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Physics Numericals
STD. XII
No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio
Video Cassettes or electronic, mechanical including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.
Edition: June 2014
Prof. Umakant N. Kondapure (M.Sc., B.Ed., Solapur)
Mr. Collin Fernandes (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, Mithagar Road, Mulund (E), Mumbai - 400 081 Off.Tel: 022 – 6551 6551 Email: mail@targetpublications.org Website: www.targetpublications.org
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Publications Pvt. Ltd.
Physics
Numericals STD. XII
Written according to the revised syllabus (2012-2013) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.
TEID : 737
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
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 sub-topic-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 : mail@targetpublications.org
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. No.
Unit Page No.
1 Circular Motion 1
2 Gravitation 38
3 Rotational Motion 67
4 Oscillations 100
5 Elasticity 129
6 Surface Tension 149
7 Wave Motion 165
8 Stationary Waves 185
9 Kinetic Theory of Gases and
Radiation 212
10 Wave Theory of light 253
Sr. No. Unit Page
No.
11 Interference and Diffraction 270
12 Electrostatics 297
13 Current Electricity 329
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
1
Target Publications Pvt. Ltd. Chapter 01: Circular Motion
Formulae Section 1: Angular Displacement, Relation
Between Linear Velocity and Angular Velocity
1. Angular velocity:
i. ω = rv
where, v = linear velocity r = radius of the circle along which
particle performs circular motion.
ii. ω = t
∆θ∆
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.
iv. ω = T2π
where, T = period of revolution of particle performing circular motion.
2. Angular displacement: θ = ωt 3. Time period:
i. T = 2 rvπ ii. T = 2π
ω
4. Frequency of revolution:
n = 1T 2
ω=
π
5. Linear velocity: i. v = rω ii. v = 2πnr Section 2: Angular Acceleration 1. Angular acceleration:
i. α = t
∆ω∆
where, ∆ω = change in the angular velocity of a particle in circular motion during a time interval ∆t.
ii. α = 2π ⋅ nt
∆∆
where, ∆n = change in frequency of the particle in circular motion during a time interval ∆t.
Section 3: Centripetal and Tangential Acceleration 1. Centripetal (or radial) acceleration:
ar = r
v2
= ωv = rω2 2. Tangential acceleration: aT = α r 3. Resultant or total acceleration: a = 2 2
t r t ra a 2a a cos+ + θ where, θ = angle made by ar with at
a = 2 2t ra a+ when θ = 90°.
4. For U.C.M.:
a = ar = 2vr
= ωv
∵ at = 0 Section 4: Centripetal and Centrifugal Forces 1. Centripetal force:
i. Fc = r
mv2
ii. Fc = mvω
iii. Fc = mrω2 iv. Fc = 4π2mrn2
v. Fc = 2
24 mr
Tπ
where, m = mass of particle performing circular motion
Section 5: Motion of a Vehicle along a Curved Unbanked Road
1. The necessary centripetal force:
Fc = µmg = 2mv
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.
01 Circular Motion
Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals
2
2. The maximum velocity with which a vehicle can take a turn safely without skidding:
v = rgµ 3. The maximum angular velocity with which
a vehicle can take a turn safely without skidding:
ω = grµ
Section 6: Banking of Roads For motion of vehicles along a banked curve road: 1. The proper velocity or optimum velocity: v = rg tanθ where, θ = angle of banking 2. The maximum velocity without skidding:
vmax = s
s
tanrg1 tan⎡ ⎤µ + θ⎢ ⎥−µ θ⎣ ⎦
where, µs = coefficient of friction between the tyres
of the vehicle and surface of the road 3. Angle of banking:
θ = tan−12v
rg⎛ ⎞⎜ ⎟⎝ ⎠
or tan θ =2v
rg⎛ ⎞⎜ ⎟⎝ ⎠
4. 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 = drg2H
where, H = height of centre of gravity (C.G.) of the vehicle from the road.
Section 7: Conical Pendulum 1. Linear speed of bob: v = rg tanθ 2. Angular velocity:
ω = gcosθl
= g tanrθ
3. Periodic time:
T = ωπ2 = 2π cos
gθl
= 2π hg
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. 4. Tension in the string:
T = mgcosθ
Section 8: Vertical Circular Motion 1. Velocity at any point in vertical circular
motion: i. vP = 2
Lv − 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.
2. Relation between velocities at different points in vertical circular motion:
i. 2Lv = 2
Hv + 4gr ii. 2
Mv = 2Hv + 2gr
3
Target Publications Pvt. Ltd. Chapter 01: Circular Motion
3. Tension at: i. Any point P,
TP = 2mv
r+ mg cosθ
ii. Lowest point L,
TL = 2Lmv
r + mg =
2Hmv
r + 5mg
iii. Highest point H,
TH = 2Hmv
r− mg =
2Lmv
r− 5 mg
iv. Midway point M,
TM = 2Mmv
r=
2Lmv
r− 2 mg
4. Total energy at any point:
E = 12
mv2 + mgr (1 − cos θ)
= 52
mgr
Section 9: Kinematical Equations Analogy between translatory motion and circular motion
No. Translatory Motion Circular Motion
1. Linear velocity →v = d r
dt
→
Angular velocity →ω= d
dt
→
θ
2. Linear acceleration
→a = 2
d v d rdt dt
→ →2
= Angular acceleration
→α = 2
d ddt dt
→ →2ω θ
=
3. Linear momentum p→
= m v→
Angular momentum →L = I
→ω
4. Linear impulse = F→
(∆t) = ∆→p Angular impulse =
→τ (∆t) = ∆L
→
5. Force →F = m
→a Torque
→τ = I
→α
6. Work W = F→
. →r Work W =
→
τ . →θ
7. Kinetic energy of translation
Et = 12
mv2
Kinetic energy of rotation
Er = 12
Iω2
8. Equations of linear motion Equations of rotational motion
i. v = u + at i. ω2 = ω1 + αt
ii. s = ut + 12
at2 ii. θ = ω1t + 21αt2
iii. v2 − u2 = 2 as iii. ω 22 −ω
21 = 2αθ
iv. sn = u + a2
(2n − 1) iv. θn = ω1 + 2α (2n −1)
Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals
4
Shortcuts 1. Angular acceleration (α): If number of rotations or revolutions is given,
then
α = 2tπ (n2 − n1) rad/s2
2. If linear velocity is given, apply ar = r
v2
and
if angular velocity is given, apply ar = ω2r to find 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,
2mv
r= µmg or
µ = 2v
rg =
2rgω =
2 24 n rg
π
5. Breaking tension is the maximum centripetal force which is given by the relation,
F = 2mv
r = mω2r
6. Speed of a vehicle on a banked road or circular turn depends upon the curvature of the road.
tan θ = 2v
rg
7. To avoid skidding, v ≤ rgµ where µ is the
coefficient of friction between the tyres and road.
µ = h2
d where d is the distance between two
wheels. 8. 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 = 2 2r−l
9. Use the expressions, Lv = 5gr and Tv = 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.
10. For looping the loop of radius r, the minimum
height from which the body should be released is given by,
h = 5r2
11. i. For a vehicle moving over a convex
bridge which is in the shape of an arc of a circle,
N = mg – 2mv
r
ii. For the motor cyclist at the upper most point in the globe of death in a circus,
N = 2mv
r – mg
where, N = normal reaction acting on the
vehicle or the motorcycle. 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 = 2 m 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.
5
Target Publications Pvt. Ltd. Chapter 01: Circular Motion
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,
θ = 2 153π×
∴ 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)
Formulae: i. ω = 2Tπ ii. v = rω
Calculation: From formula (i),
ω = 2Tπ =
6014.32×
∴ 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: Given: r = 15 cm = 0.15 m 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.
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: Given: r = 0.5 m, v = 0.2 m/s To find: Angular velocity (ω) Formula: v = rω Calculation: From formula,
ω = vr
= 0.20.5
∴ 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)
Formulae: i. T = 2 rvπ
ii. n = 1T
Calculation: From formula (i),
T = 22 3.14 10 10
20
−× × ×
= 3.14 × 10−2 ∴ T = 0.0314 s From formula (ii),
n = 1T
= 10.0314
∴ n = 31.85 s–1 Ans: i. The period of revolving particle is
0.0314 s ii. 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
Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals
6
To find: i. Angular velocity (ω) ii. Linear velocity of the tip (v)
Formulae: i. ω = 2Tπ
ii. v = rω Calculation: From formula (i),
ω = 2Tπ
= 2 3.143600×
∴ 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. 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)
Formulae: i. ω = vr
ii. v = 2 r / 2t
π
Calculation: From formula (i),
ω = 5002000
∴ ω = 0.25 rad/s From formula (ii),
t = 2 3.14 20002 500
× ××
∴ t = 12.56 s Ans: i. The angular speed of the aircraft is
0.25 rad/s. ii. The time required by the aircraft to
complete half the circular path is 12.56 s.
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 = 7060
/s,
n2 = 130/min = 60
130 /s, t = 18 s
To find: Angular acceleration (α)
Formula: α = 2 1
tω −ω = ( )2 12 n n
tπ −
Calculation: From formula,
α = 2 12 n 2 nt
π − π
=
130 70260 6018
⎛ ⎞π −⎜ ⎟⎝ ⎠
=
130 70260
18
−⎛ ⎞π⎜ ⎟⎝ ⎠
= 2 6060 18π××
3.1429 9π
= =
∴ α = 0.349 rad/s2 Ans: The angular acceleration of the particle is
0.349 rad/s2. 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:
Given: n1 = 60 r.p.m = 6060
= 1 rev/s,
n2 = 180 r.p.m = 18060
= 3 rev/s,
t = 20 s To find: Angular acceleration (α)
Formula: α = 2 1
tω −ω
Calculation: From formula,
α = 2 12 n 2 nt
π − π = ( )2 3 120
π −
7
Target Publications Pvt. Ltd. Chapter 01: Circular Motion
= 2 3.142 220
× ×
∴ α = 3.1425
∴ 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:
Given: n1 = 420 r.p.m. = 42060
r.p.s.
= 7 r.p.s α = −2 rad/s2, 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
t = 142π = 14
2 × 22
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.
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
n = 30 rpm = 3060
rps = 0.5 rps
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)
Formulae: i. ω = 2Tπ
ii. v = rω iii. ar = rω2 Calculation: From formula (i),
ω = 2 3.147200×
∴ 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.
Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals
8
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 = 2Tπ ii. v = rω
iii. ar = rω2 Calculation: From formula (i),
ω = 7
2 3.143.156 10
××
∴ 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 = ( ) ( ){ }2antilog log 1.5 log 1.99⎡ ⎤+⎣ ⎦
× 1011 × 10–14 = [ ]{ }antilog 0.1716 2 0.2989+ × × 10–3
= [ ]{ }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.
Section 3: Centripetal and Tangential Acceleration 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
ii. ar = r
v2
Calculation: From formula (i), v = rω = 0.2 × 0.6 m/s From formula (ii),
ar = 2(0.2 0.6)
0.2×
∴ ar = 0.072 m/s2 Ans: The centripetal acceleration of the mass is
0.072 m/s2. 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]
= 4 × (3.142)2 ×214
25⎛ ⎞⎜ ⎟⎝ ⎠
× 0.8
= ( ) ( )( )
2 2
2
4 3.142 14 0.8
25
× × ×
= 9.908 ∴ ar ≈ 9.91 m/s2 Ans: The magnitude of centripetal acceleration is
9.91 m/s2.
9
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 = 2v
r
But v = 2 r (5)t
π = 10 rtπ
∴ π2 = 2 2
2
100 rrtπ
∴ r = 120 120100× = 144 m
Ans: The radius of the track is 144 m. 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 = 2 2r ta a+
Calculation: From formula,
a = 2 2(473.9) (47.13)+
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 = ( ) ( )2 2473.9 47.13+
= 52.246 10 2221× +
= 226821
= antilog ( )1 log 2268212⎡ ⎤⎢ ⎥⎣ ⎦
= antilog 1 5.35562⎡ ⎤×⎢ ⎥⎣ ⎦
= 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
= r 2 1
tω −ω⎛ ⎞
⎜ ⎟⎝ ⎠
= 0.2 40 219−⎛ ⎞
⎜ ⎟⎝ ⎠
= 0.2 3819⎛ ⎞⎜ ⎟⎝ ⎠
= 0.2 × 2 = 0.4 m /s2
Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals
10
∴ r
t
aa
= 3200.4
= 32004
∴ 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 = 2 2t ra a+ or a2 = 2
ta + 2ra
or ar = 2 2ta a−
Calculation: From formula,
ar = 2 2(52.3) (29.48)−
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 = ( ) ( )2 252.3 29.48−
= 2735 869−
= 1866
= antilog ( )1 log 18662⎡ ⎤⎢ ⎥⎣ ⎦
= antilog 1 3.27092⎡ ⎤×⎢ ⎥⎣ ⎦
= antilog [1.63545] ∴ ar = 43.1 m/s2 Ans: The radial acceleration of the body is
43.1 m/s2.
Section 4: Centripetal and Centrifugal Forces 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,
v = 90 km/h = 59018
× = 25 m/s
To find: Centripetal force (FCP)
Formula: FCP = 2mv
r
Calculation: From formula,
FCP = ( )21500 25250×
∴ 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 = 2mv
r ii. v = rg tanθ
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Target Publications Pvt. Ltd. Chapter 01: Circular Motion
Calculation: Substituting formula (ii) in (i),
FCP = ( )2
m rg tan
r
θ
= mg tan θ = 1 × 9.8 × tan 30
= 9.8 × 13
= 9.81.732
= 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
n1 = 60 r.p.m = 6060
= 1 r.p.s.
∴ ω1 = 2πn1 = 2π rad/s = 2 × 3.14 = 6.28 rad/s
n2 = 75 r.p.m. = 7560
r.p.s
= 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
21ω = mr2
22ω
Calculation: From formula,
r2 = 2
1 122
rωω
= ( )( )
2
2
0.15 6.28
7.85
×
∴ 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)
Formulae: i. Tmax = 2maxmv
r
ii. v = 2 rTπ
Calculation: From formula (i),
2maxv =
mrTmax ×
= 2.1
5.098× = 40.833
∴ vmax = 40.83 = 6.39 m/s. From formula (ii),
T = 39.6
5.014.32 ×× = 3.146.39
∴ T = 0.4914 s Ans: i. The maximum speed with which the
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 To find: i. Maximum speed (vmax) ii. Period (T)
Formulae: i. T′ = 2maxmv
r
ii. T = 2 rvπ
Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals
12
Calculation: From formula (i),
2maxv = T ' r
m× = 147 0.45
1.5×
= 44.1 ∴ vmax = 44.1 ∴ vmax = 6.64 m/s From formula (ii),
T = 2 3.14 0.456.64
× ×
∴ 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,
n = 40 rpm = 4060
rps = 23
rps,
ω = 2πn = 2π × 23
= 43π rad s−1,
Tmax = 200 N To find: i. Tension (T) ii. Maximum speed (vmax) Formulae: i. Tension, T = mrω2
ii. Tmax = 2maxmv
r
Calculation: From formula (i),
T = 0.3 × 1 × 24
3π⎛ ⎞
⎜ ⎟⎝ ⎠
∴ T = 5.258 N
Now, for maximum velocity, tension is maximum.
From formula (ii),
vmax = maxT rm× = 200 1
0.3× = 2000
3
∴ vmax = antilog ( ) ( )1 log 2000 log 32
⎧ ⎫⎡ − ⎤⎨ ⎬⎣ ⎦⎩ ⎭
= antilog [ ]1 3.3010 0.47712
⎧ ⎫−⎨ ⎬⎩ ⎭
= antilog [ ]1 2.82392
⎧ ⎫⎨ ⎬⎩ ⎭
= 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)
Formulae: i. F = T = 2mv
r
ii. v = r ω = r (2πn) Calculation: From formula (i) and (ii),
T = 2m(2 r n)
rπ = m × 4π2 n2r
∴ n2 = 2
T4 r mπ
∴ n = 2
45 9.84 (3.14) 5 0.1
×× × ×
= ( )2
45 4.93.14×
= 45 4.93.14×
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Target Publications Pvt. Ltd. Chapter 01: Circular Motion
= antilog ( ) ( ) ( )1 log 45 log 4.9 log 3.142
⎧ ⎫+ −⎡ ⎤⎨ ⎬⎣ ⎦⎩ ⎭
= antilog [ ]1 1.6532 0.6902 0.49692
⎧ ⎫+ −⎨ ⎬⎩ ⎭
= antilog ( )1 2.3434 0.49692
⎧ ⎫−⎨ ⎬⎩ ⎭
= 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)
Formula: F = 0
14πε
. e p2
q qr
= 4π2 mrn2
Calculation: From formula,
n = e p2 3
0
q q 1.4 mr 4π πε
= e e2 2
0
q q 14 mr r 4
×π × πε
= e2
0
q 1 1r 4 mr 4× ×
π πε
= ( )
19 9
11 2 31 11
1.6 10 9 105.3 10 4 3.14 9 10 5.3 10
−
− − −
× ××
× × × × × ×
n = 1.65.3
× 10–8 ×( )
129
2 30 11
104 3.14 0.53 10 10− −
⎡ ⎤⎢ ⎥
× × × ×⎢ ⎥⎣ ⎦
= 1.65.3
× 10–8 × ( )
1250
2
104 3.14 0.53
+⎡ ⎤⎢ ⎥
× ×⎢ ⎥⎣ ⎦
= 1.65.3
×( )
12
2
14 3.14 0.53
⎡ ⎤⎢ ⎥
× ×⎢ ⎥⎣ ⎦× 10+17
= ( ) ( )antilog log 1.6 log 5.3⎧ ⎡⎪ −⎨ ⎢⎪ ⎣⎩
( ) ( ) ( )1 1log 4 log 3.14 log 0.532 2
⎫⎤− − − ⎬⎥⎦⎭× 10+17
= 1antilog 0.2041 0.7243 0.60212
⎧ ⎡ − − ×⎨ ⎢⎣⎩
10.4969 1.72432
⎫⎤− − × ⎬⎥⎦⎭× 10+17
= antilog 1.4798 0.3011 0.4969⎧ ⎡⎪ − −⎨ ⎢⎪ ⎣⎩
( )1 2 1.72432
⎫⎤− + ⎬⎥⎦⎭× 10+17
= { }antilog 1.1787 0.4969 1.8621⎡ ⎤− −⎣ ⎦ × 10+17
= { } 17antilog 2.6818 1 0.8621 10+⎡ ⎤+ − ×⎣ ⎦
= { } 17antilog 2.6818 0.1379 10+⎡ ⎤+ ×⎣ ⎦
= { }antilog 2.8197⎡ ⎤⎣ ⎦ × 1017
∴ 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
= 54 518× = 15 m/s
To find: Coefficient of friction (µ) Formula: v = rgµ Calculation: From formula,
µ = 2(15)
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.
Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals
14
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:
Given: v = 30 km/h = 30 100060 60××
m/s
= 253
m/s,
r = 60 m To find: Minimum value of coefficient of
friction (µs)
Formula: µsmg = 2mv
r
Calculation: From formula,
µs = rgv2
= 2
2
253 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:
Given: v = 18 km/h = 18 × 518
= 5 m/s,
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 = rgµ Calculation: From formula, vmax = 0.1 3 9.8× ×
= 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,
µ = 2rgω
= 2(2 n) r
gπ × =
2 24 n rg
π ….[∵ ω = 2πn]
∴ µ = 2 2 24 (3.14) (2) 1.5 10
9.8
−× × × ×
∴ µ = 2 216 (3.14) 1.5 10
9.8
−× × ×
∴ µ = ( ) ( ){antilog log 16 2log 3.14⎡ +⎣
( ) ( ) }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 2.3740 0.9912− × 10–2
= [ ]{ }antilog 1.3828 × 10–2
= 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. Std. XII Sci.: Physics Numericals
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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 = 2 m 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)
Formula: (T.E.)H = 52
mrg
Calculation: From formula,
(T.E.)H = 52
× 10 × 0.5 × 9.8
= 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: Given: m = 500 g = 0.5 kg, circumference = 14 m = 2πr
∴ r = 142π
m,
T = 2 N To find: Speed (v)
Formula: T = 2mv
r – mg
Calculation: From formula,
2mv
r = T + mg
v = ( )1/2r T mg
m⎡ ⎤+⎢ ⎥⎣ ⎦
= 1/ 214 (2 0.5 9.8)
2 0.5⎡ ⎤× + ×⎢ ⎥π×⎣ ⎦
= [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: Given : r = 55m, h = 0.4 m Total distance R = 55 + 0.4 = 55.4 m To find: Limiting speed (v)
Formula: F = 2mv
R = mg
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)
Formula: θ = ω1t + 21 αt2
Calculation: From formula,
400 = 0 + 21× α × (10)2
∴ α = 2 400100×
∴ α = 8 rad s−2. Ans: The angular acceleration of the wheel is
8 rad/s2.
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Target Publications Pvt. Ltd. Chapter 01: Circular Motion
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)
Formula: θ = ω1t + 12
αt2
Calculation: From formula,
200 = 4 × 10 + 12× α × 10 × 10
∴ 200 = 40 + 50α
α = 16050
∴ 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, α = 4 rad/s2, ω1 = 5 rad/s, t = 5s To find: Angular displacement (θ)
Formula: θ = ω1t + 12
αt2
Calculation: From formula,
θ = 5 × 5 + 12
× 4 × (5)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)
Formula: 22ω = 2
1ω + 2αθ Calculation: From formula, 0 = 32 + 2α × 2π ∴ 0 = 9 + 4 απ ∴ 4 α π = − 9
∴ α = 94−π
α = 94 3.142
−×
= − 912.568
∴ 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:
Given: n1 = 0 rpm = 060
rps = 0 rps,
n2 = 240 rpm = 24060
= 4 rps,
t = 3s To find: Change in angular speed (∆ω) Formula: ∆w = ω2 – ω1= 2π (n2 – n1) Calculation: By using formula, ∆ω = 2 (4) 2 (0)π − π = 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:
Given: n1 = 500 r.p.m = 50060
r.p.s.,
n2 = 400 r.p.m = 40060
r.p.s.,
α = 5 rad/s2 To find: Angular displacement (θ)
Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals
28
Formula: 22ω – 2
1ω = 2αθ Calculation: ω1 = 2πn1
= 2π × 50060
rad/s
ω1 = 52.33 rad/s ω2 = 2πn2
= 2π × 40060
rad/s
ω2 = 41.87 rad/s From formula,
θ = 2 22 1
2ω −ω
α
= 2 2(52.33) (41.87)2 5−×
….(1)
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
θ = 98510
∴ θ = 98.5 rad Ans: The angular displacement of the wheel is
98.5 rad.
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.
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?
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.
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Target Publications Pvt. Ltd. Chapter 01: Circular Motion
Section 3: Centripetal and Tangential
Acceleration 14. 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?
15. 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.
16. 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?
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 4: Centripetal and Centrifugal Forces 20. 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.
21. 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.
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.
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 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 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 5: Motion of a Vehicle along a Curved Unbanked Road
28. 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. 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?
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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. 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]
37. 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.
38. 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?
39. 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?
40. 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?
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?
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49. 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]
50. 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]
51. 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.
52. 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. 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?
56. 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.
57. 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?
58. 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. 59. 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.
60. 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.
61. 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?
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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. [g = 9.8 m/s2] [Oct 84]
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 point. [g = 9.8 m/s2] [Mar 87]
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 is 0.6 rad/s. [Mar 88]
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 point.[g = 9.8 m/s2] [Oct 89]
7. The minute hand of a clock is 10 cm long.
Calculate linear speed of the tip of the minute hand. [Oct 92]
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 force acting on it. [Mar 93]
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 body. [Mar 94]
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 horizontal circle. [Oct 94]
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. [g = 9.8 m/s2] [Oct 94]
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 of the bob. [Oct 94]
13. What is the angular velocity of an hour hand
of length 1 cm of a watch ? [Mar 95] 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 keep the balance. [g = 9.8 m/sec2] [Oct 95]
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 rails is 1 m. [Oct 96]
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 on it. [Oct 96, Feb 01]
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.
[g = 9.8 m/s2] [Oct 98]
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Target Publications Pvt. Ltd. Chapter 01: Circular Motion
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 m/s2] [Oct 01]
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 weight. [Feb 03]
22. The minute hand of a clock is 8 cm long. Calculate the linear speed of an ant sitting on its tip. [Mar 05]
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 180 km/hr. [g = 9.8 m/s2] [Feb 06]
25. 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)
[g = 9.8 m/s2] [Mar 07] 26. 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 d. centripetal force [Mar 09] 27. 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?
[Given: g = 9.8 m/s2] [Oct 10] 28. 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]
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
(A) 60π rad/s (B)
3600π rad/s
(C) 260π rad/s (D) 2
3600π rad/s
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 (C) 3.2 × 10–4 m/s (D) 3.6 × 10–2 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
(A) 0.125 rad/s2 (B) 0.15 rad/s2 (C) 1.5 rad/s2 (D) 12.5 rad/s2 7. A body moves from rest through 100 rad in
15 s. Its angular acceleration is (A) 1.15 rad/s2 (B) 0.89 rad/s2 (C) 0.75 rad/s2 (D) 0.42 rad/s2 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
(A) 1.8 rad/s2 (B) 1.2 rad/s2 (C) 0.32 rad/s2 (D) 0.23 rad/s2
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9. A body performing UCM changes from 50 r.p.m. to 100 r.p.m. in 20 s. Its angular acceleration is
(A) 0.52 rad/s2 (B) 0.26 rad/s2 (C) 0.12 rad/s2 (D) 0.02 rad/s2 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?
(A) 1.2 m/s2 (B) 0.8 m/s2 (C) 5.14 m/s2 (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?
(A) 165.23 m/s2 (B) 154.59 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?
(A) 31.38 m/s2 (B) 29.62 m/s2 (C) 24.19 m/s2 (D) 19.26 m/s2 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 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
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 5: Motion of a Vehicle along a Curved
Unbanked Road 22. 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
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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. 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 26. 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 27. 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
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
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37. 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 38. 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 39. 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 40. 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
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? (A) – 3.14 rad/s2 (B) – 31.4 rad/s2 (C) – 314 rad/s2 (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?
(A) 1.5 rad/s2 (B) 0.55 rad/s2 (C) 0.44 rad/s2 (D) 0.15 rad/s2 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. 1.047 × 10−1 rad/s, 4.188 × 10−3 m/s 2. 62.84 rad/s, 31.42 m/s 3. 25.12 rad/s, 5.024 π m/s 4. 2.093 rad 5. 1.45 rad 6. 3.0 rad 7. 7.27 × 10−5 rad/s 8. 0.2 rad/s, 6.28 s 9. 2.5 m/s 10. 5.23 rad/s2 11. 3.14 rad s−2, 300 12. 12.5 13. − 25.5 rad s−2 14. 153.74 m/s2 15. 40 π rad/s2 ; π m/s2 16. 50 m/s2 17. 0.128 m/s2 18. 2.828 rad/s 19. 6.403 m/s2 20. 3.14 s, 40 N 21. 1.22 rev/s; 15.33 m/s 22. 5 m/s 23. 394.38 N 24. 2.8 rad/s 25. 30 26. vmax = 27.39 m/s, T = 12 N 27. 63.25 r.p.m. 28. 10.84 m/s 29. 10.2 m 30. vmax = 36.56 m/s 31. 0.71 32. 11°32′, 0.2041 33. 58°38′ 34. 26° 34′ 35. 17° 41′ 36. 45° 37. 0.0454 m 38. 25.56 ms−1, 0.67 39. vmax = 17.12 m/s 40. r = 15,785 m 41. θ = 2°2′, l = 1.412 m 42. vmax = 12.325 m/s 43. θ = 2° 58′ 44. 119 cm/s; 1.32 s 45. 0.78 m; 0.57 N 46. T = 0.517 N
37
Target Publications Pvt. Ltd. Chapter 01: Circular Motion
47. Period = 2.13 s Tension = 3 N 48. i. 5.422 m/s ii 12.124 m/s iii. 9.391 m/s 49. 14072 N; 17.49 m/s 50. 12.124 m/s 51. 14 m/s. 52. 1.4 rad/s 53. 2.8 ms−1, 17.64 N 54. 10.57 rev/min 55. 35 ms−1 56. 3.92 N 57. 6.332 m/s. 58. i. 71.08 N ii. 78.92 N 59. 196 J 60. 47.04 J 61. 237.12 N, 248.88 N 62. 12.56 rad/s2, 5 sec 63. i. ω2 = 25 rad/s ii. θ =150 rad iii. at = 0.2 m/s2 64. t = 4 s 65. θ = 29° 53′
Answers to Board Problems 1. 0.2012 2. 5.422 m/s 3. 14° 19′ 4. 0.036 N 5. 14.0 m/s 6. 7.0 m/s 7. 1.744 × 10−4 m/s 8. 47.1 m/s2, 1.479 × 103 m/s2, F = 2.958 × 103N 9. 20 m/s 10. 14 rad/s 11. 66.88 r.p.m. 12. 36° 52′; 1.268 N; 1.485 m/s 13. 1.454 × 10−4 rad/s 14. 22° 12′ 15. 15°13′, 0.2625 m 16. 31.55 N 17. 29° 51′ 18. 2° 11′; 6.08 cm 19. 6.93 m/s 20. 6.28 rad/s2
21. 1.237 × 10–3 rad/s 22. 1.395 ×10–2 cm/s 23. 10 rad/s2 24. 23°2′ 25. 24.5 m/s 26. a. 6.28 rad/s b. 31.4 m/s c. 197.192 m/s2 d. 394.384 N. 27. 15.65 m/s 28. 1.47 NPractice
Answers to Multiple Choice Questions 1. (B) 2. (B) 3. (D) 4. (B) 5. (B) 6. (A) 7. (B) 8. (D) 9. (B) 10. (D) 11. (B) 12. (A) 13. (C) 14. (B) 15. (D) 16. (C) 17. (A) 18. (B) 19. (A) 20. (C) 21. (B) 22. (D) 23. (D) 24. (A) 25. (C) 26. (C) 27. (A) 28. (B) 29. (A) 30. (B) 31. (A) 32. (B) 33. (A) 34. (B) 35. (B) 36. (C) 37. (A) 38. (D) 39. (C) 40. (A) 41. (B) 42. (A) 43. (A) 44. (B) 45. (C) 46. (B) 47. (C)
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