Physics
Numericals STD. XII
Edition: June 2014
Prof. Umakant N. Kondapure
Mr. Collin Fernandes
(M.Sc., B.Ed., Solapur)
(M.Sc., Mumbai) Mr. Vivek Ghonasgi
(M.Sc., B.Ed. Mumbai) Mrs. Meenal Iyer
(M.Sc., Mumbai) Published by
Target PUBLICATIONS PVT. LTD.
Shiv Mandir Sabhagriha,
Mhatre Nagar, Near LIC Colony,
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storage and retrieval system without permission in writing from the Publisher.
Written according to the revised syllabus (2012-2013) published by the Maharashtra State
Board of Secondary and Higher Secondary Education, Pune.
Physics
Numericals
STD. XII Salient Features :
• Subtopic wise numericals with solutions.
• Shortcuts to enable quick problem solving.
• Practice problems for every subtopic.
• Includes solved board numerical.
• Numerical based multiple choice questions for effective preparation.
Solutions/hints to practice problems and multiple choice questions available
in downloadable PDF format at www. targetpublications.org
TEID : 737
Preface
In the case of good books, the point is not how many you can get through, but rather how many can get
through to you.
“STD XII Sci.: PHYSICS NUMERICALS” is a complete and thorough guide to the numerical aspect of
the HSC preparation. The book is prepared as per the Maharashtra State Board syllabus .Subtopic wise
segregation of Solved Numericals in each chapter help the student to gain knowledge of the broad spectrum
of problems in each subtopic Formulae which form a vital part of problem-solving are provided in every
chapter. Solutions and calculations have been broken down to the simplest form possible (with
log calculation provided wherever needed) so that the student can tackle each and every problem with ease.
Problems for practice are provided to test the vigilance and alertness of the students and build their
confidence. Board Numericals till the latest year have been provided to help the student get accustomed to
the different standards of board numericals. Numerical based multiple choice questions are covered subtopic-wise to prepare the student on a competitive level.
Solution/hints to practice problems and multiple choice questions can be downloaded in PDF format from
our website www. targetpublications.org
The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve
nearly missed something or want to applaud us for our triumphs, we’d love to hear from you.
Please write to us on : [email protected]
A book affects eternity; one can never tell where its influence stops.
Best of luck to all the aspirants!
Yours faithfully
Authors
R
Contents
Sr.
Unit
No.
Page
No.
Sr.
No.
Unit
Page
No.
1
Circular Motion
1
11
Interference and Diffraction
270
2
Gravitation
38
12
Electrostatics
297
3
Rotational Motion
67
13
Current Electricity
329
4
Oscillations
100
5
Elasticity
129
6
Surface Tension
149
7
Wave Motion
165
8
Stationary Waves
185
9
10
Kinetic Theory of Gases and
Radiation
Wave Theory of light
212
253
14
Magnetic Effect of Electric
Current
350
15
Magnetism
373
16
Electromagnetic Induction
389
17
Electrons and Photons
418
18
Atoms, Molecules and Nuclei
433
19
Semiconductors
458
20
Communication System
467
Target Publications Pvt. Ltd.
Chapter 01: Circular Motion
01
Circular Motion
Formulae
ii.
Section 1: Angular
Displacement,
Relation
Between Linear Velocity and Angular
Velocity
1.
Angular velocity:
v
i.
ω=
r
where,
v = linear velocity
r = radius of the circle along which
particle performs circular motion.
∆θ
ii.
ω=
∆t
where, ∆θ = angular displacement of the
particle in circular motion during time
interval ∆t.
iii. ω = 2πn
where,
n = frequency of revolution of particle
in circular motion.
2π
iv. ω =
T
where, T = period of revolution of
particle performing circular motion.
2.
Angular displacement:
θ = ωt
3.
Time period:
2πr
i.
T=
v
4.
Frequency of revolution:
1 ω
n= =
T 2π
5.
Linear velocity:
i.
v = rω
ii.
T=
1.
2.
3.
v = 2πnr
Angular acceleration:
∆ω
i.
α=
∆t
where, ∆ω = change in the angular
velocity of a particle in circular motion
during a time interval ∆t.
Centripetal (or radial) acceleration:
v2
ar =
= ωv = rω2
r
Tangential acceleration:
aT = α r
Resultant or total acceleration:
a = a 2t + a 2r + 2a t a r cos θ
where,
θ = angle made by ar with at
a=
4.
∵
a 2t + a 2r when θ = 90°.
For U.C.M.:
v2
a = ar =
= ωv
r
at = 0
Section 4: Centripetal and Centrifugal Forces
2π
ω
Section 2: Angular Acceleration
1.
Section 3: Centripetal and Tangential Acceleration
1.
ii.
∆n
∆t
where, ∆n = change in frequency of the
particle in circular motion during a time
interval ∆t.
α = 2π ⋅
Centripetal force:
mv 2
i.
Fc =
r
iii. Fc = mrω2
v.
Fc =
ii.
Fc = mvω
iv.
Fc = 4π2mrn2
2
4π mr
T2
where, m = mass of particle performing
circular motion
Section 5: Motion of a Vehicle along a Curved
Unbanked Road
1.
The necessary centripetal force:
mv 2
Fc = µmg =
r
where,
m = mass of vehicle
v = velocity of the vehicle
r = radius of the curve road
µ = coefficient of friction between the tyres
of the vehicle and the surface of the road.
1
Target Publications Pvt. Ltd.
2.
3.
The maximum velocity with which a vehicle
can take a turn safely without skidding:
v = µrg
Std. XII Sci.: Physics Numericals
3.
T=
The maximum angular velocity with which
a vehicle can take a turn safely without
skidding:
µg
ω=
r
where, θ = angle of banking
2.
The maximum velocity without skidding:
vmax =
⎡ µ + tan θ ⎤
rg ⎢ s
⎥
⎣1 − µs tan θ ⎦
4.
1.
Angle of banking:
⎛ v2 ⎞
⎛ v2 ⎞
θ = tan−1 ⎜ ⎟ or tan θ = ⎜ ⎟
⎝ rg ⎠
⎝ rg ⎠
Height of inclined road:
h = d sin θ
where,
d = distance between the two front or rear
wheels.
5.
The maximum velocity with which a vehicle
can go on a banked curved road without
toppling:
v=
rg
Tension in the string:
mg
T=
cos θ
Section 8: Vertical Circular Motion
µs = coefficient of friction between the tyres
of the vehicle and surface of the road
4.
Velocity at any point in vertical circular
motion:
i.
vP = v 2L − 2gr (1 − cosθ)
ii.
vL =
5rg
iii.
vH =
rg
iv.
vM = 3rg
where,
vP = velocity of the particle at any
point P along the circle.
vL = Minimum velocity at the lowest
point on the circle so that it can
safely travel along the vertical
circle (looping the loop).
vH = Minimum velocity of the particle at
the highest point on the circle so
that the string will not be
slackened.
vM = Minimum velocity of the particle
at a mid-way point so that it can
travel along the circle.
r = radius of the vertical circle.
θ = angle between the position vectors
at the given position of particle
and that of the lowest point on the
vertical circle.
d
2H
where, H = height of centre of gravity (C.G.)
of the vehicle from the road.
Section 7: Conical Pendulum
1.
2.
Linear speed of bob:
v = rg tan θ
Angular velocity:
g
g tan θ
ω=
=
l cos θ
r
2
h
g
where,
l = length of conical pendulum
h = the height of the fixed support from the
centre of the circle or axial height of the
cone
θ = semivertical angle of the cone.
where,
3.
l cos θ
2π
= 2π
ω
g
= 2π
Section 6: Banking of Roads
For motion of vehicles along a banked curve
road:
1.
The proper velocity or optimum velocity:
v = rg tan θ
Periodic time:
2.
Relation between velocities at different
points in vertical circular motion:
i.
v 2L = v 2H + 4gr
ii.
v 2M = v 2H + 2gr
Target Publications Pvt. Ltd.
3.
Chapter 01: Circular Motion
iv.
Tension at:
i.
Midway point M,
Any point P,
TP =
ii.
TM =
mv 2
+ mg cosθ
r
4.
Lowest point L,
Total energy at any point:
=
Highest point H,
TH =
1
mv2 + mgr (1 − cos θ)
2
E=
mv 2L
mv 2H
+ mg =
+ 5mg
TL =
r
r
iii.
mv 2M mv 2L
− 2 mg
=
r
r
5
mgr
2
mv 2H
mv 2L
− mg =
− 5 mg
r
r
Section 9: Kinematical Equations
Analogy between translatory motion and circular motion
No.
1.
Translatory Motion
→
→
→
→
dθ
Angular velocity ω =
dt
dr
Linear velocity v =
dt
→
Circular Motion
→
→
Angular acceleration
2.
d v d2 r
= 2
Linear acceleration a =
dt
dt
3.
Linear momentum p = m v
4.
Linear impulse = F (∆t) = ∆ p
5.
Force F = m a
6.
Work W = F . r
Work W = τ . θ
Kinetic energy of translation
Kinetic energy of rotation
7.
8.
→
→
→
→
Et =
→
1
mv2
2
→
→
d ω d2 θ
= 2
α=
dt
dt
→
→
→
→
→
Angular momentum L = I ω
→
→
→
Angular impulse = τ (∆t) = ∆ L
→
→
Torque τ = I α
→
Er =
→
1 2
Iω
2
Equations of linear motion
Equations of rotational motion
i.
i.
ω2 = ω1 + αt
ii.
θ = ω1t +
v = u + at
ii. s = ut +
1 2
at
2
iii. v2 − u2 = 2 as
iv.
sn = u +
a
(2n − 1)
2
1 2
αt
2
iii. ω 22 −ω 12 = 2αθ
iv.
θn = ω1 +
α
(2n −1)
2
3
Target Publications Pvt. Ltd.
Std. XII Sci.: Physics Numericals
Shortcuts
1.
v2
and
r
if angular velocity is given, apply ar = ω2r to
find ar.
If linear velocity is given, apply ar =
3.
If number of revolutions in a particular time is
given, apply ar = 4π2n2r.
4.
If some mass is placed on a rotating body,
then angular motion changes due to frictional
force between the mass and the body.
Coefficient of friction can be calculated by
applying the formula,
mv 2
= µmg or
r
v2
ω2 r
4π2 n 2 r
µ=
=
=
rg
g
g
7.
Breaking tension is the maximum centripetal
force which is given by the relation,
mv 2
F=
= mω2r
r
Speed of a vehicle on a banked road or
circular turn depends upon the curvature of the
road.
v2
tan θ =
rg
To avoid skidding, v ≤
µrg where µ is the
coefficient of friction between the tyres and
road.
d
where d is the distance between two
2h
wheels.
µ=
8.
10.
For looping the loop of radius r, the minimum
height from which the body should be released
is given by,
5r
h=
2
11.
i.
2π
(n2 − n1) rad/s2
t
2.
6.
Use the expressions, v L = 5gr and vT = gr
for the following cases:
i.
Bucket full of water whirled in a vertical
circle.
ii.
Motor cyclist riding in a vertical circle
in hollow sphere.
Angular acceleration (α):
If number of rotations or revolutions is given,
then
α=
5.
9.
In case of conical pendulum if l be the length
and r be the radius of the horizontal circle then
height of the rigid point of suspension is
calculated by the formula,
h=
4
l 2 − r2
For a vehicle moving over a convex
bridge which is in the shape of an arc of
a circle,
mv 2
r
For the motor cyclist at the upper most
point in the globe of death in a circus,
N = mg –
ii.
mv 2
– mg
r
where,
N = normal reaction acting on the
vehicle or the motorcycle.
N=
Solved Examples
Section 1: Angular
Displacement,
Relation
Between Linear Velocity and Angular
Velocity
Example 1.1
A wheel of radius 2 metre is making
60 revolutions per minute. Calculate the angular
velocity of any point on the rim.
Solution:
Given:
n = 60 r.p.m. = 60/60 = 1 rps,
r=2m
To find:
Angular velocity (ω)
Formula:
ω = 2πn
Calculation: From formula,
∴
ω = 2π × 1
∴
ω = 2π rad/s
Ans: The angular velocity of any point on the rim
of the wheel is 2π rad/s.
Target Publications Pvt. Ltd.
Example 1.2
Find the angular displacement of a particle
moving on a circle with angular velocity
(2π/3) rad/s in 15 s.
Solution:
Given:
t = 15 s, ω = (2π/3) rad/s
To find:
Angular displacement (θ)
Formula: θ = ωt
Calculation: From formula,
2π
θ=
× 15
3
∴
θ = 10 π rad
Ans: The angular displacement of the particle is
10π rad.
Example 1.3
What is the angular speed of the second hand of a
clock? If the second hand is 10 cm long, then find
the linear speed of its tip.
Solution:
Given:
r = 10 cm = 10−1 m, T = 60 s
To find:
i.
Angular speed (ω)
ii.
Linear speed (v)
2π
Formulae: i.
ω=
ii.
v = rω
T
Calculation: From formula (i),
2π
2 × 3.14
ω=
=
60
T
∴
ω = 1.047 × 10−1 rad/s
From formula (ii),
v = rω
v = 1.047 × 10−1 × 0.1
∴
v = 1.047 × 10−2 m/s.
Ans: i.
The angular speed of the second hand of
a clock is 1.047 × 10−1 rad/s.
ii.
The linear speed of the tip of the second
hand is 1.047 × 10−2 m/s.
Example 1.4
If a body rotates in a horizontal circle of radius
15 cm with an angular velocity of 0.8 rad/s, then
what is its linear velocity?
Solution:
r = 15 cm = 0.15 m
Given:
ω = 0.8 rad/s
To find:
Linear velocity (v)
Formula: v = rω
Calculation: From formula,
v = 0.15 × 0.8
∴
v = 0.12 m/s.
Ans: The linear velocity of the rotating body is
0.12 m/s.
Chapter 01: Circular Motion
Example 1.5
The linear velocity of a body is 0.2 m/s. If it
rotates in a horizontal circle having radius 0.5 m,
what is its angular velocity?
Solution:
r = 0.5 m,
Given:
v = 0.2 m/s
To find:
Angular velocity (ω)
Formula: v = rω
Calculation: From formula,
v
0.2
ω=
=
r
0.5
∴
ω = 0.4 rad/s
Ans: The angular velocity of the rotating body is
0.4 rad/s.
Example 1.6
A particle is revolving in a circle of radius 10 cm
with linear speed of 20 m/s. Find (i) its period of
revolution and (ii) frequency.
Solution:
Given:
r = 10 cm = 10 × 10−2 m,
v = 20 m/s
To find:
i.
Period of revolution (T)
ii.
Frequency (n)
2πr
Formulae: i.
T=
v
1
ii.
n=
T
Calculation: From formula (i),
2 × 3.14 × 10 × 10−2
T=
20
= 3.14 × 10−2
∴
T = 0.0314 s
∴
Ans: i.
ii.
From formula (ii),
1
1
n= =
T 0.0314
n = 31.85 s–1
The period of revolving particle is
0.0314 s
The frequency of the particle is
n = 31.85 s–1.
Example 1.7
What is the angular velocity of the minute hand
of a clock? If the minute hand is 5 cm long, what
is the linear velocity of its tip?
[Oct 04]
Solution:
Given:
R = 5 cm = 5 × 10−2 m,
T = 1h = 3600 s
5
Target Publications Pvt. Ltd.
Angular velocity (ω)
Linear velocity of the tip (v)
2π
Formulae: i.
ω=
T
ii.
v = rω
Calculation: From formula (i),
2π
ω =
T
2 × 3.14
=
3600
∴
ω = 1.74 × 10−3 rad/s
From formula (ii),
v = 5 × 10−2 × 1.74 × 10−3
∴
v = 8.7 × 10−5 m/s
Ans: For the minute hand of the clock,
i.
The angular velocity is 1.74 × 10–3 rad/s.
ii.
The linear velocity of the tip is
8.7 × 10–5 m/s.
To find:
i.
ii.
Example 1.8
An aircraft takes a turn along a circular path of
radius 2000 metre. If the linear speed of the
aircraft is 500 m/s, find the angular speed and the
time required by it to complete half the circular
path.
Solution:
Given:
r = 2000 m, v = 500 m/s
To find:
i.
Angular speed (ω)
ii.
Time required to complete half
the circular path (t)
v
Formulae: i.
ω=
r
2πr / 2
ii. v =
t
Calculation: From formula (i),
500
ω=
2000
∴
ω = 0.25 rad/s
From formula (ii),
2 × 3.14 × 2000
t=
2 × 500
∴
Ans: i.
ii.
6
t = 12.56 s
The angular speed of the aircraft is
0.25 rad/s.
The time required by the aircraft to
complete half the circular path is
12.56 s.
Std. XII Sci.: Physics Numericals
Section 2: Angular Acceleration
Example 2.1
A particle performing UCM changes its angular
velocity from 70 r.p.m to 130 r.p.m in 18 s. Find
the angular acceleration of the particle.
Solution:
Given:
n1 = 70/min =
70
/s,
60
130
/s, t = 18 s
60
To find:
Angular acceleration (α)
ω − ω 2 π ( n 2 − n1 )
Formula: α = 2 1 =
t
t
Calculation: From formula,
2 π n 2 − 2 π n1
α =
t
⎛ 130 70 ⎞
2π ⎜
− ⎟
60 60 ⎠
⎝
=
18
⎛ 130 − 70 ⎞
2π ⎜
60 ⎟⎠
= ⎝
18
2π × 60
=
60 × 18
π 3.142
= =
9
9
∴
α = 0.349 rad/s2
Ans: The angular acceleration of the particle is
0.349 rad/s2.
n2 = 130/min =
Example 2.2
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
n2 = 180 r.p.m =
= 3 rev/s,
60
t = 20 s
To find:
Angular acceleration (α)
ω − ω1
Formula:
α= 2
t
Calculation: From formula,
2πn 2 − 2πn1 2π ( 3 − 1)
α=
=
20
t
Target Publications Pvt. Ltd.
=
∴
2 × 3.142 × 2
20
α = 3.142
5
∴
α = 0.6284 rad/s2
Ans: Angular acceleration of the particle is
0.6284 rad/s2.
Example 2.3
A fly wheel rotating at 420 r.p.m. slows down at a
constant rate of 2 rad/s2. What time is required to
stop the fly wheel ?
Solution:
420
Given:
n1 = 420 r.p.m. =
r.p.s.
60
= 7 r.p.s
2
α = −2 rad/s , ω2 = 0
To find:
Time required (t)
Formulae: i.
ω = 2πn
ii.
ω2 = ω1 + αt
Calculation: From formula (i),
ω1 = 2πn1
= 2π × 7
= 14 π rad/sec
From formula (ii),
0 = 14π + (−2) × t
0 = 14π − 2t
2t = 14π
14 π
14
22
t=
=
×
2
2
7
∴
t = 22 s
Ans: The time required to stop the flywheel is 22 s.
Example 2.4
A particle performs circular motion with a
constant angular acceleration of 4 rad/s2. If the
radius of the circular path is 20 cm and the initial
angular speed of the particle is 2 rad/s, find the
angular speed of the particle after 0.5 sec.
Solution:
Given:
α = 4 rad/s2, r = 20 cm = 0.2 m,
ω1 = 2 rad/s, t = 0.5 s
To find:
Angular speed (ω2)
Formula: ω2 = ω1 + αt
Calculation: From formula,
ω2 = 2 + 4 × 0.5
=2+2
∴
ω2 = 4 rad/s
Ans: The angular speed of the particle after 0.5 s is
4 rad/s.
Chapter 01: Circular Motion
Example 2.5
A stone tied to one end of a string is whirled in a
horizontal circle of radius 40 cm with a frequency
of 30 r.p.m. Find the angular velocity and linear
velocity of the stone.
Solution:
Given:
r = 40 cm = 40 × 10−2 m
30
n = 30 rpm =
rps = 0.5 rps
60
To find:
i.
Angular velocity (ω)
ii.
Linear velocity (v)
Formulae: i.
ω = 2πn
ii.
v = rω
Calculation: From formula (i),
ω = 2 × 3.14 × 0.5
= 1.0 × 3.14
∴
ω = 3.14 rad/s
From formula (ii),
v = 40 × 10−2 × 3.14
= 4 × 3.14 × 10−2
= 125.6 × 10−2
∴
v = 1.256 m/s
Ans: For the stone:
i.
the angular velocity is 3.14 rad/s
ii.
the linear velocity is 1.256 m/s.
Example 2.6
A satellite revolves around the earth in a circular
orbit of radius 7000 km. If its period of
revolution is 2 h, calculate its angular speed,
linear speed and its centripetal acceleration.
Solution:
Given:
r = 7000 km = 7000 × 103 m,
T = 2h = 2 × 3600 = 7200 s
To find:
i.
Angular speed (ω)
ii.
Linear speed (v)
iii. Centripetal acceleration (ar)
2π
Formulae: i.
ω=
T
ii.
v = rω
iii. ar = rω2
Calculation: From formula (i),
2 × 3.14
ω=
7200
∴
ω = 8.72 × 10−4 rad/s
From formula (ii),
v = 7000 × 103 × 8.72 × 10−4
v = 61.04 × 102
v = 6.104 × 103 m/s.
∴
v = 6.104 km/s.
7
Target Publications Pvt. Ltd.
Std. XII Sci.: Physics Numericals
From formula (iii),
ar = 7000 × 103 × (8.72 × 10−4)2
= 7 × (8.72)2 × 10–5
∴
ar = 5.32 m/s2
Ans: For the revolving satellite:
i.
the angular speed is 8.72 × 10−4 rad/s,
ii.
the linear speed is 6.104 km/s
iii. the centripetal acceleration is 5.32 m/s2
Example 2.7
The earth moves round the sun in an almost
circular orbit of radius 1.5 × 1011 m with constant
angular speed. Calculate its
i.
angular velocity
ii.
linear velocity
iii. centripetal acceleration
[Given: 1 year = 3.156 × 107 sec,
Mass of earth = 5.98 × 1024 kg]
Solution:
Given:
r = 1.5 × 1011 m,
m = 5.98 × 1024 kg,
T = 1 yr = 3.156 × 107 s
To find:
i.
Angular velocity (ω)
ii.
Linear velocity (v)
iii. Centripetal acceleration (ar)
Formulae:
i.
ω=
2π
T
ii.
Calculation:
From formula (i),
∴
∴
∴
2 × 3.14
3.156 × 107
ω ≈ 1.99 × 10–7 rad/s.
From formula (ii),
v = 1.5 × 1011 × 1.99 × 10−7
v = 2.985 × 104 m/s
From formula (iii),
ar = 1.5 × 1011 × (1.99 × 10−7)2
{
}
ar = antilog ⎡log (1.5 ) + log (1.99 ) ⎤
⎣
⎦
× 1011 × 10–14
= {antilog [ 0.1716 + 2 × 0.2989]} × 10–3
2
= {antilog [ 0.1761 + 0.5978]} × 10–3
= {antilog [ 0.7739]} × 10–3
∴
ar = 5.941 × 10–3 m/s2
Ans: For the earth moving round the sun:
i.
the angular velocity is 1.99 × 10–7 rad/s,
ii.
the linear velocity is 2.985 × 104 m/s and
iii. the
centripetal
acceleration
is
5.941 × 10–3 m/s2.
8
Example 3.1
A 0.5 kg mass is rotated in a horizontal circle
of radius 20 cm. Calculate the centripetal
acceleration if the angular velocity is 0.6 rad/s
Solution:
Given:
m = 0.5 kg, r = 20 cm = 0.2 m,
ω = 0.6 rad/s
To find:
Centripetal acceleration (ar)
Formulae: i.
v = rω
v2
r
Calculation: From formula (i),
v = rω = 0.2 × 0.6 m/s
From formula (ii),
(0.2 × 0.6) 2
ar =
0.2
∴
ar = 0.072 m/s2
Ans: The centripetal acceleration of the mass is
0.072 m/s2.
ar =
ii.
v = rω
iii. ar = rω2
ω=
Section 3: Centripetal and Tangential Acceleration
Example 3.2
A stone tied to the end of a string which is 80 cm
long is whirled in a horizontal circle with a
constant speed. If the stone makes 14 revolutions
in 25 s, what is the magnitude of centripetal
acceleration?
Solution:
Given:
r = 80 cm = 0.80 m, n = 14/25 s−1
To find:
Centripetal acceleration (ar)
Formula: Centripetal acceleration,
ar = rω2
Calculation: From formula,
ar = 4π2n2r
….[∵ω = 2πn]
2
⎛ 14 ⎞
= 4 × (3.142)2 × ⎜ ⎟ × 0.8
⎝ 25 ⎠
4 × ( 3.142 ) × (14 ) × 0.8
2
=
( 25)
2
2
= 9.908
∴
ar ≈ 9.91 m/s2
Ans: The magnitude of centripetal acceleration is
9.91 m/s2.
Target Publications Pvt. Ltd.
Chapter 01: Circular Motion
Example 3.3
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
π2 m/s2.
[Oct 13]
Solution:
Given:
5 rounds = 2πr(5), t = 2 minutes = 120 s
To find:
Radius (R)
Formulae: acp = ω2r
Calculation: From formula,
acp = ω2r
∴
π2 =
v2
r
But v =
2πr (5) 10πr
=
t
t
100π2 r 2
rt 2
120 ×120
∴ r=
= 144 m
100
Ans: The radius of the track is 144 m.
∴
π2 =
Example 3.4
The tangential acceleration of the tip of a blade is
47.13 m/s2 and its centripetal acceleration is
473.9 m/s2. Find the value of linear acceleration of
the tip of the blade.
Solution:
Given:
at = 47.13 m/s2, ar = 473.9 m/s2
To find:
Linear acceleration (a)
Formula:
a=
a 2r + a 2t
Calculation: From formula,
a=
(473.9) 2 + (47.13) 2
Now,
(473.9)2 = antilog [2 log(473.9)]
= antilog [2 × 2.6757]
= antilog [5.3514]
= 2.246 × 105
(47.13)2 = antilog [2 log(47.13)]
= antilog [2 × 1.6733]
= antilog [3.3466]
= 2221
∴
ar =
( 473.9 )
2
+ ( 47.13)
2
=
2.246 × 105 + 2221
=
226821
⎡1
⎤
= antilog ⎢ log ( 226821) ⎥
2
⎣
⎦
⎡1
⎤
= antilog ⎢ × 5.3556 ⎥
⎣2
⎦
= antilog [2.6778]
∴
ar = 476.2
= 4.762 × 102 m/s2
Ans: The value of linear acceleration of the tip of
the blade is 4.762 × 102 m/s2.
Example 3.5
A particle is revolving in a circle. Its angular
speed increases from 2 rad/s to 40 rad/s in 19 sec.
The radius of the circle is 20 cm. Compare the
ratio of centripetal acceleration to tangential
acceleration.
Solution:
Given:
ω1 = 2 rad /s
ω2 = 40 rad /s
t = 19 s
r = 20 cm
= 20 × 10−2 m
= 0.2 m
To find:
Ratio of centripetal to tangential
acceleration (ar : at)
Formulae: i. ar = ω2r
ii. at = rα
Calculation: From formula (i),
ar = r ω2
= 0.2 × (40)2
= 0.2 × 1600
∴
ar = 320 m /s2
From formula (ii)
at = r α
⎛ ω − ω1 ⎞
=r ⎜ 2
⎟
t
⎝
⎠
⎛ 40 − 2 ⎞
= 0.2 ⎜
⎟
⎝ 19 ⎠
⎛ 38 ⎞
= 0.2 ⎜ ⎟
⎝ 19 ⎠
= 0.2 × 2
= 0.4 m /s2
9
Target Publications Pvt. Ltd.
∴
Std. XII Sci.: Physics Numericals
Section 4: Centripetal and Centrifugal Forces
ar
320
=
0.4
at
3200
4
∴
ar : at = 800 : 1
Ans: The ratio of centripetal acceleration to
tangential acceleration of the particle is 800 : 1.
=
Example 3.6
The tangential acceleration of a body is
29.48 m/s2 and its linear acceleration is 52.3 m/s2.
Find its radial acceleration.
Solution:
Given :
a = 52.3 m/s2,
at = 29.48 m/s2
To find:
Radial acceleration (ar)
Formula:
a 2t + a 2r or a2 = a 2t + a 2r
a =
a −a
2
or ar =
2
t
Calculation:
From formula,
ar =
(52.3) 2 − (29.48) 2
Now,
(52.3)2 = antilog [2 log (52.3)]
= antilog [2 × 1.7185]
= antilog [3.4370]
= 2735
(29.48)2= antilog [2 log (29.48)]
= antilog [2 × 1.4695]
= antilog [2.9390]
= 869
∴
ar =
=
( 52.3)
2
− ( 29.48 )
2
2735 − 869
= 1866
⎡1
⎤
= antilog ⎢ log (1866 ) ⎥
⎣2
⎦
⎡1
⎤
= antilog ⎢ × 3.2709 ⎥
2
⎣
⎦
= antilog [1.63545]
∴
ar = 43.1 m/s2
Ans: The radial acceleration of the body is
43.1 m/s2.
10
Example 4.1
A 0.2 kg mass is rotated in a horizontal circle of
radius 10 cm. If its angular speed is 0.4 rad/s, find
the centripetal force acting on it.
Given:
m = 0.2 kg,
r = 10 cm = 10 × 10−2 m,
ω = 0.4 rad /s
To find:
Centripetal force (Fc)
Formula: Fc = mrω2
Calculation: From formula,
Fc = 0.2 × 10 × 10−2 × (0.4)2
= 0.32 × 10−2 N
∴
Fc = 0.0032 N
Ans: The centripetal force acting on the rotating
mass is 0.0032 N.
Example 4.2
A car of mass 1500 kg rounds a curve of radius
250m at 90 km/hour. Calculate the centripetal
force acting on it.
[Feb 2013]
Solution:
Given:
m = 1500 kg, r = 250 m,
5
v = 90 km/h = 90 × = 25 m/s
18
To find:
Centripetal force (FCP)
Formula:
FCP =
mv 2
r
Calculation: From formula,
FCP =
1500 × ( 25 )
2
250
∴ FCP = 3750 N
Ans: The centripetal force acting on the car is
3750 N.
Example 4.3
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.
(g = 9.8 m/s2).
[Mar 14]
Solution:
Given:
m = 1 kg, l = 1 m, θ = 30°,
g = 9.8 m/s2
To find:
Centripetal force (FCP)
Formulae:
i.
FCP =
mv 2
r
ii.
v=
rg tan θ
Target Publications Pvt. Ltd.
Chapter 01: Circular Motion
Calculation: Substituting formula (ii) in (i),
FCP =
m
(
rg tan θ
)
2
r
= mg tan θ
= 1 × 9.8 × tan 30
1
9.8
=
= 9.8 ×
3 1.732
= 5.658 N
Ans: The centripetal force acting on stone is 5.658 N.
Example 4.4
A coin is placed on a revolving disc which revolves
at 60 r.p.m. It does not slip-off when it is at 15 cm
from the axis of rotation. What should be the
distance of the coin from the axis of rotation so
that it does not slip-off, when the speed of the
revolving disc is changed to 75 r.p.m?
Solution:
Given:
r1 = 15 cm = 0.15 m
60
n1 = 60 r.p.m =
= 1 r.p.s.
60
∴
ω1 = 2πn1 = 2π rad/s
= 2 × 3.14
= 6.28 rad/s
75
n2 = 75 r.p.m. =
r.p.s
60
= 1.25 r.p.s
∴
ω2 = 1.25 × 2 × 3.14
= 7.85 rad/sec
To find:
Distance of the coin from the axis (r2)
Formula: mr1 ω12 = mr2 ω22
Calculation:
From formula,
r ω2
r2 = 1 21
ω2
=
0.15 × ( 6.28 )
( 7.85)
2
2
r2 = antilog [log (0.15) + log(6.28)2 – log(7.85)2]
= antilog [log(0.15) + 2 log(6.28)
– 2 log (7.85)]
= antilog [ 1.1761+ 2 × 0.7980 – 2 × 0.8949]
= antilog [ 1 .1761 + 1.5960 – 1.7898]
= antilog [0.7726 + 2 .2102]
= antilog [ 2 .9828]
= 0.09612 m ≈ 0.096 m
∴
r2 = 9.6 cm
Ans: The distance of the coin from the axis so that
it does not slip-off is 9.6 cm.
∴
Example 4.5
A string breaks under tension of 10 kgwt. If the
string is used to revolve a body of mass 1.2 kg in a
horizontal circle of radius 50 cm, what is the
maximum speed with which the body can be
revolved? What is its period then?
Solution:
Given:
Tension, Tmax = 10 × 9.8 = 98N,
m = 1.2 kg, r = 0.5 m
To find:
i.
Maximum speed (vmax)
ii.
Time Period (T)
mv 2max
Formulae: i.
Tmax =
r
2πr
ii.
v=
T
Calculation: From formula (i),
T ×r
v 2max = max
m
98 × 0.5
= 40.833
=
1 .2
∴
vmax = 40.83
= 6.39 m/s.
From formula (ii),
2 × 3.14 × 0.5 3.14
=
T=
6.39
6.39
∴
T = 0.4914 s
The maximum speed with which the
Ans: i.
body can be revolved is 6.39 m/s.
ii.
The period of revolution of the body is
0.4914 s.
Example 4.6
A tension of 15 kg-wt breaks a string that is used
to revolve a bob of mass 1.5 kg in a horizontal
circle of radius 45 cm. Find the maximum speed
with which the bob can be revolved and with
what period?
Solution:
Given:
Tension T′ = 15 kg-wt
= 15 × 9.8 = 147 N,
m = 1.5 kg,
r = 45 cm = 0.45 m
i.
Maximum speed (vmax)
To find:
ii.
Period (T)
mv 2max
Formulae: i.
T′ =
r
2πr
ii. T =
v
11
Target Publications Pvt. Ltd.
Calculation:
From formula (i),
T '× r
147 × 0.45
v 2max =
=
m
1.5
= 44.1
∴
vmax = 44.1
∴
vmax = 6.64 m/s
From formula (ii),
2 × 3.14 × 0.45
T=
6.64
∴
T = antilog [log (2) + log (3.14)
+ log (0.45) – log (6.64)]
= antilog [0.3010 + 0.4969
+ 1 .6532 – 0.8222]
= antilog [0.4511 – 0.8222]
= antilog [0.4511 + 1 .1778]
= antilog [ 1 .6289]
= 0.4255
≈ 0.43
∴
T = 0.43 s
Ans: For the given bob:
i.
the maximum speed is 6.64 m/s.
ii.
the period of revolution is 0.43 s.
Example 4.7
A stone of mass 0.3 kg is tied to the end of a
string. It is whirled in a circle of radius 1m with a
speed of 40 rev min−1. If the string can withstand
a maximum tension of 200 N, then find the
tension and maximum speed with which the stone
can be whirled.
Solution:
Given:
m = 0.3 kg, r = 1 m,
40
2
n = 40 rpm =
rps = rps,
3
60
2
4π
ω = 2πn = 2π × =
rad s−1,
3
3
Tmax = 200 N
To find:
i.
Tension (T)
ii.
Maximum speed (vmax)
Formulae: i.
Tension, T = mrω2
mv 2max
ii.
Tmax =
r
Calculation:
From formula (i),
∴
⎛ 4π ⎞
T = 0.3 × 1 × ⎜ ⎟
⎝ 3 ⎠
T = 5.258 N
12
Std. XII Sci.: Physics Numericals
∴
Now, for maximum velocity, tension is
maximum.
From formula (ii),
Tmax × r
200 ×1
2000
=
=
vmax =
0.3
3
m
⎧1
⎫
vmax = antilog ⎨ ⎡⎣log ( 2000 ) − log ( 3) ⎤⎦ ⎬
⎩2
⎭
⎧1
⎫
= antilog ⎨ [3.3010 − 0.4771]⎬
⎩2
⎭
⎧1
⎫
= antilog ⎨ [ 2.8239]⎬
⎩2
⎭
= antilog {1.41195}
= antilog {1.4120}
∴
vmax = 25.82 ms−1
Ans: For the stone whirled in a circle:
i.
the tension in the string is 5.258 N.
ii.
the maximum speed is 25.82 m/s.
Example 4.8
A certain string breaks under a tension of
45 kg-wt. A mass of 100 g is attached to this
string of length 500 cm and whirled in a
horizontal circle. Find the maximum number of
revolutions per second without breaking the
string.
[Mar 92]
Solution:
Given:
T = 45 kg-wt = 45 × 9.8 N,
m = 100 g = 100 × 10−3 g = 0.1 kg,
r = 500 cm = 5 m
To find:
Maximum number of revolution per
second (n)
mv 2
Formulae: i.
F=T=
r
ii. v = r ω = r (2πn)
Calculation:
From formula (i) and (ii),
m(2 π r n) 2
= m × 4π2 n2r
T=
r
T
∴
n2 =
4 π2 r m
∴
n=
=
2
=
45 × 9.8
4 × (3.14) 2 × 5 × 0.1
45 × 4.9
( 3.14 )
2
45 × 4.9
3.14
Target Publications Pvt. Ltd.
Chapter 01: Circular Motion
⎧1
⎩2
⎫
⎭
1
1
⎤⎫
− log ( 4 ) − log ( 3.14 ) − log ( 0.53) ⎥ ⎬ × 10+17
2
2
⎦⎭
= antilog ⎨ ⎡⎣log ( 45 ) + log ( 4.9 ) ⎤⎦ − log ( 3.14 ) ⎬
⎧1
⎫
= antilog ⎨ [1.6532 + 0.6902] − 0.4969 ⎬
⎩2
⎭
⎧1
⎫
= antilog ⎨ ( 2.3434 ) − 0.4969 ⎬
⎩2
⎭
= antilog {1.1717 – 0.4969}
= antilog (0.6748)
∴
n = 4.73 Hz
Ans: The maximum number of revolutions of the
mass without breaking the string are 4.729.
Example 4.9
An electron of mass 9 × 10−31 kg moves in a
circular orbit of radius 5.3 × 10−11 m around a
proton in a hydrogen atom. If the charges are
1.6 × 10−19 C each, find the frequency of
revolution.
Solution:
Given:
me = 9 × 10−31 kg,
r = 5.3 × 10−11 m,
qe = qp = 1.6 × 10−19 C
(magnitudes only)
To find:
Frequency of revolution (n)
1 qeqp
Formula: F =
.
= 4π2 mrn2
4πε0 r 2
Calculation:
From formula,
qeqp
qeqe
1
1
.
n=
=
×
2
3
2
2
4π mr 4πε0
4π mr × r 4πε 0
=
qe
1
1
×
×
2
r
4π mr 4πε 0
=
1.6 ×10 −19
9 ×109
×
2
5.3×10−11
4 × ( 3.14 ) × 9 ×10−31 × 5.3×10−11
1
⎡
⎤2
1.6
109
n =
× 10–8 × ⎢
⎥
2
5.3
⎢⎣ 4 × ( 3.14 ) × 0.53 × 10−30 × 10−11 ⎥⎦
1
⎡
⎤2
1.6
10+50
× 10–8 × ⎢
=
⎥
2
5.3
⎢⎣ 4 × ( 3.14 ) × 0.53 ⎥⎦
1
⎤2
1
1.6 ⎡
=
×⎢
⎥ × 10+17
5.3 ⎢⎣ 4 × ( 3.14 )2 × 0.53 ⎥⎦
⎡
⎪⎧
= ⎨antilog ⎢ log (1.6 ) − log ( 5.3)
⎪⎩
⎣
⎧
1
⎡
= ⎨antilog ⎢ 0.2041 − 0.7243 − × 0.6021
2
⎣
⎩
1
⎤⎫
−0.4969 − × 1.7243⎥ ⎬ × 10+17
2
⎦⎭
⎡
⎪⎧
= ⎨antilog ⎢ 1.4798 − 0.3011 − 0.4969
⎪⎩
⎣
1
⎤⎫
− 2 + 1.7243 ⎥ ⎬ × 10+17
2
⎦⎭
(
)
{
}
= antilog ⎡⎣ 1.1787 − 0.4969 − 1.8621⎤⎦ × 10+17
{
}
= {antilog ⎡⎣ 2.6818 + 0.1379 ⎤⎦} × 10
= {antilog ⎡⎣ 2.8197 ⎤⎦} × 10
= antilog ⎡⎣ 2.6818 + 1 − 0.8621⎤⎦ × 10+17
+17
17
∴ n = 6.604 × 10–2 × 1017
∴ n = 6.604 × 1015 Hz
Ans: The frequency of revolution of electron is
6.604 × 1015 Hz.
Section 5: Motion of a Vehicle along a Curved
Unbanked Road
Example 5.1
A vehicle driving at 54 km/hr can safely negotiate
a curved road of 50 m radius. If the road is
unbanked, find the co-efficient of friction
between the road surface and the tyres.
Solution:
Given:
r = 50 m, v = 54 km/h
54 × 5
=
= 15 m/s
18
To find:
Coefficient of friction (µ)
Formula: v = µrg
Calculation:
From formula,
(15) 2
µ=
50 × 9.8
∴
µ = antilog [2 log (15) – log (50) – log (9.8)]
= antilog [2 × 1.1761 – 1.6990 – 0.9912]
= antilog [2.3522 – 2.6902]
= antilog ⎡⎣ 1.6620 ⎤⎦
∴
µ = 0.459
Ans: The coefficient of friction between the road
surface and the tyres of the vehicle is 0.459.
13
Target Publications Pvt. Ltd.
Example 5.2
A car is travelling at 30 km/h in a circle of radius
60 m. What is the minimum value of µs for the
car to make the turn without skidding?
Solution:
30 ×1000
Given:
v = 30 km/h =
m/s
60 × 60
=
To find:
Formula:
25
m/s,
3
r = 60 m
Minimum value of coefficient of
friction (µs)
mv 2
µsmg =
r
Calculation:
From formula,
252
v2
µs =
= 2
rg 3 × 60 × 9.8
∴
µs = antilog [2 log (25) – 2 log (3)
– log (60) – log (9.8)]
= antilog [2 × 1.3979 – 2 × 0.4771
– 1.7782 – 0.9912]
= antilog [2.7958 – 0.9542 – 1.7782
– 0.9912]
= antilog [2.7958 – 3.7236]
= antilog ⎡⎣ 1.0722 ⎤⎦
∴
µs = 0.1181
Ans: The minimum value of coefficient of friction
is 0.1181.
Example 5.3
A cyclist speeding at 18 km h−1 on a level road
takes a sharp circular turn of radius 3 m without
reducing the speed and without bending towards
the centre of the circular path. The coefficient of
static friction between the tyres and the road is
0.1. Will the cyclist slip while taking the turn?
Solution:
5
Given:
v = 18 km/h = 18 ×
= 5 m/s,
18
r = 3 m, µ = 0.1
To find:
Whether the cyclist can take the turn or
not without slipping
Formula: Maximum safe speed of cyclist
vmax = µrg
Calculation:
From formula,
vmax = 0.1× 3× 9.8
14
Std. XII Sci.: Physics Numericals
= 2.94
∴
vmax = 1.715 m/s ∴
V > Vmax
Ans: As the actual speed is greater than the
maximum safe speed, the cyclist will slip
while taking the turn.
Example 5.4
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.
Solution:
Given:
n = 120 r.p.m
= 120 rev/60 s
= 2 rev /s
= 2 Hz
r = 1.5 cm
= 1.5 × 10−2 m
g = 9.8 m/ s2
To find:
Coefficient of frictions (µ)
Formula: mrω2 = µ.m.g
OR
rω2 = µg
Calculation:
From formula,
ω2 r
µ=
g
=
∴
∴
∴
4π 2 n 2 r
(2πn) 2 × r
=
g
g
….[∵ ω = 2πn]
4 × (3.14) 2 × (2) 2 × 1.5 × 10−2
9.8
2
16 × (3.14) × 1.5 × 10−2
µ =
9.8
µ = antilog ⎡⎣ log (16 ) + 2log ( 3.14 )
µ=
{
}
+ log (1.5 ) − log ( 9.8 ) ⎤⎦ × 10–2
= {antilog [1.2041 + 2 × 0.4969
+0.1761 − 0.9912]} × 10–2
{
= antilog [1.2041 + 0.998 + 0.1761
−0.9912]} × 10–2
{
}
= {antilog [1.3828]} × 10
= antilog [ 2.3740 − 0.9912] × 10–2
–2
= 2.414 × 101 × 10–2
∴
µ = 0.2414
Ans: The coeffcient of friction between the coin
and disc is 0.2414.
Target Publications Pvt. Ltd.
Example 8.9
A small stone tied to an inextensible string of
negligible mass is rotated in a circle of radius 2 m
in a vertical plane. Find the speed at a horizontal
point on the circle.
Solution:
Given:
r=2m
To find:
Speed of stone (vM)
Formula: vM = 3rg
Calculation: From formula,
vM = 3 × 2 × 9.8
=
58.8 = 7.668
∴
vM ≈ 7.67 m/s.
Ans: The speed of the stone at a horizontal point on
the circle is 7.67 m/s.
Example 8.10
A stone of mass 10 kg tied with a string of length
0.5 m is rotated in a vertical circle. Find the total
energy of the stone at the highest position.
Solution:
Given:
m = 10 kg, r = 0.5 m
To find:
Total energy (TE)
5
Formula: (T.E.)H = mrg
2
Calculation: From formula,
5
(T.E.)H =
× 10 × 0.5 × 9.8
2
= 5 × 5 × 4.9
∴
(T.E.)H = 122.5 J
Ans: The total energy of the stone at the highest
position is 122.5 J.
Example 8.11
A 500 g particle tied to one end of a string is
whirled in a vertical circle of circumference 14 m.
If the tension at the highest point of its path is
2 N, what is its speed?
Solution:
m = 500 g = 0.5 kg,
Given:
circumference = 14 m = 2πr
14
∴
r=
m,
2π
T=2N
To find:
Speed (v)
mv 2
– mg
Formula: T =
r
26
Std. XII Sci.: Physics Numericals
Calculation:
From formula,
mv 2
= T + mg
r
1/ 2
⎡r
⎤
v = ⎢ ( T + mg ) ⎥
⎣m
⎦
⎡
14
⎤
1/ 2
= ⎢
× (2 + 0.5 × 9.8) ⎥
⎣ 2π× 0.5
⎦
= [4.4586 × 6.9]1/2
∴
v = 5.546 m/s
Ans: The speed of the particle at the highest point is
5.546 m/s.
Example 8.12
A bridge over a railway track is in the form of a
circular arc of radius 55 m. What is the limiting
speed with which a car can cross the bridge so
that no contact is lost if the centre of gravity of
the car is 0.4 m above the road?
Solution:
r = 55m, h = 0.4 m
Given :
Total distance R = 55 + 0.4
= 55.4 m
To find:
Limiting speed (v)
mv 2
= mg
Formula: F =
R
Calculation: From formula,
v = Rg = 55.4 × 9.8
∴
v = 23.3 m/s
Ans: The limiting speed of the car is 23.3 m/s.
Section 9: Kinematical Equations
Example 9.1
On the application of a constant torque, a wheel
is turned from rest through 400 radian in 10 s.
Calculate its angular acceleration.
Solution:
Given:
θ = 400 rad, ω1 = 0, t = 10 s
To find:
Angular acceleration (α)
1 2
Formula: θ = ω1t +
αt
2
Calculation: From formula,
1
400 = 0 + × α × (10)2
2
2 × 400
∴
α=
100
∴
α = 8 rad s−2.
Ans: The angular acceleration of the wheel is
8 rad/s2.
Target Publications Pvt. Ltd.
Example 9.2
The initial angular speed of a wheel is 4 rad/s. If
its angular displacement is 200 rad, find its
angular acceleration after 10 s.
Solution:
Given:
ω1 = 4 rad/s,
θ = 200 rad,
t = 10 s
To find:
Angular acceleration (α)
1 2
Formula: θ = ω1t +
αt
2
Calculation: From formula,
1
200 = 4 × 10 + × α × 10 × 10
2
∴
200 = 40 + 50α
160
α=
50
∴
α = 3.2 rad/s2
Ans: The angular acceleration of the wheel is
3.2 rad/s2.
Example 9.3
A particle moves along a circular path of length
15 cm with a constant angular acceleration of
4 rad/s2. If the initial angular speed of the particle
is 5 rad/s, find the angular displacement of the
particle in 5 sec.
Solution:
Given:
Circumference = 15 cm
= 0.15 m,
2
α = 4 rad/s , ω1 = 5 rad/s, t = 5s
To find:
Angular displacement (θ)
1 2
αt
Formula: θ = ω1t +
2
Calculation: From formula,
1
θ=5×5+
× 4 × (5)2
2
= 25 + 50
∴
θ = 75 rad
Ans: The angular displacement of the particle is
75 rad.
Example 9.4
A gramophone turntable rotating at an angular
velocity of 3 rad/s stops after one revolution. Find
the angular retardation.
Solution:
Given:
ω1 = 3 rad/sec, θ = 1 rev. = 2π rad,
ω2 = 0
To find:
Angular retardation (α)
Chapter 01: Circular Motion
Formula:
ω22 = ω12 + 2αθ
Calculation: From formula,
0 = 32 + 2α × 2π
∴
0 = 9 + 4 απ
∴
4απ=−9
−9
∴
α=
4π
−9
α=
4 × 3.142
9
=−
12.568
∴
α = − 0.716 rad/s2
Ans: The angular retardation of the turn table is
− 0.716 rad/s2.
Example 9.5
A fly wheel gains a speed of 240 rpm in 3 s.
Calculate the change in its angular speed in three
seconds.
Solution:
0
rps = 0 rps,
Given:
n1 = 0 rpm =
60
240
n2 = 240 rpm =
= 4 rps,
60
t = 3s
To find:
Change in angular speed (∆ω)
Formula: ∆w = ω2 – ω1= 2π (n2 – n1)
Calculation: By using formula,
∆ω = 2π(4) − 2π(0)
= 8π rad/s
= 8 × 3.14
∴
∆ω = 25.12 rad/s
Ans: The change in the angular speed of the
flywheel is 25.12 rad/s.
Example 9.6
A wheel rotating at 500 r.p.m. slows down to 400
r.p.m. at a constant rate of 5 rad/s2. What is the
angular displacement of the wheel?
Solution:
500
r.p.s.,
Given:
n1 = 500 r.p.m =
60
400
n2 = 400 r.p.m =
r.p.s.,
60
α = 5 rad/s2
To find:
Angular displacement (θ)
27
Target Publications Pvt. Ltd.
Formula:
ω22 – ω12 = 2αθ
Calculation:
ω1 = 2πn1
500
rad/s
60
ω1 = 52.33 rad/s
ω2 = 2πn2
400
rad/s
= 2π ×
60
ω2 = 41.87 rad/s
From formula,
Std. XII Sci.: Physics Numericals
4.
What is the angular displacement of the
minute hand of a clock in 20 minutes?
5.
Find the angular displacement of the tip of
second hand of clock whose length is 5 cm
and sweeps an arc of length 7.25 cm.
6.
If a body moves on a circular path of radius
10 m with a linear velocity of 2 m/s, find its
angular displacement in 15 s.
7.
Calculate the angular velocity of earth due to
it’s spin motion.
8.
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 by it to complete (1/5)th of the
circular path.
9.
A body is fixed to one end of a rope and the
other end of the rope is fixed to a peg on the
ground. The body rotates with a uniform
angular velocity of
5 rad/s around the peg.
If the radius of the circle in which the body
rotates is 50 cm, what is its linear velocity?
= 2π ×
ω2 − ω12
θ= 2
2α
(52.33) 2 − (41.87) 2
….(1)
2×5
Now, (52.33)2 − (41.87)2
= antilog [2 log(52.33)] − antilog[2 log(41.87)]
= antilog[2 × 1.7187] − antilog [2 × 1.6219]
= antilog[3.4374] − antilog[3.2438]
= 2738 – 1753
= 985
Substituting the value in eq (1), we get
985
θ=
10
∴
θ = 98.5 rad
Ans: The angular displacement of the wheel is
98.5 rad.
=
Section 2: Angular Acceleration
10.
Determine the angular acceleration of a
rotating body which slows down from 500
r.p.m. to rest in 10 second.
11.
The speed of a wheel increases from 600 rpm
to 1200 rpm in 20 s. What is its angular
acceleration and how many revolutions does it
make during this time?
12.
The motor of an engine is rotating about its
axis with an angular velocity of 100 rpm. It
comes to rest in 15 s after being switched-off.
Assuming constant angular retardation,
calculate the number of revolutions made by it
before coming to rest.
13.
A car is moving at a speed of 72 kmh−1. The
diameter of its wheels is 0.50 m. If the wheels
are stopped in 20 rotations by applying brakes,
calculate the angular retardation produced by
the brakes.
Problems for Practice
Section 1: Angular
Displacement,
Relation
Between Linear Velocity and Angular
Velocity
1.
Calculate the angular speed and linear speed
of the tip of a second hand of a clock, if the
second hand is 4 cm long.
2.
A flywheel turns at 600 rpm. Compute the
angular speed at any point on the wheel and
the tangential speed 0.5 m from the centre.
3.
A particle moves along a circular path of
radius 20 cm making 240 revolutions per
minute. Find the angular and linear velocities
of the particle.
28
Target Publications Pvt. Ltd.
23.
Find the centripetal force required to revolve a
body of mass 0.2 kg along a circular path of
radius 2 metre at a uniform rate of 300
revolutions per minute in the horizontal plane.
24.
A particle moves in a circle of radius 5 cm and
has its velocity of rotation increased by 100
rotations in 5 seconds. Calculate its angular
acceleration and tangential acceleration.
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 µ = 0.2, find the
angular velocity of the disc at which the coin
is about to slip-off. [g = 9.8 m/s2.]
25.
A car is moving along a circular road at a
speed of 20 m/s. The radius of the circular
road is 10 m. If the speed is increased at the
rate of 30 m/s2, what is the resultant
acceleration?
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.
26.
A 0.5 kg mass tied to the end of a string is
whirled in a horizontal circle of radius 1.5 m
with
a
angular
speed
of
4 rad/s. The maximum tension that the string
can withstand is 250 N. What is the maximum
speed with which the stone can be whirled?
What is the tension in the string?
27.
A coin placed on a revolving disc at a distance
of 25 cm from the axis does not slip-off when
the disc revolves at 80 r.p.m. The coin is then
placed 40 cm from the axis. How fast can the
disc revolve without the coin slipping-off?
Section 3: Centripetal
Acceleration
14.
15.
16.
Chapter 01: Circular Motion
and
Tangential
If a particle has a radial acceleration of
123.62 m/s2 and a tangential acceleration of
91.41 m/s2, then what is its linear
acceleration?
17.
A 0.5 kg mass is rotated in a horizontal circle
of radius 20 cm. Calculate the centripetal
acceleration acting on it, if its angular speed of
revolution is 0.8 rad/s.
18.
To simulate acceleration of large rockets,
astronauts are spun at the end of a long
rotating beam of length 9.8 m. What angular
velocity is required to generate a centripetal
acceleration
8 times the acceleration due to gravity?
(g = 9.8 m/s2)
19.
A motor car is travelling at 20 m/s on a
circular curve of radius 100 m. It is increasing
its speed at the rate of 5 m/s2. What is its
acceleration?
Section 5: Motion of a Vehicle along a Curved
Unbanked Road
28.
A body of mass 10 kg moves in a circle of
radius 1 m with constant angular speed of 2
rad/sec. Find the period of revolution and the
centripetal force.
With what maximum speed can 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]
29.
The breaking tension of a string of length 2
metre is 24 kg wt. A body of mass 2 kg is
attached to its one end and whirled in a
horizontal circle with the end fixed. Find the
maximum frequency of revolution possible.
Also find the velocity of the body when the
string breaks.
A car travelling at 18 km/hr just rounds a
curve without skidding. If the road is plane
and the coefficient of friction between the
road surface and the tyres is 0.25, find the
radius of the curve.
30.
A car moves on a level turn having a radius of
62 m. What is the maximum speed the car can
take without skidding if the coefficient of
static friction between the tyre and the road is
2.2?
31.
What should be the coefficient of friction
between the tyres and the road, when a car
travelling at 60 km h−1 makes a level turn of
radius 40 m?
Section 4: Centripetal and Centrifugal Forces
20.
21.
22.
A 1 kg mass tied at the end of a 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.
29
Target Publications Pvt. Ltd.
Section 6: Banking of Roads
32.
A cyclist speeding at 6 ms−1 in a circle of 18 m
radius makes an angle θ with the vertical.
Calculate θ. Also determine the minimum
possible value of the coefficient of friction
between the tyres and the ground.
33.
A bicycle and rider together have a mass of
90 kg. Find the angle which the rider must
make with the horizontal in travelling around
a curve of 36 m radius at 44/3 m/s.
34.
A curved road of radius 90 m is to be banked
so that a vehicle may move along the curved
road with a uniform speed of 75.6 km/hr
without any tendency to slip-off. What must
be the angle of banking? [g = 9.8 m/s2]
35.
The circumference of a track is 1.256 km.
Find the angle of banking of the track if the
maximum speed at which a car can be driven
safely along it is 25 m/s.
36.
37.
38.
39.
40.
A motorcyclist goes round a circular race
course of diameter 320 m at 144 km h−1. How
far from the vertical must he lean inwards to
keep his balance? [Take g = 10 ms−2]
A train has to negotiate a curve of radius
400 m. By how much should the outer rail be
raised with respect to the inner rail for a speed
of 48 kmh−1? The distance between the rails is
1 m.
The C.G. of a taxi is 1.5 m above the round
and the distance between its wheels is 2 m.
What is the maximum speed with which it can
go round an unbanked curve of radius 100 m
without being turned upside down? What
minimum value of coefficient of friction
would be required at this speed?
A turn on a road has a radius of 55 m and is
banked at an angle of 15° with the horizontal.
If the coefficient of friction between a car tyre
and the road surface is 0.28, what is the
maximum speed of the car on the turn?
An aircraft with its wings banked at 22°
performs a horizontal loop at a speed of
250 m/s. What is the radius of the loop?
30
Std. XII Sci.: Physics Numericals
41.
A railway track has a radius of curvature of
1.8 km. If the optimum velocity of a train on
the track is 25 m/s, what is the angle of
banking? If the elevation of the outer track
above the inner track is 0.05 m, what is the
distance between the two tracks?
42.
What is the maximum speed at which a car
can turn on a road which is banked at an angle
of 10° if the radius of the turn is 25 m and the
coefficient of friction between the tyres and
the road surface is 0.4?
43.
The maximum speed of a bus on turn of radius
50 m is 60 km/hr. If the coefficient of friction
between the tyres and the road is 0.5, what is
the angle of banking?
Section 7: Conical Pendulum
44.
A conical pendulum has a bob of mass 200 g
and a length of 50 cm. If the radius of the
circle traced by the bob is 25 cm, find the
velocity of the bob and the period of
pendulum. [g = 9.8 m/s2]
45.
A conical pendulum has a length of 1.5 m and
a bob of mass 50 g. The bob completes 20
revolutions in 45 s. Find the radius of the
circular path traced by the bob and the tension
in the thread. [g = 9.8 m/s2]
46.
The bob of a conical pendulum has mass 50 g
and it moves in a horizontal circle whose
radius is 24 cm. If the length of the string is
75 cm, what is the tension in the string?
47.
A conical pendulum has a bob of mass 300 g
and string of length 115 cm. If the angle made
by the string with the vertical is 12°, what is
the period of circular motion of the bob? What
is the tension in the string?
Section 8: Vertical Circular Motion
48.
A 1.2 kg body attached to a string is whirled
in a vertical circle of radius 3 m. What
minimum speed, vm must it have at the top of
the circle so as not to depart from the circular
path? Find its velocity at lowest point and at a
midway position?
Target Publications Pvt. Ltd.
49.
50.
51.
52.
A vehicle weighing 4000 kg is going over a
convex bridge, the radius of curvature of
which is 30 m. The height of the centre of
gravity of the vehicle from the ground is
1.2 m. If the velocity of the vehicle is 50.4
km/h, calculate the thrust of the vehicle on the
road at the highest point. Also find the greatest
speed at which the vehicle can cross the bridge
without losing contact with the road at the
highest point. [g = 9.8 m/s2]
A roadway bridge over a canal is in the form
of an arc of radius 15m. What is the maximum
speed with which a car can cross the bridge
without leaving the ground at the highest
point? [g = 9.8 m/s2]
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 19.5 m. Find the
limiting velocity at which a car can cross the
bridge without losing contact with the road at
the highest point, if the centre of gravity of the
car is 0.5 m from the ground.
A motorcyclist rides in a vertical circle in a
hollow sphere of radius 5m. Find the
minimum angular speed required so that he
does not lose contact with the sphere at the
highest point.
[g = 9.8 m/s2]
53.
A stone of mass 0.3 kg is tied to one end of a
string 0.8 m long and rotated in a vertical
circle. At what speed of the ball will the
tension in the string be zero at the highest
point of the circle? What would be the tension
at the lowest point in this case?
[Given g = 9.8 ms−2]
54.
A bucket containing water is tied to one end of
a rope 8 m long and rotated about the other
end in vertical circle. Find the minimum
number of rotations per minute in order that
water in the bucket may not spill.
[g = 9.8 m/s2]
55.
56.
What is the minimum speed required at the
bottom to perform a vertical loop, if the radius
of the death-well in a circus is 25 m?
A particle of 0.1 kg is whirled at the end of
string in a vertical circle of radius 1m at
constant speed of 7 m/s. Find the tension in
the string at the highest position.
Chapter 01: Circular Motion
57.
58.
59.
60.
61.
A stone weighing 0.5 kg tied to a rope of
length 0.5 m revolves along a circular path in
a vertical circle. What maximum speed does it
possess at the bottom where tension is 45 N?
A body weighing 0.4 kg tied to a string is
projected with a velocity of 15 ms−1. The body
starts whirling in a vertical circle. If the radius
of the circle is 1.2 m, find tension in the string
when the body is
i.
at the top of the circle and
ii.
at the bottom of the circle.
A stone of mass 8 kg tied with a string of
length 1m is rotated in vertical circle. Find the
total energy of stone at the lowest point.
A body of mass 4 kg is rotating in a vertical
circle at the end of a string of length 0.6 m.
Calulate the difference in K.E. at the top and
bottom of the circle.
A 0.6 kg bob attached to a string is whirled in
a vertical circle at a speed of 15 rad/s. If the
radius of the circle is 1.8 m., what is the
tension in the string when the bob is at the
bottom and at the top of the circle?
Section 9: Kinematical Equations
62.
The spin drier of a washing machine revolving
at 15 r.p.s slows down to 5 r.p.s. while making
50 revolutions. Find the angular acceleration
and time taken to complete the revolutions.
63.
A particle moves along a circular path of
radius 10 cm with a constant angular
acceleration of 2 rad/s2. If the initial angular
speed of the particle is 5 rad/s,
find the
i.
angular speed of the particle after 10 s.
ii.
angular displacement of the particle in
10 s.
iii. tangential acceleration of the particle.
64.
A particle moves in a circular path with an
angular velocity of 9 rad/s. The particle then
accelerates at a constant rate of 3 rad/s2. In
what time will the particle be displaced by an
angle of 60 rad ?
65.
A curved road having diameter 0.04 km is
banked at an angle θ. If a car can travel at a
maximum speed of 50 km/hr on the curved
road and if the coefficient of friction between
the tyres and the road is 0.26, what is the angle
of banking?
31
Target Publications Pvt. Ltd.
Board Problems
1.
A coin kept on a rotating gramophone disc
with its centre 5 cm away from the centre of
the disc, just begins to slip when the frequency
of rotation of the disc reaches 60 r.p.m.
Calculate the coefficient of static friction.
[Oct 84]
[g = 9.8 m/s2]
2.
A motor cyclist rides in vertical circle in a
hollow sphere of radius 3 m. Find the
minimum speed required, so that he does not
lose contact with the sphere at the highest
[Mar 87]
point. [g = 9.8 m/s2]
3.
Find the angle which the bicycle and its rider
make with the vertical, when going at
18 km/hr around a curved road of radius 10m
on level road. [Given g = 9.8 m/s2] [Oct 87]
4.
A 0.5 kg mass is rotated in a horizontal circle
of radius 20 cm. Calculate the centripetal force
acting on it, if its angular speed of revolution
[Mar 88]
is 0.6 rad/s.
5.
Find the maximum speed of a car which can
be safely driven along a curve of radius
100 m, if the coefficient of friction between
tyres and road is 0.2. [g = 9.8 m/s2]
[Oct 88]
6.
A motor cyclist rides in vertical circles in a
hollow sphere of radius 5m. Find the
minimum speed required, so that he does not
lose contact with the sphere at the highest
[Oct 89]
point.[g = 9.8 m/s2]
7.
The minute hand of a clock is 10 cm long.
Calculate linear speed of the tip of the minute
[Oct 92]
hand.
8.
A body of mass 2.0 kg is tied to the free end
of a string of length 1.5 m and is revolved
along a horizontal circle with the other end
fixed. The body makes 300 r.p.m. Calculate
the linear velocity, centripetal acceleration and
[Mar 93]
force acting on it.
9.
A body of mass 2 kg is tied to the end of a
string 2m long and revolved in a horizontial
circle. If the breaking tension of the string is
400 N, calculate the maximum velocity of the
[Mar 94]
body.
32
Std. XII Sci.: Physics Numericals
10.
One end of a string 1m long is fixed and a
body of mass 500 g is tied to the other end. If
the breaking tension is 98N, find the
maximum angular velocity of the body that
the string can withstand when rotated in a
[Oct 94]
horizontal circle.
11.
Two bodies of equal masses are connected by
a string passing through a hole at the top of a
smooth table, one of them resting on the table
and the other hanging underneath. Find the
number of revolutions per minute that the
body on the table should perform in a circle of
radius 0.2m in order to balance the other body.
[Oct 94]
[g = 9.8 m/s2]
12.
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 the angle made by
the string with the vertical. Also find the
tension in the supporting thread and the speed
[Oct 94]
of the bob.
13.
What is the angular velocity of an hour hand
[Mar 95]
of length 1 cm of a watch ?
14.
A motor cyclist moves around a circular track
of length 628 m with a speed of 72 km/hr.
Find to what extent he must lean inwards to
[Oct 95]
keep the balance. [g = 9.8 m/sec2]
15.
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.
16.
An object of mass 400 gm is whirled in a
horizontal circle of radius 2 m. If it performs
60 r.p.m., calculate the centripetal force acting
[Oct 96, Feb 01]
on it.
17.
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]
18.
Find the angle of banking of a 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]
Target Publications Pvt. Ltd.
19.
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 centre of the vehicle
is 0.5 m from the ground. [Given: g = 9.8
[Oct 01]
m/s2]
20.
If the frequency of revolution of an object
changes from 2Hz to 4Hz in 2 seconds,
calculate its angular acceleration.
[Oct 03]
21.
Find the angular speed of rotation of the earth
so that bodies on the equator would feel no
[Feb 03]
weight.
22.
The minute hand of a clock is 8 cm long.
Calculate the linear speed of an ant sitting on
[Mar 05]
its tip.
23.
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]
24.
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
[Feb 06]
180 km/hr. [g = 9.8 m/s2]
25.
26.
27.
28.
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]
An object of mass 2 kg attached to a wire of
length 5 m is revolved in a horizontal circle. If
it makes 60 r.p.m., find its
a.
angular speed
b.
linear speed
c.
centripetal acceleration
[Mar 09]
d.
centripetal force
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]
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
[Oct 13]
string. (g = 9.8 m/s2)
Chapter 01: Circular Motion
Multiple Choice Questions
Section
1:
Angular Displacement, Relation
Between Linear Velocity and
Angular Velocity
1.
A body rotating in a horizontal circle has a
linear velocity of 25 m/s and its angular
velocity is 0.5 rad/s. What is the radius of the
circle in which it rotates?
(A) 0.05 m
(B) 0.5 m
(C) 0.25 m
(D) 2.5 m
2.
The angular velocity of a point on the rim of a
wheel is 8π rad/s. How many revolution per
minute (r.p.m) does the wheel make ?
(A) 250
(B) 240
(C) 100
(D) 4
3.
The angular velocity of the minute hand of a
clock is
π
π
rad/s
(B)
rad/s
(A)
60
3600
2π
2π
(C)
rad/s
(D)
rad/s
60
3600
4.
The angular displacement of the second’s
hand of a clock in 30 s is
(A) 6.28 rad
(B) 3.14 rad
(C) 1.07 rad
(D) 60 rad
5.
If the length of the minute hand of a clock is
15 cm, then its linear velocity is
(A) 1.2 × 10–4 m/s
(B) 2.62 × 10–4 m/s
(D) 3.6 × 10–2 m/s
(C) 3.2 × 10–4 m/s
Section 2: Angular Acceleration
6.
If the angular speed of a particle changes from
2.5 rad/s to 5 rad/s in 20 s, then its angular
acceleration is
(B) 0.15 rad/s2
(A) 0.125 rad/s2
2
(C) 1.5 rad/s
(D) 12.5 rad/s2
7.
A body moves from rest through 100 rad in
15 s. Its angular acceleration is
(B) 0.89 rad/s2
(A) 1.15 rad/s2
2
(D) 0.42 rad/s2
(C) 0.75 rad/s
8.
A particle moving with an angular speed of
4.2 rad/s accelerates to move with an angular
speed of 6.4 rad/s. If its angular displacement
is 50 rad, then its angular acceleration is
(B) 1.2 rad/s2
(A) 1.8 rad/s2
2
(D) 0.23 rad/s2
(C) 0.32 rad/s
33
Target Publications Pvt. Ltd.
9.
A body performing UCM changes from 50
r.p.m. to 100 r.p.m. in 20 s. Its angular
acceleration is
(B) 0.26 rad/s2
(A) 0.52 rad/s2
2
(D) 0.02 rad/s2
(C) 0.12 rad/s
Std. XII Sci.: Physics Numericals
16.
A 250 g mass tied to the end of a string is
whirled in a circle of radius 115 cm with a speed
of 3 r.p.s. What is the tension in the string?
(A) 150.6 N
(B) 112.50 N
(C) 102.05 N
(D) 89.65 N
17.
A stone of mass 0.2 kg is tied to the end of a
string and whirled in a horizontal circle. The
maximum tension in the string is 300 N. If the
maximum velocity is 4.2 m/s, what is the
radius of the circle in which it is whirled?
(A) 1.18 cm
(B) 1.82 cm
(C) 2.11 cm
(D) 4.36 cm
18.
A 0.2 kg bob is tied to the end of a string and
whirled in a horizontal circle of radius 60 cm. The
string breaks when the tension is 200 N. What is
the time period corresponding to the maximum
velocity with which it can be whirled?
(A) 0.189 s
(B) 0.154 s
(C) 0.116 s
(D) 0.009 s
19.
A coin placed on a rotating turntable at a
distance of 12 cm from the axis does not slip
when the turntable rotates steadily at a speed
of α r.p.m. If the coefficient of friction
between the coin and the turntable is 0.4, what
is the value of α?
(A) 54.6 r.p.m.
(B) 60.3 r.p.m.
(C) 72.9 r.p.m.
(D) 90.1 r.p.m.
20.
A 0.3 kg stone is tied to the end of a string and
whirled in a horizontal circle of radius 0.6 m.
The string just breaks when the velocity of
rotation is 10 m/s. What is the maximum
tension which the string can withstand?
(A) 25 N
(B) 40 N
(C) 50 N
(D) 100 N
21.
A 0.5 kg mass is tied to the end of a string and
whirled in a circle of radius 75 cm. If the tension
in the string is 15 kg wt, what is the linear
velocity of rotation?
(A) 16.23 m/s
(B) 14.85 m/s
(C) 12.91 m/s
(D) 8.23 m/s
Section 3: Centripetal and Tangential Acceleration
10.
What is the centripetal acceleration of a
0.25 kg body which rotates in a horizontal
circle of radius 12 cm with an angular velocity
of 1.1 rad/s?
(B) 0.8 m/s2
(A) 1.2 m/s2
2
(C) 5.14 m/s
(D) 0.15 m/s2
11.
A particle moves in a horizontal circle of
radius 2 cm with a constant angular speed of
14 r.p.s. What is its centripetal acceleration?
(B) 154.59 m/s2
(A) 165.23 m/s2
(C) 138.26 m/s2
(D) 101.19 m/s2
12.
If the radial acceleration of a body is 29 m/s2
and its tangential acceleration is 12 m/s2, what
is its linear acceleration?
(B) 29.62 m/s2
(A) 31.38 m/s2
2
(D) 19.26 m/s2
(C) 24.19 m/s
13.
A body revolves in a circle of radius 14 cm
with an angular velocity of 30 rad/s. If its
tangential acceleration is 0.24 m/s2, then what
is the ratio of its centripetal acceleration to its
tangential acceleration?
(A) 600 : 1
(B) 550 : 1
(C) 525 : 1
(D) 1 : 400
14.
A particle rotates in a horizontal circle with an
angular velocity of 20 rad/s. If its tangential
acceleration is 0.2 m/s2 and the ratio of its
centripetal acceleration to its tangential
acceleration is 600 : 1, what is the radius of
the circle in which the particle rotates?
(A) 20 cm
(B) 30 cm
(C) 15 cm
(D) 10 cm
Section 4 : Centripetal and Centrifugal Forces
Section 5: Motion of a Vehicle along a Curved
Unbanked Road
22.
15.
A coin just slips when placed at a distance of
2 cm from the axis on a rotating turntable. At
what distance from the axis will it just slip
when the angular velocity of the turntable is
halved?
(A) 1 cm
(B) 2 cm
(C) 4 cm
(D) 8 cm
34
A vehicle takes a turn on an unbanked road,
the radius of the turn being 42 m. If the
coefficient of friction between the tyres and
the road surface is 0.6, then the car travelling
at which of the following velocities will slip
while taking the turn?
(A) 8 m/s
(B) 12 m/s
(C) 15 m/s
(D) 18 m/s
Target Publications Pvt. Ltd.
23.
A car travels at a speed of 60 km/h on a level
road on which there is an unbanked turn of
radius 50 m. For which of the following values
of coefficient of friction between the tyres and
the road surface will the car not skid while
turning?
(A) 0.1
(B) 0.2
(C) 0.4
(D) 0.6
24.
A road is being designed for vehicles to travel
at 80 km/h. The coefficient of friction between
the road surface and the tyres is 0.2. What
should be the radius of an unbanked level turn
on the road for vehicles to move without
skidding?
(A) 251.90 m
(B) 500 m
(C) 128.2 m
(D) 202.95 m
Section 6: Banking of Road
25.
26.
27.
The turn on a banked road has a radius of
25 m. If the coefficient of friction between a
car tyre and the road surface is 0.32 and if the
angle of banking is 17°, what is the maximum
velocity at which a car can turn?
(A) 12.61 m/s
(B) 11.47 m/s
(C) 13.04 m/s
(D) 10.93 m/s
The centre of gravity of a car is 0.62 m above
the ground. It can turn along a track which is
1.24 m wide and has radius r. If the greatest
speed at which the car can take the turn is
22.02 m/s, what is the value of r?
(A) 62.32 m
(B) 52.01 m
(C) 49.48 m
(D) 38.62 m
A motorcyclist executes a horizontal loop at a
speed of 65 km/h while himself making an
angle of 12° with the horizontal. What is the
radius of the loop?
(A) 156.6 m
(B) 161.2 m
(C) 173.4 m
(D) 180.9 m
28.
A car executes a turn of radius 22 m on a
banked road while travelling at a speed of
45 km/h. If the height of the outer edge above
the inner edge of the road is 1.1 m, what is the
breadth of the road?
(A) 2.104 m
(B) 1.875 m
(C) 1.626 m
(D) 1.213 m
29.
The angle of banking of a turn of radius 75 m
on a road is 30°. What is the speed at which a
car can turn along this curve?
(A) 20.6 m/s
(B) 22.3 m/s
(C) 24.6 m/s
(D) 28.3 m/s
Chapter 01: Circular Motion
30.
If the angle of banking of a road is 32° and if a
turn has radius 72 m, what is the maximum
speed at which a car can turn, given the
coefficient of friction between the car tyres
and the road is 0.34?
(A) 32.3 m/s
(B) 29.4 m/s
(C) 28.6 m/s
(D) 25.1 m/s
Section 7: Conical Pendulum
31.
The mass of the bob of a conical pendulum is
100 g and the length of the string is 150 cm. If
the radius of the circle in which the bob rotates
is 22 cm and if the thread makes an angle of
15° with the vertical, calculate the velocity of
the bob.
(A) 0.76 m/s
(B) 0.84 m/s
(C) 1.2 m/s
(D) 2.4 m/s
32.
The length of the string of a conical pendulum
is 90 cm and its bob moves in a circular path
of radius 25 cm. What is the tension in the
string if the bob has mass 150 g?
(A) 2.81 N
(B) 1.53 N
(C) 1.25 N
(D) 0.92 N
33.
A conical pendulum of length 120 cm moves
making an angle of 16° with the vertical. What
is the period of circular motion of the bob?
(A) 2.155 s
(B) 2.523 s
(C) 3 s
(D) 4.009 s
34.
A conical pendulum has a bob of mass 200 g
and it moves in horizontal circle making an
angle of 8° with the vertical. What is the
tension in the string?
(A) 2.113 N
(B) 1.979 N
(C) 1.504 N
(D) 1.216 N
Section 8: Vertical Circular Motion
35.
A gymnast hangs from one end of a rope and
executes vertical circular motion. The gymnast
has a mass of
40 kg and the radius of the
circle is 2.5 m. If the gymnast whirls himself
at a constant speed of 6 m/s, what is the
tension in the rope at the lowest point?
(A) 1200 N
(B) 968 N
(C) 782 N
(D) 500 N
36.
A 100 g mass attached to the end of a string is
rotated in a vertical circle of radius 40 cm.
What is the total energy of the mass at the
highest point?
(A) 1.5 J
(B) 1.1 J
(C) 0.98 J
(D) 0.76 J
35
Target Publications Pvt. Ltd.
37.
38.
39.
40.
A body of negligible mass tied to the end of a
string is rotated in a vertical circle of radius
80 cm. At a horizontal point on the circle,
what is the speed of the body?
(A) 4.85 m/s
(B) 5.25 m/s
(C) 6.58 m/s
(D) 8.01 m/s
A fighter aircraft flying at 400 km/h executes
a vertical circular loop of radius 100 m. The
pilot has a weight of 60 kg. What is the force
with which the pilot presses his seat when the
aircraft is at the highest point?
(A) 8500.9 N
(B) 7200.34 N
(C) 6900.2 N
(D) 6819.40 N
A stone of mass m tied to a string of length
60 cm is whirled in a vertical circle. If the total
energy of the stone at the highest position is
250 J, what is the value of m?
(A) 20 kg
(B) 18 kg
(C) 17 kg
(D) 15 kg
A bucket containing water is tied to one end of
a rope of length 1.8 m. It is rotated about the
other end so that water does not spill out.
What is the minimum velocity of the bucket at
which this can happen?
(A) 4.2 m/s
(B) 5 m/s
(C) 6.2 m/s
(D) 7.4 m/s
41.
A motorcyclist riding in a vertical sphere does
not lose contact with the sphere at the highest
point. If the minimum angular velocity at
which the motorcyclist achieves this is
1.5 rad/s, what is the radius of the sphere?
(A) 3.21 m
(B) 4.36 m
(C) 4.63 m
(D) 5.56 m
42.
A 200 g mass is whirled in a vertical circle
making 60 revolutions per minute. What is the
tension in the string at the top of the circle if
the radius of the circle is 0.8 m?
(A) 4.35 N
(B) 4.51 N
(C) 5.4 N
(D) 6.22 N
43.
A road bridge is in the form of a circular arc of
radius 18 m. What is the limiting speed with
which a car can traverse the bridge without
losing contact at the highest point if the centre
of gravity of the car is 0.4 m above the
ground?
(A) 13.428 m/s
(B) 14.314 m/s
(C) 15.206 m/s
(D) 16.009 m/s
36
Std. XII Sci.: Physics Numericals
Section 9: Kinematical Equations
44.
A disc starts from rest and then accelerates at a
constant rate of 12 rad/s2. If the angular
displacement of the disc is 30 rad, then what is
the final velocity with which the disc rotates?
(A) 32.71 rad/s
(B) 26.83 rad/s
(C) 24.01 rad/s
(D) 18.23 rad/s
45.
A flywheel executing 600 r.p.m. stops after one
complete rotation. What is its angular retardation?
(B) – 31.4 rad/s2
(A) – 3.14 rad/s2
2
(C) – 314 rad/s
(D) – 3140 rad/s2
46.
A wheel rotating at 10 rad/s accelerates to
12 rad/s. If its angular displacement is 40 rad,
what is its angular acceleration?
(B) 0.55 rad/s2
(A) 1.5 rad/s2
2
(D) 0.15 rad/s2
(C) 0.44 rad/s
47.
An engine required 4 s to go from a speed of
600 rpm to 1200 rpm with a constant
acceleration. The number of revolutions made
by it in this time is
(A) 6.0
(B) 600
(C) 60
(D) 6000
Answers to Problems for Practice
1.
2.
3.
4.
6.
8.
10.
12.
14.
16.
18.
20.
21.
22.
24.
26.
27.
29.
31.
33.
35.
37.
39.
41.
43.
45.
1.047 × 10−1 rad/s, 4.188 × 10−3 m/s
62.84 rad/s, 31.42 m/s
25.12 rad/s, 5.024 π m/s
2.093 rad
5.
1.45 rad
3.0 rad
7.
7.27 × 10−5 rad/s
0.2 rad/s, 6.28 s
9.
2.5 m/s
11. 3.14 rad s−2, 300
5.23 rad/s2
12.5
13. − 25.5 rad s−2
15. 40 π rad/s2 ; π m/s2
153.74 m/s2
2
50 m/s
17. 0.128 m/s2
2.828 rad/s
19. 6.403 m/s2
3.14 s, 40 N
1.22 rev/s; 15.33 m/s
5 m/s
23. 394.38 N
2.8 rad/s
25. 30
vmax = 27.39 m/s, T = 12 N
63.25 r.p.m.
28. 10.84 m/s
10.2 m
30. vmax = 36.56 m/s
0.71
32. 11°32′, 0.2041
58°38′
34. 26° 34′
17° 41′
36. 45°
0.0454 m
38. 25.56 ms−1, 0.67
vmax = 17.12 m/s
40. r = 15,785 m
θ = 2°2′, l = 1.412 m 42. vmax = 12.325 m/s
θ = 2° 58′
44. 119 cm/s; 1.32 s
0.78 m; 0.57 N
46. T = 0.517 N
Target Publications Pvt. Ltd.
47.
48.
49.
51.
53.
55.
57.
58.
59.
61.
63.
64.
Period = 2.13 s
Tension = 3 N
i.
5.422 m/s
iii. 9.391 m/s
14072 N; 17.49 m/s
14 m/s.
2.8 ms−1, 17.64 N
35 ms−1
6.332 m/s.
i.
71.08 N
196 J
237.12 N, 248.88 N
i.
ω2 = 25 rad/s
iii. at = 0.2 m/s2
t=4s
Chapter 01: Circular Motion
ii
12.124 m/s
50.
52.
54.
56.
12.124 m/s
1.4 rad/s
10.57 rev/min
3.92 N
ii.
60.
62.
ii.
78.92 N
47.04 J
12.56 rad/s2, 5 sec
θ =150 rad
65.
θ = 29° 53′
Answers to Board Problems
1.
3.
5.
7.
8.
9.
10.
11.
12.
13.
15.
17.
19.
21.
23.
25.
26.
27.
0.2012
2.
5.422 m/s
14° 19′
4.
0.036 N
14.0 m/s
6.
7.0 m/s
1.744 × 10−4 m/s
47.1 m/s2, 1.479 × 103 m/s2, F = 2.958 × 103N
20 m/s
14 rad/s
66.88 r.p.m.
36° 52′; 1.268 N; 1.485 m/s
14. 22° 12′
1.454 × 10−4 rad/s
15°13′, 0.2625 m
16. 31.55 N
29° 51′
18. 2° 11′; 6.08 cm
6.93 m/s
20. 6.28 rad/s2
–3
1.237 × 10 rad/s
22. 1.395 ×10–2 cm/s
24. 23°2′
10 rad/s2
24.5 m/s
a.
6.28 rad/s
b.
31.4 m/s
2
d.
394.384 N.
c.
197.192 m/s
15.65 m/s
28. 1.47 NPractice
Answers to Multiple Choice Questions
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
41.
45.
(B)
(B)
(B)
(C)
(A)
(B)
(C)
(A)
(A)
(A)
(B)
(C)
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
42.
46.
(B)
(A)
(D)
(B)
(B)
(D)
(C)
(B)
(B)
(D)
(A)
(B)
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
(D)
(B)
(B)
(D)
(A)
(D)
(A)
(A)
(B)
(C)
(A)
(C)
4.
8.
12.
16.
20.
24.
28.
32.
36.
40.
44.
(B)
(D)
(A)
(C)
(C)
(A)
(B)
(B)
(C)
(A)
(B)
37