MHT CET Physics

50 Pages·2016·1.24 MB·English

by  Target Publications

Useful for all Agricultural, Medical, Pharma cy and Engineering Entrance Examinations
held across India.

STD XII Sci.
.

Triumph Physics

Based on Maharashtra Board Syllabus

  Fifth Edition: February 2016

Salient Features

• Exhaustive subtopic wise coverage of MCQs.

• Important formulae provided in each chapter.

• Hints included for relevant questions.

• Various competitive exam questions updated till the latest year.
  • Includes solved MCQs from JEE (Main), AIPMT, MH CET 2014, 2015.
  • Evaluation test provided at the end of each chapter.
• Two Model Question Papers with answers at the end of the book.

Solutions/hints to Evaluation Test available in downloadable PDF format at
  www.targetpublications.org/tp10089

Printed at: Print House India Pvt Ltd., Navi Mumbai
No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical
including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

P.O. No. 11730

10089_10390_JUP
Preface

“Std. XII: Sci. Triumph Physics” is a complete and thorough guide to prepare students for a competitive
level examination. The book will not only assist students with MCQs of Std. XII but will also help them to prepare for
JEE, AIPMT, CET and various other competitive examinations.
The content of this book is based on the Maharashtra State Board Syllabus. Formulae that form a vital part of
MCQ solving are provided in each chapter. Notes provide important information about the topic. Shortcuts provide
easy and less tedious solving methods. Mindbenders have been introduced to bridge the gap between a text book
topic and the student’s understanding of the same. A quick reference to the notes, shortcuts and mindbenders has been
provided wherever possible.
MCQs in each chapter are divided into three sections:
Classical Thinking: consists of straight forward questions including knowledge based questions.

Critical Thinking: consists of questions that require some understanding of the concept.

Competitive Thinking: consists of questions from various competitive examinations like JEE, AIPMT, CET,
CPMT, G CET, KCET, Assam CEE, BCECE, EAMCET (Engineering, Medical) etc.
Hints have been provided to the MCQs which are broken down to the simplest form possible.
An Evaluation Test has been provided at the end of each chapter and two Model Question Papers to assess
the level of preparation of the student on a competitive level.
An additional feature called “The physics of .....” has been included in the book to foster a keen interest in
the subject of physics.
Informative Table of “Various Physical Quantities and Conversion Factors” has been provided at the end
for a quick glance.
The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve
nearly missed something or want to applaud us for our triumphs, we’d love to hear from you.
Please write to us on : [email protected]

Best of luck to all the aspirants!

Yours faithfully
Authors

Sr. Sr.
Topic Name Page No. Topic Name Page No.
No. No.
13 Current Electricity 536
1 Circular Motion 1
Magnetic Effects of Electric
2 Gravitation 48 14 567
Current
3 Rotational Motion 94
15 Magnetism 605
4 Oscillations 143
16 Electromagnetic Induction 623
5 Elasticity 198
17 Electrons and Photons 677
6 Surface Tension 231 18 Atoms, Molecules and Nuclei 705
7 Wave Motion 264 19 Semiconductors 752
8 Stationary Waves 301 20 Communication Systems 788
Kinetic Theory of Gases and Model Question Paper - I 807
9 343
Radiation 811
Model Question Paper - II
10 Wave Theory of Light 412
11 Interference and Diffraction 437 Various Physical Quantities
815
12 Electrostatics 480 and Conversion Factors
0 1 
Target PublicaCtionis rPvct. Lutd. lar Motion Chapter 01: Circular Motion

Syllabus

1.0 Introduction

1.1 Angular displacement

1.2 Angular velocity and angular
acceleration

1.3 Relation between linear velocity and
angular velocity

1.4 Uniform circular motion

1.5 Acceleration in U.C.M
(Radial acceleration)

1.6 Centripetal and centrifugal forces

Riding on a vertical circular arc, this roller
1.7 Banking of roads coaster fans experience a net force and
acceleration that point towards the centre of the

circle
1.8 Conical pendulum

1.9 Vertical circular motion

1.10 Kinematical equation for circular motion in analogy with linear motion

1
Std. XII : Triumph Physics 

iii. Net (linear) acceleration,
Formulae

a = a2+a2 ….(Magnitude only)
r t
1. Uniform Circular Motion (U.C.M.):
i. Instantaneous anguvlar velocity2,  = æççççèvr2ö÷÷÷÷ø+æçççèddvtö÷÷÷ø2
 = lim= = = 2n =
t0 t r T iv. Relation between tangential and angular
acceleration,
ii. Average angular velocity,
  
θ -θ Dθ a r = r
 = 2 1 = T
av
t -t Dt   = 0
2 1
where,
3. Centripetal force:
 = angular position of the body at time
1 mv2
t i. Centripetal force, F =
1 cp
r
 = angular position of the body at time
2 = mr2 = mv = mr (2f)2
t
2
iii.  =  for U.C.M. 22 42mr
av = mr =
 
iv. If a particle makes n rotations in t  T  T2
second, then ii. Magnitude of Centrifugal force
2πn = Magnitude of Centripetal force
 =
av t mv2
i.e F = (in magnitude)
v. Angular acceleration =  = 0 cf r
vi. Instantaneous angular acceleration, iii. When an electron moves round the
Dω dω d2θ nucleus of an atom along a circular path,
 = lim = =
inst Dt0 Dt dt dt2 we have
Ze2 mv2
vii. Average angular acceleration, = = m2r
ω -ω Dω 4 r2 r
 = 2 1 = 0
ave. t -t Dt 42r
2 1 = m 42 n2r = m
where,  = instantaneous angular speed T2
1
at time t where, Z = atomic number of the
1
nucleus.
 = instantaneous angular speed at time
2
t2. 4. Motion of a vehicle on a curve road:
viii. Linear acceleration The maximum velocity v, with which a
  vehicle can take a safe turn so that there is no
= centripetal acceleration =  v
skidding, is v = rg
v2 42r
= a = v = = r2 = 42f2r =
where,  = coefficient of limiting friction
r T2
between the wheels and the road.
1 2
ix. Time period = T = =
5. Banking of roads:
frequency(f) 
The proper velocity or optimum v on a road
x. Relation between linear and angular
banked by an angle  with the horizontal is
  
velocity: v=  r = r as  = 90 given by,

  tan 
2. Non-uniform circular motion:
v = rg s 
i. Radial component of acceleration, 1 tan
s
v2 where r = radius of curvature of road
a =  2r = 
r r g = acceleration due to gravity
ii. Tangential component of acceleration, s = coefficient of friction between road and
tyres
dv
a =
t when  = 0, v = rgtan
dt s

2
Chapter 01 : Circular Motion
6. Vertical Circular Motion: 8. Kinematical equations in circular motion in
analog with linear motion:
i. Velocity at highest point v  rg
H
i. =  + t
0
ii. Velocity at the lowest point v  5rg
L ii.  =  t + 1 t2
iii. Velocity at a point along horizontal 0 2
(midway position) v  3rg iii. 2 = 2 + 2 
M
0
iv. Acceleration at the highest point a = g
H
Notes

v. Acceleration at the bottom point a = 5g
L

vi. Acceleration along horizontal a = 3g
M
1. Radian measure must be used in equations
vii. Tension at top most point,
that contain linear and angular quantities.
mv2
TH = B mg  0 2. Finite angular displacement is a scalar
r
quantity because it does not obey the laws of
viii. Tension at the lowest point,
vector addition.
mv2
T = A +mg  6 mg
L 
r 3. In U.C.M., angular velocity  is only
ix. Tension at a point where the string  
makes an angle  with the lower vertical 
constant vector but angular acceleration 
line
 
mv2 
T = mgcos and angular displacement  are variable
r  
x. Tension at midway position where vectors.

 = 90 (i.e. along horizontal)
4. All the points on a rotating body in U.C.M.
mv2 have same  except centre as it is not
T = [cos 90 = 0]
M r rotating.

xi. Total energy at different points at the
5. Instantaneous angular displacement is a
top, bottom and horizontal,
vector quantity.

5
EH = EL = EM = mrg 6. Angular speed is a scalar quantity but angular
2
velocity is a vector quantity but both have
xii. Total energy at any point,
same units i.e rad/s.

1
E = mv2 mgr(1cos)   
2 7. The direction of , ,  is given by the right

hand thumb rule.
7. Conical Pendulum:
i. Angular velocity, 8. The value of  of earth about its axis is
7  105 rad/s or 360 per day.
g
a.  =
lcos 9. When a particle moves in a circle with
constant speed, its velocity is variable because
gtan of changing direction.
b.  =
r
10. Circular motion is a two-dimensional motion
2 lcos in which the linear velocity and linear
ii. Periodic time = = 2
 g acceleration vectors lie in the plane of the
circle but the angular velocity and angular
lsin
=  acceleration vectors are perpendicular to the
gtan plane of the circle.

iii. Radius of horizontal circle,
11. Centrifugal force is a fictitious force and
r = l sin  holds good in a rotating frame of reference.

3
Std. XII : Triumph Physics 
12. An observer on the moving particle 24. If a body moves in a cylindrical well (well of
experiences only the centrifugal force, but an death) the velocity required will be minimum
observer stationary with respect to the centre safest velocity and in this case the weight of
can experience or measure only the the body will be balanced by component of
centripetal force. normal reaction and the minimum safest

velocity is given by the formula rg.
13. Whenever a particle is in a U.C.M. or non

U.C.M., centripetal and centrifugal forces act
25. Cyclist leans his cycle to make an angle to
simultaneously. They are both equal and
avoid topling; not to provide centripetal force.
opposite but do not cancel each other.

14. Centripetal force and Centrifugal force are 26. If a body is kept at rest at the highest point of
not action-reaction forces as action-reaction convex road and pushed along the surface to
forces act on different bodies. perform circular motion, the body will fall

r
15. The direction of centripetal force is same
after travelling a vertical distance of from
whether the rotation of the circular path is 3
clockwise or anticlockwise. the highest point where r is the radius of the

circular path.
16. Centripetal force is not responsible for

rotational motion of a body because only
27. When a body moves in a circular path with
torque can produce rotational motion.
constant speed, its linear momentum changes
17. Since the centripetal force acting on a particle at every point, but its kinetic energy remains
in circular motion acts perpendicular to its
constant.
displacement (and also its velocity), the work
done by it is always zero. 28. In horizontal uniform circular motion, kinetic

energy and magnitude of linear momentum
18. Centrifuge is an apparatus used to separate
remains constant, but the direction of linear
heavier particles from the lighter particles in
a liquid. momentum keeps on changing.

19. Range of acceleration in circular motion 2 9 . Since the centripetal force is not zero for a
90 <  ≤ 180. particle in circular motion, the torque acting


20. The radius of the curved path is the distance is zero i.e., τ= 0 (as the force is central)
from the centre of curved path to the centre of Hence the angular momentum is constant i.e.
gravity of the body. It is to be considered 
L= constant.
when the centre of gravity of body is at a
height from the surface of road or surface of 30. Whenever the body moves, the force
spherical body. responsible for motion is the vector sum of all

the forces acting at that point.
21. Whenever a car is taking a horizontal turn, the
For example, Lift going up and down with
normal reaction is at the inner wheel. acceleration ‘a’.

22. While taking a turn, when car overturns, its 31. If a particle performing circular motion comes
inner wheels leave the ground first. 
to rest momentarily, i.e. v= 0, then it will
23. For a vehicle negotiating a turn along a move along the radius towards the centre and
circular path, if its speed is very high, then the if its radial acceleration is zero, i.e. a = 0,
r
vehicle starts skidding outwards. This causes then the body will move along the tangent
the radius of the circle to increase resulting in drawn at that point.

the decrease in the centripetal force.
32. For non uniform circular motion
1
[ F  ]     
cp T ar v

4
Chapter 01 : Circular Motion
33. When a bucket full of water is rotated in a 5. To find out number of revolutions, always
vertical circle, water will not spill only if apply the formula,
velocity of bucket at the highest point is  t 2nt
Number of revolutions = =  = nt
2 2 2
 gr .

 
34. If velocity imparted to body at the lowest 6. Since F  v, therefore, no work is done by
c
position is equal to 2rg, then it will oscillate the centripetal force. [2 rad = 360 = 1 rev.]

in a semicircle. 7. Angle which, a cyclist should make with the
vertical is the angle of banking along a curved
road.
Mindbenders

8. On frictional surface, for a body performing
1. The maximum velocity with which a vehicle circular motion, the centripetal force is
can go without toppling, is given by provided by the force of friction.
f = N but on horizontal surface N = mg
d S
v = rg = rgtan
2h 9. The minimum safe velocity for not
d gdr
where, tan  = overturning is v =
2h 2h

d = distance between the wheels
10. While rounding a curve on a level road,
h = height of centre of gravity from the road
centripetal force required by the vehicle is
g = acceleration due to gravity
provided by force of friction between the tyres

and the road.
2. Skidding of an object placed on a rotating
mv2
platform:
= F = R = mg
The maximum angular velocity of rotation of r

the platform so that object will not skid on it is
11. To avoid dependence on friction for the
 = (g/r) supply of necessary centripetal force, curved
max

roads are usually banked by raising outer edge
3. The angle made by the resultant acceleration of the road above the inner edge.

with the radius,
12. The angle of banking () is given by,
 = tan1æççççèaatö÷÷÷÷ø tan  = v2  h
r rg l2h2

where h is height of the outer edge above the
Shortcuts
inner edge and l is length of the road.

1. The basic formula for acceleration is a = v. 13. On the same basis, a cyclist has to bend

through an angle  from his vertical position
2. In U.C.M., if central angle or angular
while rounding a curve of radius r with
displacement is given, then simply apply
v2
 velocity v such that tan  =
dv = 2v sin to determine change in velocity. rg

2
If  is very very small, then
3. There are two types of acceleration; ar (radial) v2
and a (tangential) acceleration. tan  = sin  =
t rg
dv
Formula for ar = 2r and at = or r v2 h
dt =
rg l
4. To calculate angular displacement, apply the where h is height of the outer edge from the
1 inner edge and l is the distance between the
formula,  = t + t2
2 tracks or width of the road.

5
Std. XII : Triumph Physics 
14. Always remember the formulae for velocity of 19. The total energy of any body revolving in a
the body at the top, bottom and at the middle vertical circle is (5/2) mgr.

of a circle with two distinct cases:
20. The distance travelled by the particle
i. path is convex
performing uniform circular motion in t
ii. path is concave
2r
Remember in both the cases, formula will be seconds is given by the formula, d = t.
T
different.

mv2 21. To find out any unknown quantity, if body
i. = mg  N where N is normal
r moves in vertical circle, resolve mg and if the
reaction. body moves in horizontal circle, resolve
mv2 tension or normal reaction.
ii. = N  mg

r
22. Centripetal Force in Different Situations:
Remember if in the question, it is given that

body falls from a certain point then at that
The centripetal
point N = 0. Situation
force

15. Effect of rotation of earth about its axis: i. A particle tied to a Tension in the string
The apparent loss in weight of a body on its string and whirled
surface = m2 R cos2  where in a horizontal
m = mass of body circle
 = angular velocity of earth ii. Vehicle taking a Frictional force
turn on a level road exerted by the road
R = radius of earth
on the tyres
 = latitude
iii. A vehicle on a Weight of the body or
16. In horizontal circle, tension will be equal to
speed breaker a component of
mv2 weight
centripetal force i.e. T =
r
iv. Revolution of earth Gravitational force
i. The minimum velocity of projection at
around the sun exerted by the sun
the lowest point of vertical circle so that
v. Electron revolving Coulomb attraction
the string slacken at the highest point, is
around the nucleus exerted by the
given by v = 5gr
L in an atom protons on electrons
ii. velocity at the highest point is v = gr vi. A charged particle Magnetic force
H
describing a exerted by the
17. If TL is the tension at the lowest point and TH circular path in a magnetic field
is the tension at the highest point then magnetic field
T  T = 6 mg
L H vii. Coin placed on disk In this case frictional

force gives necessary
18. When
centripetal force.
i. v = 2gr , the body moves in a vertical
L
viii. Car moving on a N sin  gives
semicircle about the lowest point L,
smooth banked
necessary centripetal
ii. vL < 2gr , then the body oscillates in a road force.
circular arc smaller than the semicircle.
ix. Passenger sitting in Necessary centripetal
iii. For a motor cyclist to loop a vertical a turning car force is provided by
loop, v > 5gr and v > gr seat and passenger.
L H

6
Chapter 01 : Circular Motion
9. An electric motor of 12 horse-power generates
Classical Thinking
an angular velocity of 125 rad/s. What will be

the frequency of rotation?
1.1 Angular displacement
(A) 20 (B) 20/
1. The angular displacement in circular motion is (C) 20/2 (D) 40

(A) dimensional quantity.
10. The ratio of angular speeds of seconds hand
(B) dimensionless quantity.
and hour hand of a watch is
(C) unitless and dimensionless quantity. (A) 1 : 720 (B) 60 : 1
(D) unitless quantity. (C) 1 : 60 (D) 720 : 1

2. Angular displacement is measured in 11. A body moves with constant angular velocity
(A) metre. (B) time. on a circle. Magnitude of angular acceleration
is
(C) radian. (D) steradian.
(A) r2 (B) constant
3. A flywheel rotates at a constant speed of
(C) zero (D) r
3000 r.p.m. The angle described by the shaft
in one second is 12. A wheel having a diameter of 3 m starts from
rest and accelerates uniformly to an angular
(A) 3  rad (B) 30  rad
velocity of 210 r.p.m. in 5 seconds. Angular
(C) 100  rad (D) 3000  rad
acceleration of the wheel is
1.2 Angular velocity and angular (A) 4.4 rad s2 (B) 3.3 rad s2
acceleration (C) 2.2 rad s2 (D) 1.1 rad s2

 
4. Direction of r is 1.3 Relation between linear velocity and
angular velocity
(A) tangent to path.

(B) perpendicular to path.
13. The vector relation between linear velocity v ,
(C) parallel to the path.  
angular velocity  and radius vector r is
(D) along the path.
given by
5. What is the angular speed of the seconds hand      
of a watch? (A) v =  r (B) v = r + 
     
(A) 60 rad/s (B)  rad/s
(C) v = . r (D) v = r  

(C) /30 rad/s (D) 2 rad/s
14. A wheel has circumference C. If it makes
6. The angular velocity of a particle rotating in a f r.p.s., the linear speed of a point on the
circular orbit 100 times per minute is circumference is
(A) 1.66 rad/s (B) 10.47 rad/s (A) 2fC (B) fC
(C) 10.47 deg/s (D) 60 deg/s (C) fC/2 (D) fC/60

7. A body of mass 100 g is revolving in a 15. A body is whirled in a horizontal circle of
horizontal circle. If its frequency of rotation is radius 20 cm. It has angular velocity of
3.5 r.p.s. and radius of circular path is 0.5 m, 10 rad/s. What is its linear velocity at any
the angular speed of the body is point on circular path?
(A) 18 rad/s (B) 20 rad/s (A) 10 m/s (B) 2 m/s
(C) 22 rad/s (D) 24 rad/s (C) 20 m/s (D) 2 m/s

8. What is the angular velocity of the earth? 16. A particle moves in a circular path, 0.4 m in
radius, with constant speed. If particle makes
2 2
(A) rad/s (B) rad/s 5 revolutions in each second of its motion, the
86400 3600
speed of the particle is
2 2
(A) 10.6 m/s (B) 11.2 m/s
(C) rad/s (D) rad/s
24 6400 (C) 12.6 m/s (D) 13.6 m/s

7