) Telescope (C) Projector (D) All of the above

5.5. A small block of mass m is released from rest from point A inside a smooth hemisphere bowl of radius R,

which is fixed on ground such that OA is horizontal. The ratio (x) of magnitude of centripetal force and normal

reaction on the block at any point B varies with # as :

(A) (B) (C) (D)

6.6. A solid ice block (of any shape) is floating remaining in equilibrium in water. Some part of it is outside water

because its density is less than the density of water. Prove that the level of water does not ascend or

descend if the ice melts completely. Neglect the changes in volume due to temperature changes.


Two small spheres of mass m 1 and m 2 are moving towards each other with constant velocities 1u"

and

2u"

respectively and undergo head on inelastic collision. If the coefficient of restitution is e and

0umum 2211 *"

""

,

7.7. The velocity of sphere of mass m 1 after collision is :

(A) – 2ue"

(B) – 1ue"

(C) 1ue"

(D) None of these

8.8. The velocity of sphere of mass m 2 after collision is :

(A) – 1ue"

(B) – 2ue"

(C) 2ue"

(D) None of these

9.9. For the given situation, pick up the incorrect statementincorrect statement :

(A) During the collision, least kinetic energy of system of both spheres is non-zero.

(B) During the collision, least kinetic energy of system of both spheres is zero.

(C) Velocity of separation of both spheres after collision has magnitude = |uu|e 21

""!

(D) At the instant of maximum deformation during collision, speed of each sphere is zero.


Topics : Topics : RotationRotation , F, F luid, luid, Current Current ElectriciElectrici ty, Magnetity, Magnetic Effect c Effect of Currof Current and Magneent and Magnetic Fortic Force on Charce on Charge/ge/current, Electromagnet Inductioncurrent, Electromagnet Induction

PHYSICSPHYSICS







1.1. A uniform disc of mass m and radius R is free to rotate about its fixed

horizontal axis without friction. There is sufficient friction between the

inextensible light string and disc to prevent slipping of string over disc.

At the shown instant extension in light spring isK

mg3, where m is

mass of block, g is acceleration due to gravity and K is spring constant.

Then at the shown moment, magnitude of acceleration of block is :

M,R

Km

(A)3

g2(B)

3

g4(C)

3

g(D)

4

g

2.2. The focal length of the objective of a microscope is

(A) arbitrary (B) less than the focal length of eyepiece

(C) equal to the focal length of eyepiece (D) greater than the focal length of eyepiece

3.3. For a fluid which is flowing steadily , the level in the vertical tubes is best represented by

(A) (B)

(C) (D)

4.4. In the motorcycle stunt called "the well of death" the track is a vertical cylindrical surface of 18 m

radius. Take the motorcycle to be a point mass and 5 = 0.8. The minimum angular speed of the

motorcycle to prevent him from sliding down should be:

(A) 6/5 rad/s (B) 5/6 rad/s (C) 25/3 rad/s (D) none of these

5.5. In the figure shown a conducting rod of length !, resistance R & mass

m can m ove vertically dow nward due to gravity . Other parts are kept

fixed. B = constant = B0. MN and PQ are vertical, smooth, conducting

rails. The capacitance of the capacitor is C. T he rod is released from

rest. Find the maximum current in the circuit



Tangent Galvanometer :Tangent Galvanometer : In case of tangent galvanometer (Figure) a magnetic compass needle is placed

horizontally at the centre of a vertical fixed current-carrying coil whose plane is in the magnetic meridian (the

plane in which the earth’s magnetic field is present in vertical and horizontal directions). So if the needle in

equilibrium subtends an angle ; with the earth’s horizontal magnetic field, then for equilibrium we have

|BM| H

""K = |BM| C

""K … (1)

(direction of the torque on the needle due to field of the earth (BH) and due to field of the coil (BC) must be

opposite)

Here M "

is the magnetic moment of the needle

H B"

is the earth’s horizontal magnetic field

C B"

is the magnetic field at centre due to coil

Now from equation-(1), we have

or, MB H sin ; = MB

C sin (90 – ;) or, B

C = B

H tan ;

or,R2

N0 35= B

H tan ; here R is the radius of the coil and N is the number of turns.

i.e., 3 = K tan ; with K = N

RB2

0

H

5 M Reduction factor of the tangent galvanometer, i.e., in case of a tangent

galvanometer when the plane of coil is in magnetic meridian, current in the coil is directly proportional to the

tangent of deflection of magnetic needle.

Currentcarryingcoil (fixed invertical plane)

Compass

Tangent Galvanometer

Compassneedle

6.6. If at a place horizontal component of Earth’s magnetic field is B H = 2 × 10 –5 T, No. of turns in the coil N = 100,

current I = 10 mA, coil radius = & cm. The angle of dip at this position will be :

(A)6

&(B)

4

&(C)

3

&(D) Data insufficient

7.7. If no. of turns in coil are doubled, the reduction factor of tangent galvanometer will be :

(A) 0.001 (B) 0.002 (C) 0.005 (D) None of these

8.8. If at an instant in equilibrium after doubling the number of turns compass needle points in the direction 30º

north of east then current in coil is :

(A)200

3(B)

500

3(C)

3200

1(D)

3500

1


Topics : Electromagnet Induction, Rotation, Center of Mass, Magnetic Effect of Current and MagneticTopics : Electromagnet Induction, Rotation, Center of Mass, Magnetic Effect of Current and MagneticForce on Charge/currentForce on Charge/current

PHYSICSPHYSICS



TTyyppe e oof f QQuueessttiioonnss MM..MM.., , MMiinn..Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.51' negative marking) Q.1 to Q.5 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1155, , 1155]]Comprehension ('Comprehension (' –1' negative marking) Q.6 to Q.81' negative marking) Q.6 to Q.8 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. In the circuit shown switch S is connected to position 2 for a long timeand then joined to position 1. The total heat produced in resistance R 1

is :

(A) 22

2

R2

EL(B) 2

1

2

R2

EL

(C)21

2

RR2

EL(D) 2

22

1

221

2

RR2

)RR(EL "

2.2. A tugsten bulb radiates 2 W of energy.The filament of the bulb has surface area 2 mm2 and an emissivity of0.9.The temperature of the bulb is: (Stefan’s ConstantB = 5.6 × 10 –8 S.I. units)(A) 3500K (B) 4210K (C) 2110K (D) 211K

3.3. A person with a defective sight is using a lens having a power of +2D. The lens he is using is(A) concave lens with f = 0.5 m (B) convex lens with f = 2.0 m(C) concave lens with f = 0.2 m (D) convex lens with f = 0.5 m

4.4. A uniform solid cylinder is given an angular speed ' and placed on a

rough plate of negligible thickness. The horizontal surface below theplate is smooth. Then the angular speed of the cylinder when it startspure rolling on the plate will be: [ Assume suf ficient length of pl ate ] /////////// ////////// ////////// ///////// //////////// //////// ///

'

m'

m

(A)2' (B)

3' (C)

32' (D) none of these

5.5. Four blocks of masses M1, M

2, M

3 and M

4 are placed on a

smooth horizontal surface along a straight line as shown. It isgiven that M

1 >> M

2 >> M

3 >> M

4. All the blocks are initially at

rest. M1 is given initial velocity v

0 towards right such that it will

collide with M2. Consider all collisions to be perf ectly elastic.

The speed of M4 a fter all collision are over is

M1

M2M3 M4

(A) v0

(B) 4 v0

(C) 8 v0

(D) 16 v0


A uniform and constant magnetic field )k50 j30i20(B "!*"

Tesla exists in space. A charged particle with

charge to mass ratio19

10

m

q 3

*- .

/01

2 C/kg enters this region at time t = 0 with a velocity )k30 j50i20(V ""*

" m/

s. Assume that the charged particle always remains in space having the given magnetic field. (Use2 = 1.4)

6.6. During the further motion of the particle in the magnetic field, the angle between the magnetic field B"

and

velocity of the particle(A) remains constant (B) increases(C) decreases (D) may increase or decrease.

7.7. The frequency of the revolution of the particle in cycles per second will be

(A)19

103

&(B)

38

104

&(C)

19

104

&(D)

192

104

&8.8. The pitch of the helical path of the motion of the particle will be

(A)100

&m (B)

125

&m (C)

215

&m (D)

250

&m


Topics Topics : He: Heat, Cat, Centeenter of r of MassMass, Magnetic Effect of C, Magnetic Effect of Current aurrent and Magnetic Forcnd Magnetic Force on Chare on Charge/currenge/current, Rotatit, Rotation,on,Geometrical OpticsGeometrical Optics

PHYSICSPHYSICS









1.1. The graph shown gives the temperature along an x axis that

extends directly through a wall consisting of three layers A, B

and C. The air temperature on one side of the wall is 150 °C

and on the other side is 80 °C. Thermal conduction through the

wall is steady . Out of the three layers A, B and C, thermal

conductivity is greatest of the layer

t(ºC)

150140

120

80

A

40 60 70 x(cm)

B C(A) A

(B) B

(C) C

(D) Thermal conductivity of A = Thermal conductivity of B.

2.2. Which of the following statements is true concerning the elastic collision of two objects ?

(It is given that no net external f orce acts on the system of two object; and the objects do not exert

force on each other ex cept during collision)(A) No net work is do ne on any o f the two objects, since there is no external force on the system of

given two object.

(B) The net work done by the first object on the second is equal to the net work done by the second on

the first.

(C) The net work done by the first object on the second is exactly t he opposite of the net w ork done by

the second on the first.

(D) The net work done on the system depends on the angle of co llision.

3.3. Choose the correct statements :

(A) For a closed surface, the surface integration Q ds.B is always zero, where B is magnetic field

(B) A current carrying circular loop is in a uniform external magnetic field and is free to rotate about its diametrical

axis will be in stable equilibrium when flux of total magnetic field (external field + field due to the loop itself) is

maximum.

(C)

Spectral energy distributed graph of a black body is shown in figure. If temperature (in K) of the black body is

doubled and surface area is halved, the area under the graph will be eight times.

(D) In keplers III law,3

2

R

T depends on the m ass of the Sun, around which a planet is revolving.


4.4. A cylinder and a variable mass M are arranged on a fixed wedge using a light string and a massless pulley.

There is enough friction between cylinder and the wedge to prevent any slipping.

Cylinder

Fixed Wedge

Mass less pulley

M

(A) Only one value of M is possible for which cylinder can remain in equilibrium.(B) There is a range of value of M for which cylinder can remain in equilibrium.

(C) For a certain value of M, the cylinder starts to roll up the plane. In this situation, magnitude of friction force

on the cylinder by the wedge will be greater than tension in the string

(D) For a certain value of M, the cylinder starts to roll down the plane. In this situation, magnitude of friction

force on the cylinder by the wedge will be greater than tension in the string

5.5. In the circuit shown S1 & S

2 are switches. S

2 remains closed for a long tim e and S

1 open. Now S

1 is also

closed. Just after S1 is closed, find the potential difference (V) across R and

di

d t (with sign) in L.


Figure shows a plano-convex lens of refractive index 3 placed in air. The maximum thickness of the lens

is 3 mm and its aperture diameter is 8 mm. Point A lies on the curved surface on the principal axis. A light ray

is incident at the point A as shown in the figure making 60° with the normal.

60°A

6.6. The angle of deviation caused by the lens is :

(A) 60° (B) 30° (C) 0° (D) 15°

7.7. The lateral displacement of the light ray in passing through the lens is :

(A) 3 mm (B) 3 3 m m (C)3 mm (D) Zero

8.8. The focal length of the lens if treated as a thin lens is :

(A)21

5cm (B)

42

5cm (C)

12

25)13( " cm (D)

24

5)13( " cm


Topics : Electromagnet Induction, Geometrical Optics, Center of Mass, Heat, Magnetic Effect of CurrentTopics : Electromagnet Induction, Geometrical Optics, Center of Mass, Heat, Magnetic Effect of Currentand Magnetic Force on Charge/current,and Magnetic Force on Charge/current,

PHYSICSPHYSICS







1.1. The current through the solenoid is changing in such way that flux through it is given by ; = Ht. Then the

reading of the two voltmeters V1 and V

2 differ by :

(A) zero (B) H

(C)21

21

RR

)RR(

"

!H(D)

21

21

RR

RR

"H

2.2. A car is fitted with a convex side –view mirror of focal length 20 cm. A second car 2.8 m behind the first car is

overtaking the first car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the

mirror of the first one is :

(A)10

1m/s (B)

15

1m/s (C) 10 m/s (D) 15 m/s

3.3. In the figure shown of a block A moving with velocity 10m/s on a horizontal surface collides with another block

B at rest initially. The coefficient of restitution is2

1. Neglect friction every where. The distance between the

blocks at 5s after the collision takes place is :

BBAA

10m/s

(A) 20 m (B) 10 m

(C) 25 m (D) Cannot be determined because masses are not given.

4.4. The ends of a rod of uniform thermal conductivity are maintained at different (constant) temperatures. After

the steady state is achieved :

(A) heat flows in the rod from high temperature to low temperature even if the rod has nonuniform cross

sectional area.

(B) temperature gradient along length is same even if the rod has non uniform cross sectional area.

(C) heat current is same even if the rod has non-uniform cross sectional area.

(D) if the rod has uniform cross sectional area the temperature is same at all points of the rod.


5.5. Figure shows the path of an electron in a region of uniform magnetic field. The path consists of two

straight sections, each between a pair of uniformly charged plates, and two half circles. The electric

field exists only between the plates.

Pair-A

Pair-B

3

33

33

3

(A) Plate I of pair A is at higher potential than plate-II of the sam e pair.

(B) Plate I of pair B is at higher potential than plate I I of the sam e pair.

(C) D irection of the ma gnetic field is out of th e page [ ].

(D) D irection of t he mag netic field in to the page [ ].


A uniform rod is hinged at the ceiling of a cart and is free to rotate as shown in diagram. Hinge is smooth.

Initially the cart is at rest. Mass of the rod is 'M' and length 'L'. Now the cart starts moving with constant

acceleration in forward direction.

6.6. The minimum acceleration of the cart for which rod will become horizontal at some moment during motion is

(A) g

(B)2

g

(C) 2g

(D) Rod cannot become horizontal whatever may be acceleration

7.7. The normal reaction on the hinge at the initial instant when the cart starts moving with above minimum

acceleration is

(A) Mg (B) 2 Mg (C)4

Mg(D) 17 4

Mg

8.8. If the mass of the cart is '2M' (without rod) then for the above condition the frictional force acting on the wheels

of the cart at initial instant will be

(A) 2Mg (B) 3 Mg (C)4

Mg5(D)

4

Mg9


TopicTopics : Kins : Kinemaematictics, Electromagnet Induction, Magnetic Effect s, Electromagnet Induction, Magnetic Effect of Current of Current and Magnetic Force and Magnetic Force on Charge/on Charge/current, Center of Mass, Rotationcurrent, Center of Mass, Rotation

PHYSICSPHYSICS







1.1. For a particle moving along x-axis, which of the velocity versus position graphs given in options below

is/are possible : (position is represented b y x-coordinate of the particle)

(A) (B) (C) (D) None of these

2.2. A uniform magnetic field B increasing with time exists in a cylindrical region of centre O and radius R. The

direction of magnetic field is inwards the paper as shown. The work done by external agent in taking a unit

positive charge slowly from A to C via paths APC, AOC and AQC be WAPC

, WAOC

and WAQC

respectively. Then-

(A) WAPC

= W

AOC = W

AQC(B) W

APC> W

AOC > W

AQC

(C) WAPC

< W

AOC < W

AQC(D) W

APC=

W

AQC< W

AOC

3.3. A semi –circular current carrying wire having radius R is placed in x –y plane with its centre at srcin O. There

is a position x dependent non –uniform magnetic field kR2

xBB 0*"

(here B0 is positive constant) existing in the

region. The force due to magnetic field acting on the semi –circular wire will be along :

(A) negative x-axis (B) positive x-axis

(C) negative y-axis (D) positive y-axis


4.4. A ball is rolling without slipping in a spiral path down the inner surface of a hollow fixed cone whose axis is

vertical. The work done by the inner surface of the cone on the ball is

(A) positive (B) zero (C) negative (D) Impossible to determine

5.5. Two blocks of m ass m and 2m are fixed to the ends of a spring. The spring is initially compressed &then the system is r eleased in air (neglecting the air resistance). After time t

(A) the momentum of the system will be ze ro

(B) the momentum of the system will be 3 m g

t

(C) the momentum of the system will be m g

t

(D) the momentum of the system will not dep end on the value of spring constant.


A disc of mass ‘m’ and radius R is free to rotate in horizontal plane about a vertical smooth fixed axis passing

through its centre. There is a smooth groove along the diameter of the disc and two small balls of mass2

m

each are placed in it on either side of the centre of the disc as shown in fig. The disc is given initial angular

velocity '0and released.

•

m

m2

6.6. The angular speed of the disc when the balls reach the end of the disc is :

(A)20'

(B)3

0'(C)

3

2 0'(D)

40'

7.7. The speed of each ball relative to ground just after they leave the disc is :

•

m

m2

(A)3

R 0'(B)

2

R 0'(C)

3

R2 0'(D) none of these

8.8. The net work done by forces exerted by disc on one of the ball for the duration ball remains on the disc is

(A)9

mR2 20

2'(B)

18

mR 20

2'(C)

6

mR 20

2'(D)

9

mR 20

2'


Topics : Center of Mass, Electromagnet Induction, Magnetic Effect of Current and Magnetic Force onTopics : Center of Mass, Electromagnet Induction, Magnetic Effect of Current and Magnetic Force onCharge/current, RotationCharge/current, Rotation

PHYSICSPHYSICS



TTyyppe e oof f QQuueessttiioonnss MM..MM.., , MMiinn..Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.51' negative marking) Q.1 to Q.5 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1155, , 1155]]


1.1. Ball A of mass m after sliding from an inclined plane, strikes elastically another ball B of same mass at

rest. Find the minimum height H so that ball B just completes the circular motion.

(A) H = 3R (B) H = 2R (C) H =2

R5(D) H = 4R

2.2. The velocity time graph of a linear motion is shown in the figure. The distance from the starting pointafter 8 seconds will be:

(A) 18 m (B) 6 m (C) 8 m (D) none of these

3.3. Figure shows three regions of magnetic field, each of area A, and in each region magnitude of magnetic

field decreases at a constant rate =. If E"

is induced electric field then value of line integral Q rd.E""

along the given loop is equal to

X XX X

X XX

X XX X

X XX

(A) =A (B) –=A (C) 3 =A (D) –3=A

4.4. Wire bent as ABOCD as shown, carries current 3 entering at A and leaving at D. Three uniform magnetic

fields each B0 exist in the region as shown. The force on the wire is

(A)0BR3 3 (B)

0BR5 3 (C)0BR8 3 (D)

0BR6 3


5.5. As shown in figure, S is a point on a uniform disc rolling with uniform angular velocity on a fixed roughhorizontal surface. The only forces acting on the disc are its weight and contact forces exerted byhorizontal surface. Which graph best represents the magnitude of the acceleration of point S as afunction of time

(A) (B) (C) (D)


A uniform disc of mass M and radius R initially stands verticall y on the right end of a horizontal plankof mass M and length L, as shown.The plank res ts on smooth horizontal floor and frict ion between discand plank is sufficiently high such that disc rolls on plank without slipping. The plank is pulled to rightwith a constant horizontal force of magnitude F .

6.6. The magnitude of accelera tion of plank

(A)M8

F(B)

M4

F(C)

M2

F3(D)

M4

F3

7.7. The magnitude of angular acceleration of the disc

(A)mR4

F(B)

mR8

F(C)

mR2

F(D)

mR2

F3

8.8. The distance travelled by centre of disc from its initial position till the left end of plank comes verticallybelow the centre of disc is

(A)2

L(B)

4

L(C)

8

L(D) L


Topics : Rotation, Electromagnet Induction, Simple Harmonic MotionTopics : Rotation, Electromagnet Induction, Simple Harmonic Motion

PHYSICSPHYSICS



TTyyppe e oof f QQuueessttiioonnss MM..MM.., , MMiinn..Single choice Objective ('Single choice Objective (' –1' negative marking) Q.1 to Q.41' negative marking) Q.1 to Q.4 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[1122, , 1122]]Subjective Questions ('Subjective Questions (' –1' negative marking) Q.51' negative marking) Q.5 ((4 4 mmaarrkkss, , 5 5 mmiinn..)) [[44, , 55]]Comprehension ('Comprehension (' –1' negative marking) Q.6 to Q.81' negative marking) Q.6 to Q.8 ((3 3 mmaarrkkss, , 3 3 mmiinn..)) [[99, , 99]]

1.1. A sphere of mass ' m

' is given some angular velocity about a horizontal

axis through its centre and gently placed on a plank of mass ' m '. The co-efficient of friction between the two is 5. The plank rests on a smoothhorizontal surface. The initial acceleration of the centre of sphere relativeto the plank will be:

(A) zero (B) 5g (C) (7/5) 5g (D) 2 5g

2.2. All electrons ejected from a surface by incident light of wavelength 200 nm can be stopped before travelling1 m in the direction of uniform electric field of 4 N/C. The work function of the surface is:(A) 4 eV (B) 6.2 eV (C) 2 eV (D) 2.2 eV

3.3. Both the inductors and the cell are ideal. Find the current (in Amperes) 2Hinductance in steady state.

(A) zero (B) 1A(C) 2A (D) 3A

4.4. A system is shown in the figure. The time period for small oscillations of the two blocks will be.

(A) 2 &

k

m3(B) 2

&

k 2

m3(C) 2 &

k 4

m3(D) 2

&

k 8

m3

5.5. A spherical cavity is formed from a solid sphere by removing mass from it. Theresultant configuration is shown in figure. Find out the moment of inertia of thisconfiguration about the axis through centre of the solid sphere as shown. Takemass M (uniform) for the configuration and radius R for solid sphere and radiusR/2 for cavity.

COMPREHENSIONCOMPREHENSIONA tank of height 'H' and base area 'A' is half filled with water and there is a very small orifice at the bottom and

there is a heavy solid cylinder having base area3

A. The water is flowing out of the orifice. Here cylinder is put

into the tank to increase the speed of water flowing out. It is given that height of the cylinder is same as that

of the tank.

6.6. The speed of water flowing out of the orifice before the cylinder kept inside the tank

(A) gH (B) 1.414 gH (C)2

gh(D)

2

gh

7.7. The speed of water f lowing out of orifice after the cylinder is kept inside it

(A) gH3 (B) gH2 (C)2

gH3(D)

2

gH

8.8. After long time, when the height of water inside the tank again becomes equal to2

H. The solid cylinder is

taken out. Then the velocity of liquid flowing out of orifice will be

(A) - .

/01

2 2

Hg2 (B) -

.

/01

2 3

Hg2 (C)

3

gH(D)

2

gH3


Topics : Topics : RotationRotation , Sim, Sim ple ple Harmonic Harmonic MotMot ion, ion, ElecElectromagntromagnet Inductet Induction, Aion, Alternalternating Curting Currentrent

PHYSICSPHYSICS







1.1. A plank P is placed on a solid cylinder S, which rolls on a horizontal surface. The two are of equal

mass. There is no slipping at any of the surfaces in contact. T he ratio of the k inetic energy of P to the

kinetic energy of S is:

(A) 1: 1 (B) 2: 1 (C) 8: 3 (D) 1: 4

2.2. There is a uniform solid hemisphere. On its upper plane x and y axis are drawn which are mutually perpendicular

as shown. Z –axis is perpendicular to the upper plane and passing through the centre O. If moment of inertia

of the hemisphere about x, y and z-axis areIx, I

y and I

z respectively then :

(A) Iz = I

x + I

y(B) I

z = I

x – I

y

(C) Iz =

2

yx II "(D) I

z =

2

yx I –I

3.3. A block of mass ‘m’ is suspended from a spring and executes vertical SHM of time period T as shown in

figure. The amplitude of the SHM is A and spring is never in compressed state during the oscillation. The

magnitude of minimum force exerted by spring on the block is

(A) AmT

4mg

2

2&! (B) Am

T

4mg

2

2&"

(C) AmT

mg2

2&! (D) Am

Tmg

2

2&"


4.4. Photons of energy 5 eV are incident on cathode. Electrons reaching the anode have kinetic energies varying

from 6eV to 8eV.

(A) Work function of the metal is 2 eV

(B) Work function of the metal is 3 eV

(C) Current in the circuit is equal to saturation value.

(D) Current in the circuit is less than saturation value.

5.5. An ideal inductor, (having initial current zero) a resistor and an ideal battery are connected in series at

time t = 0. At any time t, the battery supplies energy at the rate PB, the resistor dissipates energy at

the rate PR and the inductor stores energy at the rate P

L.

(A) P B = PR + PL f or all times t. (B) P R < PL for all times t.(C) P

R < P

L only near th e starting of the circuit. (D) P

R > P

L only near the starting of the c ircuit.


A steady current 4 A flows in an inductor coil when connected to a 12 V dc source as shown in figure

1. If the same coil is connected to an ac sourc e of 12 V, 50 rad/s, a current of 2.4 A flows in the circuit

as shown in figure 2. Now after these observations, a capacitor of capacitance50

1F is connected in

series with the coil and with the same AC source as shown in figure 3 :

6.6. The inductance of the coil is nearly equal to

(A) 0.01 H (B) 0.02 H (C) 0.04 H (D) 0.08 H

7.7. The resistance of the coil is :

(A) 1 O (B) 2 O (C) 3 O (D) 4 O

8.8. The average power supplied to the c ircuit after connecting capacitance in ser ies is approximately equal

to:

(A) 24 W (B) 72 W (C) 144 W (D) None of these


Topics : Fluid, Electromagnet Induction, Rotation, Magnetic Effect of Current and Magnetic Force onTopics : Fluid, Electromagnet Induction, Rotation, Magnetic Effect of Current and Magnetic Force onCharge/currentCharge/current

PHYSICSPHYSICS







1.1. A container of a large uniform cross-sectional area A resting on a horizontal surface holds two immiscible,

non-viscous and incompressible liquids of densities ' d ' and ' 2 d ' each of height (1/2)H as shown. The

smaller density liquid is open to atmosphere. A homogeneous solid cylinder of length L ( )H21L cross-

sectional area (1/5) A is immersed such that it floats with its axis vertical to the liquid-liquid interface with

length (1/4) L in denser liquid. If D is the density of the solid cylinder then :

(A) D =2

d3(B) D =

2

d(C) D =

3

d2(D) D =

4

d5

2.2. In an L-R circuit connected to a battery of constant e.m.f. E switch S is closed at time t = 0. If e denotes the

induced e.m.f. across inductor and i the current in the circuit at any time t. Then which of the following

graphs shows the variation of e with i ?

(A) (B) (C) (D)

3.3. The effective value of current i = 2 sin 100 & t + 2 cos (100 & t + 30 º) is:

(A) 2 AA (B) 2 A (C) 4 A (D) 2 2 AA


4.4. The angular momentum of an electron in first orbit of Li++ ion is :

(A) &2

h3(B) &2

h9(C) &2

h(D) &6

h


A horizontal uniform rod of mass 'm' has its left end hinged to the fixed

incline plane, while its right end rests on the top of a uniform cylinder of

mass 'm' which in turn is at rest on the fixed inclined plane as shown. The

coefficient of friction between the cylinder and rod, and between the cylinder

and inclined plane, is sufficient to keep the cylinder at rest.

5.5. The magnitude of normal reaction exerted by the rod on the cylinder is

(A)4

mg(B)

3

mg(C)

2

mg(D)

3

mg2

6.6. The ratio of magnitude of frictional force on the cylinder due to the rod and the magnitude of frictional force on

the cylinder due to the inclined plane is:

(A) 1 : 1 (B) 3:2 (C) 2 : 1 (D) 1:2

7.7. The magnitude of normal reaction exerted by the inclined plane on the cylinder is:

(A) mg (B)2

mg3(C) 2mg (D)

4

mg5

88.. SSTTAATTEEMMEENNTT--1 1 :: A pendulum made of an insulated rigid massless rod of length ! is attached to a small

sphere of mass m and charge q. The pendulum is undergoing oscillations of small amplitude having time

period T. Now a uniform horizontal magnetic field B"

out of plane of page is switched on. As a result of this

change, the time period of oscillations does not change.

STATEMENT-2 :STATEMENT-2 : A force acting along the string on the bob of a simple pendulum (such that tension in string

is never zero) does not produce any restoring torque on the bob about the hinge.






Topics Topics : F: Fluiluid, Electromagnet Induction, Alted, Electromagnet Induction, Alternating Current, Morden Phyrnating Current, Morden Physics, Rotation, Magnetic Effecsics, Rotation, Magnetic Effect oft ofCurrent and Magnetic Force on Charge/currentCurrent and Magnetic Force on Charge/current

PHYSICSPHYSICS







1.1. A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72. If relative

density of silver is 10, then tension in the string will be:[ take g = 10 m/s2 ]

(A) 37.12 N (B) 42 N

(C) 73 N (D) 21 N

2.2. In the circuit switch S is closed at time t = 0, the current through C and L would be equal after a time 't' equal

to : (Given : R =C

L) :

(A) RC (B) RL

(C) RC !n2 (D) R/L !n2

3.3. Current in an A.C. circuit is given by i = 22 sin ( &t + & /4), then the average value of current during

time t = 0 to t = 1 sec is:

(A) 0 (B)&4

AA

(C)&

24A (D) 22 AA

4.4. A hydrogen atom is in the 4 th excited state, then:

(A) the maximum number of emitted photons will be 10.

(B) the maximum num ber of emitted photons will be 6.

(C) it can emit three photons in ultraviolet region.

(D) if an infrared photon is generated, then a visible photon may follow this infrared photon.



A uniform solid cylinder of mass M and radius R is placed on two fixed wedges as shown. The inclined

surface of each wedge makes an angle # = 45o with horizontal. The coefficient of friction between

cylinder and each wedge is 5 (5 < 1). The angular velocity of cylinder at shown instant is non zero and

sense of rotation of cylinder about its axis is clockwise.( g is acceleration due to gravity)

fixedwedge

fixedwedge

A B

R

M

5.5. At the shown instant, magnitude of frictional force on cylinder exerted by left wedge A is :

(A))1(2

)1(Mg25"

5"5(B)

)1(2

)1(Mg25"

5!5(C)

)1(2

)1(Mg25"

5"5(D)none of these

6.6. At the shown instant, magnitude of frictional force on cylinder exerted by right wedge B is -

(A))1(2

)1(Mg25"

5"5(B)

)1(2

)1(Mg25"

5!5(C)

)1(2

)1(Mg25"

5"5(D) none of these

7.7. At the shown instant, magnitude of angular acceleration of cylinder is :

(A) 0 (B) )1(R

g4

2

2

5"

5

(C) )1(R

g22

25"

5

(D)none of these

88.. SSTTAATTEEMMEENNTT –1 :1 : No electric current will be present within a region having uniform and constant magnetic

field.

STATEMENTSTATEMENT –2 :2 : Within a region of uniform and constant magnetic field B"

, the path integral of

magnetic field Q R !""

dB along any closed path is zero. Hence from Ampere circuital law IodB 5*RQ !""

(where the given terms have usual meaning), no current can be present within a region having

uniform and constant magnetic field.

(A) Statement-1 is True, Statement-2 is True; Statement-2 isis a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOTis NOT a correct explanation for Statement-1


(D) Statement-1 is False, Statement-2 is True.


1.1. (C) 2.2. (C) 3.3. (D) 4.4. (A)

5.5. (A) 6.6.(A)(B)(C)(D) 7.7. (A)(B)(C)(D)

8.8. (C) 9.9. (B) 10.10. (C)

1.1. (B) 2.2. (C) 3.3. (C) 4.4. (C)

5.5. (C) 6.6. (D) 7.7. (C) 8.8. (D)

9.9. (C) 10.10. (A) p,q (B) q,r (C) q,r (D) s

1.1. (A) 2.2. (B) 3.3. (A) 4.4. (D)

5.5. (BC 6.6. (ABD) 7.7. (BC) 8.8. 56 J

9.9. (A) s, (B) p, (C) s, (D) q

1.1. (A) 2.2. (A) 3.3. (A) 4.4. (A,D)

5.5. 70 cm 6.6. (A) 7.7. (C) 8.8. (D)

1.1. (B) 2.2. (C) 3.3. (AC) 4.4. 4 5.5. zero

6.6. 1 7.7. (A) 8.8. (D) 9.9. (B)

10.10. (A) p (B) q (C) p,q (D) s

1.1. (A) 2.2. (C) 3.3. (C) 4.4. (D) 5.5. (A)

6.6. 5 cm 7.7. (B) 8.8. (D) 9.9. (A)

10.10. (A) s (B) q (C) r (D) q

1.1. (B) 2.2. (A) 3.3. (C) 4.4. (D) 5.5.4

125 m

6.6. d = 4000 mm 7.7. (D) 8.8. (A) 9.9. (D)

1.1. (D) 2.2. (A) 3.3. (C) 4.4. (A) 5.5. (A)

66.. OOn n tthhe e oobbjjeecct t iittsseellff 77.. 13 8!3

8.8. (B) 9.9. (D) 10.10. (D)

1.1. (D) 2.2. (B) 3.3. (A,C) 44 11 cm

5.5. 3 6.6. (D) 7.7. (C) 8.8. (A)

9.9. (B)

1.1. (C) 2.2. (C) 3.3. (D) 4.4. (C)

5.5. (C) 6.6. (C) 7.7. Will not change

8.8. (C) 9.9. (C) 10.10. (C)

1.1. (D) 2.2. (C) 3.3. (D) 4.4. (A) 5.5. (A)

6.6. (B), (C) 7. 7. (A) 8.8. R = 75 cm. 9.9. (B)

10.10. (C)

1.1. (A) 2.2. (B) 3.3. (A) 4.4. (D)

5.5. h =3

125m above point of projection

6.6. 2 F2b/ma 7.7. 8 cm 8.8. (C)

9.9. (B) 10.10. (C)

1.1. (D) 2.2. (C) 3.3. (B) 4.4. (A)

5.5. (C) 6.6. (C) 7.7. (A), (B), (D)

8.8. 25 cm. 9.9. (B) 10.10. (A) 111.1. (C)

1.1. (C) 2.2. (B) 3.3. (D) 4.4. (A)

5.5. (B), (C), (D) 6.6. (A), (C)

7. 7. t =a3

A2 )12 !g

H

8.8. sin - .

/01

2 &"

&"

&t

2

5

4

3x

2 + sin -

.

/01

2 &!

&"

&t

2

5

4

3x

2

9.9. (D) 10.10. (C)

1.1. (D) 2.2. (C) 3.3. (C)

4.4. (B), (C) 5.5. (A), (C), (D) 6.6. 90 J

7.7. a1= 10

7 m/s2

, a

2= 30

7 m/s2

8.8. (C) 9.9. (C) 10.10. (A)

1.1. (A) 2.2. (C) 3.3. (D) 4.4. (B) 5.5. (C)

6.6. (A), (C), (D) 7.7. 1:1 8.8. (D) 9.9. (D)

10.10. (D)


1.1. (D) 2.2. (B) 3.3. (A), (B), (C), (D)

4.4. (A), (D) 5.5. ; = & /3

6.6. 20

2

a4

q6

C&7.7. (C) 8.8. (A)

9.9. (B) 10.10. (A) p, s ; (B)r ; (C) r ; (D) p, s

1.1. (A) 2.2. (C) 3.3. (B) 4.4. (B)

5.5. (B) 6.6. 6 7.7. (C) 8.8. (A)

9.9. (A) 10.10. (A) M ( p, r, s), (B) M (p, s),(C) M (r, s), (D) M (q, s)

1.1. (A) 2.2. (C) 3.3. (B) 4.4. (B)

5.5. (B), (C), (D) 6.6. a 7.7. (B)

8.8. (B) 9.9. (B)

10.10. (A) q,r (B) p,s (C) p,s (D) p,s

1.1. (C) 2.2. (D) 3.3. (D)

4.4. (B) 5.5. (B), (D)

6.6. (a) 6m, 0.25 Hz, 1.5m/s

(b) 1.5& mm/s,0.75&2mm/s (c) & × 10 –3

7.7. 40

2

x8

QR3

C&

8.8. (A) 9.9. (A) 10.10. (B)

1.1. (C) 2.2. (B) 3.3. (B) 4.4. (A), (B), (C)

5.5. (A), (C) 6.6. (B), (D) 7.7. (A) 8.8. (B)

9.9. (A) 10.10. (A) p (B) q,s (C) p (D) q,s

1.1. (C) 2.2. (A) 3.3. (C) 4.4. (D)

5.5. (C) 6.6. (A) 7.7. (C) 8.8. (A), (B), (D)

9.9. (A), (B), (D) 10.10. (A), (C)

1.1. (A) 2.2. (C) 3.3. (D) 4.4. (C)

5.5. (A) 66.. ((aa)) VP =

4

5 V (b)(b) ' =

R

VP =

R4

V5

7.7. (A) 8.8. (B) 9.9. (C)

10.10. (A) p (B) q (C) p,r (D) q,s

1.1. (B) 2.2. (A) 3.3. (D) 4.4. (B)

5.5. (A) 6.6. (C) 7.7. (A,C) 8.8. 8

9.9. 300 gm

1100.. (A(A)) – p,r,t, ; (B) (B) – q, s ; (C) (C) – p,s,t ; (D) – p,s,t

1.1. (B) 2.2. (A) 3.3. (B) 4.4. (B)

5.5. (B) 6.6. (B,C,D) 7.7. V =16 7

3m s /

8.8. (B) 9.9. (D) 10.10. (C)

1.1. (B) 2.2. (B) 3.3. (D) 4.4. (A)

5.5. (C) 6.6. (D) 7.7. (A, B, C, D)

8.8. (B) 9.9. (B) 10.10. (C)

1.1. (C) 2.2. (B) 3.3. (B) 4.4. (A)

5.5. (A, B, C, D) 6.6. ' = v/3R

7.7. (B) 8.8. (A) 9.9. (D)

10.10. (A) p,r (B) q,s (C) p,r (D) q,s

1.1. (A) 2.2. (D) 3.3. (B, D) 4.4. (A, C)

5.5. (i) uy = 5 m/s (ii) u x = 4 m/s6.6. 4 py ( y2 + 4 x2)1/2 7.7. (C) 8.8. (A)

9.9. (B) 10.10. (A) p (B) q,s (C) q,s (D) q,s

1.1. (C) 2.2. (B) 3.3. (D) 4.4. (D)

5.5. (A), (C), (D) 6.6. (C) 7.7. (D)

8.8. (A)

1.1. (D) 2.2. (B) 3.3. (A), (D) 4.4. 0.3 m

5.5. (a) 4.5 m/s (b) 1.5 m/s (c) 3.75 cm

6.6. (B) 7.7. (C) 8.8. (D)

1.1. (C) 2.2. 5

3.3. (i) (a) v =v =r

)qq(K ba "; E =E = 2

ba

r

)qq(K "

(b) E =E = 2

b

r

Kq(ii)

bb = = 2

b

b4

q

&

!(iii) 0


4.4. (a) ' = 2av2, R =2a2a

11(b) ' = S bv2 / a2, R = a2 / b

5.5. (A) 6.6. (B) 7.7. (B)

8.8. (A) p,q (B) p,q (C) q,r (D) q,r

1.1. (B) 2.2. (A) 3.3. (C)

4.4. 2

11

2

GM

a!

2 1 0

/ .-

5.5. 3

Q

,which is same as that before earthing

6.6. 2 7.7. (A) 8.8. (D) 9.9. (B)

1.1. (A) 2.2. (B) 3.3. (A) 4.4. (A)

5.5. (A) 6.6. (C) 7.7. (B)

8.8. 27.04 N 9.9. (A) p,s (B) p,s (C) q,s (D) r

1.1. (C) 2.2. (A,B,C) 3.3.PC

AC =

x

x!! = ( 2 – 1)

4.4.1111

9955.. x x + + 22y y = = 44 66.. )dh(

d

mg33

"

7.7. (D) 8.8. (A) 9.9. (D)

1.1. (B) 2.2. (B) 3.3. (C) 4.4. 52 m

5.5. (B) 6.6. (A) 7.7. (B)

8.8. (A) p (B) q (C) p (D) s

1.1. (C) 2.2. (D) 3.3. (D) 4.4. (C)

5.5. (B)(D) 6.6. (A) 7.7. (B) 8.8. (D)

1.1. (C) 2.2. (B) 3.3. (B) 4.4. (B)(C)

5.5. (C) 6.6. (A)

7.7. (D) 8.8. (A) p, q, r, s,t; (B) p, q, r, s,t;

(C) p, s,t; (D) p, q, r, s,t

1.1. (D) 2.2. (A) 3.3. (A) 4.4. (D)

5.5. (C) (D) 6.6. 1.05 × 109 Pa.

7.7. (A) 8.8. (D) 9.9. (C)

1.1. (B) 2.2. (A) 3.3. (D)

4.4. [13] 5.5. &

K !

10

106 2

m

6.6. (a) T = 40 (3 cos # – 2 cos #0) kg f. (b) #

0 = 60= 60°°

7.7. (D) 8.8. (A) 9.9. (A)

1.1. (B) 2.2. (D) 3.3. (B) 4.4. (A)

5.5. (D) 6.6. 1 K 109 N/m2 7.7. (C)8.8. (B) 9.9. (B)

1.1. (B) 2.2. (C) 3.3. (A) 4.4. (B)

5.5. 3 2 / 6.6. (D) 7.7. (C) 8.8. (C)

1.1. (C) 2.2. (C)

3.3. v1 =30, v 2 = – 25, v 3 = -35/3, v 4 = -25 cm

4.4. 64 5.5. 4 radians 6.6. (B)

7.7. (C) (D) 8.8. (A)(B) (C)

1.1. (B) 2.2. (C) 3.3. (A) (C) 4.4. 1

5.5.10

r96.6. (B) 7.7. (A) 8.8. (C)

1.1. (C) 2.2. (A) 3.3. (A) (C) 4.4. 8000

5.5. (C) 6.6. (C) 7.7. (B)

8.8. (A) p,r (B) q, r (C) q, r (D) q, r

1.1. (A) 2.2. (B) 3.3. 8 4.4. x = 5

5.5. (B) 6.6. (A) 7.7. (C)

8.8. (A) – q,t; (B) – p,r,s ; (C) – p,r,s ; (D) – q,r,t

1.1. (C) 2.2. (D) 3.3. (B) 4.4. (C)

5.5. (B) 6.6. (A) 7.7. (C)

8.8. (A) – p,q,s, (B) – r, (C) – r,t ; (D) – p,q,s,t.


1.1. (D) 2.2. (C) 3.3. (D) 4.4. (A)

5.5. (C) 6.6. (A) 7.7. (A) 8.8. (C)

1.1. (C) 2.2. (A) 3.3. (A,C) 4.4. (B)

5.5. (A) 6.6. (C) 7.7. (C) 8.8. (A)

1.1. (B) 2.2. (C) 3.3. (A) 4.4. (B)

5.5. t1 > t

26.6. (C) 7.7. (A) 8.8. (C)

11.. (B) 2.2. (A) 3.3. (A) 4.4. ' =!4

g9

5.5. Flag will flutter in south direction. 6.6. (C)

7.7. (D) 8.8. (C)

1.1. (A) 2.2. (D) 3.3. (A) 4.4. (A) (C) (D)

5.5. (B) 6.6. (C) 7.7. (C) 8.8. (D)

1.1. (B) 2.2. (C) 3.3. (A) 4.4. (A)

5. 5. VVABAB

= = ( )231 I"I"H

= = 1100VV 66.. (A)

7.7. (C) 8.8. (A)

1.1. (B) 2.2. (C) 3.3. (B) (C) 4.4. 3000

55.. R = r 6.6. (B) 7.7. (B) 8.8. (A)

1.1. (A) 2.2. (B) 3.3. (B) 4.4. (C)

5.5. 04 6.6. (C) 7.7. (C) 8.8. (A)

1.1. (B) 2.2. (A) 3.3. (C) 4.4. 2

5.5. v = 5gR 6.6. (A) 7.7. (C)

8.8. (C) 9.9. (A) 10.10. (B)

1.1. (B) 2.2. (B) 3.3. (C) 4.4. (A)

5.5. (B) 6.6. (B,D) 7.7. (D) 8.8. (B)

9.9. (C)

1.1. (C) 2.2. (B) 3.3. 600 4.4. 4 times

5.5. (A) 6.6. (C) 7.7. (A) 8.8. (D)

1.1. (D) 2.2. (A) 3.3. (B) 4.4. (B)

5. (B) 6.6.d

A

10

13 0C7.7. (C)

8.8. (B)

1.1. (C) 2.2. (D) 3.3. (A) 4.4. (A) (C)

5.5. (A) (C) 6.6. (C) 77. (A) 8.8. (C)

1.1. (D) 2.2. (D) 3.3. (A) 4.4. (A) (C) (D)

5.5. (A) (C) 6.6. (A) 7.7. (A) 8.8. (D)

1.1. (B) 2.2. (A) 3.3. (A) (C) 4.4. (A) (C)

5.5. x = 3 m 6.6. (C) 7.7. (B) 8.8. (A)

1.1. (D) 2.2. (B) 3.3. (A) 4.4. (A)

5.5. (D) 6.6. (C) 7.7. (C) 8.8. (C)

1.1. (C) 2.2. (A) 3.3. (B) 4.4. (D)

5.5. 96 m 6.6. (D) 7.7. (C) 8.8. (B)

1.1. (C) 2.2. (B) 3.3. (A) 4.4. 12

5.5. 1 6.6. (B) 7.7. (B) 8.8. (A)

1.1. (C) 2.2. (A) 3.3. (C) 4.4. (B) (D)

5.5. 7 6.6. (A) 7.7. (D) 8.8. (B)

1.1. (B) 2.2. (B) 3.3. (A) 4.4. (A) (B)

5.5. 4.8 Ma 2 6.6. (B) 7.7. (D)

8.8. (C)


1.1. (A) 2.2. (C) 3.3. (B) 4.4. (A)

5.5. (D) 6.6. (D) 7.7. (C) 8.8. (B)

1.1. (D) 2.2. (D) 3.3. (A) 4.4. (C)

5.5. (C) 6.6. (C) 7.7.&

*&

K*K*M22M

r2m'M l

l

8.8. (A) p,r (B) p,q,r,s (C) p,q,r,s (D) q,s

9.9. (B)

1.1. (C) 2.2. (C) 3.3. (D)

4.4. (A)(B)(C)(D) 5.5. K2 6.6. 2 M B, M B

8.8. (A) p, s; (B) q; (C) r, s; (D) q 9.9. (C)

1.1. (C) 2.2. (A) 3.3. (C) 4.4. 9

5.5. t =g

6sec. 6.6. 0.44 T 7.7. 105 m

8.8. (A) – (q) ; (B) – (r) ; (C) – (s) ; (D) – (p)

9.9. (C)

1.1. (B) 2.2. (B) 3.3. (D) 4.4. (A)

55.. ((aa) ) V V == g! / 3 ,, = = 4 3g / ! ] ]

6.6. y = 2 7.7. (A) 8.8. (D) 9.9. (A)

1.1. (C) 2.2. (A) 3.3. (A) 4.4. (D)

5.5. (A) 7.7. (B) 8.8. (B) 9.9. (A)

1.1. (B) 2.2. (B) 3.3. (A) 4.4. (B)

5.5. imax

=

cBm

cBgm

22 !

!

"6.6. (D) 7.7. (C)

8.8. (A)

1.1. (A) 2.2. (C) 3.3. (D) 4.4. (D)

5.5. (C) 6.6. (A) 7.7. (B) 8.8. (D)

1.1. (A) 2.2. (C) 3.3. (A)(B) (D) 4.4. (A) (D)

5.5. H

36.6. (C) 7.7. (C) 8.8. (C)

1.1. (C) 2.2. (B) 3.3. (C) 4.4. (A) (C)

5.5. (A) (B) (C) 6.6. (A) 7.7. (D)

8.8. (D)

1.1. (D) 2.2. (C) 3.3. (A) 4.4. (B)

5.5. (B) (D) 6.6. (B) 7.7. (C) 8.8. (D)

1.1. (C) 2.2. (B) 3.3. (B) 4.4. (D)

5.5. (C) 6.6. (D) 7.7. (C) 8.8. (A)

1.1. (D) 2.2. (D) 3.3. (D) 4.4. (C)

5.5. 3 =140

57MR2 6.6. (A) 7.7. (C)

8.8. (B)

1.1. (C) 2.2. (C) 3.3. (A) 4.4. (A) (D)

5.5. (A) (C) 6.6. (D) 7.7. (C)

1.1. (D) 2.2. (A) 3.3. (A)

4.4. (C) 5.5. (C) 6.6. (A)

7.7. (B) 8.8. (A)

1.1. (A) 2.2. (C) 3.3. (B)

4.4. (D) 5.5. (B) 6.6. (A)7.7. (C) 8.8. (A)


DPP NO. - 1DPP NO. - 1

2.2. y = ax 2

dt

dy = c = 2ax

dt

dx

2

2

dt

yd = 0 = 2a

2

dt

dx!

" #$

% &

+ 2ax 2

2

dt

xd

2

2

dt

xd = –

x

1

ax2

c

x

1

dt

dx22

! " #

$% & '(!

" #

$% &

= 32

2

xa4

c'

= 2

2

a4

c'

3.3. (DD)) The extension is spring is x = 2R cos 30° – R =

) *R13 '

Applying Newton's second law to the bead normal to

circular ring at point B

N = k ) *R13 ' cos 30° + mg cos 30°

=R

13 + mg ) *13 ' R cos 30° + mg cos 30°

N =2

mg33.

4.4. !! "

#$$%

& ,,

(s v v

v v f f 00

when approaching : f a = -./

012

'+10300

2300150

when receding : -./

012

+'

(10300

2300150r f

3 f a – f r 4 12 Hence (A).

5.5. Suppose particle strikes wedge at height‘S’ after time

t. S = 15t – 2

110 t2 = 15t – 5 t2. During this time

distance travelled by particle in horizontal direction =

5 3 t. Also wedge has travelled travelled extra distance

x = 530tan

S=

3 / 1

t5t15 2'

Total distance travelled by wedge in time t = 10 3 t.

= 5 3 t + 3 (15 – 5t2) 3 t = 2 sec.

Alternate Sol.Alternate Sol.

(by Relative Motion)

35

15

30°

310

T = 55

30cosg

30sinu2=

3

1

10

310x2 7 = 2 sec.

3 t = 2 sec.

6.6. It T be the time period ; time to go from O to Q is12

T

and from M to P is6

T.

The displacement is2A when particle goes from O

to Q, from O to N to Q, from O to N to O to P, and

so on

8 t =12

T or t =

12

T5

6

T

4

T (+

or t =12

T7

12

T

2

T(+


Hence possible time period T is

T = 12 s or T =5

112 7 = 2.4 s

or T =7

112 7 s

similarly displacement is2

A when particle goes from

M to P or M to N to P

Hence the possible time period T is

T = 1 × 6 = 6 s or T =

5

16 7s = 1.2 s

Ans.Ans. T = 1.2 s, 6s, 2.4s, 12s

7.7. There is no horizontal force on block A, therefore it

does not move in x-direction, whereas there is net

downward force (mg – N) is acting on it, making its

acceleration along negative y-direction.

Block B moves downward as well as in negative x-

direction. Downward acceleration of A and B will be

equal due to constrain, thus w.r.t. B, A moves in posi-

tive x-direction.

B

Normal reaction due to C

Due to the component of normal exterted by C on B,

it moves in negative x-direction.

NA

NC

Mg

B

The force acting vertically downward on block B are

mg and NA(normal reaction due to block A). Hence

the component of net force on block B along the

inclined surface of B is greater than mg sin9. Therefore

the acceleration of 'B' relative to ground directed along

the inclined surface of 'C' is greater than g sin 9:

8.8.

9.9. The density of liquid is four times that of cylinder,hence in equlibrium postion one fourth of the cylinderis submerged.So as the cylinder is released from initial postion, it

moves by4

3! to reach its equlibrium position. The

upward motion in this time is SHM. Therefore required

velocity is vmax

= ; A. ; =!

g4 and A =

4

3!. Therefore

vmax

= !g2

3

10.10. The require time is one fourth of time period of SHM.

Therefore t = ;<

2=

g4

!<


1.1.

AN = 3v0 cos2 37º

PB = !

" #

$% & 7+

25

16v3v

v

P00

0

0

= ! " #

$% & +

25

481 = P

0(73/25) Ans.Ans. (B)

2.2.2mmV

2

1 = 15 × 10 –3

Vm = s / m150.0

A; = s / m150.0

L qm.

L

g = s / m150.0

gL = 310100

150.0'7 3 L =

1.0

150.0 = 1.5 m

3.3. Angular acceleration ( =) = r

a t

Since, ta"

=dt

vd"

= constant

8 magnitude of = is constant

Also its direction is always constant (perpendicu-

lar to the plane of circular motion).

whereas, direction of at changes continuously ta"

is not constant.


4.4. x2 = 4ay

Differentiating w.r.t. y, we get

dx

dy =

a2

x

8 At (2a, a),dx

dy = 1

3 hence 9 = 45°

the component of weight along tangential

direction is mg sin 9.

hence tangential acceleration is g sin 9 =2

g

55.. ((CC)) FBD

T

f = 2max

2kg1N

8

f = 6max

3kgT

Net force without friction on system is ‘7N’ in right

side so first maximum friction will come on 3 kg

block.

So f2 = 1 N, f 3 = 6 N, T = 2N

6.6. As point of application of force is not moving,therefore work done by the force is zero.

9.9. Wave velocity in string is

v = >T

=1.0

40 = 20 m/s

Fundamental frequency of string oscillations is

n0

=e

v

2 =

6.0

20 =

3

100Hz

Thus string will be in resonance with a turning for k

of frequency.

nf

=3

100Hz,

3

200Hz, 100 Hz,

3

400Hz, … .

Here rider will not oscillate at all only if it is at a

node of stationary wave in all other cases ofresonance and non-resonanc e it will vibrate at the

frequency of tuning fork. At a distance3

l from one

end node will appear at 3 rd, 6th, 9th or similar higher

Harmonics i.e. at frequencies 100 Hz, 200 Hz, ...

If string is divided in odd no. of segments, these

segments can never resonate simultaneously

hence at the location of rider, antinode is never

obtained at any fre quency.

10.10. (A) p,q (B) q,r (C) q,r (D) s

(A) !!$$%

& ;';( tcos

2

1tsin

2

12x

3 x = 2 sin (; t – 4

<) is periodic with SHM.

(B) x = sin3 ;t can not be writtenas x = A sin(;'

t + ?) so it is not SHM

but periodic motion.(C) Linear combination of different periodic functionis also periodic function.

2

2

dt

xd is not directly proportional to x i.e. this motion

is not SHM(D) x continuously decreases with time. So x is notperiodic function.


1.1. With respect to the cart, equilibrium position ofthe pendulum is shown.If displaced by small angle 9 from this position,then it will execute SHM about this equilibrium

position, time period of which is given by :

T = 2<effg

L ; g

eff = 22 )g3(g +

3 geff

= 2g 3 T = 1.0 second

2.2. As ; cos9 =a2

a

9 = 60º

8 N sin60 º = mg

N cos60 º = m2

a2;

w.r.t. wire

8 tan60 º =a

g22;

;2 =3a

g2


3.3. U(x) = x2 – 4x

F = 0

dx

)x(dU = 0

2x – 4 = 0 x = 2

2

2

dx

Ud = 2 > 0

i.e. U is minimum hence x = 2 is a point of stable

equilibrium.

4.4. Acceleration of block AB = mm3

mg3

+ = 4

3

g ;

acceleration of block CD =mm2

mg2

+ =3

g2

Acceleration of image in m irror AB

= 2 acceleration of mirror

= ! " #

$% & '

4

g3 .2 =

2

3'g

Acceleration of image in mirror CD = ! " #

$% &

3

g2.2

=3

g4

8 Acceleration of the two im age w.r.t. each other

= ! " #$

% & ''

2

g3

3

g4 =

6

g17.

5.5. For equilibrium NA cos 60° + NB cos 30° = Mg

and Na sin 60° = NB sin 30°

On solving NB = 3 NA ; NA =2

Mg

6.6. r = RTM0

@3 T

M

RP

0

(@

Slope of the curve = Temperature

Hence cd and ab are isothermal processes.

@ A V

1

P

a

V2 V1

V

d

c

d

Equivalent PV diagram.

i.e. bc and da are constant volume process(A) and (B) are true.Temp. in cd process is greater than ab.Net work done by the gas in the cycle isnegative, as is clear by the PV-diagram.

@ = RTM0

@

3 TM

RP

0

(@

7.7. Frequency of horn directly heard by observer

fvv

vv

c0+

+

Frequency of echo = fvv

v

c+

Frequency of echo of horn as heard by observer.

! "

#$%

& +' v

vv.f

vv

v 0

c

Frquency of Beats :

= (v + v0) f

BCD

EFG

+'

' cc vv

1

vv

1

= f)vv(

)vv(v22c

2

0c

'+

8.8. Applying work energy theorem t o body

HKE = work done by forces delivering power P

= I(

4

2t

Pdt = I4

2

2 dtt3 = 56 J

Ans.Ans. 56 J

9.9. (A) s, (B) p, (C) s, (D) q

aiv A

""+( t = )2() ji2(i ++

= j2i5 +

j2i5v A +'(J"

A,Av J"

= i10vv AA '('J""

) j3i(vB +'("

, j3ivB +(J"

so i2v B,B (J"

For particle C (d.k C ds fy, )

dt

dv y = 2t

3 vy – 6 = t2 3 v

y = 6 + 4 = 10

j10i5vC +("

, j10i5vC +'(J"

so i10v C,C '(J"

ji3vD '("

, ji3vD ''(J"

, i6v D,D '(J"



11.. ((AA))

The slope of isothermal curve at point of intersection

isdV

dP = –

V

P = tan 135° ...(1)

The slope of adiabatic curve at point of intersection is

dV

dP= –

V

PK = tan 121º ....(2)

from (1) and (2)

K =tan 59° = 1.66 = 5/3

8 gas is monoatomic

2.2. Range = 10 m.

For point where particle strikes line PQ

8 x coordinate = 10 cos 37 ° + 2 = 10m

y coordinate = 10 sin 37 ° = 6m

z coordinate = 0m

4.4.

fmax

= > × 3g

= 0.5 × 30 = 15 N

block A starts sliding when friction force becomes

max. i.e. fmax

= 15

at that instant (F.B. D.)

both will move with same acceleration

So 15 = 3a 3 a = 5m/s 2

F – 15 = 7a

10t – 15 = 7 × 5

10t = 50

3 t = 5 sec

Work done by friction in 5 seconds

W = I ds.F

= I ds.t10 (a = t10

t10

m

F (( )

= I5

0

Vdt.t10 (ds = vdt)

= I5

0

2

dt2

t.t10 (V = II ((

2

ttdtadt

2

)

= dtt5

5

0

3I

=

5

0

4

4

t5

--.

/

001

2 = L M0625

4

5' =

4

5625 7

5.5.

From figure if man moves from source to point A

! " #

$% & (+ cm70

2

70

2

70. Then he can see image

If man moves from source to point B

! " #

$% & (+ cm70

2

70

2

70. then he can not loose sight of

image.

6.6. The equation of wave moving in negative x-direction,assuming srcin of position at x = 2 and srcin of time(i.e. initial time) at t = 1 sec.y = 0.1 sin (4<t + 8x)Shifting the srcin of position to left by 2m, that is, tox = 0. Also shifting the srcin of time backwards by 1sec, that is to t = 0 sec.y = 0.1 sin [(4<t + 8(x – 2)]

7.7. As given the particle at x = 2 is at mean position at t= 1 sec.8 its velocity v = ;A = 4< × 0.1 = 0.4 < m/s.

8.8. Time period of oscillation T =2

1

4

22 (<<(

;<

sec.

Hence at t = 1.125 sec, that is, at4

T seconds after

t = 1 second, the particle is at rest at extreme position.Hence instantaneous power at x = 2 at t = 1.125 secis zero.



1.1.

M.I. about ‘O’ is2

MR2

By parallel-axis theorem : 2

MR2

= Ncm +

2

2.3

R4M !

" #

$% &

<

3 Ncm =

22

3

R4.2M

2

MR!

" #$

% &

<' 1

2.2.

i

C

i

i

A

B

In the figure i + i = 90°8 i = 45°

3.3. Slope of graph is greater in the solid state i.e.,

temperature is rising faster, hence lower heat

capacity.

The transition from solid to liquid state takes lesser

time, hence latent heat is smaller.

4.4. W = Px

= -./

012 '

º60sin

h

º30sin

hP

= -.

/01

2'

3

11Ph2 = Ph

3

]13[2 '

=3

)3)(13)(13(2 +'

= 4 J Ans.Ans.

5.5. Angular speed of reflected light = 0 rps

There is no change in angular of incidence due to

rotation of mirror. Ans.Ans. zero

6.6.

Vefflux

= gh2

time of fall t =g

2)h4( '

x = Vefflux

t = )h4(h2 '

the roots of x are (0,4) and the maximum of x is at h

= 2.

The permitted value of h is 0 to 1 clearly h = 1 will

give the

maximum value of x in this interval.

Aliter Solution:Aliter Solution:

If the column of water itself were from ground upto a

height of 4m, h = 2m would give the maximum range

x. Farther the hole is from this midpoint, lower the

range. Here the nearest point possible to this midpoint

is the base of the container. Hence h = 1m.

7.7. W = Area under the curve =2

3 P1V1

12

12

P2P

V2V

(

(

and P1V1 = nRT 1

Therefore1nRT

w =

11

11

VP

VP.2

3

8.8. Q = dU + W

dU = nC v dT

For final state P 2V2 = 2P1 2V1

= 4P1V1 = nR(4T 1)

Hence final temp. is 4T 1

dU = n .2

3 R . 3T 1 =

2

9nRT1

Q =2

3 . nRT 1 +

2

9 nRT1 = 6nRT 1

1nRT

Q = 6


9.9. nC HT = Q 3 nCHT = 6n RT 1

dT = 4T 1 – T1 = 3T1

n . C . 3T 1 = 6nRT 1

R

C = 2

10.10. (A) p (B) q (C) p,q (D) s(A) If resultant force is zero,

systemP"

will be constant.

(B) If resultant torque is zero,

systemL"

will be constant.

(C) If external forces are absent, both systemP"

and

systemL

"

will be constant.(D) If no non conservative force acts, totalmechanical energy of system will be constant.


1.1.

v = 0 + gt3 t = 0.5 secAfter first collision :Speed becomes 5 (0.5) = 2.5 m/st1 = 2 (0.25) = 0.5

t2 = 2 (0.125) = 0.25

t3 = 0.125 and so on

[where ti is the time taken to complete the i th to andfro motion after collision]Total time = 0.5 + [0.5 + 0.25 + 0.125 + ...]

= 0.5 +5.01

5.0

' (Since above is a G.P. with

a = 0.5 and r = 0.5)= 0.5 + 1 = 1.5 sec.

2.2.

53º

37º

R

AR –R cos53=2R/5 B R –R cos37=

R/5

g

37º

g cos37

O

Reference line

By energy conservation between A & B

3 Mg5

R2 + 0 =

5

MgR +

2

1 MV2

V =5

gR2

Now, radius of curvature r

=2

R

37cosg

5 / gR2

a

V

r

2

((O

3.3. (C) For anti-clockwise motion, speed at the highest

point should be gR . Conserving energy at (1) & (2) :

2amv

2

1

= )gR(m2

1

2

Rmg +

3 va2 = gR + gR = 2gR

3 va

= gR2

For clock-wise motion, the bob must have atleast

that much speed initially, so that the string must not

become loose any where until it reaches the peg B.

At the initial position :

T + mgcos600 =R

mv2c ;

VC being the initial speed in clockwise direction.

For VC min

: Put T = 0 ;

3 VC =

2

gR

3 VC /Va =gR22

gR

=21

3 VC : V

a = 1 : 2 Ans.Ans.

6.6. Using newton’s formula

xy = f2

3 20 y = (10)2

3 y = 5 cm.


Sol. 7 to 9Sol. 7 to 9

The shift due to slab is ! " #

$% & '(!!

"

#$$%

& >

'5.1

113

11t

= 1cm towards left. Hence the object appears to mirror

at a distance 61 – 1 = 60 cm.

From mirror formulaf

1

u

1

v

1 (+ we get

v =20 cm.

x0

f=15cm

t=3cm

>>=1.5

object

1cm

1cm I

Hence the mirror forms the image at v = 20 cms

towards right. The slab again causes a shift of 1cm

towards right. hence the final image is formed at a

distance of 21 cm from pole.

Shifting of slab towards left does no cause any change

to position of final image .

The slab only causes apparent shift, but does not

cause any change to velocity of image. Hence the

velocity of image is only due to mirror. The object

appears at a distance u= 60 cm from mirror and mirror

forms its image at v=20 cm. Hence the velocity of

image is

=

2

! " #

$% & '

u

v× velocity of object= 18

60

202

7! " #

$% & '

= 2 m/s towards right

10.10. (A) s (B) q (C) r (D) q

in (A), V is on vertical axis.

Part-I

As V is icreasing, W is positive.

P

V

Part-II

V is decreasing, W is negative.

As negative work in part-II is greater t han positive

work in part-I, net work during the process is

negative.

Using PV = nRT and as V remains same for initial

and final points of the process, it is obvious that

final temp. is greater than initial temperature as

pressure has increased. Therefore dU is positive.

Hence option (S) is connected with (A).

Similar arguments can be applied to other graphs.


1.1. Particle is starting from rest, i.e. from one of itsextreme position.

99P

x

y

A/54A/5

A

Q

As particle moves a distance5

A, we can represent

it on a circle as shown.

cos 9 =5

4

A

5 / A4 (

9 = cos –1 !$% 5

4

;t = cos –1 ! " #

$% &

5

4

t = ;1

cos –1 ! " #

$% &

5

4

= <2

T cos –1 !

" #

$% &

5

4

Method :Method :

As starts from rest i.e. from extrem e position x =

A sin ( ;t + ? )

At t = 0 ; x = A

3 ? =2

<

8 A –

5

A

= A cos ;t

5

4 = cos ;t

3 ;t = cos –1 5

4

t = <2

T cos –1 !

" #

$% &

5

4


2.2. Frictional force a long the in upw ard direction

= 10 g sin 9 – 30 = 30 Nt

N = log cos 9 = 80 Nt

Direction of R is along OA.

3.3. Let v1 and v2 be the velocity of efflux from square andcircular hole respectively. S1 and S2 be cross-section

areas of square and circular holes.

4yv1

v2L

2Ry

v1 = gy8 and v

2 = )y(g2

The volume of water coming out of square and circular

hole per second is

Q1 = v1S1 = gy8 L2 ; Q2 = v2S2 = gy2 <R2

# Q1 = Q2

8 R = <2

. L

44.. ((DD)) F + f = ma .... (1)

Also ; FR – fR = N R

a

F – f = ma .... (2)

[N = mR2 ]

From (1) & (2)

f = 0.

5.5. tAB

= t

tBC

= 2t

So, for ABC part,

20m 20m

10m hC

Bu

A

uy

Time of flight,

tAC

= 3t =g

u2 y

3 uy =

2

3gt

Also, 10 = uyt –

2

1gt2 = gt2

3 t = 1s

8 uy = 15 m/s

8 h =g2

u y2

=20

225 =

4

45m.

8 Maximum height attained = 20 + 4

45

=4

125m.

66.. AAnnss.. d d = = 4400000 0 mmmm

sin 60º =2

3 sinr 3 r = 45 º

8 S = h = 1 m

y = H tan600 = 3m

8 x= S + y = 4m = 4000 mm

7.7. There is only one point image corresponding to a point

object, as long as the object lies on the water surface

(principal axis of the mirror). Any object lying at some

distance from the princpal axis results in multiple

image points.

8.8. If light rays diverge outward (forming a virtual image

behind the mirror) after reflection, there is no refraction

at water surface after reflection. This is the case when

the object lies between the focus and the pole.



1.1. average speed v =t

dt.dt

dxt

0

I =

t

dx

t

0

I

=) * ) *

t

0xtx ' =

) *;<

<6 /

1 –6 / cosA = ) *2 –3

A3

<;

since particle does not change it's direction in thegiven interval , average speed

= v = ) *3 –2A3

<;

2.2. Since the block slides down the incline with uniformvelocity, net force on it must be zero. Hence mg sin9must balance the frictional force ‘f ’ on the block.

Therefore f = mg sin9 = 5 7 10 7 ½ = 25 N.

33.. ((CC)) Angular momentum will be conserved if the nettorque is zero .Now for the sphere to move down:mg sin 9 > > mg cos 9

Let x be the perpendicular distance of the point(as shown in figure) about which torque remainszero.

for P = 0 ; x > R as shown

Note: As mgsin 9 > > mgcos 9, the point should be

inside the sphere.

4.4. Q = 2! = 3mEquation of standing wavey = 2A sin kx cos ;ty = A as amplitude is 2A.A = 2A sin kx

Q

<2

x = 6

<

3 x1 =

4

1m

and Q<2

. x =6

5<

3 x2 = 1.25 m 3 x

2 – x

1 = 1m

5.5. From figure , the velocity of approach (Vcos 9)

decrease as the source comes closer (as 9increases).And the velocity of separation also

increases as ? will decrease.

Hence the frequency of sound as heared by the

observed decreases continuously :

6.6.

ROC =150cm

S1 S2 S3

>>=3/2 >>=1 >>=3/2 >>=4/3>>=4/3

20cm

O

45cm 24cm 54cm10cm

Apparent shift in the object O dut to three slabes S1,S

2

and S3 with respect to the medium of > =

3

4 is given by

:

Shift

=!!!!

"

#

$$$$

%

&

'+!!!!

"

#

$$$$

%

&

'+!!!!

"

#

$$$$

%

&

'

3 / 4

2 / 3

1154

3 / 4

1

1124

3 / 4

2 / 3

1145

Shift = ! " #

$% & +!

" #

$% & +!

" #

$% &

9

8-154

3

4-124

9

8-145 .

Shift = 5 + ( –8) + 6 = 3 cm

8 Unet

= 150 cm and ROC = 150 cm.

Hence image will be formed on the object itself.

7.7. The moment of inertia of all seven rods parallel to

AB and not lying on AB is the moment of inertia o f

all five rods lying on AB = 0

The moment of inertia of all 18 rods perpendicular

to AB is = 18 ( Q!)3

2!

= 6 Q!3

Hence net MI of rod about

AB = 7 Q!3 + 6 Q!3 = 1313 !33 Ans.Ans.


8.8. ! " #

$% & >'>

+>

(>

Ruv1212

(>2 – >

1) is +ve and R is – ve if u is –ve, v will always

be –ve

i.e. for real object image is always virtual.

Sol. 9. to 10.Sol. 9. to 10.

Consider object on left side of spherical surface

seperating two media.

If real object is in rarer media i.e., n1 < n

2

Then)R(

n)u(nn

vn 1122

'+''( = – ve

Hence image shall be virtual for a real objectlyingon concave side with rarer media ....(1)If real object is in denser media i.e., n

1 > n

2

)R(

n

)u(

)nn(

v

n 1212

'+

'''

( =R

n

u

nn 121 ''

8 Image is real ifu

nn 21 ' >

R

n1 or u

<1

21

n

R)nn( '.... (2)

and image is virtual if u > Rn

nn

1

21!!

"

#$$%

& '.... (3)

From statements 1, 2 and 3 we can easily

conclude the answers.


1.1. As RP = 0, angular momentum rem ains conseved:

8 L = !!

"

#$$%

& +

2

R3000

2

;0 = !!

"

#$$%

& + 2

2

R302

R300.;

3 150 ;0 = 180 ; 3 ; = 5/6 ;0 Ans.Ans.

2.2.

Let v = velo. of rainPossible values of = are

–30º < = < 90º .

3.3. For first resonance with 400 Hz tuning fork

!eq

=0f4

V=

)400(4

V = (19 + 1) = 20 cm

If we use 1600 Hz tuning fork

4

20

)1600(4

V

f4

V

0

(7

( = 5 cm

for Resonance

!eq

=0f4

V,

0f4

V3 ,

0f4

V5,

0f4

V7, ....

- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -- - - - -

20 cm

400 Hz

1 cm + ! = 5 cm , 15 cm , 25 cm , 35 cm ,

45 cm .....

! = 4 cm , 14 cm , 24 cm , 34 cm , 44 cm .....

water level should be further lowered by

24 – 19 = 5 cm 3 34 – 19 = 15 cm

44 As rays are parallel to the principal axis, image is

created by lens at the focus.By placing of glass-slab,

Shift = !! "

#$$%

& '

µ

11 .t

= ! " #

$% & '

5.1

11 3 = 1 cm.

Irrespective of separation,Image is shifted to the right by 1 cm.Total distan ce from lens 10 + 1 = 11 cm 11 cmAns.Ans.

5.5.

u

n

v

n 12

' = R

nn 12 '

R2

5.1

v

1

'' =

R

5.11

''

3 v = –4R

m =un

vn

2

1 =

)R2(1

)R4(5.1

'7'7

m = 3.


6.6. If F = 20 N, 10 k g block will not move and it would

not press 5 kg block So N = 0.

8.8. If F = 50 N, force on 5 kg block = 10 N

So friction force = 10 N

9.9. Until the 10 kg block is sticked with ground

(... F = 40 N),

No force will be flet by 5 kg block. After F = 40 N,

the friction

force on 5 kg increases, till F = 60 N, and after

that, the kinetic friction start acting on 5 k g block,

which will be constant (20N)


1.1. (C) 0ss

fvv

v

vv

v

--.

/

001

2!! "

#$$%

& +

'!! "

#$$%

& '

= 2 Hz

vs = 0.5 m/s

2.2. Let velocity of c.m. of sphere be v. The velocity of

the plank = 2v.

Kinetic energy of plank =2

1× m × (2v) 2 = 2mv 2

Kinetic energy of cylinder =2

1mv2

+2

1+ !

" #$

% & ;22mR

2

1

= ! " #

$% & +

2

11mv

2

1 2=

2mv2

1.

2

3

8sphere of .E.K

plank of .E.K=

2

2

mv4

3

mv2 =

3

8.

3.3. Time of flight of projectile depends on vertical

component of velocity and not on the horizontal

component. Collision of the stone with the vertical

wall changes only the horizontal component of

velocity of stone.

Thus the total time of flight in absence of wall is

also T = 1 + 3 = 4sec

8g

u2 y = 4

or uy = 20 m/s

or Hmax

=g2

u2y

=20

400 = 20 metres.

4.4.

9min

> C

sin 9min

> sin C

R

dR ' >

n

1

3 R n – dn > R

3 R >1n

nd

'

R >12

mm4.2

'R > 8 mm

5.5.

A

B C

i

99c

60° 60°

60°

99cr

i

r

Total Deviation = (i – r) + (180 – 29c) + (i – r) = 112°

as r = 60 – 9c

2i – 120 + 2 9c + 180 – 29c = 112°

3 2i = 52 °, i = 26 °

7.7. F = a sin ;t

dt

dvm = a sin ;t 3

dt

dv =

m

asin ;tsin ;t

I I ;(

v

0

t

0

dt.tsindv

v =m

a

m

a

;'

; cos ;t =m

a

; [1 – cos ;t]

since cos ;t S 1

hence direction of velocity will not change.

AAnnss.. WWiilll nol no t t cchhaannggee


8.8. From formula for refraction at curved surface

R

nn

u

n

v

n 1212 '

(' v = – 4 cm

8 image is formed in denser medium at a distance

4 cm from pole.

9.9. Size of image =2

1

n

n

u

v7 × size of object

= 21

3

20

477 =

5

6 cm.

10.10.

i r

C

n=3 n=1

40 cm from pole in the medium of refractive index 1,

virtual, erect and 4 cm in size.

A ray incident from object O is in denser medium and

is refracted into rarer medium.

8 r > i Hence always virtual image is formed.


1.1. When object moves normal to the mirror, image

velocity wi ll be opposite to it. When object moves

parallel to the mirror, image velocity will be in the

same direction.

2.2. As the object moves from infinity to centre of

curvature, the distance between object and image

reduces from infinity to zero.

As the object moves from centre of curvature to

focus, the distance between object and image

increases from zero to infinity .

As the object moves from focus to pole, the

distance between object and its image reduces

from infinity to zero. Hence the distance between

object and its image shall be 40 cm three times.

3.3. Force on bottom surface = @gH × A

4.4. At any time

v"

= [(u cos 9 ) i + (u sin 9 – gt) j ]

P = L M j)gtsinu(icosu). jmg(v.F '9+9'(""

= mg2 t – mgu sin 9Hence, power varies linearly with time.

6.6.

Friction force acting on car will be resultant of the

components shown in the diagram.

fg and fC are components of friction force that balancegravitational pull and provides centripetal force

respectively

The resultant of fg and f

c can be horizontal only for a

point at BC and CD

and not for AB and BD

7.7. As work done in state (NN) is more than in state (N)

8.8. Image due to plane mirror will form at a distance of20 cm left of the mirror.

Since image formed by two mirrrors lie adjacent to

each other.

For convex mirror, image position is 15 cm towards

left.

u = – 25 cm

v = + 15 cm

usingv

1 +

u

1 =

f

1 =

R

2

15

1 –

25

1 =

R

2

R = 75 cm.

Ans. R = 75 cm.Ans. R = 75 cm.

9-10.9-10. Suppose velocity of ring

When it s tarts pure rolling is v. Angular momentum

can be constant about point of contact as there is

no external torque acting about it. T hus


m.2.R = m.R 2 ! " #

$% & R

v+ mvR

3 v = 1m/susing v = u + at

1 = 2 + ( –4) t

3 t = 1/4 sec. [ Ans.: 1 m/s, 1/4 sec. ][ Ans.: 1 m/s, 1/4 sec. ]


1.1.i

i'

9

T = T1 + T2 = (180 – 2i) + (180 – 2i'*

= 360 – 2 (i + i')

T = 360º – 29T = 360º – 2(90º) = 180º .

2.2. The only possibility is by reflection from concavemirror as shown.

3.3. (A) Area under P – x graph = I dxp = dxvdt

dvmI !

" #

$% &

= Iv

1

2 dVmv =

v

1

3

3

mv

--.

/

001

2 =

37

10

7 (v3 – 1)

from graph ; area =2

1 (2 + 4) × 10 = 30

837

10

7 (v3 – 1) = 30

8 v = 4 m/s

ALITERALITER ::

from graphP = 0.2 x + 2

or mv dx

dv v = 0.2 x + 2

or mv2 dv = (0.2 x + 2) dx

Now integrate both sides,

Iv

1

2dvmv = I +10

1

dx)2x2.0(

3 v = 4 m/s.

4.4. Resultant Displacementy = y

1 + y

2

for y to be zero

y = 0

(2x – 3t)2 = (2x + 3t – 6)2

on solving (x – 2

3) (t – 1) = 0

Therefore,

at x =2

3, resultant displacement is zero for all

values of t.

5.5.

30°°

h45°

v

y

x

u=50

o

h = height of the point where velocity makes 30º with

horizontal.As the horizontal component of velocity remain same50 cos45° = v cos30°

v = 503

2

Now by equationv2 = u2 + 2a

yy

2

3

250 !

! "

#$$%

& 7 = 502 – 2gxh

3 2gh = 502 – 502 ×3

2

3 2gh =3

1 × 502

3 h =60

2500 =

3

125

h =3

125m above point of projection

6.6. As Rod is in linear motion only (there ’s no rotation

of the rod), Net torque about COM must be zero.Hence

F1.

2

! – F

2 !

" #

$% & ' b

2

! = 0 ........(1)

also for linear motion.F

2 – F

1 = ma ........(2)

solving (1) and (2)

! =ma

bF2 2 Ans.Ans.


7.7. for M1 : U u = – 30, f = – 20

8v

1 +

30

1

' =20

1

' 3 v = – 60 cm

m1 = – u

v = –

30

60

''

= – 2.

The image By M 1 is PJQJfor M 2 : u = + 10 , f = + 20

8v

1 +

10

1 =

20

1 3 v = – 20 cm

m2 = – 10

20' = + 2

now the size of final image is = 8 cm. Ans.8 cm. Ans.

distance distance of of PP from Afrom AB = 1 cm.B = 1 cm. AAnsns..

distance of Q distance of Q frofrom Am AB = 7 cB = 7 cm.m. AAnsns..

8, 9, 10.8, 9, 10. From The P –V graph, the relation between P

and V is 0o

o PVV

PP +'( .... (1)

Also the ideal gas state equation for one m ole is

PV =RT .... (2)

From equation (1) and (2) is !! "

#$$%

& '(

o

o

V

V1V

R

PT

Hence the graph of T vs. V is a parabola given by

Obviously T is maximum at V=2

Vo. There

maximum value of T is R4

VP 00

Q = HU + W

where HU is the change in the internal energy ofthe gas; and W is work, done by the gas. For onemole of the monatomic ideal gas HU = 3/2R HT.Work equals the area under the graph P vs. VTherefore, for the process from the initial state wit hP1V1 = 3/2 RT 1 to the state with P,V,T the heat

given to system isQ = (3/2) R (T – T1) + (1/2) (P + P 1 (V – V1)

=2

3 (PV – P1V1) +

2

1(PV + P 1V + PV1

– P1V1) .... (3)

= 2PV +2

1P1V –

2

1PV1 – 2P1V1

from equation 1 and 3 we get

Q = !! "

#$$%

& ''+

0

1100

2

0

0

V

V

4

5VP2VP

2

5V

V

P2

The process switches from endothermic to exothermic

asdV

dQ changes from positive to negative, that is

atdV

dQ = 0. Solving we get V =

8

5V

0


1.1.

In 20 seconds , distance travelled by the Sourceis :

S =2

1(1.1) (20) 2 = 220 m.

Let, t = 0 be the sta rting time.The sound wave started at t = 0 from 'A' r eaches

the observer at 'B' after ! " #

$% &

330

330 = 1sec. ie. observer

started hearing the sound at t = 1 sec. At t = 20sec. the source reaches at 'C' 220 m from 'A'.The last sound wave starting from 'C' reaches 'B'

afrer ! " #

$% &

330

110 =

3

1 sec < 1 sec.

Hence , the observer do not hear the sound forwhole 20 sec.but for:

T = ! " #

$% & +

3

119 sec =

3

58 sec.


2.2. The retardation is given by

dt

dv = – av2

integrating between proper limits

3 – Iv

u2v

dv = I

t

0

dta

orv

1 = at +

u

1

3 dx

dt

= at + u

1

3 dx =aut1

dtu

+

integrating between proper limits

3 Is

0

dx = I +

t

0aut1

dtu

3 S =a

1 !n (1 + aut)

3.3. In the frame of ring (inertial w.r.t. earth), the initial

velocity of the bead is v at the lowest position.

The condition for bead to complete the vertical circle

is, its speed at top position

vtop

V 0

From conservation of energy

2

1 m

2topv + mg (2R) =

2

1mv2

or v = gR4

5.5. Since angular velocity is constant, acceleration of

centre of mass of disc is zero. Hence the

magnitude of acceleration of point S is ;2x where

; is angular speed of disc and x is the distance of

S from centre. Therefore the graph is

6.6. T.K.E. of disc [M.Bank_Rotation_7.120][M.Bank_Rotation_7.120]

=22

2

1mv

2

1;N+

=

222

r

v

2

mr

2

1mv

2

1!

" #$

% & 77+

=2mv

4

3

Velocity of particles of upper half is more than that of

lower half hence kinetic energy of upper half will be

more than8

3 mv2 .

7.7. ; =m

K = 10 rad/s

T = s10

22 <(;<

Maximum speed will be at the natural length of

the spring at T/4 =410

2

7<

=20

<s.

Time taken to cover 0.1 m is204

T <( s

Time taken to cover2

1 × 0.1m is

3

2

4

T 7 = s303

2

410

2 <(77<

8.8. Compression is maximum when both blocks move

with same velocity V.

By cons. of momentum

V = 21

2211

mm

vmvm

+

+

= 5 m/s

The change in K.E. = kf – k

i = – 35 J

This is stored as spring PE

Therefore2kx

2

1 = H K 3 x =

k

K2H

on solving x = 0.25 m = 25 cm.


9 to 119 to 11

Magnification is negative, therefore lens is

convex. v = m.u

u is negative and v is positive u

10

1

20

1 + =1f

1W (i)

10

1

30

1 + =2f

1W (ii)

10

1

v

1 + =21 f

1

f

1+ =

f

1W (iii)

For combination

m =u

v =

10

11 / 60

' =11

6'


1.1. using angular momentum conservation

Li = 0

Lt = mvR – N;

mvR = N:;

; = ! " #

$% &

2

1(v + ;R)t = 2<R

222211 7<(!

" #$

% & 7+ t = 2< sec.

2.2. Tmin

= i + e – A

Tmin

= A then

2A = i + e in case of Tmin

i = e

2A = 2i r1 = r

2 =

2

A

i = A = 90°

then 1 sin i = n sin r1

sin A = n sin2

A

2 sin2

Acos

2

A = n sin

2

A

2 cos2

A = n

when A = 90° = imax

then nmin

= 2

i = A = 0 nmax

= 2

3.3. P.E. is maximum at extreme position and minimumat mean position.Time to go from extreme position to mean position

is, t =4

T ; where T is time period of SHM

5 s =4

T

3 T = 20 s.

4.4.if ddd

"""'( = j6i8 +

W = d.F""

= 24 + 24 = 48 J

Pav = t

W = 8 Watt

5.5. At max. extension both should move with equalvelocity.

k = 1120 N/m

5kg 2kg

8 By momentum conservation,(5 × 3) + (2 × 10) = (5 + 2)V

V = 5 m/sec.Now, by energy conservation

2

15 × 32 +

2

1 × 2 × 10 2 =

2

1(5 + 2)V 2 +

2

1kx2

Put V and k

8 xmax = m41 = 25 cm.

Also first maximum compression occurs at ;

t =4

T3 =

4

3

k2

><

=4

3

11207

102

7< =

56

3< sec.

(where > 3 reduced mass, > =21

21

mm

mm

+ )

6.6.

For equilibrium of all the changes net force on

each charge should be zero.

3 2d

KqQ –

2dx

KqQ4

)( ' = 0 (Net force on Q)

3 d = x/3

&2d

KqQ +2

2

x

Kq4 = 0 (Net force on A ))

3 Q = q9

4' & d =3

x


7.7.

v1 = )2 / Hh(g2 '

v2 = hg2

8 By continuity equation

A !! "

#$$%

& ' td

hd = a (v 1 + v2)

3 A !! "

#$$%

& '

td

hd = a X Yhg2)2 / Hh(g2 +'

or 'g2a

AI

2 / H

H 2 / Hhh

hd

'+ = I

t

0

d t

3 t =a3

A2 ) *12 'g

HAns.Ans.

8.8. Equation of standing wave can be written as

Y = 2 sin (kx + 9) cos ;t

because the particles at t = 0 are at extreme position.

k = Q<2

=4

2< =

2

<

From the graph it is clear that x =2

1,

Amplitude = 0

8 2 sin -./

012 9+

<2

1.

2 = 0 3 0

4(9+<

or <:

we will select <(9+<4

to suit the initial condition

8 9 =4

3<

8 y = 2 sin ! " #$% & <+< 43x

2 t

25cos <

= sin ! " #

$% & <

+<

+<

t2

5

4

3x

2

+ sin ! " #

$% & <

'<

+<

t2

5

4

3x

2

9.9. Distance of image of eye from fish in upward

direction

d =reln

d + 30 =

!!!

"

#

$$$

%

&

34

1

60 = 60 ×

3

4 + 30

= 110 cm

10.10. Distance of first image of eye from the refraction

at water surface

v =

!!!

"

#

$$$

%

&

34

1

60 = 80

Distance of image of eye from mirror due to

refraction = 80 + 60 = 140

Distance of second image in downward direction

from mirror = 140 cm

Distance of second image from fish in downward

direction = 140 + 30 = 170 cm


1.1. R = 5++ 120cos44.244 22 = 4N

2.2. The force exerted by film on wire or thread depends

only on the nature of material of the film and not on

its surface area. Hence the radius of circle formed by

elastic thread does not change.

3.3. N = Mg(1 – cos9)

Mg sin 9 > >Mg (1 – cos9)


4.4. At time t 1, velocity of the particle is negative i.e.

going towards –Xm. From the graph, at time t 1, its

speed is decreasing. Therefore particle lies in

between –Xm and 0.

At time t 2, velocity is positive and its m agnitude is

less than maximum i.e. it has yet not crossed O.

It lies in between –Xm and 0.

Phase of particle at time t 1 is (180 + 91).

Phase of particle at time t 2 is (270 + 92)

Phase difference is 90 + ( 92 – 91)

92 – 91 can be negative making H? < 90° but can

not be more than 90 °.

5.5.A

AH × 100

= ! " #

$% & H

A2

! × 100

3 % increase in Area

= 2 × 0.2 = 0.4

V

VH × 100 = 3 × 0.2 = 0.6 %

Since H! = ! = HT

!

!H × 100

= = HT × 100

= 0.23 = = 0.25 × 10 –4 / ºC

6.6. Change in velocity =mass

graphT –Funderarea

=5

)10( –40 + = 6 m/s

WF = HK.E. =

2

1(5) 6 2 = 90 J

7.7. Suppose acceleration of cylinder wrt plank is ‘b’

wrt plank.

As there ’s no slipping b = R = ....(1)Equation of Rotation motion f. R = N=.

3 f =R.2

RM 2C

.R

b =

2

bMC....(2)

Linear motion of cylinder

f = MC (a – b) ....(3)

for plank

20 – f = 4.a ......(4)

putting MC = 2kg

f = b 2f = 2 (a – b)

3 f =3

a2

using in (4) 20 – 3

a2 = 4a

3 a = 7

30

m/s2

and b =7

20m/s 2

Acceleration of cylinder = a – b =7

10 m/s2

[ Ans.: a[ Ans.: a11==

10

7 m/s m/s22

,, a a

22==

30

7 m/s m/s22 ] ]

8.8. Apperent distance

(t) =31

1

n / n

d +

32

2

n / n

d + d3

but n1 = n3 = 1 & n2 = 3/2

d = 1

d1 + 2 / 3

d2 + d3

but d2 and d3 are constant when only object is

moving.

So dJ = dJ1 +0 + 0 3 vI = v0 = 12 cm/sec.

9.9. dJ =31

1

n / n

d +

32

2

n / n

d + d

3

=3 / 41

12

/ +

3 / 42 / 3

9

/ +

1

4

= 12 ×3

4 +

1

8+ 4

= 16 + 8 + 4

= 28 cm.

10.10. dJ =3 / 4 / 1

d1 +

8 / 9

d2+ d3

vJ =3

v4 1+ 0 + 0

vJ =3

4× 12 = 16 cm/sec.



2.2. (C)

Oil

Mercury

Weight = Buoyant force

V@mg =2V @Hgg +

2V @oil

rm =2

oilHg @+@ =

2

8.06.13 +

=2

4.14= 7.2

3.3. The friction force on coin just before coin is to slip will

be :

f = µs mg

f = µs mg

Normal reaction on the coin ; N = mg

The resultant reaction by disk to the coin is

= 22 fN +

= 22s

2 )mg()mg( >+

= mg 21 >+

= 40 × 10 –3 × 10 ×16

91 + = 0.5 N= 0.5 N

4.4. Applying work –energy theorem,RW

all forces = HK.E.

Wg + W

N + W

f = HK.E.

3 ( 1 × 10 ×1 ) + 0 + Wf

=2

1(1)[(2) 2 – (0)2]

10 + Wf = 2

Wf = – 8J.

5.5. Velocity of sound in air (V) =M

RTK

3 V2=T (in kelvin)

not V2 = T (in 0C)

Hence (B) is incorrect.

Velocity of transverse wave in a string :

V = >T

= V2 = T

Hence (C) is a correct graph.

6.6. If we displace the electron slightly toward

x direction, it will thrown away toward right.

So eql. is unstable along x direction.

If we displace the electron slightly towards

y direction, No extra force will act. So eql. is

neutral along y axis

If we displace the electron toward z direction, it

will be attracted and try to come to eql. positron.

So eql. is stable along z direction.

77.. [ A[ Annss. . 11::1 ]1 ]

8.8. As it is posible only when, when light reflects from P1

after refraction from the plate P2 of light coming from

light source.

First image will form due to reflection from the right

surface of P1 . As light ray is falling on P

1 from 11 cm

so it will form image at 11 cm in left. So, distance of

first image from insect is 11 + 3 = 14 cm.

9.9. Second image will form due to reflection on left

surface of P1.

For left surface of P1 light seems to coming from the


distance =2

33 + 3 =

2

39 cm

So, light seems to coming from2

45 cm from right

surfaces of P1 .

So, final position of second image will be

=

2

32

45

= 15 cm.

So, distance of second image from insect = 18 cm.

10.10. Due to multiple reflections infinite image will be formed


1.1.

Let the radius of cylinder be R

For the cylinder to remain static, net torque

on cylinder about point P (point of contact with

inclined surface) should be zero.

8 Mg (OS) = mg (SQ)

or Mg R sin q = mg R (1 – sinq)

or m = 9'9

sin1

sinM

2.2. (B) y(x, t = 0) = 2x

6then y(x,t)

= 2)t2x(

6

'

3t

y

ZZ

= 3)t2x(

24

'

at x = 2, t = 2

Vy = 3)2(

24

' = – 3 m/s.

3.3. Area under the curve is equal to number of

molecules of the gas sample. Hence

N =2

1 . a . V 0 3 aV0 = 2N

Vavg =N

1 I[

0

dV)V(Nv

=N

1 I !! "

#$$%

& 0V

0 0

dVV.v

a.C =

3

2 V0 3

0

avg

V

V =

3

2

2msrv =

N

1 I[

0

2 dV)V(NV

=N

1!! "

#$$%

& I V.

V

aV

0

v

0

20

dV =2

V20 3

0

rms

V

V =

2

1

Area under the curve from 0.5 V 0 to V0 is4

3 of total

area.

4.4. HVL= HV

V

3 YLV

L = Y

VV

V

or V

L

Y

Y

= L

V

V

V

but VV > V

L

3 YL > Y

V

5.5. Y1 = A sin ;t

Y2 = A sin(;t + ?) ? = 2< /3

As, Y2 + Y

1 = A

Now, solve and use the condition of maxima.

Alternate :Alternate : by symmetry both should be on opposite

side of mean position at equal distance from mean

(for max. seperation)

As projection of SHM on circular path

The phase difference from figure is366

<(<+<


6.6.

3

2cos (9

8 Fnet

= 3Fcos 9 =3

2

a

Kq3

2

2

= 20

2

a4

q6

\<

[ Ans.[ Ans. 20

2

a4

q6

\< ] ]


7.7. F.B.D. of disc

3 In horizontal direction N = Nx.....(1)

In vertical direction m'g – f – Ny

= m 'a.....(2)

Torque due to friction ;

fR = A2

R'm 2

Nx

m'g

f

Ng

As a = R =

8 fR =R

a

2

R'm 2

3 f =2

a'm put in (2)

8 m'g – Ny =

2

a'm3

Ny = m'g –

2

a'm3.....(3)

Now, F.B.D. of bar ;

Ny Ny

Ny Ny

mg

4Ny + mg = ma

Put Ny from (3)

4m'g – 6m'a + mg = ma ..... (4)As m' = 2kg & m = 5 kg

8 8g – 12a + 5g = 5a

a =17

g13 ]

8.8. From equation (4) (neglectin g m') a = g ]

9.9. From equation (4) (neglecting m) a =3

g2 ]

1010.. (A(A)) p, p, s ; s ; (B)(B) r ; (C) r ; (C) r ; (D) r ; (D) p, s p, s


1.1. Loss in heat from calorimeter + water astemperture changes from 10 °C to 0°C

= m1C

110 + m

2C

210

= 1 × 1 × 10 + 1 × 0.1 × 10 = 11 kcal

Gain in heat of ice as its temperature changesfrom –11°C to 0°C

= m3C

3 × 11 = 2 × 0.5 × 11 = 11 kcal

Hence ice and water will coexist at 0 °C without

any phase change.

2.2.

tan9 = gV

Fe

@ = g)(V

k / Fe

^'@

3 k = ^'@@

=8.04.2

4.2

' = 1.5

k =k = ^'@@

= =8.04.2

4.2

' = 1.5= 1.5


3.3. Because the acceleration of wedge is zero, the normal

reaction exerted by wedge on block is

N = mg cos37° .

The acceleration of the block is g sin 37° along the

incline and initial velocity of the block is v = 10 m/s

horizontally towards right as shown in figure.

The component of velocity of the block normal to the

incline is v sin 37°. Hence the displacement of the

block normal to the incline in t = 2 second is

S = v sin 37° × 2 = 10 ×5

3 × 2 = 12 m.

8 The work done by normal reaction

W = m g cos 37° S = 100 ×5

4 × 12 = 960 J

4.4. Since, the medium has not changed, speed of wave

remains same.

3 v = fQ = constant

f1Q

1 = f

2 Q

2

3 f1Q

1 = (1.25f

1) Q

2

(# frequency increased by 25%)

3 $2 =

25.11Q

3 Q2 decreases.

3 % change in wavelength

= 1

21

QQ'Q

× 100 =1

11

25.1

Q

Q'Q × 100

=25.1

25.0 × 100 =

5

100 = 20%

6.6.

T1 = 2T0 =g

m2m

)m2(m22 -.

/012

+

T1 = g3

m8 =

3

m80................(i)

In resonance,

fwire = ftube

1

1

2

V)1(

! =

2

2

4

V)1(

!

)x(2

T1

!!

"

#$$%

&

>=

! " #

$% &

2

x 4

)400(

3 T1 = >(16 × 104

)From (i),

3

80m = 10 –4 (16 × 104)

m = 0.6 kg.

7.7.

µR = 1.52

µv = 1.6

Minimum deviation condition for red is r = 30°

3 (1) sin i = (1.52) sin30°

i = 50º,TR = (50º) 2 – 60°

= 40°

8.8. For violet light

(1) sin 50° = (1.6) sin r

8 r = 28.4°

rJ = 31.6° ( # r + rJ = A )

(1) sin e = (1.6) sin 31.6°

8 e = 56°,

3 Tv = i + e – A = 50º + 56° – 60°

= 46°

8 angular width = Tv – T

R = 6°

9.9. The length of the spectrum if it is focussed on a screen

by a lens of focal length 100 cm is :

(A*) cm3

10< (B) m3

10<

(C) cm3

5<(D) m

3

5<

Sol.Sol. if 9 = 100 × 6 ×180

<cm = cm

3

10<

10.10. (A) U (p, r, s), (B) U (p, s),(C) U (r, s), (D) U (q, s)



1.1. The FBD of rod is

Taking moment of force about centre of mass

N2 ×

4

L – N

1 sin9 ×

2

L = 0

8 9(

sin2

1

N

N

2

1

2.2. E = IdE = I(

^b

axx

dxK2 = 2 K ^ !

n a

b

V = IdV = I(

^<b

axx

dxxK = K ^ < (b – a).

3V

E=

) *) *abk

a / bnk2

'^<^!

=) *) *ab

a / bln2

'<

3.3.

B

N

Nx

Ny

V

A

The horizontal component of velocity of Q will increase

and become maximum at the top ; and will again

become same at B. Because of its greater horizontal

velocity the particle Q will reach B earlier than P

8 t1 > t2 .

4.4. Given : f0

= 200 Hz. , A = 50 cm = 0.5 m, V = 340 m/s.

According to question :

) *V

Vf2 maxs0= 20 3 20

=340

5.02002 ;777 ) *) *;( AV;cesin maxS

3 ; = 34 rad/s.

5.5. B,C,D directi on of friction do not dpend on direc-

tion of force but it depends on direction of relative

motion (velocity)

6. 6. aa

9.9. (i) By conservation of volume

4 × h = 4 × 2 + 2 × 1 = 10

h = 2.5m

Pressure at top of the object

= P0 + 0.5 × 1000 × 10

= 1.05 × 105 N/m2

F = P1A

= 1.05 × 105 × 2 = 2.1 × 105 N

A=2m2

.5m

P A2

P A1

mg

4m2

1mh

T

By F.B.D.

T + P2A = mg = P

1A

T = mg + (P1 – P

2) A

= mg – (P2 – P

1) A

= 2 × 2000 × 10 – (.2 × 105)

= .4 × 105 – 0.2 × 105 = 0.2 × 105 N

Fb = V.@

wg

= 2 × 1000 × 10 = 0.2 × 105 N

It is also equal to net contact force by

the liquid = P2A – P

1A = 0.2 × 105N

Note : Net contact force and buoyant force are

same.

10.10. (A) q,r (B) p,s (C) p,s (D) p,s

(A) Centre of mass lies in second quadrant.

(B), (C) and (D) Centre of mass lies on y-axis

and below x-axis.


1.1. (Moderate)(Moderate) Draw an incident ray along the top side

of rectangular strip,which happens to be parallel

to the principal axis. After reflection this ray passes

through focus. Hence image of all points (for e.g.

O1, O

2, O

3, .......) on top side of the strip lie on

this reflected ray (at I1, I

2, I

3, .......) in between

focus and centre of curvature. Thus the image of

this strip is a triangle as shown in figure

F

C

O1 O2 O3

I1

I2I3

I[[


2.2. Vrms

=N

V..........VV 2N

22

21 ++

=N

N..........21 222 +++

=N6

)1N2()1N(N ++

3 Vrms

=6

)1N2()1N( ++

Vavg

=N

V........VV N21 +++ =N

N........21 +++

= N2

)1N(N + = 2

1N +

avg

rms

V

V =

)1N(6

)1N2(2

++

4.4. The magnitude of phase difference between the

points separated by distance 10 metres

= k × 10 = [10 < × 0.01] × 10 = <

5.5. At E,

EE = E

A – E

B

=

0

9

.2

10 / 2

\7< '

–

0

9

.2

10 / 1

\7< '

= 18, towards right.

At D

EO = E

A + E

B

=) *

0

9

2

10 / 2

\7< '

+) *

0

9

2

10 / 1

\7< '

= 54,

towards right.

6.6. (a) from y – x graph

wavelength = Q = 6m

from y – t graphTime period = T = 4 sec

3 frequency = f =4

1 = 0.25 Hz

wave speed = fQ = 0.25 × 6 = 1.50 m/s

(b) maximum velocity = 3mm ×2

< rad/sec

= 1.5 < mm/sec

maximum accelesation = w2A = mm34

2

7<

= 0.75 <2 mm/sec2.

(c) k =3

2 <(Q<

m –1

3 w =2T

2 <(< rad/sec

y = 3 sin ! " #$

% & 9+<<

ot2

–x3

y(x = 2, t = 0 ) = 0

3 sin ! " #$

% & 9+<

03

2 = 0

33

or3

20

<<'(9

andt

y

ZZ

( t = 0 , x = 2) > 0

32

3 – < cos !

" #$

% & 9+<<

o2

t –x

3 > 0

(For x = 2, t = 0)

3 cos ! " #$

% & 9+<

o3

2 < 0 3

3o

<(9

y = (x,t) = 3 sin ! " #$

% & <+<<

32

t –

3

x

<(ZZx

y cos !

" #$

% & <+<<

32

t –

3

x

3 at x = 2 and t = 4 sec ;x

y

ZZ

< × 10 –3

77. . AAnnss.. 40

2

x8

QR3

\<

8.8. Let ‘u’ be the required minimum velocity . By mo-

mentum conservation :

mu = (m + m)v 3 v = u/2.

Energy equation :

2

1 mu2 =

2

1(2m)v2 + mgH.

Substituting v = u/2 : u = 2 gH


9.9. When the block comes to rest, the wedge continues

to move at V =2

u = gH on the smooth surface.

(since, momentum of wedge-block system remains

conserved).

10.10. By work-energy theorem of the block :

– (> mg) (BD) – mg(H –h) = 0

BD = >' hH

DPP NO. - 21DPP NO. - 211.1. Let h = height to of water column

then @wgh + @

Hg g(10 –h) = @

Cu g10

3 h + 13.6 (10 – h) = 73

3 63 = 12.6 h 3 h = 5 cm

2.2. Higher is the temperature greater is the most prob-

able veloc ity.

3.3. Just before the particle transfers to inclined sur-

face, we resolve its velocity along and normal to the

plane.

ucos

usin

Before impact

ucos

After impact

For the trajectory of the particle to sharply change

from the horizontal line to the inclined line, the impact

of the particle with inclined plane should reduce the

usin9 component of velocity to zero. Hence the particle

moves up the incline with speed u cos9.

Hence as9 increases, the height to which the particle

rises shall decrease.

44.. [[MoModdeerraatete]] T = i + e – A (for minimum derivation

i = e)

8 minimum deviation = 2i – A

60 = 2 × 60 – A 3 # A = 60°

n =!

" #

$% &

! " #$

% & T+

2

Asin

2Asin m

=!

" #$

% &

! " #$

% & +

2

60sin

26060sin

= 3

T1 = i

1 + e – A

65° = i1 + 70° – 60° or i

1 = 55°

the T versus i curve is not parabolic

5.5. As N sin = = mg

N cos = = m;2 r

tan = =r

g2;

8 T2 A tan =

a h

====

NNsin ==

Ncos ==

==

8 when = increases T also increases

Also T 2 A r tan =but r = h tan =8 T2 A h tan 2 =for constant =T2 A h

Thus when h increases T also increases

9.9. # Electric field near point b is – [8 ‘b’ should be negative electric field at x

1 is O

which possible only if ‘a’ and ‘b’ are of opposite

sign.

8 ‘a’ is positive

Charge b is negative and charge a is positive

E at A = 0

3 21

b

21

a

)x(

|Q|

)x(

|Q| (+!

8 2

1

2

1

1

b

a

x1

x

x

Q

Q!!

"

#$$%

& +(!!

"

#$$%

& +(

!!

E at a general X

)x(

|Q|K

)x(

|Q|K2

b

2

a '+!

= _B

_CD

_E

_FG

'+ 2

a

b2a

x

1

Q

Q

)x(

1|Q|K

!

If E is a maximum , 0dx

dE

(

3 0x

2

x

x

)x(

23

2

1

1

3 (!!

"

#$$%

& +

++'

!!

3)x( +! =

2

1

13

x

xx !!

"

#$$%

& +!; x+! =

3

2

1

1

x

xx !! "

#$$%

& +!


8 x2 =

1x

x 3

2

1

1 '!! "

#$$%

& +!

!

Ans: xAns: x22 = =

1x

x 3

2

1

1 '!! "

#$$%

& +!

!,,

2

1b

a

x1

Q

Q!!

"

#$$%

& +( !

,,

Charge b is negative and charge a is positiveCharge b is negative and charge a is positive

10. (A) p (B) q,s (C) p (D) q,s

(A) Speed of point P changes with time

(B) Accele ration of poin t P is equal to

;2x (; = angular speed of disc and x = OP). The

acceleration is directed from P towards O.

(C) The angle betwe en accelera tion of P

(constant in magnitude) and velocity of P changes

with time. Therefore, tangential acceleration of P

changes with time.

(D) The acceleration of lowest point is directed to-

wards centre of disc and remains constant with time


1.1. QAB = H UAB + WAB

WAB

= 0

H UAB =

2

`nR

H T

3 2

`(H PV)

H UAB =

2

5(H PV)

QAB

= 2.5 P0V

0

Process BC

QBC

= H UBC

+ WBC

QBC

= 0 + 2P0V

0 !n

2

= 1.4 P0V

0

Qnet

= QAB

+ QBC

= 3.9 P0V

0

2.2. V = V1 + V

2 + V

3

=04

1

<a .R

Q +

04

1

<a !$%

'R

Q2 +

04

1

<a !$% &

R

Q3

=04

1

<a . !$% &

R

Q2

3.3. Let the body is acted upon by a force at an angle

9 with horizontal.

FBD :

F cos 9 = > (mg – F sin 9)

3 F = 9>+9>

sincos

mg. For min. force b

(cos9 + > sin9) should be max.

3 – sin9 + > cos 9 = 0

3 tan 9 = >. or 9 = tan –1 (1/ 3 ) = 30 ° Substituting

; Fmin

= 12.5 kg f12.5 kg f

4.4. Change in momentum = Impulse

kJ jJiJP zyx ++(H"

= 30(0.1) i +2

1(80) (0.1) j + ( –50) × (0.1) k

= k5 j4i3 '+

|P|"

H = 25 kg.sec

m

6.6. Plotting velocity v against time t, we get

Area under the v –t curve gives distance.

Distance =2

1 × 2 × 2 +

2

1 × 2 × 2 = 4m


7.7. (C) c = T ; and as the re is no change in length

3 Q =T

1

'T

T' (QQ

3 'T = T'Q

Q

3 T’ = (2) 2 T = 4T.

Hence (C).

8.8.

Position 1

Principle axis

O

u v

v u

NN1

NN2

Position 2

For first & second positionu

v =

O

1N ,

v

u =

O

2N

3 2

2

u

v =

2

1

NN

= 4.84

3 2.2u

v ( and v + u = 96 3 v = 66 , u = 30

2

O

N =5

112.2

u

v (( 3 A is True

Distance between two position of lens = v - u

= 36 cm

3 B is True

Focal length of lens f =3066

3066

vu

uv

+7

(+ = 20.63

3 C is False

Distance of lens from shorter image = u = 30 cm

3 D is True

9.9. Q' =f

VV s' =

1000

32332 ' = 0.3 m

f' =s

0

VV

)VV(f

'+

= 1000 ×32332

64332

'+

= 1320 Hz

Q'' ='f

VV 0' = 0.2 m.

10.10. By angular momentum conservation ;

L = N ; 3 mv2

R+ mvR = 2mR 2;

2

3mvR = 2mR 2;

; =R4

v3

99

99

N

mg

mg cos99

Also at the time of contact ;

mgcos 9 – N =R

mv2

8 N = mg cos 9 – R

mv2

when it ascends 9 decreases so cos 9 increases

and v decreases.

8 mgcos 9 is increasing andR

mv2

is decreasing

8 we can say N increases as wheel ascends.


1.1. Q = 2! = 3m

Equation of standing wave

y = 2A sin kx cos ;t

y = A as amplitude is 2A.

A = 2A sin kx

Q

<2 x =

6

<

3 x1 =

4

1m

and Q<2

. x =6

5<

3 x2 = 1.25 m 3 x

2 – x

1 = 1m


2.2. tan9 =2

1, gcos9 = a

t

10 × 22 2010

20

+ =

400100

200

+ =5

20m/s2.

3.3. Since time of flight depends only on vertical

component of velocity and acceleration . Hence

time of flight is

T =g

u2 y

where ux = cos 9 and u

y = u sin 9

8 In horizontal (x) directiond = u

xt + ½ gt2

= u cos 9 !! "

#$$%

& 9g

sinu2 +

2

g

sinu2g

2

1!! "

#$$%

& 9

=g

u2 2

(sin9 cos9 + sin 29)

We want to maximise f( 9) = cos 9 sin9 + sin 293 f

’(9) = – sin29 + cos 29 + 2 sin 9 cos9 = 0

3 cos29 + sin2 9 = 0 3 tan29 = –1

or 29 =4

3<or 9 =

8

3<= 67.5 °

Alternate :Alternate :

As shown in figure, the net acceleration of projectile

makes on angle 45 ° with horizontal. For maximum

range on horizontal plane, the angle of projection

should be along angle bisector of horizontal and

opposite direction of net acceleration of projectile.

8 9 =2

1355 = 67.5 °

4.4. Let angle of incidence i for which deviation due to

first prism is minimum, then sin i = n sin30 ° or i =

45°.

The net deviation shall be minimum if deviation due

to each prism is minimum. From the ray diagram

in figure, it is clear that angle between AC and PQ

for net deviation to be minimum is 90 °.

i

incidentray

normal

A

B C

P

Q

R

normal

45°60°°

30°°4 5 °

4 5 °

5.5. One can create a mental view of distribution of charge

i.e. how much charge is nearer and how much is

comparatively farther away.

6.6.

As a rod AB moves, the point ‘P’ will always lie on

the circle.

8 its velocity will be along the circle as shown by

‘VP

’ in the figure. If the point P has to lie on

the rod ‘ AB’ also then it should have component in‘x’

direction as ‘V’.

8 VP sin 9 = V 3 V

P = V cosec 9

here cos9 =R

x =

R

1 .

5

R3 =

5

3

8 sin9 =5

48 cosec 9 =

4

5

8 VP =

4

5 V ...Ans....Ans.

Sol. (b)Sol. (b) ; =R

VP =

R4

V5

ALALTERNATIVE SOLUTIOTERNATIVE SOLUTION :N :

Sol. (a)Sol. (a) Let ‘P’ have coordinate (x, y)

x = R cos 9, y = R sin 9.

VX =

dt

dx = – R sin 9

dt

d9 = V

3 dt

d9

= 9

'

sinR

V

and

VY = R cos 9

dt

d9 = R cos 9 !! "

#$$%

& 9

'sinR

V = – V cot 9

8 VP = 2

y2x VV + = 9+ 222 cotVV

= V cosec 9 ...Ans....Ans.

Sol. (b)Sol. (b) ; =R

VP =

R4

V5


77.. ((AA)) Electric field at r = R

E = 2R

KQ

where Q = Total charge within the nucleus = Ze

So E = 2RKZe

So electric field is independent of a

8.8. Q = I <@ drr4 2r

for a = 0,rRR

d r

'@

(

8 )rR(R

dr '(@

or, Q = IR

0R

d (R – r) 4

< r

2 dr

= --.

/

001

2'

<IIR

0

3R

0

2 drrdrrRR

d4 =

--.

/

001

2'<

4

R

3

R

R

d4 44

=3

dR3<

8 Q = Ze =3

dR3<or d = 3R

Ze3

<

9.9. From the formula of uniformly (volume) chargedsolid sphere

E =03

r

\

@

For E A r, @ should be constant throughout thevolume of nucleusThis will be possible only when a = R.

10.10. (A) p (B) q (C) p,r (D) q,s(A) If velocity of block A is zero, from conservat ionof momentum, speed of b lock B is 2u. Then K.E.

of block B =2

1m(2u) 2 = 2mu 2 is greater than net

mechanical energy of system. Since this is notpossible , velocity of A can never be zero.(B) Since initial velocity of B is zero, it shall bezero for many other instants of time.(C) Since momentum of system is non-zero, K.E.of system cannot be zero. Also KE of system isminimum at maximum extension of spring.(D) The potential energy of spring shall be zerowhenever it comes to natural length. Also P.E. of

spring is maximum at maximum extension ofspring.


11.. ((BB)) To an observer who starts falling freely undergravity from rest at the instant stones are projected,the motion of stone A and B is seen as

dt

dx = u .......(1)

dt

d! = u .......(2)

8 x = ! and d BOA = 60°

2.2. ` = `0 !!

"

#$$%

& +

sound

ob

V

V1

3 sound

ob

0 V

V1+(

``

(straight line) ;

whensound

ob

V

V = 0 ;

0``

= 1.

and assound

ob

V

V U 1 3

0``

U 2

3.3. Distance travelled by

A = s = 10 × 2 +21 × 10 × 8 = 60

Distance travelled by

B = s = 10 × 3 +2

1 × 10 × 4 = 50

Average sp eed of A =8

60 = 7.5

Average speed of B =7

50 = 7.1


4.4. Taking C as srcin and x & y –axes as shown in

figure.

Due to symmetry about y –axis

xcm = 0

ycm = !! " #$$%

& '' 21

2211mm

ymym

=2

2

22

)2(2

)6(

(8)](2)[ –3

)6(4

2

)6(

<'<<

<! " #$

% &

<<

!!

"

#$$%

& <<

(m A Area)

=)418(

)418(82

2

'<'<

= 8 cm.

5.5.

The slab does not contribute to deviation.

For minimum deviation by prism,

r1 = r

2 = 30° as shown in figure.

3 sini = 2 sin 30°or i = 45 °

8 Minimum deviation = 2i – A = 90° – 60° = 30°

6.6. Amplitude phasor diagram :

e e

8 resultant amplitude = 26 .

7.7. T cos30º + N sin30º = mg

3 3 T + N = 2 mg ..............(i)

T sin30º – N cos30º =)2 / 3(

mv2

3 T sin30º – 3N = 4mv2T3 – 3N

= 4mv2 ..............(ii)

by (i),)(ii)

3 N =4

mv4mg2 2';

T =34

mv4mg6 2'

for N > 0

3 v < 5 m/s

at v = 2 ij T =3

38N ; N = 2N.

8.8. 88

For the given case Ray diagram will be as given

here

Here say ER will be seen by observer which appear

to becoming fr om point A. To find x, the distance

of object from A we reverse the light rays by

considering ER as incident ray & find the position

of image after two refractions.

For I refraction we use

R2v

1 34

' = R'

' 341

3 v = – 3R

For II refraction we use

R

1

x3

4

'' =

R

134 '

3 x = – 2R

Hence OA = 2 × 4 = 8 cm.


9.9. Let the cube dips further by y cm and water level

rises by 2 mm.

Then equating the volumes

(/// volume = \\\ volume in figure)

3 volume of water raised

= volume of extra depth of wood

3 100 y = (1500 – 100)10

2 = 1400 ×

10

2

= 280

8 y = 2.8 cm

8 Extra upthrust

@water

× (2.8 + 0.2) × 100 g = mg

3 m = 300 gm.

m m = = 33000 0 ggmm.. ........AAnnss..

10.10. A – p,r,t

RF = 0

So, linear momentum conservation and centre of masswill not move.

B – q,s

So, linear momentum will not be conserved and cen-

tre of mass will accelerate Wext

= HE.

C – p,s,t

D – p,s,t


1.1. U =2

1 a0 E2 =

2

1

4

220

r

QKa

V =r

KQ

2V

U =

2

22

4

22

0

r

QK

r

QK2

1

a= 2

0

r2

1 a

because 2V

UA 2r

1

so the correct option is B.

2.2. The magnitude of the electric field is maximum where

the equipotentials are close together. The direction

of the field is from high potential to low potential.

4.4. From given graphs :

ax =

4

3t and a

y = !

" #

$% & +' 1t

4

3

3 vx =

8

3t2 + C

At t = 0 : vx = – 3

3 C = – 3

8 vx =

83 t2 – 3

3 dx = dt3t8

3 2 ! " #

$% & ' .... (1)

Similarly

dy = dt4tt8

3 2 ! " #

$% & +'' .... (2)

As dw = ds.F = ) jdyidx.(F +

8 II -.

/01

2! " #$

% & +''+!

" #$

% & '-

.

/01

2! " #$

% & +'(

4

0

22

W

0

dt j4tt8

3i3t

8

3. j1t

4

3it

4

3dw

8 W = 10 JW = 10 J

Alternate Solution :Alternate Solution :

Area of the graph ;

I dtax = 6 = )3(V f)x( '' 3 V(x)f = 3.

and I dtay = –10 = )4(V f)y( ' 3 V(y)f

= – 6.

Now work done = H KE = 10 J

5.5.

In diagram angle of emergence = 60°

8 d BOC = 60°

8 r + r = 60°

3 r = 30°

8 > = 5530sin60sin 3 > = 3

6.6.b (a)

V0

nV0

V0


b (b)

V0

nV0

V0

If V0 be the f low velocity of the river, then velocity of

boat relative to water = nV0.

If the boat has to adopt the shortest path, then direc-

tion of velocity of boat relative to water should make

an angle greater than 90° with the flow direction of

river

Resultant velocity of boat = 20

20 V –)nV( . This

velocity can have a real value only when n > 1. If n =

1, then resultant velocity = 0. So the boat will follow

the shortest path only when n > 1. So options (b) and

(c) are correct.

If boat is moved normal to flow direction, then it will

cross the river in a time0nV

b, where b is the width of

the river. If n f 0, the boat will cross the river. So

option (d) is also correct.

7.7. >N = 2 , N = 8 , a =2

2 –16 = 7 (U)

v2 = 2 (7) 8 U ...........(i)

when the direction of horizontal component of the

force F is reverseda

1 = 9 m/s 2 (g)

and distance covered by the block before it stops

= s1 =

92

716

77

Again s2

= 8 + s1

= 8 +92

716

7

7

anda2 = 7 (g)

v2 = 0 +2 (a2) (s

2) = 2 (7) !

" #

$% &

77

+92

7168

3 v =3

716 m/s

8.8. Using equation of continuity A1v

1 = A

2v

2

(12 cm2)vA = (6 cm2) (8.0 m/s)

vA = 4.0 m/s

9.9. Applying Bernoulli's principle between point A and C

that are at same horizontal level

A2

A pV.2

1+@ = atm

2C pV.

2

1+@

3 pA = (1.01 × 105 N/m2) +

2

1 × 13,600 (82 – 42)

= 4.27 × 105 N/m2

10.10. By applying Bernoulli's equation between point B and

C and using equation of continuity

vB = 8.57 m/s

and pB = 3.70 × 104 Pascal

@gh = 3.70 × 104

h =600,1310

1070.3 4

77

= 272 mm


1. V = KT + C

P =V

nRT

3 P =CKT

nRT+ 3 2)CKT(

nRC

dT

dP

+(

As C < 0 by diagram

3dT

dP < 0 for all T

3 P continuously decreases.

2.2. The free body diagram of the capillary tube is as

shown in the figure. Net force F required to hold

tube is

F = force due to surface tension at cross-section

(S1 + S

2) + weight of tube.

F

Free body daigramof capillary tube

T×2 ×2 RR<<

mg

S1

S2

T×2 ×2 RR<<

= (2<RT + 2 <RT) + mg = 4 <RT + mg

= 4< × 2 × 10 –3 × 0.1 + < × 10 –3 × 10 = 10.8 <mN


3.3. From conservation of energy

mgh =2

1mv2 +

2

1 Iw2

mgh =2

1mv2 +

22mr5

2

2

1;!

" #

$% &

gh =10

7(;2

r2) # = wr

7

gh10 = ;2

r2

KE =22mr

5

2

2

1;!

" #

$% &

=7

mgh2

4.4. Let ‘v’ be the initial velocity. Tangential velocityremains same during collision and equal to

v cos60° = v/2

Let vJ be the normal component of velocity after

impact.

In H OAB : tan 60° =

v

2 / v

J 3 vJ =32

v

Then : e = 5J30cosv

v =

)2 / v3(

)32 / v( =

3

1

6.6. As no. of beats = HhFor option (A) : The frequencies are :

h1 = 550 Hz, h

2 = 552 Hz, h

3 = 553 Hz,

h4 = 560 Hz.

The beats produced will be :

H h1 = h

2 – h

1 = 2,

H h2 = h

3 – h

1 = 3,

H h3 = h

4 – h

1 = 10,

H h4 = h

3 – h

2 = 1

H h5 = h

4 – h

2 = 8,

H h6 = h

4 – h

3 = 7

Which doesnot ma tches with the given set of beatfrequencies. Hence (A) is not possible.

Similarly (B) and (C) are also not p ossible.

For option (D); frequencies w ere;

h1 = 550, h

2 = 551, h

3 = 553, h

4 = 558

H h1 = h

2 – h

1 = 1,

H h2 = h

3 – h

1 = 3,

H h3 = h

4 – h

1 = 8,

H h4 = h

3 – h

2 = 2

H h5 = h

4 – h

2 = 7,

H h6 = h

4 – h

3 = 5

which matches with the given set of beat

frequencies.

Hence (D).

7.7. Work done by kinetic friction may be positive when it

acts along motion of the body.

Friction on rigid body rolling on inclined plane is along

upward because tendency of slipping is downwards.


The time taken to reach maximum height and

maximum height are

t = g

sinu 9and H =

g2

sinu 22 9

For remaining half, the time of flight is

t' =)g2(

H2 = 2

22

g2

sinu 9 =

2

t

8 Total time of flight is t + t' = !! "

#$$%

& +

2

11t

T = !! "

#$$%

& +9

2

11

g

sinu

Also horizontal range is = u cos 9 × T

= !! "

#$$%

& +9

2

11

g2

2sinu2

Let uy and vy be initial and final vertical components

of veloc ity.

8 uy2 = 2gH and v y

2 = 4gH

8 vy = yu2

Angle ( ?) final velocity makes with horizontal is

tan? =x

y

x

y

u

u2

u

v( = 2 tan9


1.1. If M0 is molecular mass of the gas then for initial

condition PV =0M

M . RT ...(1)

After 2M mass has been added

PJ .3

V =

0M

M3 . R .

3

T...(2)

By dividing (2) by (1)

PJ = 3P


2.2. Snell’s law = >0 sin (90° – 9) = >0!

" #

$% &

d

x –1 sin90°

3 ! " #

$% &

d

x –1 = cos 9

3 x = d(1 – cos 9)

3.3. The electric field at P shall be zero if q = Q.

4.4. All energy is transfered to other particles.

5.5. (A) Absolute velocity of ball = 40 m/s (upwards)

hmax = hi = ff

= 10 +102

)40( 2

7

h = 90 m

(B) Maximum height from left =102

)30( 2

7 = 45 m

(C) The ball unless meet the elevator again when dis-

placement of ball = displacement of lift

40 t – 2

1 × 10 × t2 = 10 × t 3 t = 6s.

(D) Let t0 be the total time taken by the ball to reach

the ground then – 10 = 40 × to – 2

1 × 10 × to2

3 t0 = 8.24 s.8 time taken by the ball for each the ground after

crossing the elevator = t0 – t = 2.24 s.

6.6.

Let velocity of COM after collision is v J & angular

velocity is ;.

conserving linear momentum

mv = 2mv J 3 vJ =2

v............(1)

conserving angular momentum about COM

mv.2

R= N:;

= (NRing COM

+ Nmass

)

; =--.

/

001

2+!

! "

#$$%

& +

2

MR

4

MRMR

222

;

= !!

"

#$$%

& +

4

MRMR

22

; = mv.2

R =

2

3MR2 ;

; =R3

v

7.7. As speed of ball is variable, so motion is non uniform

circular motion.

8.8. At the highest position of ball, net tangential force is

zero, hence tangential acceleration of ball is zero,

9.9. Tension in the string is minimum when ball is at the

highest position. By conservation of energy

2

1mv2 + mg (2!) =

2

1m(20 g!)

v2 = 16 g! where v is the velocity of ball at the highest

point.

So T + mg =!

2mv

T =!

!g16m – mg = 15 mg

10.10. (A) p,r (B) q,s (C) p,r (D) q,s

(A) The fundamen tal frequency in the string,

f0 =5.02

1

101

4.102

2

/ T3 7

77

(>

'!Hz = 320 Hz.

Other possible resonance frequencies are f A and

f0 = 320 Hz, 640 Hz, 960 Hz.

(B) The fundamen tal frequency in the string.

f0 =5.04

320

4

/ T

7(

>!

= 160 Hz.

Other possible resonance frequencies are

fB = 160 Hz, 480 Hz, 800 Hz.

(C) The fundamen tal freque ncy in both ends ope n

organ pipe is

f0 =5.02

320

2

v

7(

! = 320 Hz.


fc = 320 Hz, 640 Hz, 960 Hz

(D) The fundamental frequency in one end open

organ pipe is

f0 =5.04

320

4

v

7(

! = 160 Hz.


fD = 160 Hz, 480 Hz, 800 Hz.



1.1. for object O1O

1

f

1

20

1

v

1 (+ .... (1)

for object O2O

2

f

1

40

1

v

1 '(' .... (2)

from equation 1 and 2 we get

f = 80/3

2.2. As the reflector approaches OS, the beat

frequency will decrease to zero. After this, the

reflector moves away from OS thereby increasing

the beat frequency but after a long time the beat

frequency will to become constant. Hence the

correct option is (D).

3.3. Equation of process

3 @

2P

= constant = C .... (1)

Equation of State TM

RP(

@ .... (2)

From 1 and 2

PT = constant

3 C is false, D is true.

As @-changes to2

@

3 P changes to2

Pfrom equation (1)

3 A is false.

Hence T changes to T2 . 3 B is true.

4.4. w.r.t.

5.5. From graph (1) : vy = 0 att =

2

1sec.

i.e., time taken to reach maximum height H is

t =g

uy =

2

1

3 uy = 5 m/s .. ....AAnnss..((ii))

from graph (2) : vy = 0 at x = 2m

i.e., when the particle is at maximum height, its dis-

placement along horizontal x = 2 m

x = ux × t

3 2 = ux ×

2

1

3 ux = 4 m/s ....Ans (ii)....Ans (ii)

66.. 4 4 ppyy y y22 + 4 x + 4 x22))1/21/2

Force = pdx

dE

{y is not changing since p is directed along x axis}

= p i ] jxy8iy4[ 2 + i

= p4y | ] jx2iy[ + |

= 4py

22

x4y +Ans. 4 pyAns. 4 py y y22 + 4 x + 4 x22))1/21/2

Sol.7 to 9.Sol.7 to 9.

FBD of rod and cylinder is as shown.

Net torque on rod about hinge 'O' = 0

8 N1 × L = mg ×

2

L


or2

mgN1 (

Net torque on cylinder about its centre C is zero.

8 f1R = f

2 R

or 21 ff (

Net torque on cylinder about hinge O is zero.

8 N2 × L = N

1 × L + mgL

or N2 =

2

mg3

10.10. (A) p (B) q,s (C) q,s (D) q,s

(A) Work done by an ideal gas during free expansion

is zero.

(B) The angle between normal reaction on block and

velocity of block is acute (whether the block moves

up or down the incline). Hence work done by this

force is non-zero and positive.

(C) Net electrostatic potential energy

= S1 + S

2 + M

12 =

b4

Q

a8

Q

a8

Q

0

2

0

2

0

2

\<'

\<+

\<

=b4

Q

a4

Q

0

2

0

2

\<'

\< = non-zero and positive.

(# b > > a)(D) The kinetic energy of cylinder is increasing and

work is done on cylinder by only force of friction.

Therefore work done by force of friction on cylinder is

non-zero and positive.


1.1. The free body diagram of cylinder is as shown.

Since net acceleration of cylinder is horizontal,

NAB

cos30° = mg

or NAB

=3

2mg .... (1)

and NBC

– NAB

sin30° = ma

or NBC

= ma + NAB

sin 30° .... (2)

Hence NAB

remains constant and NBC

increases with

increase in a.

22.. ((BB)) The line of impact for duration of collision is

parallel to x-axis.

The situation of striker and coin just before the

collision is given as

u

3

Figure (A)

before collision

line of impactcoin striker

u

3

Figure (B)after collision

rest

coin striker

Because masses of coin and striker are same,

their components of velocities along line of im pact

shall exchange. Hence the striker comes to rest

and the x-y component of velocities of coin are uand 3 m/s as shown in figure.

Pu99

36

xO 4

y

coin

For coin to enter hole,

its velocity must be along PO

8 tan 9 =4

6 =

u

3or

u = 2 m/s Ans. (2, 0)

3.3. For no ray to emerge out of side PR

A > 2C 3 sin2

A > sinC

3 sin2

A >

2

3

or A > 120 °


4.4. @gh <r2 = 2<rS cos 9

3 r =gh

cosS2

@9

=101010

5.0123 77

77 = 10

–6 m

5.5. F = 22 )mg(f +

==f

F

mg

(f = m r);;2

Now when the angular speed of the rod isincreasing at const. rate the resultant force

will be more inclined towards f"

.

Hence the angle between F"

and horizontal plane

decreasesso as with the rod.

6.6. Consider a spherical shell of radius r(r > R) andthickness dr. Then potential at centre due to it,

dV =r

dqK= I

[U

(

<r

Rr

23

drr

r4)r / C(K

= (const.) I[U

(

r

Rr2

drr

1

= (const.) !$% & R1

7.7. Using gauss theorem

I sd.E""

=0

inq

a

E × 4 < x2 =

0

xR

Rr

2

3

0

drr4r

C

dv

a

<

(a

@ II

(

(

E × 4 < x2 = !

" #$

% &

a R

xn

x

)C(2

0

!

3 E = ! " #

$% &

a<

R

xn

)4C(

0

!

8.8.

Electric field will be radially outwards.Electric potential decreases as we move in thedirection of electric field.


1.1. Flux will be maximum when maximum length of ring

is inside the sphere.

This will occur when the chord AB is maximum. Now

maximum length of chord AB = diameter of sphere.

In this case the arc of ring inside the sphere sub-

tends an angle of3

< at the centre of ring.

8 charge on this arc =3

R<.Q

8 ? =0

3

R

\

Q<

=03

R

\<Q

2.2. The change in length of rod due to increase in

temperature in absence of walls is

H! = ! = HT= 1000 × 10 –4 × 20 mm = 2 mm

But the rod can expand upto 1001 m m only.

At that temperature its natural length is= 1002 mm.

8 compression = 1mm

8 mechanical stress = Y!

!H = 10111 ×

1000

1

= 108 N/m2

3.3. v1 = 4 cos 53 º i + 4 sin 53 º j =

5

12i +

5

16 j

v2 = 3 cos 37 º i + 3 sin 37 º j =

5

12i +

5

9 j

v12

=5

7 j = 1.4 j

Relative velocity in horizontal direction is zero.

4.4. 0.3 m

(Range)1 = (Range)

2

g

2.02)2.0 –(g2

g

1.02)1.0 –(g2

7(7!!

! = 0.3


5.5. (a) 4.5 m/s (b) 1.5 m/s (c) 3.75 cm

By conservation of angular momentu m

0.5 × V × 0.4 = 0.5 × 4 × 1.2

v = 3u

also by energy conservation

222 u5.02

1)3.0(100

2

1v5.0

2

1 77+77(77

34

u

2

9

4

v 22

+(

4

2

9

4

u8 2

(

3 u =

2

3

4

9 (

So v = 3u

= 3 × 1.5

= 4.5

(c) an =

r

u2

3 r =n

2

a

u ! " #

$% & (

7(( 2h

n s / m605.0

3.0100

m

Ka

3 r =)60(

)5.1( 2

= m0375.060

25.2 (

= 3.75 cm

6.6. Q = 4m and f = 50 Hz.

8 V = fQ = 200 m/s

# V = >T

8 T = > v2 = (0.1) × (200) 2

= 400 N

7.7. Since integral number of waves shall cross a pointis 5 seconds, therefore power transmitted in 5seconds is= <P> × 5 = 2<2 f2 A2 > v × 5

= 2 × <2 × (50) 2 × (2 × 10 –3)2 × (0.01) × 200 × 5

=5

2<

8.8. The equation of waves isy = A sin(kx – ;t + ?0)

8 where K =2

2 <(Q<

, ; = 2<f = 100 < and A = 2

at x = 2 and t = 2 y = 1 mm8 1 = 2 sin( < – 200< + ?0)

solving ?0 = –30°

8 y = 2 sin ! " #

$% & 5'<'

<30t100

2

x


1.1. The velocity of profile of each elementary sectionof the pulse is shown in figure 1 and figure 2.

pulse movingtowards left

pulse movingtowards right

(Figure-1) (Figure-2)

Velocityprofile

Velocityprofile

elementarysection

velocityvector ofelementarysection

When both the pulses completely overlaps, thevelocity profiles of both the pulses in overlap regionare identical. By superposition, velocity of eachelementary section doubles. Therefore K.E. ofeach section becomes four time s. Hence the K.E.in the complete width of overlap becomes fourtimes, i.e., 4k.

2.2.

2 160° 30°

30°60°

N2N1

N sin 60°2

N sin 30°1

N cos 60°2

N cos 30°1

N1 sin 30° = N2 sin 60°

N1 cos 30° + N2 cos 60° = mg

Solving above equation

N2 = 502

1010

2

mg (7(


3.3. (i) (a) The charge on the outer most surface will be

(qa + q

b) and it will be uniformly distributed

8 v =v =r

)qq(K ba + Ans.Ans. ;

E =E = 2

ba

r

)qq(K + Ans. Ans.

where K =04

1

\<(b) At a point inside the cavity of radius ‘b’ the

potential will be due to qb, – q

b induced on its inner

surface and due to (qa + q

b) on the outer surface of

the sphere.

v =v =r

Kqb – b

Kqb + +R

)qq(K ba + Ans. Ans.

and E at that point will be only due to qb (which is

placed at B)

E =E = 2

b

r

Kq Ans. Ans.

(ii)RR = = 2

ba

R4

qq

<+

,,aa== 2

a

a4

q

<'

,,bb = = 2

b

b4

q

<'

Ans. Ans.

(iii) 0 Ans.0 Ans.

4. (a) Parabola y = ax 2 is shown. It is clear from diagram

that at x = 0 velocity is along x-axis and constant aN

is along y-axis. So,

aN = 2

2

dt

yd

dt

dy = 2a ×

dt

dx = 2aVx

2

2

dt

yd = 2av

dt

dx = 2av2 (# 0

dt

xd2

2

( )

aN = 2av2

R = 2

2

av2

v =

a2

1.

(b) 1b

y

a

x2

2

2

(! " #$

% & +!

" #$

% &

Here again at x = 0 particle is at (0,± b) moving along

positive or negative x-axis hence aN is along y-axis

only.

aN = 2

2

dt

yd

0dt

dy

b

y2

dt

dx

a

x222

(+

0dtdy

by2

avx2 22 (! " #$% & +

0dt

yd

b

y2

dt

dy

b

2

dt

dx

a

v22

2

2

2

22 (!

! "

#$$%

& +!

" #$

% & +

[# v = const. along x-axis onlydt

dy = 0]

!!

"

#$$%

& '(

2

2

22

2

dt

yd

b

)b(2

a

v23 a

N = 2

2

a

bv'

R =N

2

a

v' =b

a2

7.7. The difference in K.E. at positions A and B is

B

O

gsin

gsinVLA

2

VLB

2

A

KA – K

B = 2

B2A mv

2

1mv

2

1' = mg (2L

cos9)

= 2mgLcos 9 .... (1) Ans.Ans.

TA =

L

mv 2A + mg cos 9

TB =

L

mv 2B – mg cos 9

8 TA – T

B =

L

mvmv 2B

2A '

+ 2 mg cos 9 .... (2)

from equation (1) and (2)T

A – T

B = 6 mg cos 9 Ans.Ans.

The comp onent of accele rations of ball at A and Bare as shown in figure.

22B

2A2

BAL

v

L

v)sing2(|aa| !

! "

#$$%

& '+9(+

""

= 9+9 222 cosg16sing4 = 9+ 2cos124g Ans.Ans.


8.8. (A) p,q (B) p,q (C) q,r (D) q,rSol.Sol. In all cases speed of balls after collision will be same.

In case of elastic collision speed of both balls aftercollision will be u, otherwise it will be less than u.


1.1. At the instant 3m is about to slip, tensionin all the strings are as shown

60°°

mg

3 mg

T

8 3 >mg = T cos 60 ° .... (1)and mg = T sin 60 ° .... (2)

8 > =33

1

2.2. |P| xH = mv sin 60° =

2

3 mv

|P| yH =2

mv + mv =

2

3 mv

3

2

y

2

xnet PP|P| H+H(H =!

"

#$%

& +4

3

4

9

mv

mv3|P| net (H

!!!

"

#

$$$

%

&

@(3

@(

vAdt

dm

)dtv(Adm,Since

3 v.dt

dm3|F| net !

" #

$% & (H = 2vA3 @ Ans.Ans.

3.3. P0 + @

1gh – @

2gh +

r

T2 = P

0

3 T =2

r(@

2 – @

1) gh

4.4.

KE = V1 – V

2 2

1 mh2 = !

" #

$% &

a

GMm – –

a2

GMm

h = !! "

#$$%

&

2

1 –1

a

GM2

5.5. Let the charge on intermediate shell be q (after

earthing)

Potential of the intermediate shell = 0

3 R2

KQ+

R2

Kq -

R3

Q2K = 0

2

Q+

2

q –

3

Q2 = 0

q = ! " #

$% & '

2

Q

3

Q22 = 2

6

Q3Q4!

" #

$% & '

= 3

Q

, which is same as that before earthing

8 No charge will flowNo charge will flow.

6.6. 22

> = Kx =dx

dM

IM

O

dM = I!

O

dxKx and K = 2

M2

!

V = >F

=Kx

F =

dt

dx I

!

O

dxx =K

F I

t

o

dt

t =fK.

94 3! =

2

3

m2.F9

4!

!

=F9

M8 !=

159

5.1458

777

= 2.

Sol. 7 to 9.Sol. 7 to 9.

The FBD of A and B are

Applying Newton's second law to block A

and B along normal to inclined surface

NB – mg cos 53° = ma sin 53°

mg cos 37° – NA = ma sin 37°

Solving NA =

5

m(4g – 3a) and N

B =

5

m(3g + 4a)

For NA to be non zero

4g – 3a > 0

or a <3

g4



1.1. The distribution of charge on the outer surface, de-pends only on the charges outside, and it distributesitself such that the net, electric field inside the outersurface due to the charge on outer surface and allthe outer charges is zero. Similarly the distribution ofcharge on the inner surface, depends only on thecharges inside the inner surface, and it distributesitself such that the net, electric field outside the innersurface due to the charge on inner surface and all theinner charges is zero.Also the force on charge inside the cavity is due tothe charge on the inner surface. Hence answer is

option (A)(A).

2.2. The distance between the orbiting stars is

d = 2rcos30° = 3 r. The net inward force on orbitingstars is

r

mv30cos

d

Gm

r

GMm30cos

d

Gm 2

2

2

22

2

!"##"

$ G 2

32

T

r4M

3

m %!&

'

()*

+#

or T =2%,,

-

.//0

1 #

3

mMG

r3

d

d dr

Mm

3.3.2

1k

2

0x + Mgh =2

1k(x

0+h)2 + 0

2 h =k

Mg2 – 2x

0

Maximum downward displacement

= [k

Mg2 – 2x

0]

4.4. H = 3

3

a2

u2

a3 is same for all the three cases.

HA =3

4a2

)sinu( 2

, HB =3a2

u2

and HC =

3

4a2

)cosu( 2

$ HB = H

A + H

C

5.5. (A)

When the ball is just released, the net force on ball isWeff (= mg – buoyant force)The terminal velocity ‘vf’ of the ball is attained whennet force on the ball is zero.$ Viscous force 6%5r vf = Weff

When the ball acquires3

2rd of its maximum velocity v

f

the viscous force is =3

2 Weff.

Hence net force is Weff – 3

2Weff =

3

1Weff

$ required acceleration is =3

a

6.6. If we complete the trapezium as shown It becomesan equilateral triangle2 A = 60°

6!&'

()*

+ 7#

2

A

Smn

2

Asin min

A

5 5

10

2

2

60sin

2

60sin min

!&'

()*

+ 7#

, 7min = 30°

7.7.

Clearly, PM = 2

3cm

37º > sin –1 )2 / 3(an

1

0 #

5

3 >

2

a3n

1

0 #


3n0 +

2

a9 > 5

2

a9 > 1

a >9

2

8.8.

25

52.!6 == 0.1 g / cm = 10 –2 Kg/m

Ist overtone

s8 = 25 cm = 0.25 m

fs = 68

T1

s

pipe in fundamental freq

p8 = 160 cm = 1.6 m

fp =

p

V

8

! by decreasing the tension , beat freq is

decreased

$ fs > f

p

2 fs –f

p = 8

2 61

320

10

T

250

12 ..

99 = 8

2 T = 27.04 N

9.9. (A) p,s (B) p,s (C) q,s (D) r

The minimum horizontal force required to push the

two block system towards left= 0.2 × 20 × 10 + 0.2 × 10 × 10 = 60.

Hence the two block system is at rest. The FBD

of both of blocks is as shown. The friction force f

and normal reaction N for each block is as shown.

F =20N1 F =60N2

f =40Nmax f =max

F =20N1 F =62 0NN=4020kg 10kg

f=20N

FBD of both blocks

N=40

f=20N

Hence magnitude of friction force on both blocks

is 20 N and is directed to right for both blocks.

Normal reaction exerted by 20 kg block on 10 kg

block has magnitude 40 N and is directed towards

right. Net force on system of both blocks is zero.


2.2. U = 3x + 4y

ay = m

Fy =

: ;m

x / U <<9 = – 3

ax = m

Fy =

: ;m

y / U <<9= – 4

2 a"

= 5 m/s5 m/s22

Let at time 't' particle crosses y-axis

then – 6 =2

1 ( – 3) t2

2 t = 2 sec.

Along y-direction :

=y =2

1 ( – 4) (2)2 = – 8

2 particle crosses y-axis at y =y = – 4 4

At (6, 4) : U = 34 & KE = 0

At (0, – 4) : U = – 162 KE = 50

or,2

1 mv2 = 50

2 v = 10 m/sv = 10 m/s while crossing y-axis

3.3.

Tension in elementary section of width dx is

T = 8 xg (8 = mass / length)

$ extension of length x (= BC) of wire is

=x = > 8

x

0

dxYA

)gx( =

YA2

x28... (1)

2 extension in total length of wire #(=AB) is 2=x

2=x =YA2

g2#8

... (2)


$ from equation (1) and (2)

x =2

#

Ans.Ans.PC

AC =

x

x9# = ( 2 – 1)

4.4. 8i = wavelength of the incident sound

=f

2

uu10 9

=f2

u19

fi = frequency of the incident sound

=

2uu10

uu10

9

9f =

f19

18 = f

r = frequency of the

reflected sound8

r = wavelength of the reflected sound

=rf

uu10 # =

f18

u11 × 19 ==

18

1911?.

f

u

r

i

88

=u1911

f18

f2

u19

?? =

1111

99 Ans.Ans.

5.5. Pr"

= ( i + j ) t

Qr"

= (2 i + j ) + ( – i + 2 j )t

QPr"

= Qr"

– Pr"

= 2 i + j + ( –2 i + t j )

QPr"

= (2 – 2t)iˆ + (1 + t) j

x = 2 – 2t y = 1 + t2 x = 2 – 2 (y – 1)x + 2y = 4 x + 2y = 4 Ans.Ans.

6.6. Force on cone while it is penetrating the sand is shownin F.B.D. belowApplying work energy theorem to the cone as xchanges from 0 to d=KE = work done by mg + work done byresistive force R

KFinal

– KInitial

= mgd – dxkx

d

0

2>

0 – mgh = mgd – dxkx

d

0

2>

$3

kd3

= (mgd + gh)

2 k = )dh(d

mg33

#

7.7. (A) Charge distri bution on syste m is shown

below

So electric lines of forces are as shown below

Since number of lines of force are proportional to

charge so no. of lines of forces emerging from inner

sphere should be equal to the no. of lines of fo rces

emerging from outer shell.

8.8. VB =

&&'

(

))*

+

?

?#

?

?9

?

?9

9

9

9

9

9

2

9

2

9

2

9

1018

103

1018

103

1027

106K = 200 V

VA =

&&'

(

))*

+

?

?#

?

?9

?

?9

9

9

9

9

9

2

9

2

9

2

9

106

103

1018

103

1027

106K

= 200 V – 150 V + 450 V= 500 V

VC = V

B (as shell is conducting)

Therefore, VC – V

B = 0

VB – V

A = – 300 V

9.9. Potential at point B and point C increases by same

value, keeping their difference unchanged.


1. F.B.D. of man and plank are -


For plank be at rest, applying Newtons second law

to plank along the incline

Mg sin 4 = f ...............(1)

and applying Newton ’s second law to man along

the incline.

mg sin 4 + f = ma ...............(2)

a = g sin 4 , -

./0

1 #

m

M1 down the incline


If the friction force is taken up the incline on man,

then application of Newton ’s second law to man

and plank along incline yields.

f + Mg sin 4 = 0 ..........(1)

mg sin 4 – f = ma ..........(2)

Solving (1) and (2)

a = g sin 4 , -

./0

1 #

m

M1 down the incline


Application of Newton ’s seconds law to system of

man + plank along the incline yields

mg sin 4 + Mg sin 4 = ma

a = g sin 4 , -

./0

1 #

m

M1 down the incline

2.2. As ON = MN = OM = r

So it is equilateral t riangle :

$ Potential at N due to two dipoles ;

V = V1 + V2

@@@ 60°

r r

r

P (0,0,0) (r,0,0)P@ @

N

OM

% - 60°

,,

//0

1 0,

2

3r,

2

r

= 2r

60cosKp " + 2r

)60(cosKp "9%= 0

3.3. 6! relx

1

xx

rel = 6

x

2

2

2

rel2

dt

xd

dt

xd6!

arel

= 6 g

4.4. At position A balloon drops first particle

So, uA = 0, a

A = – g, t = 3.5 sec.

SA = ,

-

./0

1 2gt2

1...........(i)

Balloon is going upward from A to B in 2 sec.so

distance travelled by balloon in 2 second.

, -

./0

1 ! 2

BB ta2

1S ..........(ii)

aB = 0.4 m/s 2 , t = 2 sec.

S1 = BC = (SB + SA) ...........(iii)

Distance travell by second stone which is dropedfrom balloon at B

u2 = u

B = a

Bt = 0.4 × 2 = 0.8 m/s

t = 1.5 sec.

, -

./0

1 9! 2

22 gt2

1tuS ...........(iv)

A

B

SB

SA

C

>

>

>

>

Distance between two stone

=S = S1 – S

2 .

5.5.2RR

F F

a

f

f = ma ...(i)

F2R – FR – fR = I4 ...(ii)

a = Ra ...(iii)

FR – fR = R

a I .

F – ma = 2R

Ia

F = , -

./0

1 #

2R

Im a.

a = 2R / Im

F

#

f = ma =2R

Im

mF

#

f =2R

Im

mF

#

f =3

10N


6.6. a =

2R

Im

F

# =

3

5

4 =2

a =

6

5

v = 0 + 36

5? =

2

5

B = Bo + 4 t = 0 + 33

5? = 5

KE =2

1mv2 +

2I

2

1B

=2

5

2

54

2

1552

2

1???#???

= 25 +2

25 =

2

75J

7.7. F2R – FR = I4

4 =I

FR

B = 0 + tI

FR,

-

./0

1

KE =2

2

1 Iw

=2

2

22

2

1t

I

RF I ,

, - .

//0 1 ?

=I2

RF 22

=42

331100

????

=2

925 ? = 112.5 J.

8.8. (A) p (B) q (C) p (D) s

(A) In frame of lift effective acceleration due to

gravity is2

g3

2

gg !# downwards

$ T = 2% g3

2#

(B) K# = mg $ L

g

m

k!

constant acceleration of lift has no effect in time

period of oscillation.

$ T = 2%k

m = 2%

g

#

(C) T = 2 % mgd

C = 2% g3

22

2mg

3

m 2

#

#

#

%!

(D) T = 2 % Ag

m

D = 2% Ag

A2 /

DD #

= 2% g2

#


11.. ((CC)) Work done against friction mus t equal the initial

kinetic energy.

$ 2mv2

1 = >

E

61

dxmg ;2

v2

= >E

1

2dx

x

1gA ;

2

v2

= Ag

E

&'

()*

+9

1x

1

v2 = 2gA 2 v = Ag2

2.2. F.B.D. for minimum speed (w.r.t. automobile)

:

Ffy' = N – mg cos G –

R

mv2

sin G = 0.

Ffx' =

R

mv2

cos G + 6N – mg sin G = 0

2R

mv2

cos G + 6(mg cos G +R

mv2

sin G) – mg sin G = 0

2 v2 =)sin(cos

)sinRgcosRg(

G6#GG9G6

for G = 45º and 6 = 1 :

vmin

=11

RgRg

#9

= 0


F.B.D for maxim um speed (w.r.t. automobile)

Ffx'

=R

mv2

cos G – mg sin G – 6(mg cos G

+R

mv2

sin G) = 0

for G = 45º and 6 = 1

vmax

= E (infinite)

3.3.

Taking cylindrical element of radius r and thickness

dr

dm =#)RR(

M21

22 9% × (2%r # dr)

CAB

= > C #ed = > 2rdm = > 9

2

1

R

R

3

21

22

drr.)RR(

M2

= )RR(m2

1 21

22 #

Using parallel axis theorem

IXY

= )RR(m2

1 21

22 # + MR

22

4.4. Let m be minimum mass of ball.

Let mass A moves downwards by x.

From conservation of energy,

mgx =2

1kx2

x = , -

./0

1 k

mg2

For mass M to leave contact with ground,

kx = Mg

, -

./0

1 k

mg2 K = Mg

m =2

M.

5.5. In elastic collision the velocities are exchanged if

masses are same.

$ after the collision ;

VC = 0

VA = v

Now the maximum compression will occure when

both the m asses A and B move with same velocity.

$ mv = (m + m) V (for system of A – B and spring)

$ V =2

v

$ KE of the A – B system =2

1× 2m

2

2

v,

-

./0

1

=4

mv2

And at the time of maximum compress ion ;

2

1mv2 =

2

1× 2m

2

2

v,

-

./0

1 +

2

1K X2max

$ Xmax

= vK2

m

6.6.

E =3.0

1040 9 = 100 V/m

(near the plate the electric field has to be uniform

$ it is almost due to the plate).

For conducting plate

E =0H

I

2 I = H0E

Therefore %, I = 8.85 × 10 –12 × 100= 8.85 × 10 –10 C/m2

7.7. Direction of E.F. at B is towards A that will exertforce in this direction only, causing the positive

charge to move. [ E"

is perpendicular to

equipotential surface and its direction is from high

potential to low potential.]

8.8. W = q.dV

= –1 × 10 –6[20 – ( –20)]= – 4 × 10 –5 J.



1.1. For first collision

v = 10 m/s.

t1 =10

)5(%

= % /2 sec.

velocity of sep = e. velocity of opp.

v2 – v1 = 2

1(10)

v2 –

v1 = 5 m/sfor second collision

$ t2 =5

)5(2% = 2 %

$ total time t = t + t 2

= % /2 + 2 %t = 2.5 %

2.2.

The friction force will reduce v 0 , hence translational

K.E.

The friction force will increase BThere is no torque about the line of contact, angular

momentum will remain constant

The frictional force will decrease the mechanical

energy.

3.3. Equation of the component waves are :

y = A sin(B t – kx) and y = A sin ( B

t + kx)

where; B t – kx = constant or B

t + kx = cosntant

Diffeentaitin g w.r.t. 't' ;

B – k

dtdx = 0 and B

+ k

dtdx = 0

2 v =dt

dx =

k

Bandv = –

k

B

i.e.; the speed of component waves is , -

./0

1 B

k.

Hence (B)

4.4. Kx = V(2000) (10) – V (1000) (10)

=2000

10 [ 1000 × 10]

Kx = 50 N ... (b)

Ustored

=2

1 × (100)

2

100

50,

-

./0

1 =

100

2500

2

1?

= 12.5 J

5.5. At x = 5 cm, x

Ve

<

< = slope of figure 1

at x = 5 = +ve

So Fx = – qx

Ve

<

< = – ve

at y = 15 cm,y

Vg

<

< = slope of figure 2

at y = 15 = – ve

So Fy = – m y

Vg

<

< = +ve

So particle will try to move towards – x directionand +y direction.

6.6. at (25, 35),

25xxF! = – q

25x

e

x

V

!<

<

= ,, -

.//0

1 ?99

1.0

102 4

× (20 × 10 –6) = 4 N

35y

g

35yyy

VmF

!

! <

<9!

= – (200) ,,

//0

1 9

10.0

10 3

= – 2N

Fnet = j2i4 9 = (200) a

j200

2i

200

49 = a

a = ) ji2( 9 × 10 –2 m/sec 2


7.7. W (5, 15) @ (25, 35) = U(25, 35) – U(5, 15)

= (0 + (200) ( –1.5 × 10 –3)) – [(20 × 10 –6)

(1/2 × 104) + (200) ( –1.5 / 10 –3)]

= – 0.1 J

88.. (A(A) p, ) p, q, q, rr, s, s ,t,t ;; (B(B) p) p , q, q , r, r, s, s ,t,t ;; (C(C) p) p , s,, s, t;t;

(D) p, q, (D) p, q, r, s,tr, s,t

(p, s) since there is net impulse, translations motionwill occurs for all cases.

(r,q) only in C, impulse is passing through centre of

mass. Hence rotation will occur and angularmomentum will increase in all cases except (C).

(t) About all the points on the line of action of the

impulse, torque is zero. Hence angular momentumwill conserve for many points.


1.1. T =L

F...........(i)

F = 5Adx

dv J 5L2

L

V = 5LVV

5 =LV

F...........(ii)

From (i) (ii),(i) (ii)

&

'

()

*

+

5

F J [V].

2.2. – > > B!=0

T 0

2xdxm

T

#

#

2 T =2

xm 22B

#

2 Y =#

#

=A

F=# =

Ay

F#

=# =AY

dx2

xm 22B#

=# =AY6

m 32#

#

B

=# =y6

32#DB

=# = B2

B2 = 2B

1

3.3. Both blocks loose contact immediately after therelease.

TP =K4

m2% ,

TQ =K

m2%

2 $ TQ = 2TP

Q comes at lowest position at time2

TQ travelling

a distanceK

mg2 downwards.

In time2

TQ , i.e. time period of P (T P) the block P

come back to srcinal position

$ The distance between P and Q isK

mg2

4.4. At t =2

TQ both the blocks are at extreme position

and their velocity is zero. [Soln. of SSI Sir][Soln. of SSI Sir]$ VP = VQ = 0

6.6. B = – V / V

P

==

= – V

PV

==

= – 3

3

102.0

101405.19?9

??

= 1.05 × 109 Pa .

Ans. 1.05Ans. 1.05 × 10× 1099 Pa. Pa.

7.7. Moment of inertia of one rod about the axis of frame

=12

L

4

m 2

, -

./0

1 +

2

2

L

4

m,

-

./0

1 = ,

-

./0

1 #

4

1

12

1L

4

m 2

=12

Lm 2

$ Moment of i nertia of frame = ML 2 /3.

9.9. (Fo(For thr the aboe abo ve twve tw o quo quesestitionsons))Newton's law applied on C.M. gives

mg sin G – f = ma .... (1)

Writing K ! C4 about C.M., we have

f. 2

L =

3mL

2

4 .... (2)

from the condition of rolling, we have

a =2

L4 .... (3)

from (1), (2) and (3)

f =5

sinmg2 G and a =

5

3 g sin G



1.1. From the free body diagram of the sphere :

FV = 4 mg – 2 mg – F

B

2 FV = 2 mg – F

B

2 6 % 5

r V = g

2r

3

4 3 ,/0

1 D9

I%

(since 4m = I?% 3r3

4)

2 V = 5D9I g)2(

r9

2 2

2.2. In the figure shown

! =OOL P & =CCL P are similar

P

OPOO

C!

CLC

L..............(1)

also ! = OOL C & = CCL C are similar

CLC

LOO = C

OC

C .............(2)

By equation (1) and (2)

P

OP

C =

C

OC

C 2

C

P

OC

OP

CC! Ans. Ans.

3.3. As wave has been reflected from a rarer medium,

therefore there is no change in phase. Hence equation

for the opposite direction can be written as

y = 0.5A sin ( –kx – Bt + G)

= – 0.5A sin (kx + Bt – G)

4.4. RP

m = 1kg

µ = 1.5

M = 11kg

S

Q

RP

m = 1kg

µ = 1.5

M = 11kg

S

Q

RP

m = 1kg

µ = 1.5

M = 11kg

S

Q

RP

m = 1kg

µ = 1.5

M = 11kg

S

Q

RP

If the point P has an acceleration a upwards then the

acceleration of point R will be a downwards.

R

M = 11kg

M = 11kg

M = 11kg

M = 11kg

The point R has an acceleration a downwards

so the block will also have an acceleration a

downwards.

PS

M = 11kg

M = 11kg

M = 11kg

M = 11kg

The point P has an acceleration a upwards, the block

has an acceleration a downwards so the acceleration

of S will be 3a downwards. (because

blockPS a

2

aa """

!#

).

The point Q will also have an acceleration 3a towards

right.

The F.B.D. of 11kg block

2T T

a

110 N

The F.B.D. of 1kg block

3a

T15N

Using FBD of 11 kg block, which will have acceleration

a downwards.


110 – 3T = 11a ........ (1) (in downwards direction)

For 1 kg block, which will have acceleration 3a,

T – 15 = 3a (in horizontal direction)

or 3T – 45 = 9a ............. (2)

on adding equation (1) & (2) we get

20a = 65 2 4a = 13 m/s2

5.5. 1.15 × 108 =

,,

-

.//0

1 %

#

4

d

)a10(9002

2 d = %10

6 cm = %10

06.0

m = %

? 9

10

106 2

m

Ans.Ans.%

? 9

10

106 2

m m

6.6. The situation is shown in figure.

(a) From figure h = # (cos G – cos G0)

and M2 = 2gh

= 2g# (cos G – cos G0) ....... (1)

Again T – mg cos G = mM2 / # ....... (2)

Substitting the value of M2 from eq. (1) in eq. (2)

we get

GGN

Mmg

h

T#

T – mg cos G = m {2g# (cos G – cos G0) / # }

or T = mg cos G + 2mg (cos G – cos G0)

or T = mg (3 cos G – 2 cos G0)

or T = 40g (3 cos G – 2 cos G0) newton

Ans.Ans. T = 40 (3 cos G – 2 cos G0) kg f.

(b) Let G0 be the maximum amplitude. The maximum

tension T will be at mean position where G = 0.

$ Tmax

= 40 (3 – 2 cos G0)

But Tmax

= 80

Solving we get G0 = 60= 60°°

Ans.Ans. G0 = 60= 60°°

9.9. (i) Cons. linear mo mentum

– 2m.v + 2v.m = 0 = MVcm

Vcm

= 0

(ii) As ball sticks to Rod

Conserving angular momentum about C

2v.m. 2a + 2mva = CB

= ,, -

.//0

1 ## 22

2

a4.ma.m212

a36.m8

6mv.a = 30 ma 2.B

2 B =a5

v

(iii) KE =

2

1 CB2 =

2

1. 30 ma 2 × 2

2

a25

v

=5

mv3 2

.


1.1. Upward force by capillary tube on top surface of

liquid is

f up

= 4Ia cos G

If liquid is raised to a height h then we use

4Ia cos G = ha 2 Dg

or h =ga

cos4

D

GIAns.Ans.

2.2. =# =yr

F2%

#2 =# 4 2r

#

Only option 'radius 3mm, length 2m' is satisfying

the above relation.


3.3. Velocity gradient = 210

5.0

25.2 9?

as force on the plate due to viscocity is from upperas well as lower portion of the oil, equal from each

part,

Then, F = 2 5

Adz

dv

= 2 × 5 × (0.5) 21025.1

5.09?

2 5 = 2.5 × 10 –2 kg – sec/m 2

4.4.

2

5 sin i = 1 sin e 2

2

5 22 hcm4

cm2

#

= 1 ×22 hcm16

cm4

# 2 h = 4 cm

5.5.


mgh =2

1mv2

2 mg# sinG =2

1 mv2

2 2g sin G =#

2v = a

C

g cos G = at

Total acceleration a = 2t

2c aa #

= 22 )sin2(cosg G#G = 1sin3g 2 #G

6.6. µ =cm40

g3.2 m = 2

3

1040

102.39

9

?

? ==

40

2.3 = =

4000

32kg/m

# =2

8

2 8 = 2# ...........(1)

f =8v

=#2

1 µ

T

2 64

1000

= 210402

1

9?? 4000 / 32

T

2

2210402

64

1000&'

()*

+??? 9

4000

32 = T

64

1000 ×

4000

32 = T

2 T =8

10 N

vc y =2

2

6

1040

1005.

10

8 / 10

9

9

9

?

? =8

107

)05(.

40

= 109 N/m2. [ [ Ans. Ans. 1 1 101099 N/m N/m22 ] ]

7.7. Torque of f riction a bout A is zero.

8.8. Angular momentum conservation about point A.

Lin = mv0r – mk2B0

Lfin = 0

Lfin = Lin

2 v0 = B0k2 /r.

9.9. acm = –6g

02 = v02 – 26gs 2 S =

g2

v20

6


1.1. (B) VB > V

D = In a Wheatstone's bridge circuit shown

if PS = QR, VB = V

D . No current flows between B

and D.

If PS < QR , VB > V

D current flows from B to D.

If PS > QR , VB < V

D current flows from D to B.


Alternate :Alternate :B

QP

A C

SR

D

G

Let resistance P = 0 and all other resistances Q,R,S,G

are non –zero then PS < QR condition is satisfied.

JG

P = 0 P = 0

i1

VA = V

B

2 VA – i

1R = V

D

2 VA – V

D = i

1R = positive (or current flows from A to

D through G, then VA > V

D)

2 VA > V

D

2 VB > V

D

Let resistance S = 0 and all other resistances P,Q,R,G

are non –zero then PS < QR condition is also satis-fied.

J

B

QP

A C

R

D

B

QP

A C

R

D

G

i2

G

S = 0 S = 0

VD = V

C

2 VB – i

2Q = V

C = V

D

2 VB – V

D = i

2Q = positive

2 VB > V

D

Suppose resistance Q is zero and all other resis-

tances P,R,S,G are non –zero then PS > QR.

J

B

Q=0P

A C

R S

D

G

B

P

A C

R S

D

G

i3

Q=0

VB = V

C

VD – V

C = i

3S = positive

2 VD > V

C

2 VD > V

B

If resistance R is zero and all other resistances arenon –zero, then PS > QRand similarly, we get V

D > V

B

Hence if PS < QR, VB > V

D .

2.2.

K.E. system =2

1 × 2 mR 2 B0

2 +2

1m(2R) 2B0

2 +

2

1m( 2 R)2 Br2 +

2

12m( 2 R)2 B0

2

= 6 mv 02

K.E.2m =2

1 × 2m ( 2 R)2B2 = 2mv 0

2 .

3.3. a =m

T2 #

v = u + at 2 0 = u – at

t =a

u =

#T2

um

Total time T = 2t =#T

um.

4.4. According to given condition,

, -

./0

1 # e

2 4

v13

# =

, -

./0

1 # e2

2 2

v7

#

e =24

#

So, r =6

e10

r =72

5#


5.5. For TIR at B, the angle of incidence i O c

& r + i = 90 2 i = 90 – r

by snell’s law at pt A,

sin 45° = n sin r = n cos i

Now ! i > c 2 sin i > sin c

2 cos r >n

1

2 n >rcos

1

2 n >rsin1

1

29

2 n >

2

n2

11

1

9

2 n >1n2

n2

2 9

2 2n2 – 1 > 2 2 n >2

3

6.6. The time in which the planet rotates about its axis

is not given for either planet.

7.7. For geostationary satellite, time period = 1 planet

day (by def.)

Let T = 1 planet day

T0 = 1 planet year

Now T2 =3G

2

rGm

4% = ,

-

./0

1 %M

mr

Gm

4 32

=30

32

TrGM

4!

% 2 T = T0

8.8. The energy of any geostationary satellite is the

sum of kinetic energy of satellite, interaction energy

of satellite and its own planet and interaction

energy of satellite and star. Both planets have

same mass and same length of day. Geostationary

satellite - planet system will have same interaction

energy in either planet. Also kinetic energy of both

satellites will be same. But the satellite-sun system

will account for the energy difference.

Ui = –r2

GMm0 + Usatellite – planet

Uf = –)r4(2

GMm 0 + Usatellite – planet

Emin = Uf – U i = r8

GMm3 0


1.1. In a binary star system

B1 = B

2

2.2. Strain (P) =#

#= = Q =T = (10 –5) (200)

= 2 × 10 –3

Stress = Y (strain)

Stress = 1011 × 2 × 10 –3 = 2 × 108 N/m2

2 Required force = stress × Area

= (2 × 108) (2 × 10 –6) = 4 × 102 = 400 N

$ Mass to be attached =g

400 = 40 kg

3.3. [Ans. v 11 =30, v =30, v 22 = = – 25, v 25, v 33 = -35/3, v = -35/3, v 44 = -25 cm] = -25 cm]

44.. AAnnss.. 6644

C =v

p2

D

1

2

CC

= 2

1

22

p

p ×

2

1

DD

2

1

v

v

=1200

400

3

5.1

16

9?? =

2316

9

?? =

32

3

62

1

C

C =

3

326 ? = 64


5.5. By energy conservation,

mg.2

3# =

22

.3

)3(m.

2

1B

#

B =#

g

and velocity of centre of mass of rod BC

Vcm = ##

2.g

Vcm = #g2

If time taken by centre of mass of rod BC from break-

ing position to line PQ is t.

8# =2tg

2

1??

t = g4

#

G = B.t = 4 radian.

6.6. iR = R (58.3)

& iX = R (68.5)

where R is the potential gradient of the

potentiometer

$ X

R =

5.68

3.58

X

10 =

5.68

3.58 2 X = 11.7511.75

7.7. The maxim um P.D. which we can measure by this

potentiometer is V

8.8. Any change can be done which assures p.d. across

R or X less then or equal to V


1.1. Originally

VA = VD = VE

After connecting C & B. The equivalent circuit will be

[Now VA = VD =VE and VC = VB ]

J

$ Ratio = 3.

2.2. The only force acting on the body is the viscous force

Here,

dx

vdvm = –6%5rv

= – rv

2 > > 9!0

v

x

0

rdxmdv 2 x =r

mv.

3.3. The image of a point closer to the focus will be farther.

As the transverse magnification of B will be more than

A, the image of AB will be inclined to the optical axis.

4.4. a1 =

m

F = 2r

GM

It is same in both cases

$2

1

a

a = 1

5.5. Loudness S = 10 log10

0CC

$ S2 – S

1 = 10log

10

1

2

CC

&

! C =T%r4

P

$1

2

CC

=22

21

r

r

$ (S + 20) – S = 10 log10

22

2

r

r

= 20 log 10 2r

r

22r

r = 10 2 r

2 = 0.1r

$ shift = r – 0.1 r = 0.9 r.

Ans. Ans.10

r9



The angular speed of rod = B =2 /

vu B

#

9

As given v B = 0 $ B =#

u2Ans.Ans.

The time after which centre of rod reaches the

highest point is t 0 =g

u

The angular acceleration of rod is zero and in the

given time to the rod undergoes angular displacement

2

%.

$ from G = Bt

2 2

% = g

uu2?

# or u =

4

gL%


1.1. By energy conservation between A & B

2 Mg5

R2 + 0 =

5

MgR +

2

1 MV2

53º

37º

R

AR –R cos53=2R/5 B R –R cos37=

R/5

g

37º

g cos37

O

Reference line

V =5

gR2

Now, radius of curvature r

=2

R

37cosg

5 / gR2

a

V

r

2

!!3

r =2R

37cosg5 / gR2

aV

r

2 !!3

2.2. Relative displacement of glass window w.r.t. cyclist

is 20 cm time taken = 1 sec.

So, relative velocity of glass window w.r.t. cyclist =

1

20 cm/sec. = 0.2 m/sec.

3.3. a = 3t 2 + 1

dt

dv = 3

t2 + 1 2 >> #!

1

0

2v

0

dt)1t3(dv

v = : ;103 tt # = 2 m/s.

2 v = t3 + t

$ >> #!1

0

3s

0

dt)tt(ds

2 S =

2

1

4

1# = 0.75

4.4. 8000

Conserving angular momentum m.(V 1 cos60°).

4R = m.V 2.R ;1

2

V

V = 2.

Conserving energ y of the system

m.(V1 cos60 °). 4R = m.V 2.R ;1

2

V

V = 2.

21mV

2

1

R4

MmG#9 =

22mV

2

1

R

GMm#9

2

1V2

2 – 2

1V1

2 =4

3

R

GM

or V1

2 =2

1

R

GM

V1 =2

161064 ? =

2

8000 m/s Ans. 8000Ans. 8000

7.7. On the positive side of x axis, potential is zero at

distance x1 (it is between both charges), then

r8

e10.k

r

e6.k

9! 2 r = 3nm

For the left side

22 x8

e10.k

x

e6.k

#!

2 x2 = 12nm

R =2

xx 21 #=7.5nm

xc

y

xR

x2 x1

|Xc| = X

2 – R = 4.5 nm

Vc =

9105.4

e6.k9?

– 9105.12

e10.k9?

= 9×109×109×1.6×10 –19 &'

()*

+9

5

4

3

4 = 1.44 ×

15

8

= 0.77V


8.8. (A) p,r (B) q, r (C) q, r (D) q, r

Initially the image is formed at infinity.

(A) As m is increased the focal length decreases.

Hence the object is at a distance larger than focal

length. Therefore final image is real. Also final image

becomes smaller is size in comparision to size of

image before the change was made.

(B) If the radius of curvature is doubled, the focal

length decreases. Hence the object is at a distance

lesser than focal length. Therefore final image is

virtual. Also final image becomes smaller is size in

comparision to size of image before the change was

made.(C) Due to insertion of slab the effe ctive object for

lens shifts right wards. Hence final image is virtual.

Also final image becomes smaller is size in


made.

(D) The object comes to centre of curvature of right

spherical surface as a result. Hence the final image

is virtual. Also final image becomes smaller is size in


made.


1.1. The linear relationship between V and x is

V = – mx + C where m and C are positive constants.$ Acceleration

a =dx

dVV = – m( –

mx + C)

$ a = m2x – mCHence the graph relating a to x is.

2.2. Relative to lift initial velocity and acceleration of coin

are 0 m/s and 1 m/s2 downward

$ 2 = )1(2

1 t2 or t = 2 second

3.3. R1 = Gcosg

v20

v0v cos0

g

gR2 =

g

)cosv( 20 G

$2

1

R

R = 8

)(cos

13

!G

Ans. 8Ans. 8

4.4.

f' = 70016334

9334,

-

./0

1 #

9 =

350

325× 700 = 650 Hz.

5.5. Consider two small elements of ring having charges

+dq and – dq symme trically located about y-axis.The potential due to this pair at any point on y-

axis is zero. The sum of potential due to all such

possible pairs is zero at all points on y-axis.

Hence potential at P(0,

2

R) is zero.

x

y

+ + +

+ +

+ +

+ +

+ +

+ +

+ + + +

+ +

+ + +

+ +++

x´

y´

– – – – –

– – –

–

–

–

– – –

–

– –

–

–

– –

+8 –8

GGdG dG

+dq –dq

6.6. Since all charge lies in x-y plane, hence direction

of electric field at point P should be in x-y plane

Also y-axis is an equipotential (zero potential) line.

Hence direction of electric field at all point on y-

axis should be normal to y-axis.

$ The direction of e lectric field at P should be in

x-y plane and normal to y-axis. Hence directio n of

electric field i s along positive-x directi on.

7.7. Consider two small elements of r ing having charge

+dq and –dq as shown in figure.The pair constitutes a dipole of dipole moment.

dp = dq 2R = ( 8 RdG) 2R

The net dipole moment of system is vector sum

of dipole moments of all such pairs of elementary

charges.


By symmetry the resultant dipole moment is

along negative x-direction.

x

y

+ + +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ + +

+ +++

x´

y´

– – – – –

– – – –

–

–

– – –

–

– –

–

–

– –

+8 –8

G

dG

–dq

+dq

$ net dipole moment

= – i)dcosR2(i)cosdp(

2 /

2 /

2

2 /

2 /

>>%#

%9

%#

%9

GG89!G

= – 4R2 8 i

8.8. (A) %&%'(%&%'(

#

#9

,, -

.//0

1 9,,

-

.//0

1 9!

21s R

1

R

1 1

n

n

f

1

f = – ve

P =f1 = –ve q,t

(B) %&%'(%&%'(

#

##

,, -

.//0

1 9,,

-

.//0

1 9!

21s R

1

R

11

n

n

f

1

f = +ve

P =f

1 = +ve p,r,s

(C) %&%'(%&%'(

#

99

,, -

.//0

1 9,,

-

.//0

1 9!

21s R

1

R

1 1

n

n

f

1

f = + , P =f1 = +ve p,r,s

(D) f =2

R

P =f

1 = –ve r,q,t


1.1. The electric field intensity due to each uniformly

charged infinite plane is uniform. The elec tric field

intensity at points A, B, C and D due to plane 1,

plane 2 and both planes are given by E1, E

2and E

as shown in figure 1. Hence the electric lines of

forces are as given in figure 2.z

x

–I

+I

(figure 1)

E1

E2

E1

E2

E1

E2

E1

E2

EE

E E

1

2

AB

C D

z

x

(figure 2)

Aliter :Aliter : Electric lines of forces srcinate from

positively charged plane and terminate at negatively

charged plane. Hence the correct representation

of ELOF is as shown figure 2.

2.2.

Force on q = Eq = q)cos1(2

o0

G9PI

= f

Consider a ring of radius y and thickness dy. Flux

through this ring dR =(E cosG)2%ydy

$ Total flux = G%%P>

G

cosydy2r

q

4

12

0 0

0


= )cos1(2

q0

0

G9P

So,I

!Rf

3.3. (Moderate)(Moderate) a =dx

dvv = cx + d

Let at x = 0 v = u

$ >> #!x

0

v

u

dx)dcx(dvv

or v2 = cx2 + 2dx + u 2

v shall be linear function of x if cx 2 + 2dx + u 2 is

perfect square

44.. ((CC))

Velocity of approach of P and O is

– dt

dx = v cos 60° = 5 m/s

It can be seen that velocity of approach is always

constant.

$ P reaches O after =5

100 = 20 sec.

5.5. a = (kx – f)/m

f × R = mR2 / 2 × a/R

6.6. by energy conservation

2

2

22

mv2

1

R

v

2

mR

2

13.03.020

2

1#??!???

mv2 = 1.2 J

Rotational K.E. =4

mv2

= 0.3 J

7.7. by energy conservation

2

2

22

mv2

1

R

v

2

mR

2

13.03.020

2

1#??!???

mv2 = 1.2 J

Translation K.E. =2

mv2

= 0.6 J

88.. ((AA)) – p,q,s, (B) p,q,s, (B) – r, (C) r, (C) – r,t ; (D) r,t ; (D) – p,q,s,t. p,q,s,t.

(p) Parallel beam can be obtained from concave mir-

ror and convex lens when point object is at focus.

(q) Real image for real object is for concave mirror

and convex lens.(r) Virtual and diminished images are obtained for

convex mirror and concave lens.

(s) Real and magnified image is obtained for concave

mirror and convex lens.

(t) The direction of motion of image is in the same

direction as motion of object in lens and opposite in

mirror.


1.1. As seen from the figure

the displacement is 22 )FD()AF( #

= 27 m

2.2. Max. frictional force

fmax

= 6N

= 6(mg + F sin53°)

= 0.2 (20 × 10 + 30 ×5

4)

mg

Fsin53°

F

Fcos53°53°

N

6N

= 44.8 N

As applied horizontal force is Fcos53°= 18N < f

max, friction force will also be 18 N.


3.3. Potential gradient = 1.6 v/m

E.M.F. = potential gradient × balancing length1.2 = 1.6 × #

$ # =6.1

2.1 =

4

3 = 0.75 m = 75cm

4.4. mg = m B2 R , B =R

g

6.6. 2x

KQ = 2)xr3(

Q4K

9 (Since r >> R)

2 (3r – x) = 2r 2 x = r

7.7. Once the charge reaches the neutral point it will be

accelerated towards center of ring 2, will cross it, be

reatarded, come to rest and then return towards it.

Thus the motion is oscillatory, but not SHM.

8.8. Uin = R

qKQ4 + r3

qKQ

Its energy when it reaches the center of ring 1

Ufin. = R

qKQ + r3

qKQ4 + K.E. = Uin

2 K.E. is positive


1.1.

1

2 3 4+

I2 Î 0

I2 Î 0

Electric field due to both the plates will be

cancelled out for all the points. So the net electric

field at the points will be governed only by the

sphere. Farther the point from the sphere, lesser

the magnitude of electric field.

Therefore E3 = E

4 > E

2 > E

1

2.2. Since the block slides down the incline with uniform

velocity, net force on it must be zero. Hence mg sinGmust balance the frictional force ‘f ’ on the block.Therefore f = mg sinG = 5 ? 10 ? ½ = 25 N.


1.1. Four lines, perpendicular to lines of electric field

and passing through A, B, C and D are drawn.

These are equipotential lines. As potential

decreases in the direction of electric field, therefore

VA > V

B > V

D > V

C

30°A

D

C

B

E

y

x

2.2. For A,dt

ds = V

A =

3

1

For B,dt

ds = V

B = 3

B

A

V

V =

3

1.

3.3.

If pot. drop between A and B is also 2V, then no

currrent will pass through the gelvanomter.

Pot. drop across R = ,/0

1 # 5R

12R = 2

12 R = 2R + 10

R = 1 U

4.4. Let the angular speed of disc when the balls reach

the end be B.

From conservation of angular momentum

2

1

mR2

B0= 2

1

mR2

B + 2

m

R2

B + 2

m

R2

B

or B =3

0B

5.5. In second case due to psuedo force acting on the

block its acceleration will be more as compared to

the first case.

Hence t1 > t

2

Ans.Ans. t1 > t

2


6.6. V = E + ir (during charging)

= 14 V.

7.7. P = I2 r (Due to internal resistance)

= 502 × 4 × 10 –2 = 100 W

8.8. Rate of charging = E.I.

= 12 V. 50 A = 600 W


1.1. Initially : –

Vv + V

A= 6 ..(1)

; VV

& VA

being the potential across voltmeter &

ammeter respectively

after closing S1.

2

Vv + 2 V

A= 6 ...(2)

Solving (1) & (2)

Vv

= 4, VA

= 2 .

after closing S2: –

Vv = 0

VA= 6

So that value after closing S2

is 3/2 times the value

after closing S1 .

22.. [ [ AAnnss: : VV11 ==

3

V10 , V , V

22 ==

3

V5, ], ]

3.3. The speed of sound in air is v = M

RTV

M

V of H

2 is least, hence speed of sound in H

2 shall be

maximum.

4.4.

Decrease in PE = Gain in rotational K.E.

Mg2

L + MgL =

213 RRR C#C!C

2

1.

3

4 ML2. B2 =

3

ML2

ML2

22

3MgL =

3

2.ML2 . B2 =

3

ML4 2

2L4

g9 = B2 2 B =

L4

g9

[ Ans.:[ Ans.: = =#4

g9 ] ]

5.5. From given data

j25i25VM #!"

velocity of man

Velocity of wind

i25VW !"

The flag will flutter in the direction in which wind is

blowing with respect to the man holding the flag.

2 WMV"

= WV"

– MV"

WMV"

= ( i25 ) – ) j25i25( #

WMV"

= – j25 = 25 ) j(9

This implies direction of wind with respect to man in

south.

Flag will flutter Flag will flutter in south direction. in south direction. Ans.Ans.

6.6.20V

A B

1 X

P

0V

C

Let reference potential of B be zero. No current

shall flow through g alvanometer.

If VC –

Vp = 16 volts.Now V

p = 2 volts.

$ Vc should be 18 volts.

Now1

VV CA 9 =

X

VV BC 9

Solving X = 9 U.

20VA B

1 X

P

0V

C

7.7. Balance point is independent of r. It can be seen

for balance point at P, VC – V

P = E in absence of

cell, jockey and galvanomete r.

8.8. For balance point at P.

VC – V

P = E = 12

! VC = 18 , V

P should be 6 volts.

Therefore

## 99

!9

100

0VVV PPA or # = 70 cm.



1.1. E – ir 1 = 0 2 i =

1r

E and i = Rrr

E2

21 ##

Therefore R = r1 – r

2

= 2 U

2.2. S

4 F6

50V

2 F6

4 F6

2 F6q= 100 Cf 6

q= 50 Cf 6

q =100/3 C1 6

q =50/3 C2 6

1

2

q=200/3 Ci 6

q=200/3 Ci 6

Initial and final charges are marked on 46f

and 26f capacitors as shown.

Hence charge passing through segment 1 and 2 are

C3

100q

1

6! C3

50q

2

6!

$ charge through switch = q1 + q

2 = 50 6C.

S

4 F6

50V

2 F6

4 F6

2 F6q= 100 Cf 6

q= 50 Cf 6

q =100/3 C1 6

q =50/3 C2 6

1

2

q=200/3 Ci 6

q=200/3 Ci 6

3.3. The equation of pressure variation due to sound is

p = – Bdx

ds = – B

dx

d[s

0 sin2 (Bt – kx)]

= B ks0 sin (2Bt – 2kx)

4.4.

V

xx=x0

Vmax

x = x0 is the point where potential is maximum . So, if

the impulse is sufficient enough and point chargecrosses the maximum PE barrier than point chargewill move to infinity otherwise it will perform oscilla-tory motion and for very small impulse the motionmay be SHM.

Sol.(69-72)Sol.(69-72)

Resitance of wire AB is -

RAB

= , -

./0

1 D

2

0#

A

# =

%

%

2

24= 12U

WX

WYZ

W[

W\] D

! >1

0A

dxR

Current in wire AB is C =312

15

#= 1A

when switch is open, null point at C (AC = x)

RAC

= , -

./0

1 D

2

x0 , -

./0

1 A

x=

A2

x20D

=%

%

23

224

= 8U

EMF E = 1 × 8 = 8 Vwhen switch closed null point at D (AD = x)

RAD

= , -

./0

1 D

2

x0 , -

./0

1 A

x=

A2

x20D

=%

%

22124

= 6U

=Vbattery

= 6 × 1=V

cSVjh = 6 × 1

8 – r3r

8

# = 6

r = 1 U


2.2. (C) For pipe A, second resonant frequency is third

harmonic thus f =AL4

V3

For pipe B, second resonant frequency is second

harmonic thus f =BL2

V2

Equating,AL4

V3 =

BL2

V2

2 LB =

3

4 L

A=

3

4.(1.5) = 2m.2m.


55.. [ [ AAnnss: : VVABAB

= = : ;231 5#5#

P = 10V ] = 10V ]

The distribution of charges is shown in fig. In

closed loop (1)

0C

)q –q( –

C

q –

2

1

1

!P ...(i)

In closed loop (2) – 2

1

2

1

1

1

C

q –q

C

q –

C

q# = 0

or q = ,,

-

.

//0

1 #

1

21

C

CC2

q1

From Eq. (i),2

1

21

1

C

q

C

q –

C

q – #P = 0

or ,, -

.//0

1 #!#P

2121

21

CC

CCqC

q

or ,, -

.//0

1 #,,

-

.//0

1 #!#P

21

211

1

21

2

1

CC

CCq

C

CC2

C

q

$ q1 = 2

22121

212

CCC3C

CC

##

P

$2

1

2

1BA

C

q

C

q – – !!RR

= 2221

21

21

CCC3C

C

##

P

=21

22

1

2

C

C

C

C31 ##

P

= 231 5#5#

P= 10V

,, -

.//0

1 !5! 2

C

C

1

2!

6.6. Torque equation

KHinge = CHinge Q

mg # = Q,, -

.//0

1 # 2

2

m12

)4(m#

#

#7

g3 = Q

Tangential acceleration = Q# =#7

g3

Radial acceleration = B2 # = 0

Ans.Ans.7

g3

7.7. mg – N1 = m , -

./0

1 7

g3

N1 =7

mg4

N2 = 0

8.8. Energy conservation

mg # =2

1 .

3

7 m#2 B2

#7g6 = B


1.1. Moment of inertia of semicircular portion s about x

and y axes are same. But moment of inertia of

straight portions about x-axis is zero.

$ Cx < C

yor 1

y

x ^C

C

2.2. As voltage applied across capacitor is same i.e.

10V in both case. Therefore in both case

Ed = 10 2 E =d

10, as d is constant . Therefore

electric field remians the same as 10 V/m

3.3. after collision

By momentum conservation in horizontal direction


V = V1 + V

2.............(i)

and

e =2

1

V

V –V 12 ! .............(ii)

By (i) and (ii) V2 =

4

V3

So impulse on B

= m , -

./0

1 4

V3

and loss in K.E.

=16

3 mV2

44.. AAnnss.. 3000

Time period is minimum for the satellites with

minimum radius of the orbit i.e. equal to the radius

of the planet. Therefore.

R

mv

R

GMm 2

2 !

2 V =R

GM

Tmin

=

R

GM

R2%

=GM

RR2%

using M =3

4 p

R3. D

Tmin. = D%

G

3

Using values T min = 3000 s

5.5. When S 1 is closed

i =rr2

p#

P =r3pP

$ P =r3

pP. r =

3

pP ............. (1)

when S2 is closed (Let resistance of w 2 be R)

i =rR

p

#

P

P =rR

p

#

P . ,

-

./0

1 3

2.R ............... (2)

From (1) & (2) From (1) & (2) R = rR = r. A. Ans.ns.


If in second case both S 1 & S2 are closed,

P =2

.r2

.r2r

p #

##

P =

3

pP

P =3

2.

xr2

rx2.

xr2

rx2r

p

##

#

P

2 3

pP=

x2xr2

x4 p

##

P.

Sol.(57 to 59)Sol.(57 to 59)

Final potential of spheres will be same

So,R

xK =

R2

Ky =

R3

)yxQ(K 999

y = 2x and 3x = –Q – x – y$ 6x = –Q

x =6

Q9 y =

3

Q9

Charge on sphere of radius 3R is2

Q9

Change in potential energy of sphere of radius ‘R’ is

=U =R2

)6 / Q(K

R2

KQ 22 99

=U =R72

KQ35 2

.


1.1. At highest point

mu =3

m2 v – 2u

3

m

mu +3

u2 =

3

2v

2 v =2

u5

total horizontal distance =2

R

2

5 +

2

R =

4

R7


2.2. (i) Vcm

v

V

smoothB

Vcm

+ BR = VV

cm = V – BR

B depends on value of friction between plank &

cylinder, hence Vcm

is undetermined.

(ii) B = R2

v2 = R

V(iii) B = R2

V2 = R

V

Vcm

= 0

(iv) BA/C

=R

V –V3 =

R

V2 2 B =

R

V2

3.3. ! wheat stone bridge is in balanced condition

So1

100

# =

2

x100

x100

#

#

100U

x

#2#1

100U

!2

1

#

# = 2

2 x = 100 U

5.5. As gravitational force provides centripetal force

r

mv2

= 3r

GMm

i.e., v2 = 2r

GM

So that T =v

r2% =

GM

rr2

2

%

$ T2 Q r4

AAnnss.. 44

6.6. VA +

2

1 × 1× (2)2 = – 3

VA = – 5 Ans. (C)

7.7. –R

GMAns. (C)

8.8. Ex = –

x

V

<<

= – 1

4

Ez = –

3 / 1

8 –4 = 4 3

| E | = 2z

2x EE # = 64 = 8 N/kg Ans. (A)


1.1. Ei = Q.

a2

KQ

a4

KQ

a2

KQ 22

##

=a4

KQ

a

KQ 22

# =a4

KQ5 2

Ef =

a4

)Q2(K 2

=a

KQ2

Ei

– Ef

= H =a4

KQ2

2.2. In potentiometer wire potential difference is directly

proportional to length

Let potential drop unit length a potentiometer wire be

K.

For zero deflection the current will flow indepen-

dently in two circles

C R = K × 10 .... (1)

C R + C

X = K × 30 .... (2)

(2) – (1)2 C

X = k × 20 .... (3)

(1)/(2) =2

1

X

R!


3.3. Let I be the charge density of conducting plate and

V be the volume of either dielectric

$2

1

U

U =

VEk2

1

VEk2

1

2202

2101

, -

./0

1 H

, -

./0

1 H

=2

1

k

k

2

02

2

01

k

k

,

,

-

.

/

/

0

1

H

I

,, -

.//0

1

HI

=1

2

k

k

4.4. 22

PE = –4 MJTE = –2MJThe additional energy required to make the satellite

escape = +2MJ.

5.5. Applying conservation of m ech. energy.

decrease in P.E. = increase in rotational K.E.

2 (2m) g , -

./0

1 4

R32 = _ ` 2

system2

1BC

3 mgR =2

1 [C

disc. + C

mass] B2

3 mgR

=2

222

16

mR25

16

mR

4

mR

2

1

B&&'

(

))*

+

##

2

16

30

2

1

R

g3B&

'

()*

+! 2

R

g

5

16 = B

Vparticle

= gR5R4

5!B,

-

./0

1

6.6. (A) 5 m/sv

A

from linear momentum conservation

MAV = m

b5 2 v =

40

54 ? = 0.5 m/s Ans.

7.7. (C)

mA 0.5 + m

b 5 = (M

A + m

b) V

1

V1 =

44

545.040 ?#? =

44

40 =

11

10m/s Ans.Ans.

8.8. (C)

after throught the ball velocity of man A is 0.5 m/s

For man B

4 × 5 = 40 v2 – 4 × 5 2 v

2 = 1 m/s

velocity B is 1 m/s after through the ball

after through the ball second time, velocity of man

A is

4 × 5 + 40 × 0.5 = 40 × v3 – 4 × 5

v3 = 1.5 m/s

similarly for man B v4 = 2 m/s

after 5 round trip and man A hold the ball velocity

of man B is 5 m/s

velocity of man A

4.5 × 40 + 4 × 5 = (40 + 4)v5

v5 =

11

50 m/s Ans.Ans.

99.. ((AA))

When man through the ball 6 times it velocity is

greater than 5 m/s and velocity of B is 5 m/s

therefor maximum number of times man A can

through the ball is 6 .

10.10. Fext

= 0 , Centre of mass of system cannot move

Initial position of centre of mass from A.

44

(A+ball) d

40

B

Xcm

=4044

d40

# =

21

10 d


1.1. Potential drop. @ V = i (R + R A)

i

V = R + R A = Rmeasured .


4.4. When charge on plate is constant electric field

remains constant E =0A2

Q

P

In case when potential difference is constant E =

d

V

Electric field increases when 'd' decreases and

hence chances of breakdown increases.

5.5. C = k H0 A /d

Af = 4A

i

df = 2 d

i

Because, AL L linear dimensions are doubled so

capacitance become doubled.

6.6. x = 3 sin 100 t + 8 cos2 50 t

= 3 sin 100 t +2

]t100cos1[8 #

x = 4 + 3 sin 100 t + 4 cos 100 t

(x – 4) = 5 sin (100t + R)XYZ

[\]

!R3

4tan

Amplitude = 5 units

Maximum displacement = 9 units.

7.7. C = k H0 A /dformula suggest that it depends on area, separation

and surrounding medium.

9.9. Most Approp riate capacitor is a capacitors of high

capacitance & high dielectric strength. By dielectric

strength

C > B > A > D

By capacitance C =d

Ak 0P

C = D > B > A

So C is best.


2.2. 100T

T?

= = 100

2

1?

=#

# is not valid as =# is not

small.

g2T1

#%!

g

22T2

#%! % change =

1

12

T

TT 9 ×

100 = 100)1 –2( ? = 41.04.

3.3. Switches open :

E

R

AR1

R2

R3

CA = 321 RRR

E

##

Switches closed :

There will be no current through R 3.

Current through E and R 2

CL =

1

12

RR

RRR

E

##

Current through the ammeter

CLA =1RR

R

#CL

=121 RRR)RR(

ER

##

=

R

RRRR

E

1221 ##

As CA = CLA

R =3

21

R

RR =

50

300100 ? = 600 U

44.. 4 4 ttiimmeess

A1 = A

3 = 8 (area)

A2 = A

4 = 9

Position of the particle at any time t is given by

X = X0 + >

t

0

Vdt X0 = Initial position


>t

0

Vdt = Area under the curve

Now at t = 0 X = X0 = – 1

at t = 2 X = X0 + A

1 = – 1 + 8 = 7

at t = 5 X = X0 + A

1 – A

2 = – 1 + 8 – 9

= – 2t = 7 X = X

0 + A

1 – A

2 + A

3 = – 1 + 8 – 9 + 8

= 6

t = 10 X = X0 + A

1 – A

2 + A

3 – A

4 = – 1 + 8 – 9 + 8 –

7 = 3

As during 10 seconds four times the position of the

particle changed in sign. Particles passes 4 timesthe srcin

5.5. Once the switch is closed, the capacitor is charged

through resistance R1 by the battery's e.m.f. Time

constant is R1 C.

6.6. Using Vc = E (1 – e –t/RC)

2 Vc = 12 (1 – e –2)

= 10.4 V

7.7. At any moment in the circuit –

Vc +

1RV = 12 V

2 1RV = 12 V – 10.4 V

= 1.6 V

8.8. If loop law is applied to the left hand loop in clockwise

direction

E – 2RV

= 0

2RV = E = 12 V

i.e. VR2

does not change during the charging process.


1.1. By symmetry

CAB

= CBC

& CAD

= CDC

$ No current in BO and OD

$ TB

= TO

= TD

2.2. In (A) xf – x

i

0 – x = – x = – veSo average velocity is – ve.

3.3. Speed of block is m aximum at m ean position. A t

mean position upper spring is extended and lower

spring is compressed.

4.4. 5 and Y are properties of material.

These coefficients are independent of geometry of

body.

5. Relative to liquid, the velocity of sphere is 2v 0

upwards.

$ viscous force on sphere

= 6 % 5 r 2v0 downward

= 12 % 5 r v0 downward

6.6. J

J

C1 = d

2 / A0P , C2 =

2

d

k

2 / d

2 / A0

#

P =

d5

A4 0PC

= C1 + C2 = 10

13

d

A0P Ans.Ans.d

A

10

13 0H

7.7. The current through the galvanometer is ~1000

1

of total current, the S << G.

8.8. Potential difference across galvanometer =

Potential difference across S.

2 ig . G = ( C – ig) . S

2 10 × 10 –3 a 10 = (1 – 10 × 10 –3) a S

2 RS = 2

1

101

109

9

9 =

99

10U


33.. ((AA)) Moment of inertia of the rod w.r.t. the

axis through centre of the disc is : (by parallel axis

theorum).

C = 22

mR12

mL#


& K.E. of rod w.r.t. disc

=2

2

1BC

=&&'

(

))*

+#B

12

LRm

2

1 222

Ans.Ans.

5.5. Since, Heat energy radiated per se c = Ae IT4

where, e is emissivity of the surface which depends

upon its nature and A is its area.

T is its own temperature (independent of

surrounding temperature)

Hence, (A) and (C) are correct.

8.8. Let VA = 0 volts.

! Net current entering node C = 0

$ 02

V1

5.6

V2

2

V0 ccc !9

#99

#9

$ VC =

6

1volt. = p.d across wire AC.

Also5

VV BC 9 =

5.1

2VB # =

3

1

5.6

2VC !#

$ VC – V

B =

3

5 = p.d. across wire BC

$ VC > V

B > V

A

Hence potential gradient across BC =

A

1V

C

2V -2v

1v

5

1.5v

VC VBOV

3

10

2 / 1

3 / 5! V/m

also potential gradient across

AC6

5

5 / 1

6 / 1! V/m


2.2. The velocity of the body a time t is given by

2

t

4

t

dt

d

dt

dx 2

!,, -

.//0

1 !!b

$ At t = 0 , ! = u = 0 and t = 2 s, ! = 1ms -1, Now,,

work done = increase in KE

02

1

2

1

2

1222 9!9! ! ! mmum

22)1(6

2

1

2

1??!! ! m

= 3J,

Hence the correct choice is (d).

3.3. From conservation of energy, the kinetic energy of

ball at lowest portion is (vc = speed of centre of

ball)

2cmv

2

1 +

2cmv

5

2

2

1? = mgR

or2cmv

10

7 = mgR

Since net tangential force on sphere at lowest point

is zero, net force on sphere at lowest position is

= mg7

10

R

mv 2c ! upwards. upwards.

6.6. RA =

V

V

RR

R.R

# < R

7.7. RB = R + R

G > R

8.8. % error in case A.

R

RRA 9× 100 = ,,

-

.//0

1 9

#1

RR

R

V

V × 100

=VRR

R

#9

× 100 c - 1%

% error in case B

R

RRB 9× 100 =

R

RG× 100 c 10%

Hence percentage error in circuit B is m ore than

that in A.



2.2. C1B

1 = C

2B

2

Since, men move towards middle of turn table C2

decreases hence B2 increases.

$ =K =2

1C1B

12 –

2

1C

2B

22

=2

1C1B

12

&&'

(

))*

+

B

B

C

C9

21

22

1

2 .1XYZ

[\]

OB

B1

1

2

=2

112

1BC &

'

()*

+

B

B9

1

21 < 0

So kinetic energy increases.

33.. ((AA,,BB)) For steady state

indt

dQ,

-

./0

1 =

outdt

dQ,

-

./0

1

(V) (i55

) = 45(T – 20)

(500) (4.5) = 45(T – 20)

T55

= 70ºC.

(C,D)(C,D) Resistance at 20ºC is R =iv =

5500

R20

= 100 U

Resistance at 70ºC is R =i

v =

5.4

500 –~ 11111 U

Rf = R

0(1 + 4=T)

111 = 100(1 + 4(50))

4 =50

11.0 –~ 2.2 × 10 –3 / ºC.

4.4. For painter ;

R + T – mg = ma

R + T = m(g + a) ............(1)

For the system ;

2T –

(m + M)g = (m + M)a2T = (m + M) (g + a) ..............(2)

where ; m = 100 kg

M = 50 kg

a = 5 m/sec 2

$ T =2

15150 ?= 1125 N

and ; R = 375 N

5.5. By Wnet

= =K.E = 0

22

1k(x

02 – x2) = 6mgx

22

1× 200(22 – x2) =

2

1 × 60 × 10x

2 x = 1m

Also at this moment fmax

> kx

So, block will not move so total distance travelled =

2 + 1 = 3m.

6.6. In steady state, before the switch S is closed,potential difference across capacitor is 40 volts. Just

after switch S is closed, charge and hence potential

difference across the capacitor does not change

appreciably. So, the potential difference acr oss R 2

is 40 –10 = 30 volt. The curent through R 2 is 3

ampere.

7.7. The current through resi stors when t he capacitor

is in steady state with switch S closed.

I = amp1RR

1040

21

!#9

. Therefore potential difference

across R 2 is 10 × 1 = 10 volts. Hence the potential

difference across th e capacitor i s 10 + I R 2= 20

volts. So, the charge on capacitor q = CV = 200 6C.

8.8. At the given instant, p.d across capacitor is 20

Volts. Hence the current through R1 at the required

instant of time is I = amp1R

2040

1

!9


1.1. Heat current : i = – k Adx

dT

idx = – kA dT

>#

0

dxi = – A 4

>

2

1

T

T

dTT

2 i # = – A 42

)TT( 21

22 9

2 i =#2

)TT(A 22

21 94


2.2. i =R10

2

# 2 VAB =

R10

2

# × 10

2 x =10

10

R10

2?

# 2 xmax = 0.2 V/m.

3.3. 3

0

r

rv2

4B

""" ?%

6!

2

0

r

sinqv

4B

G%

6!

= 10 – 7 ×2)2(

30sin1002 "??

= 10 – 7 ×22

2

11002 ??

= 25 × 10 – 7 T

= 2.5 × 10 – 6 T

= 2.5 6T

5.5.

V0 = 5 × 2t = 10 t

S = 1500 = t3.V2

10 = t3.t10

2

1

2 t = 10 sec.

$ total time = 3t = 30 sec.

6.6. Just before collision velocity of M and

m = GsingL2 0 = 12 m/s

Since collision is elastic, let velocity of m just after

collision is v then by relative velocity of separ ation

= relative velocity of approach

v = 12 + 12 = 24 m/s Ans.Ans.

7.7. By momentum conservation during collision of m

and M.

12 M – 12 m = 24 m

m : M = 1 : 3

8.8. By mechanical energy conservation for m, just after

collision

2

1mv2 = mgL sin G

L 0

C M

3 m = M

m L

2 L =420524

sing2v

22

??!G = 36 meter..

Alternate :Alternate :

since there is no energy loss, center of mass of m

and M rises to the same initial position.

3mL0 = m(L – L0)

2 4mL0 = mL

2 L = 4L0 = 36 meter.


1.1. (B) Initially effective resistance = 2R. In parallel

effective resistance =2

R. It has reduced by a factor

of 1/4 so rate of heat transfer would be increased

by a factor of 4, keeping other parameters same.

4.4.

21

1

FF

LF

# – 21

2

FF

LF

# =4

L

24

1

F

F =

3

5 K

[ Ans.: 3: 5 ][ Ans.: 3: 5 ]

5.5. (A)

Distance travelled = 24t0 + 21 t0


= 45 t0

2 45t0 = 180 m

2 t0 = 4 seconds

$ Distance translted by A is

24t0 = 24 × 4 = 96 m

Sol(57,58,59)Sol(57,58,59)

(57).(57). Initial velocity of com

u = mm

0m20

##

= 10 m/s d

acceleration of com

= g = 10 m/s2

einitial height = 10 m

S = ut +2

1 at2

10 = – 10t +2

1(10) t2

5t2 – 10 t – 10 = 0t2 – 2t – 2 = 0

t =2

842 ##

t = 1 + 3

(58).(58). Hmax.

= 10 +g2

u2

= 10 + 5 = 15 m

(59).59). =t =a

202? –

g

202 ? = 4 – 2 = 2sec.


1.1.#2

TKA

dt

dQ =! =

KA

2

T

#

= =

120

10 J /sec.

New ratedt

Qd ) =

KA2

T

#

=

=120

40 J /sec. ;

So time taken is t =40

20 × 120 sec.

= 60 sec.

22.. ((BB))

Point A shall record zero magnetic field (due to 4-

particle) when the 4-particle is at position P and Q

as shown in figure. The time taken by4-particle to go

from P to Q is

t =3

1

B%2

or B =t3

2%

3.3. Angular velocity w =5.0

1020 9 = 20 rad/sec.

4.4. Potential difference across wire AB = 5 V

$ p.d. across 40 cm of this wire

=100

5 × 40 = 2 volt.

$ Potential difference across 20 cm of wireCD = 2 volt.

$ p.d. across wire CD =20

2 × 80 = 8 volt.

p.d. across 2 U resistor = 2 × 2 = 4 volt$ Emf of the cell = 12 volt.

5.5. > dsv

2 = > dsdt

ds

2 = 2t = 8 × 4 +

2

1 × 10 × 2

t = 21 s

t – 20 = 1 s Ans.Ans.

SSooll.. ((1 1 tto o 33))

mA × 0.8 = m

A × 0.2 + m

B × 1.0

mA × 0.6 m

B × 1.0 m

B = 0.6 m

A

e =8.0

2.0 –1 = 1 = 1.5

Cd = 6 × 0.5 – 6 × 0 = 3N – 5

= 10 × {0.8 – 0.5} = 10 × 0.3= 3 NS

=U =2

1 × 10 × (0.8)2 –

2

1 × 10 × (0.5)2

= 5 × 0.64 8 × 0.25 = 3.2 – 2.0 = 1.2 J



1.1.

Mx +3

M (x + L) = 0

3

M4 x = –

3

MLx = –

4

L

2.2. B!

= 3

0

r

rvq

4

!!!

"#

and E!

= 30 r

rq

4

1!

$"

% B!

= #0

$0 )Ev(

!!! = 2c

Ev!!

!

3.3. v =100

50Ve = R

GM2

2

1

Applying energy conservation

& 2vm2

1

R

GMm() =

)hR(

GMm

()

v2 =hR

GM2

R

GM2

()

&R

GM2.

4

1 = *

+

,-.

/ (

)hR

1

R

1GM2

&)hR(R

h

R4

1

(0

& R + h = 4h& h = R/3

4.4. Keq.

is same in all three cases. All other parameter

being same, rate of energy conduction is same in all

three cases.

Simlarly temperature difference across any material

in any wall is also same.

5.5. 1 =12

ma2

+

2

2

a m **

+

,--.

/ =

12

ma7 2

= 7.

6.6. mgh =2

1mv2 +

2

1mv2 +

2

1.

5

2mv2

2

r

v2*

+

,-.

/

=2

1mv2 [1 + 1 +

5

8] =

2

1mv2

5

18 =

5

mv9 2

& v = gh9

5

7.7. KE of the ball =2

1 mv2 +

2

1 5

2 mv2

2

r

v2*

+

,-.

/ =

18

13

mgh

8.8. X = 2vt = 2vg

h2

= 2.g

h2gh

9

5 = h

3

102


1.1. Initially the centre of mass is at

4

L

distance from the vertical rod.

**

+

,

--

.

/ 0

(

(0

4

L

mm

)0(m)(mx,As 2

1

cm

centre of mass does not move in x-direction as

2Fx = 0.

After they lie on the floor, the pin jo int should be atL/4 distance from the srcin shown inorder to ke ep

the centre of mass at rest.

% Finally x-displacement of the pin is4

L and

y-displacement of the pin is obviously L.

Hence net displacement =4

L17

16

LL

22 0(


2.2. H = – kAdx

dT&

dx

dT =

kA

H)

Now as k increases,dx

dT becomes less ( –)ve

So slope becomes less ( –)ve

So curve will be

3.3. Since

B!

= 3

0

r

rvq

4

!!!

"#

, rv!!

! must be same

where v!

= velocity of charge with respect to

observer

Let A and B are the observers

then r)vv( AC

!!!!) = r)vv( BC

!!!!)

or 0r)vv( BA 0!) !!!

or r||)vv( BA

!!!)

4.4. 1131 = 1131 (Angular momentum i s conserved)

As 12 decreases. 32 increases.

Thus T =3"2 i.e. T decreases.

Therefore the earth is completing each circle around

its own axis in lesser time.

K.E. =2

11 3

2

Therefore K.E. of rotation increases.

Duration of the year is dependent upon time tak en

to complete one revoluti on around the sun.

5.5. Using 4 axis theorem

"x = "

y

21x= 1.6

1x

= .8 Ma 2

1AB

= 1x + M(2a) 2

= 4.8 Ma 2

Ans.: 4.8 MaAns.: 4.8 Ma 22

6, 7 & 8.6, 7 & 8.

Let 5 be the angular acceleration of rod and a be

acceleration of block just after its r elease.

% mg – T = ma ..... (1)

T" – mg2

" = 5

3

m 2"

.... (2)

and a = "5 .... (3)

Solving we get

T =8

mg5 and 5 =

"8

g3

Now from free body diagram of rod, let R be the

reaction by hinge on rod

R + T – mg = m a cm = m2

15

Solving we get

R =16

mg9


1.1. Moment of inertia is more when mass is farther from

the axis. In case of axis BC, mass distribution is

closest to it and in case of axis AB mass distribution

is farthest .Hence

cm

x yC

5

A

B

3

IBC< IAC< IAB

& IP > I

B > I

HI

C= I

CM + my2

= IB

1 – mx2 + my2

= IB

1 + m (y2 –x2) = IP

+ IB + m (y2 – x2)

> IP

+ IB

> IP

Here IB

1 is moment of inertia of the plate about an

axis perpendicular to it and passing through B.

% IC

> IP

> IB

> IH


22.. ((CC)) Taking the srcin at the centre of the plank.

A B40 kg 60 kg

40 kg

60 cm

smooth

x

m16x1 + m2 6x2 + m36x3 = 0

(# 6xCM = 0)

(Assuming the centres of the two m en are exactly

at the axis shown.)

60(0) + 40(60) + 40 ( –x) = 0 , x is the displacement

of the block.

& x = 60 cm

i.e. A & B meet at t he righ t end of the p lank.

3.3. The slope of temperature variation is more in inner

dt

dQ= T.

KA6

"

6T =dtdQ.

KA"

Slope 5 K

1

Larger the conductivity, smaller is the slope.

4.4. Let Q be the charge on the ring. The electric field at

point P is

% E =04

1

$" 2 / 322 )Rx(

Qx

(

=04

1

$" 2 / 32 )R2(

QR

The rotating charged (Q) ring is equivalent to

a ring in which current 1 flows, such that

1 =R2

Qv

"

The magnetic field at point P is

B ="4

µ0 2 / 322

2

)Rx(

R2

(

1" =

"4

µ0 2 / 32 )R2(

QvR

%B

E = vµ

1

00 $ =v

c2

5.5. The orbital velocity,

v0 = r

GM

Its velocity is increased by 2 times, new velocity

v =r

GM2

r

GM 2 0 = escape velocity

The path is parabolic in case of escape velocity.

8.8. 6L = 7 8dt = 7 Fdtx = x6P (because x is

essentially constant during the quick blow)

since, the rod starts at rest, the final values

therefore satisfy L = xP.

& 2

1m"2 3 = xmv &

2

x12

v "0

3.... (1)

Another expression forv

3 is obtained from the given

information that rod makes one revolution by th etime centre reaches the dot.

3t = 2" and vt = d

& d

2

v

"0

3.... (2)

from equation 1 and 2 :d

2x122

"0

"% x =

d6

2""

x cannot be larger than2

"

2d6

2""

9"

& d3

9""


1.1. MP =*

+

,-.

/ (

f

D1

=*

+

,-.

/ (

5

251

= 6

2.2. Heat radiated (at temp same temp) : A

& Q : 4"R2 and Q' : (4"R2 + 2 × "R2)

& 2

2

R4

R6

Q

'Q

"

"0 = 1.5

Here "R2 is extra surface area of plane surface of one

of the hemisphere.


3.3.

Megnetic moment M!

= " r2 i i & B!

= j4i3 (

% 8! = M

! × B

! = " r2 (3 i – 4 i )

8

!

will be along the direction shown .Hence , the point about which the loop will be lift

up will be : (3, 4)

4.4.

Radius of Curvature =onAcceleratiNormal

)velocity( 2

=R / v

)v2(2

2

= 4R

5.5. By linear momentum conservation in horizontal di-

rection = for (bob + string + cart)mV0 = (m + m)v

v =2

V0

By mechanical energy conservation for

(bob + string + cart + earth)

2

1mV0

2 + 0 + 0 =2

1(2m)v2 + mgh + 0

2

1mV0

2 – 2

1(2m)

4

V2

0 = mgh

Solving it,

h =g4

V20

.

6.

When two drops of radius r each combine to form abig drop, the radius of big drop will be given by

3R3

4" =

32 r3

4r

3

4 "(

"

or R3 = 2r3 or R = r2 3 / 1 Now

2

r

R

r

R

V

V*

,-.

/ 0 = 3

1

3

2

42 0

% VR= 5 × 4 1/3 cm/s

7.7. If m is pole strength , then

m =l

Mm 0

When the wire is bent into a semicircular arc, the

separation between the two poles changes from l to

2l, where new magnetic moment of the steel wire,

"0

"!0!0

M22Mr2m'M l

l

8.8. (A) Real image of a real object is formed by concave

mirror and convex lens.

(B) Virtual image of a real object is formed by all four.

(C) Real image of a virtual object may be formed by

all four.(D) Virtual image of a virtual object may be formed by

convex mirror and concave lens.

(A) p,r (B) p,q,r,s (C) p,q,r,s (D) q,s

9.9. Let E1 < E

2 and a current i flows through the circuit.

Then the potential difference across cell of emf E1 is

E1 + ir

1 which is positive, hence potential difference

across this cell cannot be zero. Hence statement 1

is correct.

For current in the circuit to be zero, emf of both the

cells should be equal. But E1 ; E

2. Hence statement

2 is correct but it is not a correct explanation of state-

ment 1.

E , r1 1 E , r

2 2


3.3. Equal area means equal power output. A3 area

pertains to highest wavelength range, thus photons

with minimum range of fr equency. Thus maximum

number of photons are required from this segment

to keep the power same.

4.4. Work done by kinetic friction may be positive when it

acts along motion of the body.

Friction on rigid body rolling on inclined plane is along

upward because tendency of slipping is downwards.

5.5. The torque of s ystem = T orque on loop

[AFGH + BCPE + ABEF]

= 1SB )i() + 1SB( i )+ 1SB k (I = current,

S = area of loop, B = m agnetic field.

= 1 S B k

= 1 × 1 × 2 k = 2 k units

[Ans:[Ans: K2 ]]


6.6. The work done to rotate a bar magnet from its initial

position < = <1 to the final position < = <

2 is given by

W = M B (cos <1 – cos<

2),

(i) Here <1 = 0° and <

2 = 180°

% W = M B (cos <1 – cos180°) = M B = [1 –( –1)] = 2

M BM B

(ii) Here <1 = 0° and <

2 = 90°

% W = M B (cos0° – cos 90°) = M B = [1 – 0)]

= M BM B

7.7. If velocity of m2 is zero then

by momentum conservation

m1v= = m2 v

v= =1

2

m

vm

Now kinetic energy of m1

=2

1m1 v=2 =

2

1 m

1

2

2

1

2 vm

m**

+

,--.

/

=2

1 **

+

,--.

/

1

2

m

mm

2v2 = **

+

,--.

/

1

2

m

m

2

1m

2v2 =

1

2

m

m × initial

Kinetic energy

Kinetic energy of m1 > initial mechanical energy of

system m1

Hence provedHence proved

8.8. C = sin –1 * +

,-.

/ 1 / 2

1 = 30°

for i = 37 T 1Rso, > = " – 2 (37°) = 104°

i = 25, Refraction > <2

" – C

i = 45°, T1R

so, > = " – 2 * +

,-.

/ "4

= 90°

By applying snells law for prism :

i = 90

r1 = 30

r2 = 30e = 45

> = 90 + 45 – 60 = 75°

9.9. The points A and B are at same potential, then under

given conditions points A and B on the circuit can be

connected by a conducting wire. Hence the circuit

can be redrawn as shown in figure 2.

E

BA

E

R1

R2

R3

Fig-1

Therefore statement 1 is true. Statement 2 is obvi-

ously false.


1.1. In normal adjustment

m =e

0

f

f

so 50 = ef

100

& fe = 2 cm

(# eyepiece is concave lens)

and L = f0 – f

e = 100 – 2 = 98 cm

2.2. ?m T = const.

"n ?m + "nT = C

0T

dTd

m

m 0(??

% m

md

??

=T

dT)

Nowm

md

??

= – 1 % =100

1) ( –ve sign indicates

decrease)

dT = 1 (given)

% T = 100 K.

3.3. Emissive power = @ T4

= 6 × 10 –8 × 1004 W/m2


4.4.

to reach < = 270º, it has to cross the potential

energy barrier at < = 180º and to cross < = 180º

angular velocity at < = 180º should be 0+

ki + U

i = k

f + U

y

22 MR2

3

2

13*

+

,-.

/ + ( –Mi AB cos 0º) = 0 + ( –NiAB

cos180º)

3 = 980 A rad/sec.

5.5. Equation for linear motion

mgsin < – f = ma

for rotary motion

f. R = 1 .R

a& f = 2R

1.

a mgsin< = ma +2

2

R.2

MR.a =

2

3ma

a =3

sing2 < =

3

g

using S = ut +2

1at2 for linear motion.

1 = 0 +2

1.

3

g. t2

t = g

6sec. Ans.Ans.

6.6. Here BH

= 0.22 T ; BV = 0.38 T

Now 2V

2H BBB (0

= 1928.0)38.0()22.0( 22 0( = 0.44 T0.44 T

7.7. w.r.t. the wedge

As maximum height = 125 m

& block want by a height 20m over the wedge

& (v sin53º)2 = 2.g.20

v2 25

16 = 400

v2 = 25 × 25

v = 25 m/sec.

& block left the wedge with a relative velocity

25 m/sec.

Now, time of flight = 2 + 5 = 7 sec.

horizontal range w.r.t. wedge

= vx × T

= 25 cos 53 × 7

= 105 m.

8.8. For slab no deviation so > = 0 for any i

for slab for light from D to R

> = r – i ....(i)

nd sin i = nr sin r

& r = sin –1 BC

DEF

Gisin

n

n

r

d

> = sin –1 BC

DEF

Gisin

n

n

r

d – i

it is non-linear function and graph is

After i > C T. 1HR. will occur and graph is straight line

for D to R


Similarly for light going from R to D

= i – sin –1

isin

n

n

d

r

and for prism graph is drawn from

i = imin to i = e that is graph(s)

9.9. Statement-2 is wrong as in this case

A B

E

E r

A is at high potential and B is at low potential and

there is no current from A to B. It also justifies

Statement-1.


1.1. By right hand thumb rule, the field by both the

segments are out of the plane i.e.along

+ve z-axis+ve z-axis .

2.2. Let us compute the magnetic fi eld due to any one

segment :

B = ))180cos(0(cos)sind(4

i 00

=

cos1

)sind(4

i0 =

2tan

d4

i0

Resultan t field will be :

Bnet

= 2B =2

tand2

i0

k =

d2

i0

3.3. Due to the motion of the loop, there will be an

induced curre nt flowing in the circuit, resulting in

a force acting on each element of the loop equally

& radially. Therefore the net force on the loop is

zero.

Hence (D).

5.5. Decrease in PE =

Increase in rotation K.E

2mg. 2

– mg. 2

= 2

1

. 2

=2

1

4.m

4m2

2

2

2

mg =

2

1 .

4

m3 2

. =8

m3 2

2

=3

g4and v = r =

2

3

g4 =

3

g

[ Ans.: (a) V =[ Ans.: (a) V = g / 3 ,, = = 4 3g / ] ]

6.6.

mg cos + T1 =

21mv

for leaving circle T1 = 0

mv12 = mg cos ...(i)

and by energy conservation

0 +2

1m 2)g3( =

2

1mv

12 + mg )cos(

2

1 m )g3( =

2

1mv

12 + mg (1 + cos)

2

mg3 =

2

cosmg + mg + mg cos


(by eqation (i))

2

mg = mg

2

3cos

cos =3

1

sin =9

11 =

3

8

ac =

21v

=

cosg = g cos

at = g sin

thent

c

a

a

=

sing

cosg =

3/8

3/1 =

8

1 =

22

1 =

2y

1

so y = 2

Ans. y = 2Ans. y = 2

SSooll.. ((7 7 tto o 99))

Applying bernoulli's equation

P0 +2

1 × 2 × V2 = P0 + 2g ×

2

h + gh

v = gh2

2

1 × g × t 2 =

2

h

t =g

h

R = v × t

h2

Applying continuity equation

6 × gh2 = gh3 × A A

A = 2cm2


1.1. BOD

= 0

BOB

= 0

B AB

=2a4

0

]k45cos)i(45[cos

= )ki(a8

0

X

Y

Z

a

a

A

B

DO

upto

upto

parallel to 'y' axis

2.2.

The magnetic flux must remain constant

m = B

0ab =

kt1

B0

bx

where x is as shown

x = a(1 + kt)

or v =dt

dx = ak Ans.Ans.

33.. ((AA)) Let 'F' be the magnitude of force exerted on

the rod due to the collision.

Then : F = ma

and F.4

=

12

m 2

.

(about 'O')

a =

3

.... (1)

Using ; S = ut +2

1at2 and =

0t +

2

1

t2

S = ut +2

1at2 =

0t +

2

1

t2

S =2

1at2 and 6

=

2

1

t2

s

6 =

3

a

(from (1)) )

S = 2

Ans.Ans.

5.5.

R

mv2

= N – mg sin


N =R

mv2

+ mg sin

By energy conservation,

mgR sin =2

1mv2

R

mv2

= 2mg sin

N = 3mg sin

Ratio =

RN

mv2

=

3

2(constant)

x =3

2.

6.6. Consider a block of ice having volume V and density

i.

Let the volume of ice submerged in water (of density

w) be V

Since the ice block is in equilibrium

i V g =

wVg or V =

w

i V ... (1)

Let V volume of ice melt in to V of water. Then

i V =

w V

or V =w

iV

... (2)

from (1) and (2) V = VHence when V volume of ice melts it occupies V V volume of water.

Hence the level of water does not change on melting

of ice.

9.9. The velocity of centre of mass is always zero . At

maximum deformation during hea d on coll ision,

velocity of each sphere is equal to velocity of centre

of mass and hence zero. Therefore at maximum

deformation K.E. of system is also zero.

Velocity of separation after collision = e (v elocity

of approach before collision).

From centre of m ass frame in a head-on collision,

if1u

and

1v

be velocity of a ball before and after

collision11 uev

. Since, v

cm = 0 from ground

frame, ground frame and centre of mass frame carry

same meaning.


1.1. For disc, from t orque equation

3 mg R – TR = 2

mR2

.... (1)

By application of Newton's second law on block

we get,

T – mg = ma .... (2)

where a = R ..... (3)

F=3mgT

M,R

mg

T

a

solving a = 3

g4

3.3. (A) From continuity equation, velocity at cross-

section (1) is more than that at cross-section (2).

Hence ; P1 < P

2

Hence (A)

4.4.

N

mg

N = mr 2 .............(i)

N = mg .............(ii)

From (i) & (ii),

(mr 2) = mg

2 =r

g

=188.0

10

=6

5rad/s.


5.5. By newton's law :

mg i l B = m

td

vd (1)

By k v

l B

l v = i R +

c

q (2)

differentiate (2) w.r .t. time

B l

td

vd = R

td

id +

c

i (3)

Eliminatetd

vd by (1) & (3)

mg i l B =B

m

c

i

td

idR

m g B

l i

B2

l 2 = m

R

tdid +

cim (4)

i will be maxi mum whentd

id = 0.

Use this in (4)

m g B

l c = i (B 2 l 2 c + m)

imax

=cBm

cBgm22

Ans.Ans.

6.6. Angle of dip can not be calculated by tangent

galvanometer.

7.7. K =200104

1021027

52

= 0.005

8.8. i = K tan

using values ( = 60°)

i = 0.005 × 3 =200

3


1.1. When the key is at position (B) for a long time ;

the energy stored in the inductor is :

UB =2

1Li0

2 =2

1.L.

2

2R

E

= 2

2

2

R2

.E.L

This whole energy will be dissipated in the f orm of

heat when the inductor is connected t o R 1 and no

source is connected.

Hence (A).Hence (A).

2.2. P = 4T Ae

2 = 2 x 10-6 x0.9x 5.6 x 10-8 x T4

T4 =6.5x9.0

1014

T = 2110 k

3.3. f =2

1

p

1 metre

f = 0.5 m this is positive so lense is convex lense.

4.4.

Condition for pure rolling

V – R = V' ...........(i)


Momentum conservation

m'V' = mV

V' ='m

mV...........(ii)

From (i) and (ii) V =m'–m

Vm' 1

Torque of friction about point P is zero Angular

momentum will remain c onserved about this point

mVR2

mR

2

mR1

22

Solving this we get ' =m'–m3

m'–m .

5.5. M1

is very large as compared to M2. Hence for

collision between M1 and M

2, M

1 can be considered

equivalent to a wall and M2

as a small block. T hus

the velocity of M2 will be 2v

o after collision with M

1

. Similarly after collision betwee n M2 and M

3, the

velocity of M3 will be 2(2v

o). In sequence, the

velocity of M4 shall be 2(2(2v

o)) = 8 v

o after collision

with M3.

6.6. The component of velocity of charged particle along

the magnetic field does not change. The component

of velocity of charged par ticle normal to magnetic

field only changes in direction but always remains

normal to magnetic f ield. Hence angle betwee n

velocit y and magnetic f ield remains same.

7.7. f =m2

qB

=

2

3800

19

103

=38

104

8.8. Pitch = T.V||

=3800.10

400.38

|B|

B.V.

f

14

= 310

4m

=250

m


1.1. Rate of heat transfer is same through all walls

cm40

)10.( A.K1 =

cm20

)20( AK2 =

cm10

)40( AK2

4

K1 = K2 = 4K3 K1 = 4K2 = 16 K 3 .

cm40

)10.( A.K1 =

cm20

)20( AK2 =

cm10

)40( AK2

4

K1 = K2 = 4K3 K1 = 4K2 = 16 K 3 .

2.2. For the elastic collision, total kinetic energy is

constant.

If we consider both the colliding bodies as system,

then net work done on the system is zero (By work

energy theorem)

K1 + K

2 = K'

1 + K'

2.... (1)

K is kinetic energy of object before the collision

and K' is kinetic energy of object af ter the collision

In general masses of both the objects are different,

So speed of each object beco mes different after

collision K1 K'

1 and K

2 K'

2

By work energy theorem

W1 = K'

1 – K

1.... (2)

W2 = K'

2 – K

2.... (3)

From (1), (2) and (3)W1 0, W

2 0

W1 + W

2 = 0

W1 = – W

2

i.e. the work done by the first object on the second

object is ex actly the opposite of t he work done by

the second ob ject on the fi rst object.

33.. AA, , BB, , DD

4.4. The F.B.D. of cylinder is as shown. In equilibrium T = Mg = tension in string

In equilibrium, net force on cylinder is zero

Torque is same about any axis.

Torque on cylinder about any point is zero.

A = Mg 2R – mg sin R = 0

mgsin

f

A

T


M =2

sinm Hence for only one value of M

cylinder can remain in equilibrium.

A is true, B is false

When the cylinder rolls up the incline, sense of

rotation of cylinder about center of mass is clockwise.

Hence T > f. C is false.

When the cylinder rolls down the incline, sense of

rotation of cylinder about center of mass is

anticlockwise. Hence T < f. D is True.

5.5. When S 2 is closed current in inductor

remains, i =

2 R

V

R

1 + V

R

1

2 =

2 R

3

V2

1

Potential difference

(V) = 2

3

=

3

Ans.Ans.

And Ld i

d t =

2

3

d i

d t = +

2

3

L

Ans.Ans.

6.6. From ray diagram it is clear that ray emerges out of

lens parallel to itself. Hence the angle of deviation

caused by the lens is 0°.

60°

r r 60°

7.7. From snells law at first interface

sin 60 = 3 s inr or r = 30°

Since the emergent ray is parallel to initial incident

ray, the portion of lens used for refraction can be

assumed as slab

Hence lateral displacement

=30cos

)3060sin(3

r cos

)r isin(t

= 3 mm.

8.8. If R denotes radius of curvature of curved surface,

then from above figure

3 × (2R – 3) = 4 × 4

or R =6

25mm

From the formulae of focal length for plano-convex

lens

3

4

4

2R-3

f = 136

25

1

R

= 136

25

×

13

13

=12

25)13( cm


1.1.

i =21

21

RR

22

=

21 RR

Where =dt

d is the net emf in the circuit.

V1 – V

2= ( – iR

1) – ( – iR

2) =

21

12

RR

R –R

2.2. Mirror formula :

20

1

280

1

v

1

280

1

20

1

v

1


280

114

v

1

v =15

280

v = – om

2

v.u

v

v = – 15.

28015

2802

v =

1515

15

v =

15

1 m/s Ans.Ans.

3.3. A

10m/sB A

v1

Bv2

m × 10 = mv1 + mv2

10 = v1 + v2 ...(i)

and2

1 × 10 = v2 – v1 ...(ii)

From and

v1 = 2

5 m /s ; v2 = 2

15m/s

Distance between the two blocks

S = (– v1 + v2) . t

=

2

15

2

5 × 5 = 25 m

4.4. Heat obviously flows from higher temperature to lower

temperature in steady state. A is true.

Temperature gradient areationseccross

1 in

steady state. B is false.

Thermal current through each cross section area is

same. C is true.Temperature decreases along the length of the rod

from higher temperature end to lower temperature end.

D is false.

55.. AA,,BB,,CC

Using – e )BV(

for the region outside the plates,

direction of magnetic f ield can be found. Inside the

plates, net force on the electron is zero hence

electric force is opposite to that of magnetic f orce.

Direction of electric field between the plates is

opposite to that of direction of force on the negative

(electron) charge.

6.6.

Mg

Ma

CM

CM L/2

L/2

Hinge

From the cart's frame

Wall

= KE2 – KE

1

Ma

2

L + Mg

2

L = 0 – 0

a = g

7.7.

N1

HingeN2

Ma

Mg

Initially rod is at rest

So, N1 = Mg] N

1 = Mg

Torque =

Ma

3

ML

2

L 2


4

3a =

2

L

4

3g =

2

L

Ma + N2 = Ma

CM

Mg + N2 =

g

4

3M

N2 = –

4

Mg

N = 22

21 NN =

4

Mg17

8.8. Equation of motion for the cart

–4

Mg+ f = 2Ma

f = 2Ma +4

Mg

f =4

Mg9


1.1. It is clearly visible from all graphs that as x-increases. Velocity changes sign. Since this is

not possible, no graph represents the possible

motion.

2.2. Because of increase in magnetic field with time,

electric field is induced in the circular region and

represented by lines of forces as shown in figure.

The signs of minimum work done by external agent

in taking unit positive charge from A to C via path

APC, AOC and AQC are

W APC

= – ve, W AOC

= 0, W AQC

= + ve

(C) is the correct choice.

3.3. The direction of forces on the two elements taken

symmetrical on two sides of the y–axis are shown.

Clearly the net force will be on negative x-axis.

4.4. For a ball rolling without slipping on a fixed rough

surface, no work is done by friction.

5.5. p - Implse = Ft = 3mgt

6.6. Let the angular speed of disc when the balls reach

the end be . From conservation of angular

momentum

2

1mR2

0=

2

1 mR2 +

2

mR2 +

2

mR2 or

=3

0

7.7. The angular speed of the disc just after the balls leave

the disc is =3

0

Let the speed of each ball just after they leave the

disc be v.


2

1

2mR2

1 02 =

2

1

2mR2

12 +

2

1

2

mv2

+

2

1

2

mv2

solving we get

v =3

R2 0

NOTE :NOTE : v = 2r

2 v)R(

; vr

= radial velocity of the ball

8.8. Workdone by all forces equal Kf – K

i =

2

m

2

1v2 =

9

mR 20

2



1.1.

For the just completing the circular motion,

minimum velocity at bottom in

vB = gR5

Energy conservation b/w point A and B

MgH + 0 = 0 +2

1mv

B2

MgH =2

1m (5gR)

H =2

R5

3.3. r d.E

= –dt

d

and take the sign of f lux according to right hand

curl rule get.

r d.E

= – (– (– A) – (– A) + (– A)) = – A

4.4. F

= BF

= AD = R ) ji(

B

= B0 )k ji(

F

= RB 0 ) ji( × )k ji( = RB 0

111

011

k ji

= RB0 )k2 ji(

F = RB0 6

Aliter :Aliter :

)k ji(BB 0

= R ) ji(

B

= 0

Angle = 90°

F = B = 3 B0 2 R = 6 B0 R

5.5. Since angular velocity is constant, acceleration of

centre of mass of disc is zero. Hence the

magnitude of acceleration of point S is 2x where

is angular speed o f disc and x is the distance of

S from c entre. Therefore the graph is

Sol. 14 to 16Sol. 14 to 16

The free body diagram of plank and disc is

Applying Newton's second law

F – f = Ma1

.... (1)

f = Ma2

.... (2)

FR =2

1MR2 .... (3)

from equation 2 and 3

a2 =

2

R

From constraint a1 = a

2 + R

a1 = 3a

2.... (4)

Solving we get a1 =

M4

F3 and =

MR2

F

If sphere moves by x the plank moves by L + x.

The from equation (4)

L + x = 3x or x =2

L



1.1. FBD for sphere & block

a1

f r

m

m

f r

a2

a1 =

m

f r =

m

mg

a2 =

m

f r =

m

mg

iga1

iga2

ig2aaa 21rel

arel

= 2g.

2.2. The electron ejected with maximum speed vmax

are

stopped by electric field E = 4N/C after travelling a

distance d = 1m

2

1 mV

max2 = eE d = 4eV

The energy of incident photon =200

1240 = 6.2 eV

From equation of photo electric effect

21 mv

max2 = h –

0

0 = 6.2 – 4 = 2.2 eV.

3.3. In steady state current from battery =2

10 = 5A

In parallel inductors L1I1 = L

2I2 all the time

i1 =

21

2

LL

L

i =23

3

× 5 = 3A

4.4. Both the spring are in series

Keq

=K2K

)K2(K

=

3

K2

Time period

T = 2eqK where =

21

21

mmmm

Here =2

m

T = 2 K2

3

2

m =

K4

m32

Method IIMethod II

/////////////////////////////////////////////////////////////////////////////////////

m mk

/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

m

mk

x1 x2

mx1 = mx

2 x

1 = x

2

force equation for first block;

3k2 (x 1 + x 2 ) = –m 21

2

dtxd

Put x1 = x

2

2

12

dt

xd +

m3

k4 × x

1 = 0

2 =m3

k4

T =K4

m32

5.5. = 33 )2/R()3/4(R)3/4(

M

M1 = M7

8, M2 = M

7

1

=5

2 M1R2 –

2

2

2

22

RM

2

RM

5

2

; =140

57MR2

6.6.

P0

HH2

V P0

V = gh2 =

2

Hg2 = gH


7.7.

Let h be height of water column just after puttingcylinder,

A

2

H

3

A A´h

h´ =4

3 H

V´ = ´gh2 = gH2

3

8.8.

3

A A

2

H = h A

h =3

H

v = gh2 = gH3

2


1.1. Let velocity of c .m. of sphere be v. The velocity of

the plank = 2v.

Kinetic energy of plank =2

1× m × (2v) 2 = 2mv 2

Kinetic energy of cylinder =2

1mv2

+2

1+

22mR

2

1

=

2

11mv

2

1 2

=2mv

2

1.

2

3

sphere of .E.K

plank of .E.K=

2

2

mv4

3

mv2 =

3

8.

2.2. Ix = I

y = I

z =

5

2mR 2

3.3. The spring is never compressed. Hence spring shall

exert least force on the block when the block is at

topmost position.

Fleast

= kx0 – kA = mg – m2 A = mg – 4

2

2

T

mA

4.4. KEmax = (5 – ) eV

when these electrons are accelerated through 5V,

they will reach the anode with maximum energy

= (5 – + 5)eV

10 – = 8

= = 2eV 2eV Ans.Ans.

Current is less than saturation current because if

slowest electron also reached the plate it would have

5eV energy at the anode, but there it is given that the

minimum energy is 6eV.

5.5. By principal of energy conservation.

PB = P

R + P

L

Near the starting of the circuit

PR = i2 R and PL = L i dt

di

.

Asdt

di has greater value at the starting of the

circuit, PL > P

R


Sol. 1 to 3. :Sol. 1 to 3. : When connected with the DC source

R =4

12 = 3

When connected to ac source =Z

V

2.4 = 222 L3

12

L = 0.08 H

Using P = rms

Vrms

cos

=Z

V2rms cos

=22

2rms

)C

1 –L(R

RV

= 24 W


2.2. The potential difference across the inductor ise

= E– iR.Hence the plot of e versus i is a straight line with

negative slope.

3.3. Equation can be written as i = 2 sin 100 t + 2 sin

(100 t + 120º)

so phase difference = 120º

= cos A A2 A A 2122

21

==

2

1 –22244 == 2 so effective value will

rms. value = 2 / 2 = 2 A

4.4. Angular momentum =2

nh =

2

h( n = 1)

(mvr) = n.2

h =

2

h(n = 1)

Sol.5 to 7.Sol.5 to 7.

FBD of rod and cylinder is as shown.

Net torque on rod about hinge 'O' = 0

N1 × L = mg ×

2

L

or 2

mgN1

Net torque on cylinder about its centre C is zero.

f 1R = f

2 R or 21 f f

Net torque on cylinder about hinge O is zero.

N2 × L = N

1 × L + mgL

or N2 =

2

mg3

8.8. The magnetic force on bob does not produce any

restoring torque on bob about the hinge. Hence this

force has no effect on time period of oscillation.

Therefore both statements are correct and statement-

2 is the correct explanation.



3.3. <i> ==1

dti

1

0

= = 22 2

1

04

tsin dt = =4

4.4. The hydrogen atom is in n = 5 state.

Max. no of possible photons = 4

To emit photon in ultra violet region, it must jump to n

= 1, because only Lyman series lies in u.v. region.

Once it jumps to n = 1 photon, it reaches to its ground

state and no more photons can be emitted. So only

one photon in u. v. range can be emitted.

If H atom emits a photon and then another photon of

Balmer series, option D will be correct.

7.7. The FBD of cy linder is as shown

Resolvi ng forces along x and y axis

yx

NBN A

mgf = N A A

f = NB B

N A + f

B =

2

mg.... (1)

f B =

N

B

NB – f

A =

2

mg.... (2)

f A =

N

A

solving we get

f A =

)1(2

)1(Mg2

and

f B =

)1(2

)1(Mg2

Angular acceleration

=

2

MR

R)f f (2

B A =

)1(R

g222

8.8.

dB along any closed path within a uniform

magnetic field is always zero. Hence the closed

path can be chos en of any size, even v ery small

size enclosing a very smal l area. Hence we can

prove that net current through each area of

inf init esimall y sm all size wi thin region of uniform

magnetic field is zero. Hence we can say no current

(rather than no net current) flows through region of

uniform magnetic field. Hence statement -2 is correct

explanation of statement-1.

PREFACE - SMAROSA

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