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 Written as per the revised syllabus prescribed
by the Maharashtra State Board
of Secondary and Higher Secondary Education, Pune.
Perfect
Physics – I
STD. XII Sci.
Salient Features
• Exhaustive coverage of syllabus in Question Answer Format.
• Covers answers to all Textual Questions and Numericals.
• Covers answers to all Board Questions till date.
• Includes marking scheme for Board Questions from 2013 to 2016.
• Covers relevant NCERT Questions.
• Simple and Lucid language.
• Neat, Labelled and authentic diagrams.
• Solved & Practice numericals.
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P.O. No. 39707
10105_11131_JUP
Preface
In the case of good books, the point is not how many of them you can get through, but rather how many can get
through to you.
“Std. XII Sci. : PERFECT PHYSICS - I” is a complete and thorough guide critically analysed and
extensively drafted to boost the students confidence. The book has been prepared as per the Maharashtra State
board syllabus.
While preparing the book, our main aim was to make a student’s journey of learning and understanding the
various concepts, an effortless and pleasant process.
Our Sub-topic wise classified ‘Question and Answer’ format of the book will help the student to
understand each and every concept thoroughly. Neatly labelled diagrams have been added wherever required.
Answers to all textual and intext questions have been provided.
We have included National Council Of Educational Research And Training (NCERT) questions and
problems based on Maharashtra State board syllabus along with their solutions which will help the students to
have a better grasp of the concept and prepare him/her on a competitive level.
For the students to have a crystal clear idea about the board paper pattern and the range of questions asked,
we have included Solved Board Questions and Numericals upto the latest year. Board Questions from old
syllabus which fall under the revised syllabus are also added. Solved Board Questions from 2013 to 2016 have
been included along with its marking scheme which is indicative and is subject to change as per Maharashtra
State Board’s discretion.
To develop better understanding of concepts; we have discussed relevant points and questions in the form
of Additional Information. Any additional information about a concept is provided in the form of Note. We had
developed the concept of Brain Teasers, which are theory questions and numericals build within the frame work
of State Board syllabus to develop higher order thinking among students. Concept Builders are designed to
enable the students guage their grasp of a given concept and strengthen it further. A quick review of each chapter
is provided in the form of Summary. Definitions, statements and laws are specified with italic representation.
Formulae are provided in every chapter which are the main tools to tackle numericals. Ample amount of solved
problems are provided to understand the application of different concepts and formulae.
Practice Problems and Multiple Choice Questions will help the students to test their range of preparation
and the amount of knowledge of each topic. Selected multiple choice questions have been provided with hints to
help the students overcome conceptual or mathematical hinderances.
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 at : [email protected]
A book affects eternity; one can never tell where its influence stops.
Best of luck to all the aspirants!
Yours faithfully,
Publisher
PAPER PATTERN

There will be one single paper of 70 Marks in Physics.

Duration of the paper will be 3 hours.

Physics paper will consist of two parts viz: Part-I and Part-II.

Each part will be of 35 Marks.

Same Answer Sheet will be used for both the parts.

Each Part will consist of 4 Questions.

The sequence of the 4 Questions in each part may or may not remain same.

The paper pattern for PartI and PartII will be as follows:
Question 1:
(7 Marks)
This Question will be based on Multiple Choice Questions.
There will be 7 MCQs, each carrying one mark.
One Question will be based on calculations.
Students will have to attempt all these questions.
Question 2:
(12 Marks)
This Question will contain 8 Questions, each carrying 2 marks.
Students will have to answer any 6 out of the given 8 Questions.
4 questions will be theory-based and 4 will be numericals.
Question 3:
(9 Marks)
This Question will contain 4 Questions, each carrying 3 marks.
Students will have to answer any 3 out of the given 4 Questions.
2 questions will be theory-based and 2 will be numericals.
Question 4:
(7 Marks)
This Question will contain 2 Questions, each carrying 7 marks.
Students will have to answer any 1 out of the given 2 Questions.
4/5 marks are allocated for theory-based question and 3/2 marks for numerical.
Distribution of Marks According to Type of Questions
Type of Questions
Marks
Marks with option
Percentage (%)
Objectives
14
14
20
Short Answers
42
56
60
Brief Answers
14
28
20
Total
70
98
100
Topicwise Weightage
Topic Name
No.
Marks Without
Option
Marks With
Option
1
2
Circular Motion
Gravitation
04
03
05
05
3
Rotational Motion
04
06
4
5
6
Oscillations
Elasticity
Surface Tension
05
03
04
07
04
05
7
8
Wave Motion
Stationary Waves
03
05
04
07
9
Kinetic Theory of Gases and Radiation
04
06
Contents
Sr. No.
Unit
Page No.
1
Circular Motion
1
2
Gravitation
47
3
Rotational Motion
81
4
Oscillations
121
5
Elasticity
160
6
Surface Tension
192
7
Wave Motion
221
8
Stationary Waves
254
9
Kinetic Theory of Gases and Radiation
292
Board Question Paper – March 2015
345
Board Question Paper – October 2015
347
Board Question Paper – March 2016
349
Board Question Paper – July 2016
351
Note: 1.
* mark represents all Textual questions.
2.
# mark represents all Intext questions. 3. **mark represents marks allotted for ‘Explanation’ at the end of specific board
answers are for the detailed description of the whole answer.
Chapter 01: Circular Motion
01
Circular Motion Subtopics
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)
1.0
Centripetal
forces
1.7
Banking of roads
1.8
Vertical circular motion due to
earth’s gravitation
1.9
Equation for velocity and
energy at different positions in
vertical circular motion
1.10
Kinematical
equation
for
circular motion in analogy with
linear motion
Introduction
Q.1. Define circular motion. Give its examples.
Ans: Definition:
Motion of a particle along the circumference
of a circle is called circular motion.
Examples:
i.
The motion of a cyclist along a circular path.
ii.
Motion of the moon around the earth.
iii. Motion of the earth around the sun.
iv. Motion of the tip of hands of a clock.
v.
Motion of electrons around the nucleus
in an atom.
1.1
1.6
centrifugal
where, s = small linear distance
 = small angular displacement
iii.
It is directed radially outwards.
iv.
Unit: metre (m) in SI system and
centimetre (cm) in CGS system.
v.
Dimensions: [M0L1T0]
Q.3. *Define angular displacement.
OR
What is angular displacement?
Ans: i.
Angular displacement
Q.2. What is radius vector?
Ans: i.
A vector drawn from the centre of a
circle to position of a particle on
circumference of circle is called as
‘radius vector’.
B
ii.
It is given by,
 s

A
O 
s
|r|=
r

and
ii.
Angle traced by a radius vector in a
given time, at the centre of the circular
path is called as angular displacement.
Consider a particle
performing
circular motion in
anticlockwise
sense as shown in
the figure.
Let, A = initial
position of particle
at t = 0
Y
B
O
s

r
A
Y
1
Std. XII Sci.: Perfect Physics ‐ I B = final position of particle after time t
 = angular displacement in time t

d
r = radius of the circle
B

s = length of arc AB
iii.
=
Length of arc
radius of circle
Direction of angular displacement

s
=
r
iv.
Unit: radian
v.
Direction of angular displacement is
given by right hand thumb rule or right
handed screw rule.
Note:
1.
If a particle performing circular motion
describes an arc of length s, in short time
interval t then angular displacement is given
s
by  =
.
r

s =  r



In vector form, s =   r
2.
If a particle performing circular motion
completes one revolution then angular
displacement is given by  = 360 = 2c
where, c represents angular displacement in
radians.
3.
One radian is the angle subtended at the centre
of a circle by an arc of length equal to radius
of the circle.
*Q.4. State right hand thumb rule to find the
direction of angular displacement.
Ans: Right hand thumb rule:
Imagine the axis of rotation to be held in
right hand with the fingers curled around it
and thumb out-stretched. If the curled
fingers give the direction of motion of a
particle performing circular motion then
the direction of out-stretched thumb gives
the direction of angular displacement
vector.
2
A
Angular displacement is given by,
*Q.5. Explain right handed screw rule to find the
direction of angular displacement.
Ans: i.
Imagine the right handed screw to be
held in the place in which particle is
performing circular motion. If the right
handed screw is rotated in the direction
of particle performing circular motion
then the direction in which screw tip
advances, gives the direction of angular
displacement.
ii.
The tip of the screw advances in
downward direction, if sense of rotation
of the object is clockwise whereas the
tip of the screw advances in upward
direction, if sense of rotation of the
object is anticlockwise as shown in the
figure.

Y
d
O
B

A
Tip of screw
advancing
in upward
direction
Y
Right handed screw rule
Additional Information
Characteristics of angular displacement:
i.
Instantaneous angular displacement is a vector
quantity (true vector), so it obeys commutative
and associative laws of vector addition.
ii.
Finite angular displacement is a pseudo
vector.
iii. Direction
of
infinitesimal
angular
displacement is given by right hand thumb
rule or right handed screw rule.
iv. For anticlockwise sense, angular displacement
is taken as positive while in clockwise sense,
angular displacement is taken negative.
Chapter 01: Circular Motion
#Q.6. Are the following motions same or different?
i.
Motion of tip of second hand of a clock.
ii.
Motion of entire second hand of a clock.
Ans: Both the motions are different.
The tip of the second hand of a clock performs
uniform circular motion while the entire hand
performs rotational motion with the second
hand as a rigid body.
1.2
ii.
iii.
iv.


d

 = lim
=
t  0 t
dt
Finite angular velocity is given by,

=
t
It is a vector quantity.
Direction: The direction of angular
velocity is given by right hand rule and
is in the direction of angular
displacement.

v.
Unit: rad s1
vi. Dimensions: [M0L0T1]
Note:Magnitude of angular velocity is called angular
speed.
Q.8. *Define angular acceleration.
OR
What is angular acceleration? State its unit
and dimension.
Ans: i.
The rate of change of angular velocity
with respect to time is called angular
acceleration.

It is denoted by  .




  0

 =
=
t
t  t0


iv.
Direction: The direction of  is given
by right hand thumb rule or right handed
screw rule.
Unit: rad /s2 in SI system.
v.
Dimensions: [M0L0T2].
iii.
Angular velocity and angular acceleration
Q.7. *Define angular velocity.
OR
What is angular velocity? State its unit and
dimension.
Ans: i.
Angular velocity of a particle
performing circular motion is defined as
the time rate of change of limiting
angular displacement.
OR
The ratio of angular displacement to
time is called angular velocity.
ii.
Instantaneous angular velocity is given
by,

If 0 and  are the angular velocities
of a particle performing circular motion
at instant t0 and t, then angular
acceleration is given by,
Q.9. Define the following terms.
i.
Average angular acceleration
ii.
Instantaneous angular acceleration
Ans: i.
Average angular acceleration:
Average angular acceleration is defined
as the time rate of change of angular
velocity.

It is given by  avg
ii.



  1 
= 2
=
t 2  t1
t
Instantaneous angular acceleration:
Instantaneous angular acceleration is
defined as the limiting rate of change of
angular velocity.



d
It is given by  = lim
=
t 0 t
dt

Concept Builder
Positive angular acceleration:
When an angular acceleration will have the same
direction as angular velocity, it is termed as positive
angular acceleration.
eg: When an electric fan is switched on, the angular
velocity of the blades of the fan increases with time.
Negative angular acceleration:
When an angular acceleration will have a direction
opposite to that of angular velocity. It is termed as
negative angular acceleration.
eg: When an electric fan is switched off, the angular
velocity of the blades of fan decreases with time.
3
Std. XII Sci.: Perfect Physics ‐ I Additional Information
Q.10. What happens to the direction of angular
acceleration
i.
if a particle is speeding up?
ii.
if a particle is slowing down?
Ans: i.
Direction of  when the particle is
speeding up:

Consider a particle

moving
along
a

circular
path
in


d
anticlockwise direction
and is speeding up.


results in d to be
directed up the plane.
 increases
Figure (a)


and  are ar to the plane, they are
parallel to each other. [See figure (a)].

Direction of  when the particle is
slowing down:
Consider a particle is

moving
along
a

circular
path
in
anticlockwise direction
and is slowing down.


results in d to be
directed down the
plane.
d
dt
2.
If a body completes one revolution in time
2
= 2n,
interval T, then angular speed,  =
T
where n = frequency of revolution.
3.
d ,  and  are called axial vectors, as they



are always taken to be along axis of rotation.
Hence, direction of  is upward. As 
Magnitude of  keeps
on decreasing which
 = || =
O


d


 decreases
Figure (b)

Hence, direction of  is downward. [See
figure (b)].
Q.11. Write down the main characteristics of
angular acceleration.
Ans: Characteristics of angular acceleration:
i.
Angular acceleration is positive if
angular velocity increases with time.
ii.
Angular acceleration is negative if
angular velocity decreases with time.
iii. Angular acceleration is an axial vector.
iv. In uniform circular motion, angular
velocity is constant, so angular
acceleration is zero.
4


Magnitude of  keeps
on increasing which
ii.
Note:
1.
When a body rotates with constant angular
velocity its instantaneous angular velocity is
equal to its average angular velocity, whatever
may be the duration of the time interval. If the
angular velocity is constant, we write


4.
The direction of d and  is always given by
right handed thumb rule.
1.3
Relation between
angular velocity
linear
velocity
and
Q.12. *Show that linear speed of a particle
performing circular motion is the product
of radius of circle and angular speed of
particle.
OR
Define linear velocity. Derive the relation
between linear velocity and angular
velocity.
[Mar 02, Mar 96, 08, 12, Oct 09]
Ans: Linear velocity:
Distance travelled by a body per unit time in a
given direction is called linear velocity.
It is a vector quantity and is given by,


ds
v =
dt
Relation between linear velocity and
angular velocity:
i.
Consider a particle moving with uniform
circular motion along the circumference
of a circle in anticlockwise direction
with centre O and radius r as shown in
the figure.

v
B

r
O


r

s

v
A
Chapter 01: Circular Motion
ii.
iii.


Let the particle cover small distance s
from A to B in small interval t.
In such case, small angular displacement
is AOB = .
Magnitude of instantaneous linear
velocity of particle is given by,
s
v = lim
δt  0 t
But s = r 
 

v = r  lim 
[  r = constant]
 t 0 t 
iv.





ds
d
=
 r
dt
dt



d
=  = angular velocity
dt




v =   r
[**Explanation – ½ Mark]



Calculus method:
i.
A particle is moving in XY plane with
position vector,



r = r î cos t + r ĵsin t

Q.13. *Prove the relation v =   r , where
symbols have their usual meaning.
OR
In U.C.M. (Uniform Circular Motion),

ii.
….(1)
Angular
velocity
is
directed
perpendicular to plane, i.e., along

Z-axis. It is given by  =  k̂ ,

where, k̂ = unit vector along Z-axis.
prove the relation v   r , where symbols
have their usual meanings.
[Mar 16]
Ans: Analytical method:
i.
Consider a particle performing circular
motion in anticlockwise sense with centre
O and radius r as shown in the figure.
iii.


  r =  k̂  (r î cos t + r ĵ sin t)
[From equation (1)]
= r cos t ( k̂  î ) + r sin t ( k̂  ĵ)
= r ĵ cos t + r (– î ) sin t

 

   

 k i  j and k j   i 
Let,  = angular velocity of the particle

v = linear velocity of the particle


r = radius vector of the particle




  r = – r î sin t + r ĵ cos t


  r = r ( î sin t + ĵ cos t) .…(2)



Also v =
O
iii.
v = r ˆi sin t  ˆjcos t
v
r

s =   r
[½ Mark]
Dividing both side by t,



….(3)
From equation (2) and (3),


s  
=
 r
t t

dr
= r ˆi sin t  ˆjcos t
dt




[Diagram - ½ Mark]
Linear displacement in vector form is
given by,

[½ Mark]
ds 
But,
= v = linear velocity,
dt
In vector form, v =   r
ii.

s
 
lim
= lim
 r
t  0 t
t  0 t

Also lim
=
t  0 t
v = r

Taking limiting value in equation (1)
….(1)

1.4



v =   r
Uniform Circular Motion
Q.14. *Define uniform circular motion. OR
What is uniform circular motion?
Ans: i.
The motion of a body along the
circumference of a circle with constant
speed is called uniform circular motion.
5
Std. XII Sci.: Perfect Physics ‐ I ii.
iii.
iv.
In U.C.M, direction of velocity is along
the tangent drawn to the position of
particle on circumference of the circle.
Hence, direction of velocity goes on
changing continuously, however the
magnitude of velocity is constant.
Therefore, magnitude of angular
velocity is constant.
Examples of U.C.M:
a.
Motion of the earth around the sun.
b.
Motion of the moon around the earth.
c.
Revolution of electron around the
nucleus of atom.
Q.15. State the characteristics of uniform circular
motion.
Ans: Characteristics of U.C.M:
i.
It is a periodic motion with definite
period and frequency.
ii.
Speed of particle remains constant but
velocity changes continuously.
iii. It is an accelerated motion.
iv. Work done in one period of U.C.M is
zero.
Q.16. Define periodic motion. Why U.C.M is
called periodic motion?
Ans: i.
Definition:
A type of motion which is repeated after
equal interval of time is called periodic
motion.
ii.
The particle performing U.C.M repeats
its motion after equal intervals of time
on the same path. Hence, U.C.M is
called periodic motion.
Q.17. Define period of revolution of U.C.M. State
its unit and dimension. Derive an expression
for the period of revolution of a particle
performing uniform circular motion.
Ans: Definition:
The time taken by a particle performing
uniform circular motion to complete one
revolution is called as period of revolution.
2
It is denoted by T and is given by, T =
.

Unit: second in SI system.
Dimensions: [M0L0T1]
Expression for time period:
During period T, particle covers a distance
equal to circumference 2r of circle with
linear velocity v.
6




Time period = Distance covered in one revolution
Linear velocity
2r
v
But v = r
2r
T=
r
2
T=

T=
Q.18. What is frequency of revolution? Express
angular velocity in terms of frequency of
revolution.
The number of revolutions performed by
Ans: i.
a particle performing uniform circular
motion in unit time is called as
frequency of revolution.
ii.
Frequency of revolution (n) is the
reciprocal of period of revolution.
1
1

v
n= =
=
=
2
2r
T  2 
 
 
iii. Unit: hertz (Hz), c.p.s, r.p.s etc.
iv. Dimensions: [M0L0T1]
Angular velocity in terms of frequency of
revolution:
2
1
=
= 2 
T
T
1
=n
But
T

 = 2n
*Q.19.Define period and frequency of a particle
performing uniform circular motion. State
their SI units.
Ans: Refer Q.17 and Q.18
1.5
Acceleration in U.C.M (Radial acceleration)
Q.20. Define linear acceleration. Write down its
unit and dimensions.
Ans: i.
Definition:
The rate of change of linear velocity
with respect to time is called linear
acceleration.

It is denoted by a and is given by
ii.
iii.

dv
a =
dt
Unit: m/s2 in SI system and cm/s2 in
CGS system.
Dimensions: [M0L1T2]

Chapter 01: Circular Motion
Q.21. U.C.M is an accelerated motion. Justify this
statement.
Ans: i.
In U.C.M, the magnitude of linear
velocity (speed) remains constant but
the direction of linear velocity goes on
changing i.e., linear velocity changes.
ii.
The change in linear velocity is possible
only if the motion is accelerated. Hence,
U.C.M is an accelerated motion.
Q.22. *Obtain an expression for acceleration of
a particle performing uniform circular
motion.
OR
Define centripetal acceleration. Obtain an
expression for acceleration of a particle
performing U.C.M by analytical method.
Ans: Definition:
The acceleration of a particle performing
U.C.M which is directed towards the centre
and along the radius of circular path is called
as centripetal acceleration.
The centripetal acceleration is directed along
the radius and is also called radial
acceleration.
Expression for acceleration in U.C.M by
analytical method (Geometrical method):
i.
Consider a particle performing uniform
circular motion in a circle of centre O
and radius r with a uniform linear
velocity of magnitude v.
ii.
Let a particle travel a very short distance
s from A to B in a very short time
interval t.
iii. Let  be the angle described by the
radius vector OA in the time interval t
as shown in the figure.
D
C
v 
B 
O
iv.
v.

r
M
v
A
The velocities at A and B are directed
along the tangent.
Velocity at B is represented by BC
while the velocity at A is represented by
AM . [Assuming AM = BD]
Angle between BC and BD is equal to
 as they are perpendicular to OB and
OA respectively.
vii. Since BDC  OAB
DC
v
AB
AB

=

=
AO
BD
v
r
viii. For very small t, arc length s of
circular path between A and B can be
taken as AB
v s
v

=
or v = s
r
v
r
where, v = change in velocity
v
v s
= lim
Now, a = lim
t 0 t
t 0 r t
v
s
v2
v

a=
lim
=
v=
r
r
r t 0 t
As t  0, B approaches A and v
becomes perpendicular to the tangent
i.e., along the radius towards the centre.
ix. Also v = r
r 2 2
a=
= 2r
r
vi.
x.


In vector form, a =  2 r

Negative sign shows that direction of a

is opposite to the direction of r .
Also a =

v2 
r0 , where r0 is the unit
r
vector along the radius vector.
Q.23. Derive an expression for linear acceleration of
a particle performing U.C.M.
[Mar 98, 08]
Ans: Refer Q.22
Q.24. Derive an expression for centripetal
acceleration of a particle performing uniform
circular motion by using calculus method.
Ans: Expression for centripetal acceleration by
calculus method:
i.
Suppose a particle is performing U.C.M
in anticlockwise direction.
The co-ordinate axes are chosen as
shown in the figure.
Let,
A = initial position of the particle
which lies on positive X-axis
P = instantaneous position after time t
 = corresponding angular displacement
7
Std. XII Sci.: Perfect Physics ‐ I  = constant angular velocity
vi.
From equation (1) and (3),

ii.
a =  2 r
Negative sign shows that direction of
acceleration is opposite to the direction
of position vector.
Magnitude of centripetal acceleration is
given by,
a = 2r
v
v2
As  =

a=
r
r

r = î x + ĵy
vii.
where, î and ĵ are unit vectors along
X-axis and Y-axis respectively.
Y
v

y r
N
r
O
P(x, y)
Note:



x
A
X


  v =  k̂  ( – r î sin t + r ĵ cos t)


r = [r î cos t + r ĵ sin t]
....(1)
Velocity of the particle is given as rate
of change of position vector.

d
dr
v =
[r î cos t + r ĵ sin t]
=
dt
dt
d

d

= r  cos  t  î + r  sin  t  ĵ
 dt

 dt

v =  r î sin t + r ĵ cos t
v = r ( î sin t + ĵ cos t) ....(2)
Further,
instantaneous
linear
acceleration of the particle at instant t is
given by,


d
dv
[r ( î sin t + ĵcos t)]
=
a =
dt
dt
d

= r  ( ˆi sin  t  ˆjcos  t ) 
dt


d
d


= r  ( sin  t )iˆ  (cos  t )ˆj
dt
 dt

= r (  î cos t   ĵ sin t)
=  r2 ( î cos t + ĵ sin t)

8
= – r2 î cos t – r2 ĵ sin t

r = [r î cos  + r ĵ sin ]


v.



   

 k  i  j and k  j   i 



= – r2 sin t ĵ + r2 cos t (– î )
Also, x = r cos  and y = r sin 


2

= – r2 sin t ( k̂  î ) + r2 cos t ( k̂  ĵ)
But  = t

iv.

To show a =   v ,

iii.


r = instantaneous position vector at time t
From the figure,
a =   (r î cos t + r ĵsin t) ....(3)
= a [From equation (3)]
Q.25. Derive an expression for centripetal
acceleration of a particle performing
uniform circular motion. [Mar 02, Mar 06]
Ans: Refer Q. 24
*Q.26.Derive the relation between linear
acceleration and angular acceleration if a
particle performs U.C.M.
Ans: Relation between linear acceleration and
angular acceleration in U.C.M:
i.
Consider a particle performing U.C.M.
with constant angular velocity  with
path radius r.
ii.
Magnitude of linear acceleration is
given by,
v
a = lim
t 0 t
dv

a=
dt
iii. But, v = r
d
 d 
 dr 

a=
(r) = r 
 +  
dt
 dt 
 dt 
iv. Since, r = constant
dr

=0
dt

 d 

 dt 
a = r
Chapter 01: Circular Motion
d
=
dt
a = r
In vector form,
From figure,
Magnitude
But,












iii.
....(3)

ω  v is along the radius of the circle,
pointing towards the centre, hence it is

called radial acceleration a R .

iv.



aR =   v

Note:
1.
Resultant linear acceleration in different
cases
Situation

aR = 0, aT = 0
aR = 0, aT  0
aR  0, aT = 0
….(4)

  r is along the tangent of the
circumference of the circular path, hence

aR  0, aT  0
2.
it is called tangential acceleration a T .

v.






aT =   r
….(5)
From equation (3), (4) and (5),
3.
a = aR + aT




a

O


aR

aT
P

a and  – ½ Mark, Final expression for linear
acceleration – ½ Mark]
....(2)
a =  v +   r



Equation (2) becomes,

 – ½ Mark, Expression for relation between



dv
dr
d
= a,
= v and
= 
dt
dt
dt




But

v   r ,
resultant
terms of
[Mar 15]
Mark, Derivative of the relation between v and

ii.

[Expression for relation between v and  – ½
dv
d  
=
( r )
to t, we get
dt
dt


In circular motion, assuming
obtain an expression for the
acceleration of a particle in
tangential and radial component.
Ans: Refer Q.27
Since, v =   r ....(1)
Differentiating equation (1) with respect


dv
dr
d
=  
+
 r
dt
dt
dt

motion v = ω × r . Obtain an expression for
linear acceleration of the particle performing
non-uniform circular motion.
[Mar 14]
OR


a 2R  a T2
Q.28. For a particle performing uniform circular
Q.27. Define non-uniform circular motion. Derive
an expression for resultant acceleration in
non-uniform circular motion.
Ans: Non-uniform circular motion:
Circular motion with variable angular speed
is called as non-uniform circular motion.
Example: Motion of a body on vertical circle.
Expression for resultant acceleration in
non-U.C.M:
i.
linear
acceleration is given by | a | =


resultant

a =  r
This is required relation.

of
4.
Resultant
motion
Uniform linear
motion
Accelerated
linear motion
Uniform
circular motion
Non-uniform
circular motion
Resultant linear
acceleration
a =0
a = aT
a = aR
a =
a 2R  a T2
In non-uniform circular motion, aR is due to
change in direction of linear velocity, whereas
aT is due to change in magnitude of linear
velocity.
In uniform circular motion, particle has only
radial component aR due to change in the
direction of linear velocity. It is so because
 = constant
d
=
= 0 so, aT =   r = 0
dt
Since the magnitude of tangential velocity
does not change, there is no component of
acceleration along the tangent. This means the
acceleration must be perpendicular to the
tangent, i.e., along the radius of the circle.
9
Std. XII Sci.: Perfect Physics ‐ I ii.
Brain Teaser
Q.29. Read each statement below carefully and
state, with reasons, if it is true of false:
i.
The net acceleration of a particle in
circular motion is always along the
radius of the circle towards the centre.
ii.
The acceleration vector of a particle in
uniform circular motion averaged over
one cycle is a null vector.
(NCERT)
The statement is false. The acceleration
Ans: i.
of the particle performing circular
motion is along the radius only when
particle is moving with uniform speed.
ii.
The statement is true. When we consider
a complete cycle, for an acceleration at
any point of circular path, there is an
equal and opposite acceleration vector at
a point diameterically opposite to the
first point, resulting in a null net
acceleration vector.
iii.
iv.
Example: Electron revolves around the
nucleus of an atom. The necessary
centripetal force is provided by
electrostatic force of attraction between
positively
charged
nucleus
and
negatively charged electron.
Unit: N in SI system and dyne in CGS
system.
Dimensions: [M1L1T2]
Q.32. Derive the formula for centripetal force
experienced by a body in case of uniform
circular motion. Express the formula in
vector form.
Ans: Expression for centripetal force:
i.
Suppose a particle performs uniform
circular motion. It has an acceleration of
magnitude v2/r or 2r directed towards
the centre of the circle.

*Q.30.What is the difference between uniform
circular motion and non uniform circular
motion? Give examples.
Ans:
Sr.
U.C.M
No
i. Circular motion with
constant angular speed
is known as uniform
circular motion.
ii. For U.C.M,  = 0
iii. In U.C.M, work done
by tangential force is
zero.
iv. Example: Motion of the
earth around the sun.
1.6
Non-U.C.M
Circular motion with
variable angular speed
is called as non-uniform
circular motion.
For non-U.C.M,   0
In non-U.C.M, work
done by tangential
force is not zero.
Example: Motion of a
body on vertical circle.
Centripetal and centrifugal forces
Q.31.What is centripetal force? Write down its
unit and dimensions.
Force acting on a particle performing
Ans: i.
circular motion along the radius of
circle and directed towards the centre of
the circle is called centripetal force.
mv 2
It is given by FCP =
r
where, r = radius of circular path.
10
v
r
O

FCP
P

ii.
iii.
According to Newton’s second law of
motion, acceleration must be produced
by a force acting in the same direction.
If m is the mass of particle performing
U.C.M then the magnitude of centripetal
force is given by,
FCP = Mass of particle
 centripetal acceleration
FCP = maCP
iv.
But, aCP =

FCP =
v.

v2
= v = r2
r
mv 2
= mv = mr2
r
Also  = 2n
FCP = mr(2n)2 = 42n2mr
Centripetal force in vector form:

mv 2
FCP = 
r̂0 = mr2. r̂0
r
where r̂0 is unit vector in the direction

of radius vector r .
Chapter 01: Circular Motion
Q.33. State four examples where centripetal force
is experienced by the body.
A stone tied at the end of a string is
Ans: i.
revolved in a horizontal circle, the
tension in the string provides the
necessary centripetal force. It is given
by equation T = mr2.
ii.
The planets move around the sun in
elliptical orbits. The necessary centripetal
force is provided to the planet by the
gravitational force of attraction exerted
by the sun on the planet.
iii. A vehicle is moving along a horizontal
circular road with uniform speed. The
necessary centripetal force is provided
by frictional force between the ground
and the tyres of wheel.
iv. Satellite revolves round the earth in
circular orbit, necessary centripetal force
is provided by gravitational force of
attraction between the satellite and the
earth.
Q.34. *Define centripetal force. Give its any four
examples.
OR
Define centripetal force and give its any two
examples.
[Mar 11]
Ans: Refer Q.31 and Q.33
Q.35. Define and explain centrifugal force.
Ans: Definition:
The force acting on a particle performing
U.C.M which is along the radius and directed
away from centre of circle is called
centrifugal force.
Magnitude of centrifugal force is same as that of
centripetal force but acts in opposite direction.
Explanation:
i.
U.C.M is an accelerated motion. Thus,
a particle performing U.C.M is in an
accelerated (non-inertial) frame of
reference.
ii.
In non-inertial frame of reference, an
imaginary force or a fictitious or a
pseudo force is to be considered in order
to apply Newton’s laws of motion.
iii. The magnitude of this pseudo force is
same as that of centripetal force but its
direction is opposite to that of
centripetal force. Therefore, this pseudo
force is called centrifugal force.
iv. If m is the mass of a particle performing
U.C.M
then
centrifugal
force
experienced by the body is given by,
mv 2
r
But v = r
FCF = mr2
In vector form,
FCF =

v.



FCF = m2 r

mv 2
r̂0
FCF =
r

v
r
FCP
P
FCF

where r̂0 = unit vector in the direction of r .
Q.36. Explain applications of centrifugal force in
our daily life.
Ans: Applications:
i.
When a car in motion takes a sudden
turn towards left, passengers in car
experience an outward push to the right.
This is due to centrifugal force acting on
the passengers.
ii.
A bucket full of water is rotated in a
vertical circle at a particular speed, so
that water does not fall. This is because,
weight of water is balanced by
centrifugal force acting on it.
iii. The children sitting in merry-go-round
experience an outward pull as merry-goround rotates about vertical axis. This is
due to centrifugal force acting on the
children.
iv. A coin kept slightly away from the axis
of rotation of turn table moves away
from axis of rotation as the speed of
rotation of turn table increases. This is
due to centrifugal force acting on coin.
v.
The bulging of earth at equator and
flattening at the poles is due to
centrifugal force acting on it.
vi. Drier in washing machine consists of a
cylindrical vessel with perforated walls.
As the cylindrical vessel is rotated fast,
centrifugal force acts on wet clothes.
This centrifugal force, forces out water
through perforations thereby drying wet
cloths quickly.
vii. A centrifuge works on principle of
centrifugal force. In centrifuge, a test
tube containing liquid along with
suspended particles is whirled in a
horizontal circle. Denser particles are
acted upon by centrifugal force, hence
they get accumulated at bottom which is
on outside while rotating.
11
Std. XII Sci.: Perfect Physics ‐ I *Q.37. Define centrifugal force. Give its any four
examples.
Ans: Refer Q.35 and Q.36
iii.
Q.38. What is pseudo force? Why centrifugal
force is called pseudo force?
[Oct 99]
The force whose origin is not defined
Ans: i.
due to the known natural interactions is
called pseudo force.
ii.
The known interactions are gravitational
force, electromagnetic force, nuclear
force, frictional force, etc.
iii. It is directed opposite to the direction of
accelerated frame of reference.
iv. The centrifugal force is a fictitious force
which arises due to the acceleration of
the frame of reference. Therefore it is
called a pseudo force.
iv.
Q.39. Define:
i.
Inertial frame of reference
ii.
Non-inertial frame of reference
Ans: i.
Inertial frame of reference:
A frame of reference which is fixed or
moving with uniform velocity relative to
a fixed frame, is called as inertial frame
of reference.
Newton’s laws of motion can be directly
applied when an inertial frame of
reference is used, without inclusion of
pseudo force.
ii.
Non-inertial frame of reference:
A frame of reference which is moving
with an acceleration relative to a fixed
frame of reference is called non-inertial
frame of reference.
In non-inertial frame of reference,
Newton’s laws of motion can be applied
only by inclusion of some fictitious force
(pseudo force) acting on the bodies.
*Q.40.Distinguish between centripetal force and
centrifugal force.
[Mar 05, 09, 10, Feb 13 old course]
Ans: Difference between centripetal force and
centrifugal force:
Sr.
Centripetal force
No.
i. Centripetal force is
directed along the
radius towards the
centre of a circle.
ii.
12
It is a real force.
Centrifugal force
Centrifugal force is
directed along the
radius away from
the centre of a circle.
It is a pseudo force.
It is considered in
inertial frame of
reference.
In vector form, it is
given by

mv 2
r̂0
F = 
r
with usual notations.
It is considered in
non-inertial frame of
reference.
In vector form, it is
given by

mv 2
r̂0
F =+
r
with usual notations.
[Any two points  1 Mark each]
Note:
1.
If centripetal force, somehow vanishes at any
point on its path, the body will fly off tangentially
to its path at that point, due to inertia.
2.
The work done on the revolving particle by a
centripetal force is always zero, because the
directions of the displacement and force are
perpendicular to each other.
 

3.
4.
Thus, W = F. s = Fs cos
But  = 90
W = Fs cos 90 = 0
Any one of the real forces or their resultant
provide centripetal force.
Accelerated frame is used to attach the frame
of reference to the particle performing U.C.M.
#Q.41.Do centripetal and centrifugal force
constitute action reaction pair? Explain.
Centripetal and centrifugal force do not
Ans: i.
form action reaction pair.
ii.
The centripetal force is necessary for the
body to perform uniform circular
motion. It is real force in inertial frame
of reference. The centrifugal force is not
a real force. It is the force acting on the
same body in non-inertial frame of
reference to make Newton's laws of
motion true. As both centripetal and
centrifugal forces are acting on the same
body in different frame of reference,
action reaction pair is not possible.
1.7
Banking of roads
*Q.42. Derive the expression for maximum safety
speed with which vehicle should move along
a curve horizontal road. State the
significance of it.
Ans: Expression for maximum safety speed on
horizontal curve road:
i.
Consider a vehicle of mass m moving
with speed v along a horizontal curve of
radius r.
Chapter 01: Circular Motion
ii.
While taking a turn, vehicle performs
circular motion. Centripetal force is
provided by the frictional force between
tyres and road.
iii. Centripetal force is given by,
mv 2
FCP =
r
iv. Frictional force between road and tyres
of wheel is given by, Fs = N
where,  = coefficient of friction
between tyres of wheels and road.
N = normal reaction acting on vehicle in
upward direction.
But, N = mg

Fs = mg
v.
For safe turning of vehicle,
FCP = Fs
mv 2

= mg
r

v2 = rg

v = rg
vi. Maximum safe speed of the vehicle
without skiding is provided by
maximum centripetal force.

vmax = rg
This is maximum speed of vehicle.
Significance:
i.
The maximum safe speed of a vehicle
on a curve road depends upon friction
between tyres and road.
ii.
Friction depends on the nature of the
surface and presence of oil or water on
the road.
iii. If friction is not sufficient to provide
centripetal force, the vehicle is likely to
skid off the road.
Q.43. What force is exerted by a vehicle on the
road, when it is at the top of a convex
bridge of radius R?
Ans: Force exerted by the vehicle on the convex
bridge:
i.
Let,
m = mass of vehicle
R = radius of convex bridge
g = acceleration due to gravity
N
A
mg
ii.
In the figure, centripetal force acting on
the vehicle is given by,
mv 2
mg  N =
R
mv 2
N = mg 
R
where, N = normal reaction
iii. Normal reaction is balanced by the net
force exerted on the vehicle.
It is given by, N = F
mv 2

F = N = mg 
R
This is the required force on the vehicle.
Note: If bridge is concave then,
mv 2
F = mg +
R
*Q.44.What is banking of road?
[Mar 99, 12; Oct 01, 06]
Explain the necessity of banking of the
road.
[Mar 99, Oct 01, Oct 06]
Ans: Banking of road:
The process of raising outer edge of a road
over its inner edge through certain angle is
known as banking of road.
The angle made by the surface of road with
horizontal surface of road is called angle of
banking.
Angle of banking
h

Necessity of banking of the road:
i.
When a vehicle moves along horizontal
curved road, necessary centripetal force
is supplied by the force of friction
between the wheels of vehicle and
surface of road.
ii.
Frictional force is not enough and
unreliable every time as it changes when
road becomes oily or wet in rainy season.
iii. To increase the centripetal force the road
should be made rough. But it will cause
wear and tear of the tyres of the wheel.
iv. Thus, due to lack of centripetal force
vehicle tends to skid.
13
Std. XII Sci.: Perfect Physics ‐ I v.
vi.
When the road is banked, the horizontal
component of the normal reaction
provides the necessary centripetal force
required for circular motion of vehicle.
Thus, to provide the necessary
centripetal force at the curved road,
banking of road is necessary.
Q.45. *Show that the angle of banking is
independent of mass of vehicle.
[Mar 10, Oct 10]
OR
Obtain an expression for maximum safety
speed with which a vehicle can be safely
driven along curved banked road.
[Mar 10, 12; Oct 10]
Ans: Expression for angle of banking:
i.
The angle made by the surface of road
with horizontal surface of road is called
angle of banking. It is given by angle .
N
N cos 

G
F

A

ii.
iii.
14


vi.

vii.
C
h
F sin 
B
W = mg
AC
AB
BC
G
W
N
v.

N sin 
F cos 
iv.
:
:
:
:
:
:
inclined road surface
horizontal surface
height of road surface
centre of gravity of vehicle
(mg) weight of vehicle
normal reaction exerted on
vehicle
: angle of banking
Consider a vehicle of mass m moving
with speed v on a banked road banked at
an angle  as shown in the figure.
Let F be the frictional force between
tyres of the vehicle and road surface.
The forces acting on the vehicle are
a.
Weight mg acting vertically
downward.
b.
Normal reaction N in upward
direction through C.G.
The frictional force between tyres of
vehicle and road surface can be resolved
into,
Fcos - along horizontal direction
Fsin - along vertically downward
direction.
The normal reaction N can be resolved
into two components:
a.
Ncos along vertical direction
b.
Nsin along horizontal direction.
The component Ncos balances the
weight mg of vehicle and component
Fsin of frictional force.
Ncos = mg + Fsin
Ncos  Fsin = mg
….(1)
The horizontal component Nsin along
with the component Fcos of frictional
force provides necessary centripetal
mv 2
force
.
r
mv 2
….(2)
Nsin + Fcos =
r
Dividing equation (2) by (1),
N sin θ  Fcosθ
v2
=
….(3)
rg
N cosθ  Fsin θ
The magnitude of frictional force
depends on speed of vehicle for given
road surface and tyres of vehicle.
viii. Let vmax be the maximum speed of
vehicle, the frictional force produced at
this speed is given by,
Fmax = sN
….(4)
v 2max  N sin   Fmax cos  

= 
….(5)

rg
 N cos   Fmax sin  
Dividing the numerator and denominator
of equation (5) by Ncos , we have,
 N sin  Fmax cos  



v max = rg  N cos  N cos  
F
sin

N
cos



 max
 N cos  N cos  
2

F 

tan   max 

N
v2max = rg 
Fmax tan  
1 

N



   tan  
v2max = rg  s

1  s tan  
 Fmax  s N 

vmax =
   tan  
rg  s

1  s tan  
….(6)
Chapter 01: Circular Motion
ix.
For a curved horizontal road,  = 0,
hence equation (6) becomes,
vmax = μ s rg
….(7)
x.
Comparing equation (6) and (7) it is
concluded that maximum safe speed of
vehicle on a banked road is greater than
that of curved horizontal road/level road.
If s = 0, then equation (7) becomes,
xi.
vmax = vo =

xii.
Q.46. Draw a diagram showing all components of
forces acting on a vehicle moving on a
curved banked road. Write the necessary
equation for maximum safety, speed and
state the significance of each term involved
in it.
[Oct 14]
Ans:
N
N cos 

N sin 

G
F
A
AC
AB
BC
G
C
h
F sin 
B
:
:
:
:
i.
W = mg
: (mg) weight of vehicle
: normal reaction exerted on
vehicle
: angle of banking
[1 Mark]
Equation for maximum safety speed for
the vehicle moving on the curved
banked road is,
vmax =
xiii. Equation (9) represents angle of banking
of a banked road. Formula for angle of
banking does not involve mass of
vehicle m. Thus angle of banking is
independent of mass of the vehicle.


 0  tan θ 
rg 

1  0 tan θ 
vo = rg tan θ
….(8)
At this speed, the frictional force is not
needed to provide necessary centripetal
force. There will be a little wear and tear
of tyres, if vehicle is driven at this speed
on banked road. vo is called as optimum
speed.
From equation (8) we can write,
v2
tan  = o
rg
 vo2 
1 

 = tan
....(9)
 rg 
F cos 
W
N
inclined road surface
horizontal surface
height of road surface
centre of gravity of vehicle
ii.
a.
b.
   tan  
rg  s

1  s tan  
[1 Mark]
where, r is radius of curved road.
s is coefficient of friction
between road and tyres,
 is angle of banking.
Significance of the terms involved:
The maximum safety speed of a vehicle
on a curved road depends upon friction
between tyres and roads.
It also depends on the angle through
which road is banked. Also absence of
term ‘m’ indicates, it is independent of
[1 Mark]
mass of the vehicle.
Q.47. Draw a neat labelled diagram showing the
various forces and their components acting
on a vehicle moving along curved banked
road.
[July 16]
Ans: Refer Q.45 (only diagram)
[Diagram  1 Mark, Labelling  1 Mark]
Q.48. State the factors which affect the angle of
banking.
Ans: Factors affecting angle of banking:
i.
Speed of vehicle: Angle of banking ()
increases with maximum speed of
vehicle.
ii.
Radius of path: Angle of banking ()
decreases with increase in radius of the
path.
iii. Acceleration due to gravity: Angle of
banking () decreases with increase in
the value of ‘g’.
Q.49. Define angle of banking. Obtain an
expression for angle of banking of a curved
road and show that angle of banking is
independent of the mass of the vehicle.
[Mar 97, Mar 03, Oct 03]
Ans: Refer Q.45
15
Std. XII Sci.: Perfect Physics ‐ I h :
v :
r :
 :
mg :
#Q.50. The curved horizontal road is banked at
angle . What will happen for vehicle
moving along this road if,
 < 
ii.  > ?
i.
where  is angle of banking for given road.
Ans: i.
If  < , the necessary centripetal force
will not be provided and the vehicle will
tend to skid outward, up the inclined
road surface.
ii.
If  > , the centripetal force provided
will be more than needed and the
vehicle will tend to skid down the
banked road.
*Q.51.Define conical pendulum. Obtain an
expression for the angle made by the string
of conical pendulum with vertical. Hence
deduce the expression for linear speed of
bob of the conical pendulum.
Ans: Definition:
A simple pendulum, which is given such a
motion that bob describes a horizontal circle
and the string describes a cone is called a
conical pendulum.
Expression for angle made by string with
vertical:
i.
Consider a bob of mass m tied to one
end of a string of length ‘l’ and other
end is fixed to rigid support.
ii.
Let the bob be displaced from its mean
position and whirled around a
horizontal circle of radius ‘r’ with
constant angular velocity , then the
bob performs U.C.M.
iii. During the motion, string is inclined to
the vertical at an angle  as shown in
the figure.
S

l
h
T

O
r
T sin
T cos
P
mg
S : rigid support
T : tension in the string
l : length of string
16
iv.
v.
vi.

height of support from bob
velocity of bob
radius of horizontal circle
semi vertical angle
weight of bob
In the displaced position P, there are two
forces acting on the bob.
a.
The weight mg acting vertically
downwards.
b.
The tension T acting upward
along the string.
The tension (T) acting in the string can
be resolved into two components:
a.
T cos acting vertically upwards.
b.
T sin acting horizontally towards
centre of the circle.
Vertical component T cos balances the
weight and horizontal component T sin
provides the necessary centripetal force.
T cos  = mg
….(1)
2
T sin  = mv = mr2 .…(2)
r
vii.
Dividing equation (2) by (1),
v2
tan  =
….(3)
rg
Therefore, the angle made by the string
 v2 

 rg 
with the vertical is  = tan1 

Also, from equation (3),
v2 = rg tan 
v = rg tan 
This is the expression for the linear speed of
the bob of a conical pendulum.
Q.52.*Define period of conical pendulum and
obtain an expression for its time period.
[Oct 08, 09]
OR
Derive an expression for period of a conical
pendulum.
[Mar 08]
Ans: Definition:
Time taken by the bob of a conical pendulum
to complete one horizontal circle is called
time period of conical pendulum.
Expression for time period of conical
pendulum:
(Refer Q. 51 with diagram)
Chapter 01: Circular Motion
v=

=
i.

ii.

rg tan 
g tan 
[ v = r]
r
r
In  SOP, tan  =
h
From equation (i),
gr
 =
rh
….(1)
g
h
If the period of conical pendulum is T
then,
2
=
T
2
g
=
T
h
 =
h
g

T = 2
iii.
Also, In  SOP,
h
cos  =
l
h = l cos 
where, l = length of the string
 = angle of inclination
Substituting h in equation (2),
l cos 
T = 2
g
This is required expression for time
period of conical pendulum.

….(2)
Q.53. Draw a neat labelled diagram of conical
pendulum. State the expression for its
periodic time in terms of length. [Oct 15]
Ans: Refer Q.51(diagram only.)
Refer Q.52(Expression for time period of
conical pendulum)
[Diagram with labelling  1 Mark, Formula
for time period in terms of length  1 Mark]
Q.54. Discuss the factors on which time period of
conical pendulum depends.
Ans: Time period of conical pendulum is given by,
l cos 
.…(1)
T = 2
g
where, l = length of the string
g = acceleration due to gravity
 = angle of inclination
From equation (1), it is observed that period of
conical pendulum depends on following
factors.
i.
Length of pendulum (l): Time period of
conical pendulum increases with increase
in length of pendulum. i.e., T  l
ii.
Acceleration due to gravity (g): Time
period of conical pendulum decreases
1
with increase in g. i.e., T 
g
Angle of inclination (): As  increases,
cos  decreases, hence, time period of
conical pendulum decreases with increase
in . (For 0 <  < ) i.e., T  cos 
iii.
Q.55. Find an expression for tension in the string
of a conical pendulum.
Ans: Expression for tension in the string of a
conical pendulum:
i.
Consider a bob of mass ‘m’ tied to one
end of a string of length ‘l’ and other
end fixed to rigid support (S).
ii.
Let the bob be displaced from its mean
position and whirled around a
horizontal circle of radius ‘r’ with
constant angular velocity ‘’.
iii. During the motion, string is inclined to
the vertical at an angle  as shown in
the figure.
S
S : rigid support
T : tension in the
 l
string
l : length of string
h : height of support
from bob
h
T
T cos v : velocity of bob

r : radius
of
r
horizontal
circle
 O T sin P
 : semi vertical angle
mg
mg : weight of bob
iv.
In the displaced position P, there are two
forces acting on the bob:
a.
The weight mg acting vertically
downwards and
b.
The tension T acting upwards
along the string.
17
Std. XII Sci.: Perfect Physics ‐ I v.
vi.

The tension (T) acting in the string can
be resolved into two components:
a.
T cos  acting vertically upwards
b.
T sin  acting horizontally towards
centre of the circle
Vertical component T cos  balances the
weight of the bob and horizontal
component
Tsin 
provides the
necessary centripetal force.
T cos  = mg
….(1)
2
T sin = mv
vii.
1.8
.…(2)
r
Squaring and adding equations (1) and
(2),
 mv 2 
T cos  + T sin  = (mg) + 

 r 
2
2
2
2
2
2
2
 mv 2 
T (cos  + sin ) = (mg) + 

 r 
2
2
2
2
2
 mv 2 
T2 = (mg)2 + 
 ….(3)
 r 
[ sin2  + cos2 = 1]
viii. Dividing equation (2) by (1),
v2
….(4)
tan  =
rg
r
From figure, tan  =
h

r
v
r 2g
=


v2 =
h
rg
h
ix. From equation (3) and (4),
 m  r 2g 
T = (mg) +  

 r  h 
2
 r 
2
2 1    
T = (mg)   h  


2

2
2
2
r
T = mg 1   
h
This is required expression for tension in the
string.

#Q.56. Is there any limitation on semivertical
angle in conical pendulum?
Ans: i.
For a conical pendulum,
Period T  cos 
1
Tension F 
cos 
Speed v  tan 
18
ii.
With increase in angle , cos 
decreases and tan  increases. For
 = 90, T = 0, F =  and v =  which
cannot be possible.
Also,  depends upon breaking tension
of string, and a body tied to a string
cannot be resolved in a horizontal circle
such that the string is horizontal. Hence,
there is limitation of semivertical angle
in conical pendulum.
Vertical circular motion due to earth’s
gravitation
*Q.57.What is vertical circular motion? Show that
motion of an object revolving in vertical circle
is non uniform circular motion.
A body revolves in a vertical circle such
Ans: i.
that its motion at different points is
different then the motion of the body is
said to be vertical circular motion.
ii.
Consider an object of mass m tied at one
end of an inextensible string and whirled
in a vertical circle of radius r.
iii. Due to influence of earth’s gravitational
field, velocity and tension of the body
vary in magnitude from maximum at
bottom (lowest) point to minimum at the
top (highest) point.
iv. Hence motion of body in vertical circle
is non uniform circular motion.
1.9
Equation for velocity and energy at different
positions in vertical circular motion
*Q.58.Obtain expressions for tension at highest
position, midway position and bottom
position for an object revolving in a vertical
circle.
Ans: Expression for tension in V.C motion:
i.
Let a body of mass m be tied at the end
of a massless inextensible string and
whirled in a vertical circle of radius r in
anticlockwise direction.
ii.
At any point P the forces acting on it are:
a.
Tension T along PO
b.
Weight mg along vertically
downward direction.
iii. The weight mg can be resolved into two
rectangular components:
a.
mg cos  acting along OP.
b.
mg sin  acting tangentially in a
direction opposite to velocity at
that point.
Chapter 01: Circular Motion
vH
H
TM  mg cos 90 =
TH
T
H
N
vM
TTM
O
M

TTLL
T
L
mg sin 

M
vP
P

vL mg
mg cos 
iv.
To complete vertical circular path, the
necessary centripetal force is provided
by the difference in the tension T and
mg cos .

T  mg cos  =
v.



vi.



vii.

mv 2p
….(1)
r
where, vp = velocity at point P.
When body is at highest position,
tension in the string = TH and  = .
Using equation (1),
mv 2H
TH  mg cos  =
r
where vH = velocity at highest point
mv 2H
[ cos  = 1]
TH  mg (1) =
r
mv 2H
TH + mg =
r
mv 2H
 mg
.…(2)
TH =
r
When the body is at bottom position:
 = 0
cos  = 1
From equation (1),
mv 2L
TL  mg cos 0 =
r
where TL = tension at lowest point
vL = velocity at lowest point
mv 2L
TL  mg =
r
2
mv L
+ mg
.…(3)
TL =
r
When the body is at midway position,
(M or N)
 = 90
cos 90 = 0
If tension at horizontal position is TM
then
TM  0 =
mv 2M
r
[From (1)]
mv 2M
r
mv 2M
….(4)
r
From equation (2), (3) and (4) it is observed
that tension is maximum at lowest position
and minimum at highest position.

TM =
Q.59. *Derive expressions for linear velocity at
lowest point, midway and top position for a
particle revolving in a vertical circle if it
has to just complete circular motion
without string slackening at top.
OR
Obtain an expression for minimum velocity
of a body at different positions, so that it
just performs vertical circular motion.
Ans: Expression for velocity in vertical circular
motion:
i.
Consider a body of mass m which is tied to
one end of a string and moves in a vertical
circle of radius r as shown in the figure.
H
vH
2r
N
vN
ii.
iii.


mg
TH
O r
TL
L
vM
M
vL
Let,
vH = velocity at highest position
vL = velocity at lowest position
vM = velocity at midway position
The velocity at any point on the circle is
tangential to the circular path.
Velocity at highest position:
Tension in the string at highest position
mv 2H
TH =
 mg
.…(1)
r
In order to continue the circular motion,
TH  0
TH = 0
Equation (i) becomes
mv 2H
 mg = 0
r

mv 2H
= mg
r
19
Std. XII Sci.: Perfect Physics ‐ I 
v 2H = rg
vH =
rg





Velocity at lowest position:
According to law of conservation of
energy,
Total energy at L = Total energy at H
(K.E)L + (P.E)L = (K.E)H + (P.E)H
.…(3)
At lowest point, P.E = 0
1
K.E = mv 2L
2
At highest point,
1
P.E = mg (2r) and K.E = mv 2H
2
From equation (3)
1
1
mv 2L  0  mv 2H  mg(2r)
2
2
1
1
1
mv 2L  mv 2H  (4 mgr)
2
2
2
1
1
mv 2L  m  v 2H  4gr 
2
2
2
2
v L  v H  4gr
.…(4)
To complete vertical circular motion,
vH = rg
v 2L 
 rg 

vM =
….(2)
Equation (2) represents minimum
velocity at highest point so that string is
not slackened.
To continue vertical circular motion,
vH  rg (at top position).
iv.

5
1
mgr  mgr = mv 2M
2
2
3
1
mgr = mv 2M
2
2
2
v M = 3rg

2

Q.60. Derive an expression for the minimum
velocity of a body at any point in vertical
circle so that it can perform vertical
circular motion.
Ans: Expression for minimum velocity at any
point in V.C. motion:
i.
Consider a body of mass ‘m’,
performing vertical circular motion of
path radius r. P is any point on the circle
as shown in the figure. We have to find
velocity at P.
ii.
Let vP = velocity at P
H
N
v.
20
M
P
L
…(5)
1
mv 2M and
2
P.E = mgr
Total energy at L = Total energy at M
(P.E)L + (K.E)L = (P.E)M + (K.E)M
1
1
m  5rg = mgr + mv 2M
2
2
5mgr
1
= mgr + mv 2M
2
2
r
vP
At midway position, K.E =


h
Equation (5) represents minimum
velocity at the lowest point, so that body
can safely travel along vertical circle.
Velocity at midway position:
0+
O
rh
K
 4rg = rg + 4 rg
5rg
….(6)
Equation (6) represents minimum
velocity of a body at midway position, so
that it can safely travel along vertical
circle. To continue vertical circular
motion, vM = 3rg .
v 2L = 5 rg
vL =
3rg
iii.
In OKP,
OK = r cos 
h = r  OK
= r  r cos 
h = r (1  cos )
From principle of conservation of
energy,
Total energy at L = Total energy at P
(P.E)L + (K.E)L = (P.E)P + (K.E)P
1
1
0 + mv 2L = mgh + mv 2P
2
2
But min. vL =
5rg
Chapter 01: Circular Motion



1
1
 5mgr = mgr (1  cos ) + mv 2P
2
2
1
1
mv 2P   5 mrg  mrg (1  cos )
2
2
5

= mrg   1  cos  
2

1 2
rg(5  2  2cos )
vP =
2
2

v 2P = (3 + 2 cos ) rg

vP =
vH
(3  2cos )rg
H
TL
L
When the particle is at midway point
in V.C.M:
P.E = mgh = mgr
[ h = r]
1
1
3
m v 2M = m  3rg = mgr
2
2
2
Total energy at M = K.E + P.E
3
=
mgr + mgr
2
5
mgr
....(3)
(T.E)M =
2
Equation (3) represents total energy of
particle at midway position in V.C.M
From equation (i), (ii) and (iii),
it is observed that total energy at any
5
point in V.C.M is mgr, i.e., constant.
2
Hence, total energy of a particle
performing vertical circular motion
remains constant.
K.E =

v.
Q.62. *A particle of mass m, just completes the
vertical circular motion. Derive the
expression for the difference in tensions at
the highest and the lowest points. [Mar 13]
OR
Show that for a body performing V.C.M.,
difference in tension at the lowest and
highest point on vertical circle is 6mg.
Ans:
H
vH
vM
TH
TH
O
2r
r
iii.

iv.
Q.61. *Obtain expression for energy at different
positions in the vertical circular motion.
Hence show that total energy in vertical
circular motion is constant.
OR
Show that total energy of a body
performing vertical circular motion is
conserved.
[Mar 11]
Ans: Expression for energy at different points in
V.C.M:
i.
Consider a particle of mass m revolving
in a vertical circle of radius r in
anticlockwise direction.
ii.
When the particle is at highest point H:
1
1
K.E = mvH2 = m  rg [ vH = rg ]
2
2
P.E = mg(2r) = 2mgr
Total energy at highest point
T.E = K.E + P.E
1
5

T.E = mgr + 2mgr = mgr
2
2
5

(T.E)H = mgr
....(1)
2
Equation (1) represents energy of
particle at the highest point in V.C.M.
1
1
5
mv 2L = m  5rg = mgr
2
2
2
Total energy at lowest point = K.E + P.E
5
5
=
mgr + 0 = mgr
2
2
5
....(2)
(T.E)L = mgr
2
Equation (2) represents energy of
particle at lowest point in V.C.M
K.E =
O
M
TL

vL
When particle is at lowest point L:
P.E = 0
[ At lowest point, h = 0]
r
L
i.
vL
[½ Mark]
Suppose a body of mass ‘m’ performs
V.C.M on a circle of radius r as shown
in the figure.
21
Std. XII Sci.: Perfect Physics ‐ I ii.
iii.

Let,
TL = tension at the lowest point
TH = tension at the highest point
vL = velocity at the lowest point
vH = velocity at the highest point
[½ Mark]
At lowest point L,
mv 2L
+ mg
TL =
[½ Mark]
r
At highest point H,
mv 2H
 mg
TH =
[½ Mark]
r
 mv 2H

mv 2L
+ mg  
 mg 
TL  TH =
r
 r

m 2
 vL  v2H  + 2mg
r
m 2
v L  v 2H  + 2mg ....(1)
TL  TH =

r
By law of conservation of energy,
(P.E + K.E) at L = (P.E + K.E) at H
1
1
0 + mv 2L = mg.2r + mv 2H
2
2
1
m  v 2L  v 2H  = mg.2r
2
v 2L  v 2H = 4gr
....(2)
From equation (1) and (2),
m
(4gr) + 2mg
[½ Mark]
TL  TH =
r
= 4mg + 2mg
TL  TH = 6mg
[½ Mark]
ii.
by  = 0 t + 1 t2
2
It is analogous to the kinematic equation
of linear motion,
s = u t + 1 at2
2
iii.
=

iv.




1.10 Kinematical equation for circular motion in
analogy with linear motion
*Q.63.State the kinematical equations for circular
motion in analogy with linear motion.
Ans: The kinematical equations of circular motion
are analogue to the equations of linear motion
which is given below:
i.
Angular velocity of a particle at any
time t is given by,
 = 0 +  t,
where,
0 = initial angular velocity of the particle
 = angular acceleration of the particle
It is analogue to the kinematical
equation of linear motion,
v = u + at
where, u = initial velocity of particle
v = final velocity of particle
22
a = constant acceleration of particle
The angular displacement of a particle
in rotational motion after time t is given
where,
s = linear displacement
u = initial velocity
a = constant acceleration
t = time interval.
The angular velocity of rotating particle
after certain angular displacement  is
given by,
2 = 02 + 2  
It is analogous to the kinematic equation
of linear motion
v2 = u2 + 2as,
where,
u = initial velocity
v = final velocity
a = constant acceleration
s = linear displacement
Summary
1.
Motion of a particle along a circumference of
a circle is called circular motion.
2.
Angle described by a radius vector in a given
time at the centre of circle to other position is
called as angular displacement.
3.
Infinitesimal small angular displacement is a
vector quantity. Finite angular displacement is
a pseudo vector (scalar), as for large values of
, the commutative law of vector addition is
not valid.
4.
The rate of change of angular displacement
w.r.t time is called angular velocity.


d
.
It is given by  =
dt
Angular velocity relates with linear velocity



by the relation, v =   r or v = r.
5.
The rate of change of angular velocity w.r.t
time is called as angular acceleration.
d   0

.
It is given by relation,  =
dt
t
Chapter 01: Circular Motion
6.
non-U.C.M is given by, a =
7.
8.
9.
10.
11.
12.
13.
14.
15.
Case 1: At highest point,  = 180
There are two types of acceleration aR (radial)
and aT (tangential) in non U.C.M.
dv
= r,
Formula for aR = 2r and aT =
dt
resultant acceleration of a particle in
a 2R  a T2 .
Centripetal force is directed towards the centre
along the radius and makes the particle to
move along the circle.
Centrifugal force is directed away from the
centre along the radius and has the same
magnitude as that of centripetal force.
vH =
rg
Case 2: At lowest point,  = 0
vL =
5rg
Case 3: At horizontal point,  = 90

vM =
16.
Energy of a particle at any point in vertical
5
circular motion is given by T.E = mgr
2
3rg
Formulae
The process in which the outer edge of the
road is made slightly higher than the inner
edge is called as banking of roads.
rg
.
The formula for vmax = rg and vmin =

On frictional surface, a body performing
circular motion, the centripetal force is
provided by the force of friction given by,
Fs = mg.
1.
In U.C.M angular velocity:

v
i.
=
ii.
=
r
t
2
iii.  = 2n
iv.  =
T
2.
Angular displacement:
The angle of banking () is given by,
v2
tan  =
.
rg
3.
The period of revolution of the conical
pendulum is given by,
r
l cos 
= 2
T = 2
g tan 
g
4.
The linear speed of the bob of conical
pendulum v = rg tan 
Tension at any point P in vertical circular
motion is given by,
mv 2P
+ mg cos 
T=
r
Where, vP = velocity at any point in V.C.M
Case 1: At highest point,  = 180
mv 2H
so, TH =
 mg
r
Case 2: At lowest point,  = 0
mv 2L
so, TL =
+ mg
r
Velocity at any arbitrary point is given by,
v=
rg  3  2cos  
i.
 = t
Angular acceleration:
  1
i.
= 2
ii.
t
Linear velocity:
i.
v = r
2t
T
=
ii.
=
ii.
2
(n2  n1)
t
v = 2nr
5.
Centripetal acceleration or radial
v2
= 2r
acceleration: a =
r
6.
Tangential acceleration: a T =   r
7.
Centripetal force:

mv 2
r
i.
FCF =
iii.
FCP = 42 mrn2
v.


ii.
FCP = mr2
iv.
FCP =
42 mr
T2
2
FCP = mg = m r
 v2 

 rg 
8.
Inclination of banked road:  = tan1 
9.
Maximum velocity of vehicle to avoid
skidding on a curve unbanked road:
vmax =
rg
23
Std. XII Sci.: Perfect Physics ‐ I 10.
Maximum safe velocity on banked road:
i.
vmax =
vmax =
rg tan  (in absence of friction)
11.
Height of inclined road: h = l sin 
12.
Conical Pendulum:
i.
Angular velocity of the bob of conical
pendulum,
g
g
g tan 
= r
=
=
r
l cos
h
ii.
Linear velocity of the bob of conical
pendulum v = rg tan 
iii.
Period of conical pendulum
l cos 
a. T = 2
g
b. T = 2
l
g
c.
r
g tan 
T = 2
( is small)
13.
Minimum velocity at lowest point to
complete V.C.M: vL = 5rg
14.
Minimum velocity at highest point to
complete V.C.M: vH = rg
15.
Minimum velocity at midway point to
complete in V.C.M: vM = 3rg
16.
Tension at highest point in V.C.M:
mv 2H
 mg
TH =
r
17.
Tension at midway point in V.C.M:
mv 2m
TM =
r
18.
Tension at lowest point in V.C.M:
mv 2L
+ mg
TL =
r
19.
Total energy at any point in V.C.M:
5
T.E = mgr
2
24
Kinematic equations of linear motion:
1 2
at
i.
v = u + at
ii.
s = ut +
2
iii. v2 = u2 + 2as
21.
Kinematic equations of rotational motion:
1
i.
 = 0 + t
ii.
 = 0t + t2
2
2
2
iii.  = 0 + 2 
 μ  tan θ 
rg  s

1  μ s tan θ 
(in presence of friction)
ii.
20.
Solved Examples
Example 1
What is the angular displacement of second hand
in 5 seconds?
Solution:
Given:
T = 60 s, t = 5 s
To find:
Angular displacement ()
2t
Formula:
=
T
Calculation: From formula,
2  3.142  5
 =
60

 = 0.5237 rad
Ans: The angular displacement of second hand in
5 seconds is 0.5237 rad.
Example 2
Calculate the angular velocity of earth due to its
spin motion.
Solution:
T = 24 hour = 24  3600 s
Given:
To find:
Angular velocity ()
2
Formula:
=
T
Calculation: From formula,
2
=
24  3600
2  3.142 
=
24  3600
  = 7.27  105 rad/s
Ans: The angular velocity of earth due to its spin
motion is 7.27  105 rad/s.
Example 3
What is the angular speed of the minute hand of
a clock? If the minute hand is 5 cm long. What is
the linear speed of its tip ?
[Oct 04]
Chapter 01: Circular Motion
Solution:
Given:
Length of minute hand, r = 5 cm,
T = 60 min = 60  60 = 3600 s
To find:
i.
Angular speed ()
ii. Linear speed (v)
2
Formulae: i.
=
ii.
v = r
T
Calculation: From formula (i),
 =
2  3.142
3600
 = 1.74  103 rad/s
From formula (ii),
v = r = 5  102  1.74  103

v = 8.7  105 m/s
Ans: The minute hand of the clock has angular
speed 1.74  103 rad/s and linear speed
8.7  105 m/s.

*Example 4
Calculate the angular velocity and linear velocity
of a tip of minute hand of length 10 cm.
Solution:
Given:
T = 60 min. = 60  60 s = 3600 s,
l = 10 cm = 0.1 m
To find:
i.
Angular velocity ()
ii. Linear velocity (v)
2
Formulae: i.
=
ii.
v = r
T
Calculation: From formula (i),
2 2  3.142
 =
=
T
3600

 = 1.744  103 rad/s
From formula (ii),
v = r = 0.1  1.745  103

v = 1.745  104 m/s
Ans: The tip of the minute hand has angular
velocity 1.744  103 rad/s and linear velocity
1.745  104 m/s.
Example 5
An aircraft takes a turn along a circular path of
radius 1500 m. If the linear speed of the aircraft
is 300 m/s, find its angular speed and time taken
th
1
by it to complete   of circular path.
5
Solution:
Given:
r = 1500 m, v = 300 m/s
To find:
i. Angular speed ()
th
ii. Time taken for   of circular
5
path (t)

Formulae: i. v = r
ii.
=
t
Calculation: From formula (i),
v
300
=
=
r
1500
1
300

=
=
= 0.2 rad/s
1500
5
The angular displacement () of the
1
1
aircraft to complete  
5
th
 
circular path is  =
of the
2
rad
5
From formula (ii),

2 / 5
=
t =
0.2


t=
2  3.142
= 6.284 s
5  0.2
Ans: The angular speed of the aircraft is 0.2 rad/s
1
th
and time taken by it to complete   of
5
circular path is 6.284 s.
*Example 6
Propeller blades in aeroplane are 2 m long
i.
When propeller is rotating at 1800 rev/min,
compute the tangential velocity of tip of the
blade.
ii.
What is the tangential velocity at a point on
blade midway between tip and axis?
Solution:
1800
60
Given:
l = 2 m, n = 1800 r.p.m =
To find:
= 30 r.p.s.
r1 = l = 2m, r2 = l/2 = 1m
i.
Tangential velocity of tip of the
blade ( vT1 )
ii.
Tangential velocity at a point on
blade midway between tip and
axis ( v T2 )
Formula:
v = 2nr
Calculation: i.
From formula,
Tangential velocity of the tip of
blade,
vT1 = 2nr1 = 2  3.14  30  2

vT1 = 376.8 m/s
25
Std. XII Sci.: Perfect Physics ‐ I Ans: i.
ii.
ii.
Tangential velocity at a point
midway between tip and axis,
v T2 = 2nr2= 2  3.14  30  1

v T2 = 188.4 m/s
The tangential velocity of tip of the
blade is 376.8 m/s.
The tangential velocity at a point on blade
midway between tip and axis is 188.4 m/s.
Example 7
A particle, initially at rest, performs circular
motion with uniform angular acceleration
0.2 rad/s2. What speed will it attain in
10 seconds?
Solution:
Given:
1 = 0,  = 0.2 rad/s2, t = 10 s
To find:
Speed (2)
Formula:
2 = 1 + t
Calculation: From formula,
2 = 0 + (0.2)  10

2 = 2 rad/s
Ans: Speed attained by the particle in 10 seconds is
2 rad/s.
Example 8
The frequency of a particle performing circular
motion changes from 60 r.p.m to 180 r.p.m in
20 second. Calculate the angular acceleration.
[Oct 98]
Solution:
60
= 1 rev/s,
Given:
n1 = 60 r.p.m =
60
180
= 3 rev/s,
n2 = 180 r.p.m =
60
t = 20 s
To find:
Angular acceleration ()
  1
Formula:
= 2
t
Calculation: From formula,
2n 2  2n1 2 (3  1)
=
=
t
20
2  3.142  2
=
20
3.142

=
5

 = 0.6284 rad/s2
Ans: Angular acceleration of the particle is
0.6284 rad/s2.
26
Example 9
The spin dryer of a washing machine rotating at
15 r.p.s. slows down to 5 r.p.s. after making
50 revolutions. Find its angular acceleration.
[Mar 15]
Solution:
Given:
n0 = 15 r.p.s., n = 5 r.p.s.,
No. of revolutions = 50
To find:
Angular acceleration ()
 = 2n
Formulae: i.
ii.
2 = 02  2
[½ Mark]
Calculation: Using formula (i),
 = 2  5 = 10 
0 = 2  5 = 30 
In 1 revolution, angular displacement
= 2

in
50
revolutions,
angular
displacement =  = 100 
Using formula (ii),
(10)2 = (30 )2 + 2(100 )
[½ Mark]
900 2  100 2
=
200 

 =  4 rad/s2 =  12.56 rad/s2
Ans: Angular acceleration of spin dryer is
 12.56 rad/s2.
[1 Mark]
*Example 10
The length of hour hand of a wrist watch is
1.5 cm. Find magnitude of
i.
angular velocity
ii.
linear velocity
iii. angular acceleration
iv. radial acceleration
v.
tangential acceleration
vi. linear acceleration of a particle on tip of
hour hand.
Solution:
Given:
T = 12  60  60 = 43200 s,
r = 1.5 cm = 1.5 102 m
To find:
i.
Angular velocity ()
ii.
Linear velocity (v)
iii. Angular acceleration ()
iv. Radial acceleration (aR)
v.
Tangential acceleration (aT)
vi. Linear acceleration (a)
Formulae:
i.
=
2π
T
ii.
v = r
Chapter 01: Circular Motion
iii.
=
d
dt
iv.
aR = v
v.
aT = r
vi. a = aR + aT
Calculation:
i.
From formula (i),
2  3.142
=
43200

 = 1.454  104 rad/s
ii.
From formula (ii),
v = 1.5  102  1.46  104

v = 2.19  106 m/s
iii
Since angular velocity of hour hand is
constant.

=0
iv. From formula (iv),
aR = 2.182  106 1.454  104

aR = 3.175  1010 m/s2
v.
From formula (v),
aT = 0  1.5  102

aT = 0
vi. From formula (vi),
a = 3.175  1010 + 0

a = 3.175  1010 m/s2
Ans: i.
The hour hand of the wrist watch has
angular velocity 1.454  104 rad/s.
ii.
The hour hand of the wrist watch has
linear velocity 2.19  106 m/s.
iii. The hour hand of the wrist watch has
angular acceleration 0.
iv. The hour hand of the wrist watch has
radial acceleration 3.175  1010 m/s2.
v.
The hour hand of the wrist watch has
tangential acceleration 0.
vi. Linear acceleration of the particle on tip
of hour hand is 3.175  1010 m/s2.
*Example 11
A block of mass 1 kg is released from P on a
frictionless track which ends in quarter circular
track of radius 2 m at the bottom as shown in the
figure. What is the magnitude of radial
acceleration and total acceleration of the block
when it arrives at Q?
P
Solution:
Given:
H = 6 m, r = 2 m,
u = 0 (body starts from rest)
To find:
i.
ii.
Radial acceleration (aR)
Total Acceleration (aTotal)
Formulae:
i.
aR =
ii.
aTotal = a 2R  a T2
Calculation: Height lost by the body = 6  2 = 4 m
From equation of motion,
v2 = u2 + 2gh

v2 = 0 + 2  9.8  4 = 78.4
From formula (i),
aR =

Q
r=2m
78.4
2
aR = 39.2 m/s2
aT = g = 9.8 m/s2
From formula (ii),
aTotal =

Ans: i.
ii.
 39.2 2   9.82
= 1536.64  96.04 = 1632.68
aTotal = 40.4 m/s2
The magnitude of radial acceleration of
the block is 39.2 m/s2.
The total acceleration of the block
is 40.4 m/s2.
Example 12
A car of mass 1500 kg rounds a curve of radius
250m at 90 km/hour. Calculate the centripetal
force acting on it.
[Mar 13]
Solution:
Given:
m = 1500 kg, r = 250 m,
5
v = 90 km/h = 90  = 25 m/s
18
To find:
Centripetal force (FCP)
mv2
Formula:
FCP =
[½ Mark]
r
Calculation: From formula,
FCP =
H=6m
v2
r
1500   25 
2
[½ Mark]
250

FCP = 3750 N
Ans: The centripetal force acting on the car is
[1 Mark]
3750 N.
27
Std. XII Sci.: Perfect Physics ‐ I Example 13
A racing car completes 5 rounds of a circular
track in 2 minutes. Find the radius of the track if
the car has uniform centripetal acceleration of
2 m/s2.
[Oct 13]
Solution:
Given:
5 rounds = 2r(5), t = 2 minutes = 120 s
To find:
Radius (r)
Formula:
acp = 2r
[½ Mark]
Calculation: From formula,
acp = 2r

2 =
v2
r
But v =

2 =

r=
2r (5) 10r
=
t
t
1002 r 2
rt 2
120 120
100
= 144 m
Ans: The radius of the track is 144 m.
[½ Mark]
[1 Mark]
*Example 14
A car of mass 2000 kg moves round a curve of
radius 250 m at 90 km/hr. Compute its
i.
angular speed
ii.
centripetal acceleration
iii. centripetal force.
Solution:
Given:
m = 2000 kg, r = 250 m,
5
v = 90 km/h = 90  = 25 m/s
18
To find:
i.
Angular speed ()
ii.
Centripetal acceleration (aCP)
iii. Centripetal force (FCP)
v
Formulae: i.
=
ii.
aCP = 2r
r
iii.
Calculation: i.

28
mv 2
r
From formula (i),
25
=
250
 = 0.1 rad/s
FCP =
ii.

iii.
From formula (ii),
aCP = (0.1)2  250
aCP = 2.5 m/s2
From formula (iii),
FCP =
2000   25 
2
250

FCP = 5000 N
Ans: The car has,
i.
angular speed 0.1 rad/s.
ii.
centripetal acceleration 2.5 m/s2.
iii. centripetal force 5000 N.
Example 15
A one kg mass tied at the end of the string 0.5 m
long is whirled in a horizontal circle, the other
end of the string being fixed. The breaking
tension in the string is 50 N. Find the greatest
speed that can be given to the mass.
Solution:
Given:
Breaking tension, F = 50 N,
m = 1 kg, r = 0.5 m
To find:
Maximum speed (vmax)
Formula:
B.T = max.C.F =
mv 2max
r
Calculation: From formula,
v2max =
F r
m

v 2max =
50  0.5
1

vmax =
50  0.5

vmax = 5 m/s
Ans: The greatest speed that can be given to the
mass is 5 m/s.
Example 16
A mass of 5 kg is tied at the end of a string
1.2 m long revolving in a horizontal circle. If the
breaking tension in the string is 300 N, find the
maximum number of revolutions per minute the
mass can make.
Solution:
Given:
Length of the string, r = 1.2 m,
Mass attached, m = 5 kg,
Breaking tension, T = 300 N
Chapter 01: Circular Motion
To find:
Maximum number of revolutions per
minute (nmax)
Formula:
Tmax = mr2max
Calculation: From formula,

5  1.2  (2n)2 = 300

5  1.2  42n2 = 300

n2max =

nmax = 1.26618 = 1.125 rev/s
300
4   3.142   6.0
2
= 1.26618

nmax = 1.125  60

nmax = 67.5 rev/min
Ans: The maximum number of revolutions per
minute made by the mass is 67.5 rev/min.
Example 17
A coin placed on a revolving disc, with its centre
at a distance of 6 cm from the axis of rotation just
slips off when the speed of the revolving disc
exceeds 45 r.p.m. What should be the maximum
angular speed of the disc, so that when the coin is
at a distance of 12 cm from the axis of rotation, it
does not slip?
Solution:
Given:
r1 = 6 cm, r2 = 12 cm, n1 = 45 r.p.m
To Find:
Maximum angular speed (n2)
Formula:
Max.C.F = mr2
Calculation: Since, mr1mr222
[As mass is constant]
2
2

r11 = r22
2
=
1
r1
r2

2n 2
=
2n1

n2
=
n1

n2 = n1
r1
r2
r1
r2
r1
= 45
r2
6
r.p.m
12
45
r.p.m = 31.8 r.p.m
2
Ans: The maximum angular speed of the disc
should be 31.8 r.p.m.
Example 18
A stone of mass 0.25 kg tied to the end of a string
is whirled in a circle of radius 1.5 m with a speed
of 40 revolutions/min in a horizontal plane. What
is the tension in the string? What is the maximum
speed with which the stone can be whirled
around if the string can withstand a maximum
tension of 200 N?
(NCERT)
Solution:
Given:
m = 0.25 kg, r = 1.5 m, Tmax = 200 N,
n = 40 rev. min1 =
40
rev s1
60
To find:
i.
ii.
Tension (T)
Maximum speed (vmax)
Formulae:
i.
T = mr2
Calculation: Since,  = 2n =

ii.
Tmax =
mv 2max
r
2
60
 = 1.33  rad s1
From formula (i),
T = 0.25  1.5  (1.33)2
T = 6.55 N
From formula (ii),
T r
v2max = max
m
200  1.5
v2max =
0.25
vmax = 1200

Ans: i.
ii.
vmax = 34.64 m s1
The tension in the string is 6.55 N.
The maximum speed with which the stone
can be whirled around is 34.64 m s1.
Example 19
A coin kept on a horizontal rotating disc has its
centre at a distance of 0.1 m from the axis of the
rotating disc. If the coefficient of friction between
the coin and the disc is 0.25; find the angular
speed of disc at which the coin would be about to
[Oct 11]
slip off. (Given g = 9.8 m/s2)
Solution:
Given:
r = 0.1 m,  = 0.25
To find:
Angular speed ()
Formula:
v = r
29
Std. XII Sci.: Perfect Physics ‐ I Calculation: Using formula,
rg
=
r
1/ 2
g  0.25  9.8 
=

r
 0.1 
= 4.949 rad/s
Ans: The angular speed of the disc at which the
coin would be about to slip off is 4.949 rad/s.
=
v
=
r
Example 20
A coin kept at a distance of 5 cm from the centre
of a turntable of radius 1.5 m just begins to slip
when the turnable rotates at a speed of 90 r.p.m.
Calculate the coefficient of static friction between
the coin and the turntable. [g = 9.8 m/s2]. [Mar 16]
Solution:
Given:
r = 5 cm = 0.05 m
90
r.p.s.,
n = 90 r.p.m. =
60
g = 9.8 m/s2
To find:
Coefficient of static friction (s)
Formula:
s =
r2
g
[½ Mark]
2    90
60
= 3 rad/s
0.05  (3) 2

s =
[½ Mark]
9.8
0.45  (3.14) 2
=
= 0.4527
9.8
Ans: The coefficient of static friction between the
coin and the turntable is 0.4527.
[1 Mark]
Calculation: Since,  = 2n =
Example 21
1
rev/min. and
3
has a radius of 15 cm. Two coins are placed at
4 cm and 14 cm away from the centre of record.
If the co-efficient of friction between the coins
and the record is 0.15, which of the coins will
revolve with the record?
(NCERT)
Solution:
The coin will revolve with record if the force of friction
is enough to provide centripetal force. If this force is not
enough, then the coin will slip off the record.
To prevent slipping, the condition is
mg  mr2
g  r2
For the first coin, r = 4 cm = 0.04 m,
A disc revolves with a speed of 33
30
1
2
n = 33 rev/min =
 = 2n =
100
100
5
rev/min =
r.p.s = r.p.s
3
180
9
10 
9
g = 0.15  9.8 = 1.47 m/s2
2
and
 10  
2
r1 = 0.04  
  0.488 m/s
 g 
2
As g > r12, the coin will revolve with the record.
For the second coin,
 10  
2
2
r22 = 0.14  
 = 1.706 m/s
 9 
As r22 > g, the coin will not revolve with the record.
Brain Teaser
Example 22
A 70 kg man stands in contact against the inner
wall of a hollow cylindrical drum of radius 3 m
rotating about its vertical axis with 200 rev/min.
The coefficient of friction between the wall and
his clothing is 0.15. What is the minimum
rotational speed of the cylinder to enable the man
to remain stuck to the wall (without falling) when
the floor is suddenly removed?
(NCERT)
Solution:
The horizontal force N of the wall on the man
provides the necessary centripetal force.
mv 2
N
= mr2
r
The frictional force ‘’ acting upwards balances the
weight ‘mg’ of the man.
i.e.,   N or mg  mr2
g
g
 2 or 2 
r
r
So, the minimum angular velocity of rotation of the
drum is given by,
9.8
g
min =
=
0.15  3
r

min = 4.667 rad s1
Ans: The minimum rotational speed of the cylinder
is 4.667 rad s1.
Example 23
Calculate the maximum speed with which a car
can be safely driven along a curved road of
radius 30 m and banked at 30 with the
horizontal [g = 9.8 m/s2].
[Mar 96]
Chapter 01: Circular Motion
Solution:
Given:
To find:
Formula:
2
r = 30 m,  = 30, g = 9.8 m/s
Maximum speed (vmax)
vmax = rg tan 
Calculation: From formula,
vmax =
=
30  9.8  tan  30 
30  9.8 
1
3
=
30  9.8
1.732

vmax = 13.028 m/s
Ans: The maximum speed with which the car can
drive safely is 13.028 m/s.
Example 24
An aircraft executes a horizontal loop at a speed
of 720 km h-1 with its wings banked at 15. What
is the radius of the loop?
(NCERT)
Solution:
5
Given:
v = 720 km h1 = 720 
= 200 m/s,
18
 = 15
To find:
Radius (r)
v2
Formula:
tan  =
rg
Calculation: From formula,
v2
r 
g tan 

r=
 200 
2
= 15232 m
9.8  tan 15o

r = 15.23 km
Ans: The radius of the loop is 15.23 km.
Example 25
A train runs along an unbanked circular track of
radius 30 m at a speed of 54 km h1. The mass of
the train is 106 kg.
i.
What provides the centripetal force required
for this purpose? The engine or the rails?
ii.
What is the angle of banking required to
prevent wearing out of the rail? (NCERT)
Solution:
i.
The centripetal force is provided by the lateral
force action due to rails on the wheels of the
train.
5
= 15 m/s, r = 30 m
ii.
v = 54 km h1 = 54 
18
v2
tan  =
rg
(15) 2
30  9.8
 = tan1(0.7653 )

 = 37 25
The centripetal force required is provided
Ans: i.
by the lateral force action due to rails.
ii.
The angle of banking required is 37 25.
tan 
*Example 26
A motor cyclist at a speed of 5 m/s is describing a
circle of radius 25 m. Find his inclination with
vertical. What is the value of coefficient of
friction between tyre and ground?
Solution:
Given:
v = 5 m/s, r = 25 m, g = 9.8 m/s2
To find:
i. Inclination with vertical ()
ii. Coefficient of friction ()
v2
v2
Formulae: i. tan  =
ii.
= g
rg
r
Calculation: From formula (i),
2
1
 5
tan  =
=
= 0.1021
25  9.8 9.8

 = tan1 (0.1021) = 550
From formula (ii),
52
=
= 0.1021
25  9.8
Ans: i.
The inclination of the motor cyclist with
vertical is 550.
ii.
The value of coefficient of friction
between tyre and ground is 0.1021.
*Example 27
A motor van weighing 4400 kg rounds a level
curve of radius 200 m on an unbanked road at
60 km/hr. What should be minimum value of
coefficient of friction to prevent skidding? At what
angle the road should be banked for this velocity?
Solution:
Given:
m = 4400 kg, r = 200 m,
5
50
v = 60 km/hr = 60 
=
m/s,
3
18
g = 9.8 m/s2
To find:
i.
Coefficient of friction ()
ii.
Angle of banking ()
v2
Formulae: i.
v = rg ii. tan  =
rg
Calculation: From formula (i),
=

25
(50 / 3) 2
=
18  9.8
200  9.8
 = 0.1417
31
Std. XII Sci.: Perfect Physics ‐ I 
Ans: i.
ii.
From formula (ii),
tan  = 0.1417
 = tan1 (0.1417) = 8 4
The minimum value of coefficient of
friction to prevent skidding 0.1417.
The angle at which the road should be
banked is 8 4.
Example 28
A stone of mass 1 kg is whirled in horizontal
circle attached at the end of a 1 m long string. If
the string makes an angle of 30 with vertical,
calculate the centripetal force acting on the stone.
[Mar 14]
(g = 9.8 m/s2).
Solution:
Given:
m = 1 kg, l = 1 m,  = 30,
g = 9.8 m/s2
To find:
Centripetal force (FCP)
mv 2
ii. v = rg tan 
Formulae:
i. FCP =
r
[½ Mark]
Calculation: Substituting formula (ii) in (i),
FCP =
m

rg tan 
r

= mg tan 
Example 29
A stone of mass 2 kg is whirled in a horizontal
circle attached at the end of 1.5 m long string. If
the string makes an angle of 30 with vertical,
[July 16]
compute its period. (g = 9.8 m/s2)
Solution:
Given:
m = 2 kg, l = 1.5 m,  = 30,
g = 9.8 m/s2
To find:
Period (T)
T = 2
l cos 
g
[½ Mark]
Calculation: From formula,
T = 2  3.14
= 6.28 
1.5  cos 30
9.8
[½ Mark]
1.5  0.8660
9.8

T = 2.29 s
Ans: Period of revolution is 2.29 s.
32
iii.
2
[½ Mark]
= 1  9.8  tan 30
1
9.8
= 9.8 
=
= 5.658 N
1.732
3
Ans: The centripetal force acting on the stone is 5.658 N.
[1 Mark]
Formula:
Example 30
A motorcyclist rounds a curve of radius 25 m at the
speed of 36 km/hr. The combined mass of motorcycle
and motorcyclist is 150 kg.
(g = 9.8 m/s2)
i.
What is centripetal force exerted on the
motorcyclist?
ii.
What is upward force exerted on the
motorcyclist?
iii. What angle the motorcycle makes with
vertical?
[Feb 13 old course]
Solution:
5
Given:
r = 25 m, v = 36 
= 10 m/s,
18
m = 150 kg
To find:
i.
Centripetal force (F)
ii.
Upward force (N cos )
ii.
Angle motorcycle makes with
vertical ()
mv 2
Formulae: i.
F=
[½ Mark]
r
ii.
N cos  = mg
[½ Mark]
[1 Mark]
tan  =
v2
rg
Calculation: From formula (i),
150  (10) 2
F=
[½ Mark]
25
F = 600 N
From formula (ii),
N cos = 9.8  150 = 1470 N
From formula (iii),
 v2 
 = tan1  
[½ Mark]
 rg 

Ans: i.
ii.
iii.
 102 
[½ Mark]
= tan1 

 25  9.8 
 = 2212
Centripetal force exerted on the
[½ Mark]
motorcyclist is 600 N.
Upward force exerted on the
[½ Mark]
motorcyclist is 1470 N.
Angle the motorcycle makes with
vertical is 2212.
[½ Mark]
*Example 31
A circular race course track has a radius of
500 m and is banked to 10. If the coefficient of
friction between tyres of vehicle and the road
surface is 0.25. Compute.
i.
the maximum speed to avoid slipping.
ii.
the optimum speed to avoid wear and tear
of tyres. (g = 9.8 m/s2)
Chapter 01: Circular Motion
Solution:
Given:
To Find:
r = 500 m, θ = 10,  = 0.25
i.
Maximum speed to avoid
slipping (vmax)
ii. Optimum speed to avoid wear
and tear of tyres (vo)
Formulae:
i.
ii.
Calculation: i.
 μ  tan θ 
vmax = rg  s

1  μ s tan θ 
vo = rg tan θ
From formula (i),
vmax =
Ans: i.
ii.
 0.25  tan10 
500  9.8 

1  0.25  tan10 

vmax = 46.72 m/s
ii.
From formula (ii),
vo = 500  9.8  tan10
= 500  9.8  0.176
 vo = 29.37 m/s
The maximum speed to avoid slipping is
46.72 m/s.
The optimum speed to avoid wear and
tear of tyres is 29.37 m/s.
Example 32
The radius of curvature of meter gauge railway
line at a place where the train is moving with a
speed of 10 m/s is 50 m. If there is no side thrust
on the rails, find the elevation of the outer rail
above the inner rail.
Solution:
Given:
Radius of curve, r = 50 m,
Speed of train, v = 10 m/s
To find:
Elevation of rails (h)
Formulae:
i.
tan  =
v2
rg
ii.
h = lsin
h = 1  sin (1132)
= 1  (0.2000) = 0.2 m

h = 20 cm
Ans: The elevation of the outer rail above the inner
rail is 20 cm.
Example 33
A string of length 0.5 m carries a bob of mass
0.1 kg at its end. It is used as a conical pendulum
with a period 1.41 s. Calculate angle of
inclination of string with vertical and tension in
the string.
Solution:
Given:
l = 0.5 m, m = 0.1 kg, T = 1.41 s
To find:
i.
Angle of inclination ()
ii.
Tension in the string (T)
Formulae:
h

From formula (i),
 v2 

 rg 
 = tan1 
 100 

1
= tan1 
 = tan (0.2041)
 50  9.8 
 = 1132
From formula (ii),
h = l sin 
T = 2
ii.
Tension, T =
mg
cos 
Calculation: From formula (i),
1.41 = 2  3.142
0.5  cos 
9.8
cos 
19.6

1.41
=
2  3.142

cos 
 1.41 

 = 19.6
 2  3.142 
2
2




Calculation:
l=1m
l cos 
g
i.

Ans: i.
ii.
 1.41 
cos  = 19.6 

 2  3.142 
cos  = 0.9868
 = cos1 (0.9868)
 = 919
From formula (ii),
0.1  9.8
0.98
T =
=
cos 919
0.9868
T = 0.993 N
The angle of inclination of string with
vertical is 919.
The tension in the string is 0.993 N.
Example 34
In a conical pendulum, a string of length 120 cm
is fixed at rigid support and carries a mass of
150 g at its free end. If the mass is revolved in a
horizontal circle of radius 0.2 m around a vertical
axis, calculate tension in the string. (g = 9.8 m/s2)
[Oct 13]
33
Std. XII Sci.: Perfect Physics ‐ I Solution:
Given:
To find:
Formula:
Solution:
l = 120 cm = 1.2 m, r = 0.2 m,
m = 150 g = 0.15 kg
Tension in the string (T)
mg
Tension, T =
cos 

l
h
r
Calculation:
h
O
 T
mg
l
T cos
Given:
r
mg





By Pythagoras theorem,
l2 = r2 + h2
h2 = (1.2)2  (0.2)2
= 1.4
h = 1.183 m
[1 Mark]
The weight of bob is balanced by
vertical component of tension T.
Tcos = mg
….(i)
From figure,
h
cos  =
l
h2
h2
cos2 = 2  2
r  h2
l
h
cos  =
2
r  h2
Substituting in formula,
Formulae:
[½ Mark]
 0.04 
= 0.15  9.8  1  

 1.4 
[½ Mark]
= 1.491 N
Ans: Tension in the string is 1.491 N.
[1 Mark]
Example 35
A conical pendulum has length 50 cm. Its bob of mass
100 g performs uniform circular motion in
horizontal plane, so as to have radius of path 30 cm.
Find
i.
the angle made by the string with vertical
ii.
the tension in the supporting thread and
iii. the speed of bob.
l = 50 cm = 0.5 m,
r = 30 cm = 0.3 m,
m = 100 g = 100  103 kg = 0.1 kg
i.
Angle made by the string with
vertical ()
ii.
Tension in the supporting thread
(T)
iii. Speed of bob (v)
i.
tan  =
r
h
ii.
tan  =
v2
rg
Calculation: By Pythagoras theorem,
l2 = r2 + h2
h2 = l2  r2
h2 = 0.25  0.09 = 0.16

h = 0.4 m
i.
From formula (i),
tan  =
ii.
2
r
= mg    1
h
To find:

mg r 2  h 2
T=
h
34
T cos 



iii.


Ans: i.
ii.
iii.
0.3
= 0.75
0.4
 = tan1 (0.75)
 = 3652
The weight of bob is balanced by
vertical component of tension T
T cos  = mg
h
0.4
=
= 0.8
cos  =
l
0.5
mg
0.1  9.8
T=
=
cos
0.8
T = 1.225 N
From formula (ii),
v2 = rg tan 
v2 = 0.3  9.8  0.75 = 2.205
v = 1.485 m/s
Angle made by the string with vertical is
3652.
Tension in the supporting thread is 1.225 N.
Speed of the bob is 1.485 m/s.
Chapter 01: Circular Motion
*Example 36
A bucket containing water is whirled in a vertical
circle at arms length. Find the minimum speed at
top to ensure that no water spills out. Also find
corresponding angular speed. [Assume r = 0.75 m]
Solution:
Given:
r = 0.75 m, g = 9.8 m/s2
To find:
i.
Minimum speed (vH)
ii.
Angular speed (H)
v
Formulae: i.
vH = rg
ii.
H = H
r
Calculation: From formula (i),


Ans: i.
ii.
vH = 0.75  9.8
vH = 2.711 m/s
From formula (ii),
2.711
H =
0.75
H = 3.615 rad/s
For no water to spill out, the minimum
speed at top should be 2.711 m/s.
The angular speed of the bucket is
3.615 rad/s.
Example 37
A stone of mass 100 g attached to a string of
length 50 cm is whirled in a vertical circle by
giving velocity at lowest point as 7 m/s. Find the
velocity at the highest point.
[Acceleration due to gravity = 98 m/s2] [Oct 15]
Solution:
Given:
m = 100 g= 0.1 kg, r = 50 cm = 0.5m,
g = 9.8 m/s2, vL = 7 m/s
To find:
Velocity at the highest point (vH)
Formula:
vH =
2  T.E.(H) 
m
 4gr
[½ Mark]
Calculation: Total energy at highest point,

1
T.E.(H)= K.E. at lowest point = mv 2L
2
[½ Mark]
1
T.E.(H) =  0.1  7 2
[½ Mark]
2
= 2.45 J
[½ Mark]
From formula,
vH =
2  2.45
  4  9.8  0.5  =
0.1
Example 38
A stone of mass 5 kg, tied to one end of a rope of
length 0.8 m, is whirled in a vertical circle. Find
the minimum velocity at the highest point and at
the midway point. [g = 98 m/s2]
[Oct 14]
Solution:
Given:
m = 5 kg, r = 0.8 m, g = 9.8 m/s2
To find:
i.
Minimum velocity at the highest
point (vH)
ii.
Minimum velocity at midway
point (vM)
Formulae: i.
vH = rg
[½ Mark]
ii.
vM = 3rg
Calculation: From formula (i),
vH = 0.8  9.8 =
7.84 = 2.8 m/s
[½ Mark]
From formula (ii),
vM = 3  0.8  9.8
= 3  2.8
[½ Mark]
= 4.85 m/s
Ans: The minimum velocity at the highest point and
midway point is 2.8 m/s and 4.85 m/s
respectively.
*Example 39
A stone weighing 1 kg is whirled in a vertical
circle at the end of a rope of length 0.5 m.
Find the tension at
i.
lowest position
ii.
mid position
iii. highest position
Solution:
Given:
m = 1 kg, r = l = 0.5 m, g = 9.8 m/s2
To find:
i.
Tension at lowest position (TL)
ii.
Tension at mid position (TM)
iii. Tension at highest position (TH)
mv 2L
+ mg
Formulae: i.
TL =
r
mv 2M
ii.
TM =
r
mv 2H
iii. TH =
mg
r
Calculation: Since, v 2L = 5rg
From formula (i),
 5rg

 g  = 6mg
 r

TL = m 
29.4

vH = 5.422 m/s
Ans: The velocity at the highest point is 5.422 m/s.
[1 Mark]
[½ Mark]

= 6  1  9.8 = 58.8 N
TL = 58.8 N
Since, v2M = 3rg
35
Std. XII Sci.: Perfect Physics ‐ I From formula (ii),
ii.
 3rg 
 = 3 mg
 r 
TM = m 

= 3  1  9.8 = 29.4 N
TM = 29.4 N
Since, v 2H = rg
From formula (iii),
 rg

 g = 0
r

TH = m 

Ans: i.
ii.
iii.
TH = 0
The tension at lowest position in the
vertical circle is 58.8 N.
The tension at mid position in the
vertical circle is 29.4 N.
The tension at highest position in the
vertical circle is 0.
*Example 40
A pilot of mass 50 kg in a jet aircraft is executing
a loop-the-loop with constant speed of 250 m/s. If
the radius of circle is 5 km, compute the force
exerted by seat on the pilot
i.
at the top of loop.
ii.
at the bottom of loop.
Pilot
A
N1
mg
N2
B
Pilot
Ans: i.
ii.
From formula (ii),
50  (250) 2
Fbottom =
+ 50  9.8
5  103
= 625 + 490

Fbottom = 1115 N
The force exerted by seat on the pilot at
the top of loop is 135 N.
The force exerted by seat on the pilot at
the bottom of loop is 1115 N.
*Example 41
An object (stone) of mass 0.5 kg attached to a rod
of length 0.5 m is whirled in a vertical circle at
constant angular speed. If the maximum tension
in the string is 5 kg wt. Calculate
i.
speed of the stone
ii.
maximum number of revolutions it can
complete in a minute.
Solution:
Given:
m = 0.5 kg, r = l = 0.5 m,
Tmax = 5 kg wt. = 5  9.8 N
To find:
i.
Speed (v)
ii.
Maximum number of
revolutions (nmax)
mv 2
+ mg
Formulae: i.
Tmax =
r
v
ii.
n=
2r
Calculation: i.
From formula (i),
r
(T  mg)
v2 =
m

ii.
= 0.5 
 9.8 
 0.5

= 49  4.9 = 44.1
v = 44.1 = 6.64 m/s
From formula (ii),
v
nmax =
[ v = r]
2r
 5  9.8
Solution:
Given:
36

v2 = r   g 
m

mg
m = 50 kg, v = 250 m/s,
r = 5 km = 5  103 m
To find:
i.
Force at the top of loop (Ftop)
ii.
Force at the bottom of loop
(Fbottom)
mv 2
Formulae: i.
Ftop =
 mg
r
mv 2
+ mg
ii.
Fbottom =
r
Calculation: i.
From formula (i),
50  (250) 2
 50  9.8
Ftop =
5  103
= 625  490

Ftop = 135 N
T

=
Ans: i.
ii.

6.64
2  3.14  0.5
= 2.115 r.p.s

nmax = 2.115  60
= 126.9 r.p.m
The speed of the stone is 6.64 m/s.
The maximum number of revolutions
the stone can complete in a minute is
126.9 r.p.m.
Chapter 01: Circular Motion
*Example 42
A ball is released from height h along the slope
and moves along a circular track of radius R
without falling vertically downwards as shown in
the figure. Show that h =
acting on it, if its angular speed of rotation is
0.6 rad/s.
6.
A body of mass 1 kg is tied to a string and
revolved in a horizontal circle of radius 1 m.
Calculate the maximum number of revolutions
per minute, so that the string does not break.
Breaking tension of the string is 9.86 N.
7.
A coin just remains on a disc rotating at
120 r.p.m when kept at a distance of 1.5 cm
from the axis of rotation. Find the coefficient
of friction between the coin and the disc.
8.
A coin kept on a horizontal rotating disc has
its centre at a distance of 0.25 m from the axis
of rotation of the disc. If  = 0.2, find the
angular velocity of the disc at which the coin
is about to slip off. [g = 9.8 m/s2]
9.
With what maximum speed a car be safely
driven along a curve of radius 40 m on a
horizontal road, if the coefficient of friction
between the car tyres and road surface is 0.3?
[g = 9.8 m/s2]
10.
A vehicle is moving along a circular road
which is inclined to the horizontal at 10.
The maximum velocity with which it can
move safely is 36 km/hr. Calculate the radius
of the circular road.
11.
Calculate angle of banking for circular track
of radius 50 m as to be suitable for driving a
car with maximum speed of 72 km/hr.
12.
At what angular speed should the earth rotate
about its axis so that apparent weight of a
body on the equator will be zero? What would
be the length of the day at that periodic time?
[Radius of earth = 6400 km, g = 9.8 m/s2]
13.
A body of mass 200 gram performs circular
motion of radius 50 cm at a constant speed of
240 r.p.m. Find its linear speed.
14.
A body of mass 2 kg is tied to a string 1.5 m
long and revolved in a horizontal circle about
the other end. If it performs 300 r.p.m,
calculate its linear velocity, centripetal
acceleration and force acting on it.
15.
A string breaks under a tension of 10 kg-wt. If
the string is used to revolve a body of mass
12 g in a horizontal circle of radius 50 cm,
what is the frequency of revolution and linear
speed with which the body can be revolved?
[g = 9.8 m/s2]
5
R.
2
A
h
R
B
Solution:
The total energy of any body revolving in a vertical
5
mgR.
circle =
2
When a ball is released from a height ‘h’ along the
slope and moves along a circular track of radius R
without falling vertically downwards, its potential
energy (mgh) gets converted into kinetic energy.
5
1
2
 mv  = mgR
2
 2
Hence according to law of conservation of energy,
5
mgh =
mgR
2
5

gh = gR
2
5

h= R
2
EXERCISE
Section A: Practice Problems
1.
Calculate the angular speed of minute hand of
a clock of length 2 cm.
2.
Determine the angular acceleration of a
rotating body which slows down from
500 r.p.m to rest in 10 seconds.
3.
The minute hand of a clock is 5 cm long.
Calculate the linear speed of an ant sitting at
the tip.
4.
Calculate the angular speed and linear speed
of tip of a second hand of clock if second hand
is 5 cm long.
5.
A 0.5 kg mass is rotated in a horizontal circle
of radius 20 cm. Calculate the centripetal force
37
Std. XII Sci.: Perfect Physics ‐ I 16.
17.
18.
19.
20.
21.
22.
The breaking tension of a string is 80 kg-wt. A
mass of 1 kg is attached to the string and
rotated in a horizontal circle on a horizontal
surface of radius 2 m. Find the maximum
number of revolutions made without breaking.
[g = 9.8 m/s2]
Find the angle of banking of a railway track of
radius of curvature 250 m, if the optimum
velocity of the train is 90 km/hr. Also find the
elevation of the outer track over the inner
track if the two tracks are 1.6 m apart.
The distance between two rails of rail track is
1.6 m along a curve of radius 800 m. The
outer rail is raised about the inner rail by
10 cm. With what maximum speed can a train
be safely driven along the curve?
A particle, initially at rest, performs circular
motion with uniform angular acceleration
0.18 rad/s2. What speed, in r.p.m will it attain
in a time of 10 seconds? What is angular
displacement in this time?
38
Obtain an expression for optimum speed of a
car on a banked road.
[Oct 99]
3.
Explain
i.
Centripetal force
ii.
Centrifugal force.
A 50 g mass is attached to a string and rotated
in a vertical circle of radius 1.8 m. What is the
minimum speed the mass must have at the top
of the circle in order that the string may not
slacken? What will be the velocity of mass
and the tension in the bottom of the circle
under the above conditions? [g = 9.8 m/s2]
A body of mass 1 kg tied to string is whirled
in a vertical circle of radius 1 m. Find velocity
and tension in the string
i.
at the top of the circle
ii.
at the bottom of the circle and
iii. at a point level with the centre.
Assume that the mass just goes around the
circle at the top with minimum speed without
the string slackening.
Define angle of banking. Draw a neat labelled
diagram showing different forces and their
components acting on a vehicle moving on a
banked road.
[Oct 97]
[Mar 00]
4.
What is banking of road? Obtain an
expression for angle of banking. On what
factors does it depend?
[Oct 00, 02, 04, Mar 06]
5.
Derive the expression for the maximum speed
of a vehicle on the banked road. State the
factors on which the optimum speed depends.
[Mar 01]
6.
Derive an expression for radial acceleration of
a particle performing uniform circular motion.
[Mar 04, Oct 05]
Why is it so called?
7.
What is centripetal and centrifugal force?
[Mar 04]
8.
A certain body remains stationary on the
vertical inner wall of a cylindrical drum of
radius ‘r’, rotating at a constant speed. Show
that the minimum angular speed of the drum is
Determine the force that presses the pilot
against his seat at the upper and lower points
of a loop, if the weight of the pilot is 75 kgf.,
the radius of the loop is 200 m and the
velocity of the plane looping the loop is
constant and equal to 360 km h1.[g = 10 m/s2]
Section B: Theoretical Board Questions
1.
2.
g
, where “” is coefficient of friction
r
between the body and surface of the wall.
[Oct 06]
9.
For a conical pendulum prove that tan  =
v2
rg
[Oct 09]
10.
Obtain an expression for maximum speed with
which a vehicle can be driven safely on a
banked road. Show that the safety speed limit
is independent of the mass of the vehicle.
[Mar 10, Oct 10]
11.
Derive an expression for linear velocity at
lowest point and at highest point for a particle
revolving in vertical circular motion. [Oct 11]
Section C: Numerical Board Problems
1.
A train rounds a curve of radius 150 m at a
speed of 20 m/s. Calculate the angle of
banking so that there is no side thrust on the
rails. Also find the elevation of the outer rail
over the inner rail, if the distance between the
[Oct 96]
rails is 1 m.
Chapter 01: Circular Motion
2.
An object of mass 400 g is whirled in a
horizontal circle of radius 2 m. If it performs
60 r.p.m, calculate the centripetal force acting
[Oct 96, Mar 01]
on it.
3.
Find the angle which the bicycle and its rider
will make with the vertical when going round
a curve at 27 km/hr on a horizontal curved
road of radius 10 m. [g = 9.8 m/s2] [Mar 98]
4.
Find the angle of banking of curved railway
track of radius 600 m, if the maximum safety
speed limit is 54 km/hr. If the distance
between the rails is 1.6 m. find the elevation
of the outer track above the inner track.
[Oct 98]
[g = 9.8 m/s2]
5.
The vertical section of a road over a bridge in
the direction of its length is in the form of an
arc of a circle of radius 4.4 m. Find the
greatest velocity at which a vehicle can cross
the bridge without losing contact with the road
at the highest point, if the center of the vehicle
is 0.5 m from the ground.
[Oct 01]
[Given: g = 9.8 m/s2]
6.
If the frequency of revolution of an object
changes from 2 Hz to 4 Hz in 2 second,
[Oct 03]
calculate its angular acceleration.
7.
The minute hand of a clock is 8 cm long.
Calculate the linear speed of an ant sitting at
[Mar 05]
its tip.
8.
The frequency of a spinning top is 10 Hz. If it is
brought to rest in 6.28 sec, find the angular
acceleration of a particle on its surface. [Oct 05]
9.
Calculate the angle of banking for a circular
track of radius 600 m as to be suitable for
driving a car with maximum speed of
[Mar 06]
180 km/hr. [g = 9.8 m/s2]
10.
A vehicle is moving along a curve of radius
200 m. What should be the maximum speed
with which it can be safely driven if the angle
of banking is 17? (Neglect friction)
[Mar 07]
[g = 9.8 m/s2]
11.
An object of mass 1 kg is tied to one end of a
string of length 9 m and whirled in a verticle
circle. What is the minimum speed required at the
[Oct 08]
lowest position to complete a circle?
12.
An object of mass 2 kg attached to wire of
length 5 m is revolved in a horizontal circle. If
it makes 60 r.p.m. Find its
i.
angular speed
ii.
linear speed
iii.
iv.
13.
centripetal acceleration
centripetal force
[Mar 09]
A stone of mass one kilogram is tied to the end
of a string of length 5 m and whirled in a verticle
circle. What will be the minimum speed required
at the lowest position to complete the circle?
[Oct 10]
[Given: g = 9.8 m/s2]
Section D: Multiple Choice Questions
1.
When a particle moves in a circle with a
uniform speed
(A) its velocity and acceleration both are
constant.
(B) its velocity is constant but the
acceleration changes.
(C) its acceleration is constant but the
velocity changes.
(D) its velocity and acceleration both change.
2.
A particle is performing a U.C.M along a
circle of radius R in half the period of
revolution, its displacement and distance
covered are
(B) 2R, 2R
(A) R, R
(D)
(C) 2R, R
2 R , 2R
3.
A particle rotates in U.C.M. with tangential
velocity ‘v’ along a horizontal circle of
diameter ‘D’. Total angular displacement of
the particle in time ‘t’ is _______ [Mar 16]
v
(A) vt
(B)    t
D
vt
2vt
(D)
(C)
2D
D
4.
When a body performs a U.C.M it has
(A) a constant velocity.
(B) a constant acceleration.
(C) an acceleration of constant magnitude
but variable direction.
(D) an acceleration, which changes with time.
5.
When a body performs a U.C.M
(A) its velocity remains constant.
(B) work done on it is zero.
(C) work done on it is negative.
(D) no force acts on it.
6.
Angular speed of the second hand of a watch is
(A) /60 rad/s
(B) /30 rad/ s
(C)  rad/s
(D) 2/3 rad/s
39
Std. XII Sci.: Perfect Physics ‐ I 7.
8.
9.
When a particle moves in a uniform circular
motion. It has
(A) radial velocity and radial acceleration.
(B) tangential
velocity
and
radial
acceleration.
(C) tangential velocity and tangential
acceleration.
(D) radial velocity and tangential acceleration.
For a particle moving along a circular path,
the angular velocity vector () is directed
(A) along the radius towards the centre.
(B) along the radius but away from the centre.
(C) along the tangent to the circular path.
(D) along the axis of rotation.
The ratio of the angular speeds of the hour
hand and the minute hand of a clock is
(A) 1 : 12
(B) 1 : 6
(C) 1 : 8
(D) 12 : 1
10.
A wheel having radius one metre makes 30
revolutions per minute. The linear speed of a
particle on the circumference will be
(B)  m/s
(A) /2 m/s
(D) 62 m/s
(C) 30 m/s
11.
A particle starts from rest and moves with an
angular acceleration of 3 rad/s2 in a circle of
radius 3 m. Its linear speed after 5 seconds
will be
(A) 15 m/s
(B) 30 m/s
(C) 45 m/s
(D) 7.5 m/s
12.
13.
14.
40
Angular speed of a minute hand of a wrist
[Oct 10]
watch in rad/s is


(A)
(B)
60
900


(C)
(D)
1800
3600
To enable a particle to describe a circular path,
what should be the angle between its velocity
and acceleration?
(B) 45
(A) 0
(C) 90
(D) 180
The bulging of earth at the equator and
flattening at the poles is due to _______.
[Mar 14]
(A) centripetal force
(B) centrifugal force
(C) gravitational force
(D) electrostatic force
15.
A flywheel rotates at a constant speed of
2400 r.p.m The angle in radian described by
the shaft in one second is
(B) 80 
(A) 2400 
(C) 20 
(D) 4800 
16.
A body is revolving with a uniform speed v in a
circle of radius r. The tangential acceleration is
(A) v/r
(B) Zero
2
(C) v /r
(D) v/r2
17.
For keeping a body in uniform circular
motion, the force required is
(A) centrifugal
(B) radial
(C) tangential
(D) centripetal
18.
The magnitude of centripetal force cannot be
expressed as
19.
(A)
mr2
(B)
42 mr
T2
(C)
mv
(D)
mv/
Particle A of mass M is revolving along a
circle of radius R. Particle B of mass m is
revolving in another circle of radius r. If they
take the same time to complete one revolution,
then the ratio of their angular velocities is
(A) R/r
(B) r/R
(C)
1
(D)
R
 
r 
2
20.
If a cycle wheel of radius 0.4 m completes one
revolution in 2 seconds, then acceleration of
[Mar 11]
the cycle is _______.
2
2
(A) 0.4  m/s
(B) 0.4  m/s2
2
0.4
(C)
m/s2
(D)
m/s2
0.4
2
21.
A particle is performing circular motion. Its
frequency of revolution changes from 120 rpm
to 180 rpm in 10 s. The angular acceleration of
the particle is
(B) 0.628 rad/s2
(A) 1 rad/s2
2
(C) 0.421 rad/s
(D) 0.129 rad/s2
22.
Which of the following force is a pseudo
force?
(A) Force acting on a falling body.
(B) Force acting on a charged particle
placed in an electric field.
(C) Force experienced by a person standing
on a merry-go- round.
(D) Force which keeps the electrons moving
in circular orbits.
Chapter 01: Circular Motion
23.
24.
25.
26.
27.
In uniform circular motion, the angle between
the radius vector and centripetal acceleration is
(A) 0
(B) 90
(C) 180
(D) 45
30.
[Oct 09, Mar 10]
The centripetal force acting on a mass m
moving with a uniform velocity v on a circular
orbit of radius r will be
(A)
mv 2
2r
(B)
1
mv2
2
(C)
1
mrv2
2
(D)
mv 2
r
A body performing uniform circular motion
[Oct 08]
has _______.
(A) constant velocity
(B) constant acceleration
(C) constant kinetic energy
(D) constant displacement
Which of the following statements about the
centripetal and centrifugal forces is correct?
(A) Centripetal force balances centrifugal force.
(B) Both centripetal force and centrifugal
force act in the same frame of reference.
(C) Centripetal force is directed opposite to
centrifugal force.
(D) Centripetal force is experienced by the
observer at the centre of the circular
path described by the body.
The linear acceleration of the particle of mass
‘m’ describing a horizontal circle of radius r,
with angular speed ‘’ is
(A) /r
(B) r
2
(C) r
(D) r2
28.
An unbanked curve has a radius of 60 m. The
maximum speed at which a car can make a turn,
if the coefficient of static friction is 0.75, is
(A) 2.1 m/s
(B) 14 m/s
(C) 21 m/s
(D) 7 m/s
29.
Centrifugal force is
(A) a real force acting along the radius.
(B) a force whose magnitude is less than
that of the centripetal force.
(C) a pseudo force acting along the radius
and away from the centre.
(D) a force which keeps the body moving
along a circular path with uniform speed.
A stone is tied to a string and rotated in a
horizontal circle with constant angular velocity.
If the string is released, the stone flies _____
31.
32.
(A)
radially inward
(B)
(C)
(D)
radially outward
tangentially forward
tangentially backward
A particle performs a uniform circular motion
in a circle of radius 10 cm. What is its
centripetal acceleration if it takes 10 seconds
to complete 5 revolutions?
(A)
2.5 2 cm/s2
(B)
52 cm/s2
(C)
102 cm/s2
(D)
202 cm/s2
When a car takes a turn on a horizontal road,
the centripetal force is provided by the
(A) weight of the car.
(B) normal reaction of the road.
(C) frictional force between the surface of
the road and the tyres of the car.
(D)
centrifugal force.
33.
On being churned the butter separates out of
milk due to _______.
(A) centrifugal force (B) adhesive force
(C) cohesive force
(D) frictional force
34.
When a particle moves on a circular path then
the force that keeps it moving with uniform
velocity is
(A) centripetal force.
(B) atomic force.
(C) internal force.
(D) gravitational force.
35.
A car is moving along a horizontal curve of
radius 20 m and coefficient of friction
between the road and wheels of the car is 0.25.
If the acceleration due to gravity is 9.8 m/s2,
then its maximum speed is _______ .
[Mar 08]
(A)
(C)
3 m/s
7 m/s
(B)
(D)
5 m/s
9 m/s
41
Std. XII Sci.: Perfect Physics ‐ I 36.
A particle of mass m is observed from an
inertial frame of reference and is found to
move in a circle of radius r with a uniform
speed v. The centrifugal force on it is
mv 2
towards the centre.
(A)
r
mv 2
away from the centre.
(B)
r
mv 2
(C)
along the tangent through the particle.
r
(D) zero.
37.
If a cyclist goes round a circular path of
circumference 34.3 m in 22 s, then the angle
made by him with the vertical will be
(A) 42
(B) 43
(D) 45
(C) 49
38.
39.
A motor cycle is travelling on a curved track
of radius 500 m. If the coefficient of friction
between the tyres and road is 0.5, then the
maximum speed to avoid skidding will be
[g = 10 m/s2]
(A) 500 m/s
(B) 250 m/s
(C) 50 m/s
(D) 10 m/s
A coin placed on a rotating turntable just slips
if it is placed at a distance of 4 cm from the
centre. If the angular velocity of the turntable
is doubled, it will just slip at a distance of
(A) 1 cm
(B) 2 cm
(C) 4 cm
(D) 8 cm
40.
Two bodies of mass 10 kg and 5 kg are
moving in concentric orbits of radius R and r.
If their time periods are same, then the ratio of
their centripetal acceleration is
(A) R/r
(B) r/R
(D) r2/R2
(C) R2/r2
41.
A body is moving in a horizontal circle with
constant speed. Which one of the following
statements is correct?
(A) Its P.E is constant.
(B) Its K.E is constant.
(C) Either P.E or K.E of the body is constant.
(D) Both P.E and K.E of the body are
constant.
42.
42
A cyclist bends while taking a turn to
(A) reduce friction.
(B) generate required centripetal force.
(C) reduce apparent weight.
(D) reduce speed.
43.
A cyclist has to bend inward while taking a
turn but a passenger sitting inside a car and
taking the same turn is pushed outwards. This
is because
(A) the car is heavier than cycle.
(B) centrifugal force acting on both the
cyclist and passenger is zero.
(C) the cyclist has to balance the centrifugal
force but the passenger cannot balance
the centrifugal force hence he is pushed
outward.
(D) the speed of the car is more than the
speed of the cycle.
44.
The minimum velocity (in m s1) with which a
car driver must traverse a flat curve of radius
150 m and coefficient of friction 0.6 to avoid
skidding is (g = 10 m/s2)
(A) 60
(B) 30
(C) 15
(D) 25
45.
Maximum safe speed does not depend on
(A) mass of the vehicle.
(B) radius of curvature.
(C) angle of inclination (banking).
(D) acceleration due to gravity.
46.
A motor cyclist moving with a velocity of
72 km per hour on a flat road takes a turn on
the road at a point where the radius of
curvature of the road is 20 metres. The
acceleration due to gravity is 10 m/s2. In order
to avoid skidding, he must not bend with
respect to the vertical plane by an angle
greater than
(A)  = tan1(6)
(B)  = tan1(2)
(C)  = tan1(25.92)
(D)  = tan1 (4)
47.
A car of mass 1500 kg is moving with a speed
of 12.5 m/s on a circular path of radius 20 m
on a level road. What should be the coefficient
of friction between the car and the road, so
that the car does not slip?
(A) 0.2
(B) 0.4
(C) 0.6
(D) 0.8
48.
A particle is moving in a circle of radius r with
constant speed v. Its angular acceleration will be
(A) vr
(B) v/r
(C) zero
(D) vr2
Chapter 01: Circular Motion
49.
A hollow sphere has radius 6.4 m. Minimum
velocity required by a motor cyclist at bottom
to complete the circle will be
(A) 17.7 m/s
(B) 12.4 m/s
(C) 10.2 m/s
(D) 16.0 m/s
50.
A curved road having a radius of curvature of
30 m is banked at the correct angle. If the
speed of the car is to be doubled, then the
radius of curvature of the road should be
(A) 62 m
(B) 120 m
(C) 90 m
(D) 15 m
51.
(C)
54.
55.
l cos 
g
2
l cos 
g
(B)
(D)
2
l sin 
g
57.
When a car crosses a convex bridge, the
bridge exerts a force on it. It is given by
mv 2
mv 2
(B) F =
(A) F = mg +
r
r
2
2 2
mv
(D) F = mg +  mv 
(C) F = mg 
r
 r 
58.
Out of the following equations which is
[Mar 12]
WRONG?
l sin 
g
The period of a conical pendulum in terms of
its length (l), semivertical angle () and
[Mar 15]
acceleration due to gravity (g) is:
1 l cos 
1 l sin 
(A)
(B)
2
g
2
g
(C)
53.
The period of a conical pendulum is
(A) equal to that of a simple pendulum of
same length l.
(B) more than that of a simple pendulum of
same length l.
(C) less than that of a simple pendulum of
same length l.
(D) independent of length of pendulum.
The time period of conical pendulum is
_______.
[Oct 11]
(A)
52.
56.
4
l cos 
4g
(D)
4
(A)
(C)
A car is moving on a curved path at a speed of
20 km/ hour. If it tries to move on the same
path at a speed of 40 km/hr then the chance of
toppling will be
(A) half
(B) twice
(C) thrice
(D) four times
Consider a simple pendulum of length 1 m. Its
bob performs a circular motion in horizontal
plane with its string making an angle 60 with
the vertical. The period of rotation of the bob
is (Take g = 10 m/s2)
(A) 2 s
(B) 1.4 s
(C) 1.98 s
(D) none of these


 r F



at    r
(B)
(D)



ar    v



v r  
59.
A car is moving with a speed of 30 m/s on a
circular path of radius 500 m. Its speed is
increasing at the rate of 2 m/s2. The
acceleration of the car is
(B) 9.8 m/s2
(A) 2 m/s2
2
(C) 2.7 m/s
(D) 1.8 m/s2
60.
A ball of mass 250 gram attached to the end of a
string of length 1.96 m is moving in a horizontal
circle. The string will break if the tension is
more than 25 N. What is the maximum speed
with which the ball can be moved?
(A) 5 m/s
(B) 7 m/s
(C) 11 m/s
(D) 14 m/s
61.
A 500 kg car takes a round turn of radius 50 m
with a speed of 36 km/hr. The centripetal
force acting on the car will be
(A) 1200 N
(B) 1000 N
(C) 750 N
(D) 250 N
62.
Angle of banking does not depend upon
(A) Gravitational acceleration
(B) Mass of the moving vehicle
(C) Radius of curvature of the circular path
(D) Velocity of the vehicle
63.
What would be the maximum speed of a car
on a road turn of radius 30 m, if the coefficient
of friction between the tyres and the road is
0.4?
(A) 6.84 m/s
(B) 8.84 m/s
(C) 10.84 m/s
(D) 4.84 m/s
l tan 
g
A stone of mass m is tied to a string and is
moved in a vertical circle of radius r making
n revolutions per minute. The total tension in
the string when the stone is at its lowest point
is
(A) m(g +  nr2)
(B) m (g + nr)
(C) m (g + n2 r2)
(D) m [g + (2 n2 r)/900]

43
Std. XII Sci.: Perfect Physics ‐ I 64.
65.
In a conical pendulum, when the bob moves in
a horizontal circle of radius r, with uniform
speed v, the string of length L describes a cone
of semivertical angle . The tension in the
string is given by
(L2  r 2 )1/ 2
mgL
(B)
(A) T = 2 2
mgL
(L  r )
mgL
(D) T = mgL 2
(C) T =
L2  r 2
 L2  r 2 
(C)
mgr
68.
69.

2

mgL
(B)
71.
Let  denote the angular displacement of a
simple pendulum oscillating in a vertical
plane. If the mass of the bob is m, the tension
in the string at extreme position is
(A) mg sin 
(B) mg cos 
(C) mg tan 
(D) mg
(D)
72.
Kinetic energy of a body moving in vertical
circle is
(A) constant at all points on a circle.
(B) different at different points on a circle.
(C) zero at all the point on a circle.
(D) negative at all the points.
73.
The difference in tensions in the string at
lowest and highest points in the path of the
particle of mass ‘m’ performing vertical
[July 16]
circular motion is
(A) 2 mg
(B) 4 mg
(C) 6 mg
(D) 8 mg
74.
A body of mass 1 kg is moving in a vertical
circular path of radius 1 m. The difference
between the kinetic energies at its highest and
lowest position is
(A) 20 J
(B) 10 J
(C) 4 5 J
mgr
L2  r 2
mgL
L  r 
2
2 1/ 2
3gd
(D)
gd
Which quantity is fixed of an object which
moves in a horizontal circle at constant speed?
(A) Velocity
(B) Acceleration
(C) Kinetic energy
(D) Force
A particle of mass 0.1 kg is rotated at the end
of a string in a vertical circle of radius 1.0 m at
a constant speed of 5 m s1. The tension in the
string at the highest point of its path is
(A) 0.5 N
(B) 1.0 N
(C) 1.5 N
(D) 15 N
A stone of mass 1 kg tied to a light
inextensible string of length L = (10/3) metre
in whirling in a circular path of radius L in a
vertical plane. If the ratio of the maximum
tension in the string to the minimum tension is
4 and if g is taken to be 10 m/s2. The speed of
the stone at the highest point of the circle is
(A) 20 m/s
(B) 10 / 3 m/s
(C)
(D)
44
L r
L2  r 2
2
A metal ball tied to a string is rotated in a
vertical circle of radius d. For the thread to
remain just tightened the minimum velocity at
highest point will be
(B) gd
(A)
5gd
(C)
67.
Water in a bucket is whirled in a vertical circle
with a string attached to it. The water does not
fall down even when the bucket is inverted at the
top of its path. We conclude that in this position.
(A) mg = mv2/r
(B) mg is greater than mv2/r
(C) mg is not greater than mv2/r
(D) mg is not less than mv2/r
In a conical pendulum, the centripetal force
 mv 2 

 acting on the bob is given by
 r 
(A)
66.
70.
5 2 m/s
10 m/s
(D)
10 ( 5 1) J
75.
A circular road of radius 1000 m has banking
angle 45. The maximum safe speed of a car
having mass 2000 kg will be, (coefficient of
friction between tyre and road is 0.5)
(B) 124 m/s
(A) 172 m/s
(C) 99 m/s
(D) 86 m/s
76.
For a particle in circular motion the centripetal
acceleration is
(A) less than its tangential acceleration.
(B) equal to its tangential acceleration.
(C) more than its tangential acceleration.
(D) may be more or less than its tangential
acceleration.
Chapter 01: Circular Motion
77.
One end of a string of length l is connected to
a particle of mass m and the other to a small
peg on a smooth horizontal table. If the
particle moves in a circle with speed v the net
force on the particle (directed towards the
(NCERT)
centre) is
mv 2
(A) T
(B) T 
l
mv 2
(D) 0
(C) T +
l
9.
10.
11.
12.
13.
14.
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
57.
61.
65.
69.
73.
77.
Section A
1.74  103 rad/s
5.237 rad/s2
8.72  105 m/s
1.07  101 rad/s, 5.235  103 m/s
0.036 N
30
0.2418
2.8 rad/s
10.84 m/s
57.87 m
39 12
1.237  103 rad/s, 5080 s
12.57 m/s
47.13 m/s, 1480 m/s2, 2.960  103 N
20.34 rev/s, 63.95 m/s
3.150 rev/s
1419, 0.3955 m
22.16 m/s
17.18 r.p.m, 1.43 rad
300 kgf, 450 kgf
42 m/s, 9.39 m/s, 2.94 N
i.
3.13 m/s, zero
ii.
7 m/s, 58.8 N
iii. 5.42 m/s, 29.4 N
Section C
1.
2.
3.
4.
5.
6.
7.
8.
1513, 0.2625 m
31.59 N
2952
212, 0.061 m
6.429 m/s
6.28 rad/s2
1.396 102 cm/s
10 rad/s2
ii.
iv.
31.4 m/s
394.384 N
Section D
ANSWERS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
232
24.48 m/s
21 m/s
i.
6.28 rad/s
iii. 197.192 m/s2
15.65 m/s
1.47 N
(D)
(B)
(A)
(C)
(D)
(B)
(C)
(C)
(A)
(D)
(D)
(A)
(A)
(D)
(C)
(B)
(A)
(B)
(C)
(A)
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
58.
62.
66.
70.
74.
(C)
(B)
(B)
(B)
(D)
(C)
(C)
(C)
(A)
(C)
(B)
(B)
(B)
(D)
(D)
(B)
(D)
(C)
(A)
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
59.
63.
67.
71.
75.
(D)
(B)
(C)
(B)
(C)
(C)
(C)
(C)
(C)
(A)
(C)
(D)
(C)
(B)
(C)
(C)
(C)
(C)
(A)
4.
8.
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
60.
64.
68.
72.
76.
(C)
(D)
(C)
(B)
(B)
(D)
(C)
(C)
(D)
(A)
(B)
(C)
(C)
(C)
(D)
(C)
(C)
(B)
(D)
Hints to Multiple Choice Questions
2.
In half the period, particle is diametrically
opposite to its initial position. Hence, its
displacement is 2R. It has covered a semicircle,
hence distance covered by particle is R.
3.
=
S
r
S = vt and r =
D
2
2vt
D

=
11.
v = r = r (t) = 3  3  5 = 45 m/s
15.
 2400 
d =  dt = 2n  dt = 2 
  1 = 80 
 60 
45
Std. XII Sci.: Perfect Physics ‐ I 37.


69.
 v2 
 = tan1  
 rg 
Circumference, 2r = 34.3 m
34.3
m
r=
2
2r 34.3
and v =
m/s

t
22



 34.3 / 22 2  2 

 = tan1 


34.3  9.8


1
= tan [0.9997]
= 44.99
 45

10
 10
3
10
=
m/s
3
=
59.
Tangential acceleration aT = 2 m/s2
v2 (30) 2
=
Radial acceleration = ar =
r
500

Acceleration a =
a T2  a 2r
2
 900 
4

 500 
= 2.69 m/s2  2.7 m/s2
=
64.
Tension in the string,

T = T 2 cos 2   T 2 sin 2 
 mv 2  2

2
= 
  (mg) 
 r 

but v =


v=

1/ 2
1/ 2
rg tan  and tan  =
r
h
r2 g / h
 r 2  2 
T = mg    1
 rh 

2
2
2
but h = (L – r )
1/ 2
1/ 2

65.
46
Tmax – Tmin = 6 mg
T
Also, max = 4
….(Given)
Tmin
 Tmax = 4 Tmin
 Tmin = 2 mg
mv 2
 mg
but Tmin =
r
mv 2
= mg
r
v = rg
2



r


1
T = mg 

 L2  r 2 

mgL
=
L2  r 2
Centripetal force,
mv 2
mgL
r
= T sin  =
 =
2
2
L
r
L r
mgr
L2  r 2
77.
….(here r = L =
10
m)
3
The particle is performing circular motion and
is constantly accelerated. Hence, it is under
the action of external force. As the motion
here is confined to horizontal plane, net force
on the particle is T.