ENGINEERING PHYSICS I & II

DIPLOMA COURSE IN ENGINEERING AND TECHNOLOGY

FIRST & SECOND SEMESTER

A Publication under
Government of Tamil Nadu
Distribution of Free Textbook Programme
(NOT FOR SALE)

Untouchability is a sin
Untouchability is a crime
Untouchability is an inhuman

DIRECTORATE OF TECHNICAL EDUCATION

GOVERNMENT OF TAMILNADU
© Government of Tamilnadu
First Edition – 2020

CHAIRPERSON
Thiru. K. VIVEKANANDAN, I.A.S.,
Commissioner of Technical Education
Directorate of Technical Education
Guindy, Chennai - 600 025.

Co-ordinator Authors
Dr. M.S. Padmanabhan Thiru L. JOHN LOUIS
Principal (i/c) Lecturer (S.G) / Civil,
Central Polytechnic College Government Polytechnic College,
Chennai – 113. Nagercoil

Convener: Thiru. M. MOHANA DAS,

Thiru. AL.M. PERUMAL PILLAI, Lecturer (S.G) / Physics,
Lecturer (S.G) / Physics, Kamaraj Polytechnic College,
Government Polytechnic College, Pazhavilai, Nagercoil.
Nagercoil.
Thiru. D. NAVILAN,
Lecturer (S.G) / Physics,
Sankar Polytechnic College,
Sankar Nagar.

Thiru. S. SABAREESH,
Lecturer / Physics,
Sree Krishna Polytechnic College,
Nagercoil.

Dr. R. PAZHANI,
Lecturer / Physics,
Government Polytechnic College,
Nagercoil.

This book has been prepared by the Directorate of Technical Education

This book has been printed on 60 G.S.M Paper

Through the Tamil Nadu Text book and Educational Services Corporation

Printed by Web Offset at:

ii
 THE NATIONAL ANTHEM

FULL VERSION

Jana-gana-mana-adhinayaka jaya he
Bharata-bhagya-vidhata
Punjaba-Sindhu-Gujarata-Maratha-
Dravida-Utkala-Banga
Vindhya-Himachala-Yamuna-Ganga
Uchchhala-jaladhi-taranga
Tava Subha name jage,TavaSubhaasisa mage,
Gahe tava jaya-gatha.
Jana-gana-mangala-dayaka jaya he
Bharata-bhagya-vidhata.
Jaya he, jaya he, jaya he,
Jaya jaya jaya jaya he.

- Rabindranath Tagore

SHORT VERSION

Jana-gana-mana-adhinayaka jaya he
Bharata-bhagya-vidhata.
Jaya he, jaya he, jaya he,
Jaya jaya jaya jaya he.

AUTHENTIC ENGLISH TRANSLATION OF

THE NATIONAL ANTHEM

Thou art the ruler of the minds of all people,

Thou dispenser of India's destiny.
Thy name rouses the hearts of the Punjab, Sind,
Gujarat and Maratha, of Dravida, Orissa and Bengal
It echoes in the hills of the Vindhyas and Himalayas,
mingles in the music of the Yamuna and Ganges
and is chanted by the waves of the Indian Sea.
They pray for Thy blessings and sing Thy praise
The saving of all people waits in Thy hand,
Thou dispenser of India's destiny.
Victory, Victory, Victory to Thee

iii
 THE NATIONAL INTEGRATION PLEDGE

“I solemnly pledge to work with dedication to preserve and strengthen the

freedom and integrity of the nation.”
“I further affirm that I shall never resort to violence and that all differences
and disputes relating to religion, language, region or other political or
economic grievances should be settled by peaceful and
constitutional means.”

INVOCATION TO GODDESS TAMIL

Bharat is like the face beauteous of Earth clad in wavy seas;

Deccan is her brow crescent-like on which the fragrant 'Tilak' is the
blessed Dravidian land.
Like the fragrance of that 'Tilak' plunging the world in joy supreme
reigns Goddess Tamil with renown spread far and wide.
Praise unto ‘You, Goddess Tamil, whose majestic youthfulness, inspires
awe and ecstasy’.

iv
 FOREWORD

With the concept of Global Village after liberalisation and globalisation, our country India
has become one of the most sought after destinations by many Multi National Companies
for investment. The rapid growth in various fields of Engineering and Technology like
Information technology, Automobile engineering, Infrastructure development, etc.,
demands much needed Human Resources.

The growth of various Industries demand qualified Technicians at different categories. The
Diploma level education of Engineering and Technology offered by the various Polytechnic
Colleges cater these Technician needs of the Industries.

Every Polytechnic college is offering diploma programmes in many fields of Engineering

like Mechanical, Electrical, Civil, Electronics, Information Technology and special fields like
Aeronautics, Chemical, Printing, Leather and textile technology etc.,

For all branches of study, the first year curriculum is common. The syllabus provides the
necessary bridge between the school education and engineering education which the
students pursue from their second year of study. For successful completion of engineering
diploma with flying colours, a thorough knowledge of basics is very much essential.

The Content of this Engineering Physics I and Engineering Physics Il provide necessary
basic ideas and concepts in a bright manner. Real life applications and practical examples
are included in this text wherever required. The experiments to be performed by the
student in I and II semester Engineering Physics practical are also included in this text for
the benefit of students.

The students should give due importance in understanding the various basic concepts with
which only they can shine in their career, rather than reading them by heart.

- AUTHORS

v
 40013 ENGINEERING PHYSICS – I

DETAILED SYLLABUS

Unit Name of the Topics Hours

I S I UNITS AND STATICS

1.1 UNITS AND MEASUREMENTS: - 5
Unit – Definition - Fundamental Quantities – Definition - Seven fundamental
quantities; their SI units and symbol for the units - Supplementary quantities
- plane angle and solid angle; their SI units and symbol for the units Derived
physical quantities.
Dimensional formula for length, mass and time-derivation of dimensional
formula for area, volume, density, velocity, momentum, acceleration, force,
impulse, work or energy and power. Uses of Dimensional formula.
Conventions followed in SI Units Multiples & sub-multiples and prefixes of
units – Unit conversions (Horse Power to watt & calorie to joule) -
Applications of the method of dimensional analysis.
1.2 STATICS: -
10
Scalar and vector quantities – Definitions and examples – Concurrent forces
and coplanar forces – Definition - Resolution of a vector into two
perpendicular components - Resultant and equilibrant – Definitions -
Parallelogram law of forces-statement - Expressions for magnitude and
direction of the resultant of two forces acting at a point with an acute angle
between them - Lami’s theorem - Statement and explanation - Experimental
verification of parallelogram law of forces and Lami’s theorem. Simple
problems based on expressions for magnitude and direction of resultant.
Moment of a force - Clockwise and anti-clockwise moments - Principle of
moments – Couple – Torque acting due to a Couple – Experimental
determination of mass of the given body using principle of moments - Solved
problems.
II PROPERTIES OF MATTER
5
2.1 ELASTICITY: -
Elastic and plastic bodies – Definition - stress, strain – Definitions – Hooke’s
law – statement - three types of strain – Elastic and plastic limit – Young’s
modulus, Bulk modulus, Rigidity modulus – Definitions - Uniform and non-
uniform bending of beams - Experimental determination of the Young’s
modulus of the material of a beam by uniform bending method - Poisson’s
ratio - Simple problems based on stress, strain and Young’s modulus –
Applications of elasticity.
2.2 VISCOSITY: - 5
Viscosity – Definition - Coefficient of viscosity - Definition, SI unit and
dimensional formula - Stream line flow, turbulent flow – Explanation - Critical
velocity – Reynolds number – Definition - Experimental comparison of
coefficient of viscosity of two low viscous liquids – Terminal velocity –
vi
 Definition - Experimental determination of coefficient of viscosity of a highly
viscous liquid by Stokes method – Practical applications of viscosity –
Practical applications of Stoke’s law.
2.3 SURFACE TENSION: - 5
Surface tension & angle of contact – Definitions - Expression for surface
tension of a liquid by capillary rise method - Experimental determination of
surface tension of water by capillary rise method – Practical applications of
capillarity. Simple problems based on expression for surface tension –
Applications of surface tension – Solved problems.
III DYNAMICS–I
3.1. STRAIGHT LINE MOTION: - 2
Introduction-Newton’s Laws of motion-Fundamental Equations of motion for
objects- horizontal motion-falling freely-thrown vertically upwards.
3.2 PROJECTILE MOTION: - 5
Projectile motion, angle of projection, trajectory, maximum height, time of
flight, and horizontal range–Definitions-Expressions for maximum height,
time of flight and horizontal range–Condition for getting the maximum range
of the projectile. Path of the projectile (the trajectory) is a Parabola - Simple
problems based on expressions for maximum height, time of flight and
horizontal range – Examples of projectile motion.
3.3 CIRCULAR MOTION: -
8
Circular motion, angular velocity, period and frequency of revolutions –
Definitions – Relation between linear velocity and angular velocity – Relation
between angular velocity, period and frequency – Normal acceleration,
centripetal force and centrifugal force – Definitions – Expressions for normal
acceleration and centripetal force. Banking of curved paths – Angle of
banking – Definition – Expression for the angle of banking of a curved path.
{tanθ =v2/(r g)} - Simple harmonic motion, amplitude, frequency and period -
Definition. Simple problems based on the expressions for centripetal force
and angle of banking - Applications of centripetal force and centrifugal force -
solved problems.
IV DYNAMICS–II
4.1 ROTATIONAL MOTION OF RIGID BODIES: - 7
Rigid body – Definition - Moment of inertia of a particle about an axis -
Moment of inertia of a rigid body about an axis – expressions – Radius of
gyration – Definition – Expression for the kinetic energy of a rotating rigid
body about an axis – Angular momentum – Definition – Expression for the
angular momentum of a rotating rigid body about an axis – Law of
conservation of angular momentum – Examples.
4.2 GRAVITATION: -
3
Newton’s laws of gravitation – Acceleration due to gravity on the surface of
earth – Expression for variation of acceleration due to gravity with altitude

vii
 4.3 SATELLITES: - 4
Satellites – Natural and artificial – Escape velocity and orbital velocity –
Definitions – Expression for escape velocity and Orbital velocity – Polar and
Geostationary satellites – Uses of artificial satellites. Simple problems based
on the expressions for escape velocity and Orbital velocity.
V. SOUND AND MAGNETISM
5.1 SOUND: - 9
Wave motion – Introduction and definition – Audible range – Infrasonic –
Ultrasonics - Progressive waves, longitudinal and transverse waves –
Examples - Amplitude, Wave length, period and frequency of a wave –
Definitions – Relation between wavelength, frequency and Velocity of a
wave - Stationary or standing waves. Vibrations - Free & forced vibrations
and resonance – definitions and examples - Laws of transverse vibration of a
stretched string – Sonometer – Experimental determination of frequency of a
tuning fork. Acoustics of buildings – Echo - Reverberation, reverberation
time, Sabine’s formula for reverberation time (no derivation) – Coefficient of
absorption of sound energy – Noise pollution. Simple problems based on
expression for frequency of vibration. Doppler effect – Definition and
Applications – Ultrasonic and its uses – SONAR – Solved Problems.
5.2 MAGNETISM: - 5
Pole strength – Definitions – Magnetic moment, intensity of magnetisation,
magnetising field intensity, magnetic induction, Permeability, hysteresis,
saturation, retentivity and coercivity – Definitions - Method of drawing
hysteresis loop of a specimen using a solenoid – Uses of Hysteresis loop.
Simple problems based on intensity of magnetization – Types of magnetic
materials and their applications – Solved problems.

viii
 CONTENTS
FIRST SEMESTER

UNIT CONTENTS Page No.

UNIT-I SI UNITS AND STATICS

1.1 Units and Measurements 2

1.2 Statics 10

UNIT-II PROPERTIES OF MATTER

2.1 Elasticity 24

2.2 Viscosity 32

2.3 Surface tension 38

UNIT-III DYNAMICS- I

3.1 Straight line Motion 50

3.2 Projectile Motion 52

3.3 Circular Motion 62

UNIT-IV DYNAMICS- II

4.1 Rotational motion of rigid bodies 80

4.2 Gravitation 87

4.3 Satellites 89

UNIT-V SOUND AND MAGNETISM

5.1 Sound 97

5.2 Magnetism 108

Model Question Paper 117

Engineering Physics Practical (Semester I) 119

ix
 40023 ENGINEERING PHYSICS – II
DETAILED SYLLABUS

Name of the Topics Hours

I. HEAT
1.1 TRANSFER OF HEAT 4
Concept of Heat and Temperature - Centigrade, Fahrenheit and Kelvin scales of temperature
measurement- Conduction, convection and radiation - Definitions and explanations-Good and
Poor conductors - Examples. Coefficient of thermal conductivity-Definition and SI Unit-
Properties of thermal radiation – Heat conversions.
1.2 KINETIC THEORY OF GASES 4
Postulates –Mean square velocity and Root Mean Square (RMS) velocity of molecules -
Definitions and expressions – Expression for the pressure of a gas on the basis of postulates
of kinetic theory of gases - Relation between pressure and kinetic energy of the gas-Relation
between kinetic energy and absolute temperature–Simple problems based on the expression
for the pressure of a gas.
1.3 SPECIFIC HEAT CAPACITY 4
Specific heat capacity of a substance (solids and liquids) –Definition – Specific heat capacity
of a gas at constant volume – Specific heat capacity of a gas at constant pressure– Ratio of
specific heat capacities – Explanation for Cp is greater than Cv– Derivation of Mayer’s relation
– calculation of Universal gas constant R from the gas equation PV= RT. Simple problems
based on Mayer’s relation – Solved problems.
II. THERMODYNAMICS, LIQUEFACTION OF GASES and NON-CONVENTIONAL ENERGY
2.1 THERMODYNAMICS
5
First law of thermodynamics – Statement - Isothermal and Adiabatic changes - Explanation –
Equations for isothermal and adiabatic changes (No derivation) Simple problems based on
equations P1V1 = P2V2 and P1V1= P2V2. Second law of thermodynamics – Clausius
statement and Kelvin’s statement - Working of Carnot’s reversible engine with indicator
diagram and its efficiency – Applications of heat and thermodynamics.
4
2.2 LIQUEFACTION OF GASES
Critical temperature, critical pressure and critical volume – Definitions – Principle used in
cascade process –Cascade process of liquefaction of oxygen –Disadvantages of cascade
process - Joule Thomson effect – Temperature of inversion – Liquefaction of air by Linde’s
process.
3
2.3 NON – CONVENTIONAL ENERGY
Introduction – Non-renewable and Renewable (Alternate) energy sources – Examples – Solar
energy, wind energy, – Advantages and disadvantages of renewable energy – Tidal Energy.
III. LIGHT AND REMOTE SENSING
3.1 OPTICS 4
Refraction – Laws of refraction – Refractive index of a medium – Derivation of refractive index
of glass prism using minimum deviation - Definition –Spectrometer – Experimental
determination of refractive index using spectrometer- Phenomenon of total internal reflection –
Fiber optics - Introduction –Optical Fiber Cable as a wave guide – advantages of OFC –
problems using the refractive index – Applications of total internal reflection.

x
3.2 LASER 3
LASER – Characteristics of LASER – principle of LASER –Spontaneous emission –
Stimulated emission – population inversion – Ruby Laser-Construction and working- Uses of
LASER.
3.3 REMOTE SENSING
4
Remote sensing – Introduction – Active and passive remote sensing – Explanation and
examples – Components of remote sensing – Data acquisition and data analysis - Reference
data – RADAR – principle and working with block diagram.
IV. ELECRICITY
4.1 ELECTRICAL CIRCUITS
4
Ohm’s law – Laws of resistances – Resistivity, Conductivity, Super conductivity and Meissner
effect- Definitions – Kirchhoff’s current and voltage laws. Condition for balancing the Wheat
Stone’s bridge - Simple problems based on expression for resistivity. Capacitance of a
capacitor – Definition – ‘farad’– Definition– expressions for effective capacitance when
capacitors are connected in series and in parallel. Simple problems based on effective
capacitance for series and parallel connections of capacitors – Applications of capacitors.
4.2 EFFECTS OF CURRENT 3
Joule’s law of heating – Experimental determination of specific heat capacity of a liquid using
Joule’s calorimeter – Faraday’s laws on electrolysis – Electro chemical equivalent (e.c.e) of an
element – Definition – Experimental determination of e.c.e. of copper - Simple problems based
on expressions for e.c.e – Applications of heating effect of electric current.
4.3 MEASURING INSTRUMENTS 4
Expression for the force acting on a current carrying straight conductor placed in a uniform
magnetic field – Fleming’s Left Hand rule – Expression for the torque experienced by a
rectangular current carrying coil placed inside a uniform magnetic field – Working of a moving
coil galvanometer and its merits – Conversion of galvanometer into an Ammeter and
Voltmeter. Simple problems based on conversion of Galvanometer into Ammeter and
Voltmeter – Solved problems.
V. ELECTRONICS
3
5.1 SEMI CONDUCTORS
Semiconductors – Intrinsic semiconductors -Concept of holes - Doping – Extrinsic
semiconductors – Energy bands in solids – Energy band diagram of good conductors,
insulators and semiconductors – Concepts of Fermi level – P type and N type semiconductors.
5.2 DIODES AND TRANSISTORS
5
P-N junction diode – Forward bias and reverse bias –Rectification action of diode – Working of
full wave bridge rectifier using P N junction diodes -PNP and NPN transistors – Three different
configurations – Working of NPN and PNP transistor in common base configuration.
5.3 DIGITAL ELECTRONICS 3
Digital electronics – Introduction – Logic levels – Basic logic gates: OR, AND , NOT gates –
Universal logic gates: NAND and NOR gates – Symbolic representation, Boolean expression
and Truth table for all above logic gates – Integrated circuits– Levels of integration – SSI, MSI,
LSI and VLSI-Advantages of ICs – Applications of transistors, gates and ICs – Solved
problems.

xi
 SECOND SEMESTER

UNIT CONTENTS Page No.

UNIT-I HEAT

1.1 Transfer of Heat 140

1.2 Kinetic theory of gases 145

1.3 Specific heat capacity 150

UNIT-II THERMODYNAMICS, LIQUEFACTION OF GASES AND NON-CONVENTIONAL

ENERGY

2.1 Thermodynamics 159

2.2 Liquefaction of gases 166

2.3 Non-conventional Energy 169

UNIT-III LIGHT AND REMOTE SENSING

3.1 Optics 177

3.2 Laser 189

3.3 Remote sensing 193

UNIT-IV ELECTRICITY

4.1 Electrical circuits 200

4.2 Effects of current 213

4.3 Measuring instruments 220

UNIT-V ELECTRONICS

5.1 Semi-conductors 232

5.2 Diodes and Transistors 236

5.3 Digital Electronics 242

ENGINEERING PHYSICS PRACTICAL (Semester – II) 259

xii
 STATE BOARD OF TECHNICAL EDUCATION & TRAINING, TAMILNADU
DIPLOMA IN ENGINEERING / TECHNOLOGY SYLLABUS
N-SCHEME

(Implements from the Academic year 2020-2021 onwards

Course Name: All Branches of Diploma in Engineering and Technology and Special
Programmes except DMOP, HMCT and Film & TV

Subject Code: 40006

Subject Title: ENGINEERING PHYSICS PRACTICAL

TEACHING AND SCHEME OF EXAMINATION:

No. of Weeks per Year: 32 Weeks

Instructions Examination

Marks

Subject Hours/ Hours/ Internal

Board Duration
Week Year Assessment
Examinatio Total
/
n
Record

ENGINEERING
PHYSICS 2 64 25 75 100 3 Hours
PRACTICAL

xiii
 40006- ENGINEERING PHYSICS PRACTICAL

Semester I

LIST OF EXPERIMENTS WITH OBJECTIVES:

1. MICROMETER (SCREW GAUGE).

To measure the thickness of the given irregular glass plate using micrometer. To
determine the area of the glass plate using a graph sheet and to calculate the volume of
the glass plate.

2. VERNIER CALIPERS.
To measure the length and diameter of the given solid cylinder using vernier calipers and
to calculate the volume of the solid cylinder.

3. PARALLELOGRAM LAW.
To verify the parallelogram law using concurrent force.

4. LAMI’S THEOREM
To verify Lami’s theorem using concurrent forces.

5. COMPARISON OF VISCOSITIES
To compare the co-efficient of viscosities of two low viscous Liquids by capillary flow
method.

6. STOKES’ METHOD.
To determine the coefficient of viscosity of a highly viscous liquid.

7. SONOMETER.
To determine the frequency of the given tuning fork.

8. DEFLECTION MAGNETOMETER
To compare the magnetic moments of the two bar magnets using Deflection
Magnetometer in Tan A position, by equal distance method.

xiv
 Semester II

LIST OF EXPERIMENTS WITH OBJECTIVES:

9. REFRACTIVE INDEX
To determine the refractive index of a transparent liquid (water) using travelling
Microscope.

10. SPECTROMETER.
To measure the angle of the prism using Spectrometer.

11. SOLAR CELL.

To draw the V – I characteristics of the solar cell.

12. LAWS OF RESISTANCES.

To verify the laws of resistances by connecting the two given standard resistances
(i) in series and
(ii) in parallel, using Ohm’s law.

13. JOULE’S CALORIMETER.

To determine the specific heat capacity of water.

14. COPPER VOLTAMETER.

To determine the electro chemical equivalent (e.c.e.) of copper.

15. P-N JUNCTION DIODE.

To draw the voltage – current characteristics in forward bias and to find the ‘dynamic
Forward resistance’ & ‘knee voltage’ from the graph.

16. LOGIC GATES.

To find the output conditions for different combinations of the input for NOT gate and 2
inputs AND, OR, NAND & NOR logic gates, using IC chips. ( IC 7404 – NOT Gate, IC
7408 – AND Gate, IC 7432 – OR gate, IC 7400 – NAND Gate, IC 7402 – NOR Gate)

xv
LIST OF EQUIPMENTS:

1. MICROMETER (SCREW GAUGE).

Screw gauge, graph sheet and irregular glass plate.

2. VERNIER CALIPERS.

Vernier Calipers and Solid Cylinder

3. PARALLELOGRAM LAW.

Vertical drawing board, two Z pulleys, three sets of slotted weights (5 x 50g) and twine
thread.

4. LAMI’S THEOREM

Vertical drawing board, two Z pulleys, three sets of slotted weights (5 x 50g) and twine
thread.

5. COMPARISON OF VISCOSITIES

Burette stand, graduated burette without stopper, rubber tube, capillary Tube, beaker,
digital stop watch, two liquids and funnel.

6. STOKES’ METHOD.

Stokes’ Apparatus, highly viscous liquid (Castrol oil), glass beads of different radii, digital
stop watch and screw gauge.

7. SONOMETER.

Sonometer, screw gauge, tuning fork, rubber hammer, slotted weight hanger set (5 x
0.5kg) and paper rider.

8. DEFLECTION MAGNETOMETER

Deflection Magnetometer, meter scale and two bar magnets

9. REFRACTIVE INDEX
Travelling Microscope, Beaker with transparent liquid and Saw dust.

10. SPECTROMETER.
Spectrometer, Sodium vapour lamp, Reading lens and Glass prism

11. SOLAR CELL.

Solar cell Kit for drawing the V - I characteristics

12. LAWS OF RESISTANCES.

Battery Eliminator, key, rheostat, ammeter, voltmeter, Connecting wires and two known
standard resistances.

xvi
13. JOULE’S CALORIMETER.
Joule’s Calorimeter, Battery eliminator, Rheostat, Key, Ammeter, voltmeter, stop clock,
thermometer, digital Balance and connecting wires.

14. COPPER VOLTAMETER.

Copper Voltameter, Battery eliminator, Rheostat, Key, Ammeter, stop clock, digital
balance, emery sheet and Connecting wires.

15. P-N JUNCTION DIODE.

P-N Junction Diode forward characteristics kit.

16. LOGIC GATES.

Logic gates testing apparatus kit with bread board for Mounting ICs and Integrated circuit
chips ( IC 7404 –NOT Gate, IC 7408 – AND Gate, IC 7432 – OR gate, IC 7400 –
NAND Gate, IC 7402 – NOR Gate)

========================

ALLOCATION OF MARKS

Formula & Diagram 20 marks

Tabulation with proper units 10 marks
Observation (including taking readings) 45 marks
Calculation 15 marks
Result 10 marks
----------------
Total 100 Marks

xvii
 40013 - ENGINEERING PHYSICS – I
UNIT I

SI UNITS AND STATICS

OBJECTIVES

▪ To know about the branches of physics

▪ To understand the physical quantities
▪ To know about the dimensional formulas and its uses.
▪ To know about the conventions of SI units
▪ To convert unit (hp to watt).
▪ To understand the scalar and vector quantities
▪ Can derive the expression for vector quantity (Force)
▪ Can derive the expression for magnitude and direction of a resultant.
▪ To verify the parallelogram law and Lami’s theorem
▪ To define principle of moments
▪ To verify mass of the body using principle of moments
▪ To solve simple problems based on the expression for magnitude and
direction of the resultant of two forces.

1
UNIT

1 SI UNITS AND STATICS

1.1 UNITS AND MEASUREMENT

Introduction:
The word physics comes from the Greek word meaning “nature”. Today physics is treated as the
most fundamental branch of science and finds numerous applications of life. Physics deals with matter
in relation to energy and the accurate measurement of the same. Thus physics is inherently a science of
measurement. The fundamentals of physics form the basis for the study and the development of
engineering and technology.
Measurement consists of the comparison of an unknown quantity with a known fixed quantity. The
quantity used as the standard of measurement is called ‘unit’. For example, a vegetable vendor weighs
the vegetables in terms of units like kilogram.
Fundamental physical quantities
Fundamental quantities are the quantities which cannot be expressed in terms of any other physical
quantity.
(eg) length, mass and time.
Derived quantities
Quantities that can be expressed in terms of fundamental quantities are called derived quantities.
(eg) area, volume, density.
Unit
Unit of a physical quantity is defined as the accepted standard used for comparison of given physical
quantity.
The unit in which the fundamental quantities are measured are called fundamental unit and the units
used to measure derived quantities are called derived units.
SI Units
SI unit is the abbreviation for System International de units and is the modern form of metric
system finally agreed upon at the eleventh International conference of weights and measures, 1960. This
system of units is now being adopted throughout the world and will remain the primary system of units
of measurement. SI system possesses features that make it logically superior to any other system.
There are seven fundamental units (base units) and two supplementary units.
SI system of units
Physical quantity Unit Symbol
Fundamental quantities
1.Length metre m
2.Mass kilogram kg
3.Time second s
4.Electric current ampere A

2
 5.Temperature kelvin K
6.Luminous Intensity candela cd
7.Amount of substance mole mol
Supplementary
quantities
1. Plane angle radian rad
2. Solid angle steradian sr
Derived quantities and their units
Sl.No Physical Formula Unit Symbol
1 quantity
Area of the square side × side 2
metre or square metre m2
2 Volume of the side × side × side metre3 or cubic metre m3
3 cube
Density Mass / volume kilogram metre–3 kgm–3
4 Velocity Displacement / time metre second–1 ms–1
5 Acceleration velocity/ time metre second–2 ms–2
6 Momentum mass × velocity Kilogram metre second–1 kgms–1
7 Force mass × acceleration newton N
8 Impulse force × time newton second Ns
9 Work (or) Energy force × displacement newton metre or joule J
10 Power Work / time joule second–1 or watt W
Dimensions
The fundamental physical quantities namely length, mass and time are symbolically represented by
the capital letters L, M and T respectively.
Dimensional formula is the formula in which the given physical quantity is expressed in terms of the
fundamental quantities raised to suitable powers.
Dimensional formula for derived physical quantities.
1. Area of the square = side × side
Applying dimensions
Area of the square = L × L = L2
Dimensional formula for Area = [ L2 ]
2. Volume of the cube = side × side × side
Applying dimensions
Volume of the cube = L × L × L = L3
Dimensional formula for Volume = [L3]
3. Density is the mass per unit Volume.
mass
Density =
volume
Applying dimensions
M
Density = 3 = ML–3
L
Dimensional formula for Density = [ML–3]

3
4. Velocity is the rate of change of displacement
displacement
Velocity =
time
Applying dimensions,
L
Velocity = = LT –1
T
Dimensional formula for Velocity = [LT–1]
5. Acceleration is the rate of change of velocity
velocity
Acceleration =
time
displacement 1
= 
time time
Applying dimensions
L L
= = = LT –2
T  T T2
Dimensional formula for Acceleration = [LT–2]
6. Momentum is the product of mass and velocity.
Momentum = mass × velocity
displacement
= mass 
time
Applying dimensions
L
Momentum = M  = MLT –1
T
Dimensional formula for Momentum = [MLT–1]
7. Force is the product of mass and acceleration.
Force = mass × acceleration
velocity
= mass 
time
mass displacement
= 
time time
Applying dimensions,
M L
Force = T  T = MLT -2
Dimensional formula for force = [MLT -2]
8. Impulse is the product of force and time
Impulse = force × time
= mass × acceleration × time

velocity
= mass × time time
= mass × velocity
= mass × (Displacement/time)
4
 Applying dimensions,
ML
Impulse = T = MLT -1
Dimensional formula for Impulse= [MLT-1]
9. Work (or) Energy is the product of force and displacement
Work or Energy = force × displacement
= mass × acceleration × displacement

velocity
= mass × time × displacement
mass displacement × displacement
= 
time time
Applying dimensions
L L
Work or energy = M ×  = ML2 T-2
T T
Dimensional formula for Work or Energy = [ML2 T-2]
10. Power is the rate of doing work.
work
Power = time

force  displacement
=
time
Applying dimensions,
L
2 -3
Power = MLT X T = ML T
–2

Dimensional formula for Power = [ML2 T-3]

Uses of Dimensional formula
• To check the dimensional correctness of a given equation.
• To convert a physical quantity from one system of unit to another.
• To establish a relationship between the physical quantity in an equation.
Problem 1:
If the force is proportional to the square of the velocity, then calculate the dimension of
the proportionality constant.
F α v2
F = k v2
Applying dimension,
MLT-2 = k [ LT-1 ]2
MLT-2 = k [ L2T-2 ]
𝑀𝐿 𝑇 −2
k =
𝐿2𝑇 −2

Dimension of constant ‘k’ is = ML-1T0

5
1. To check the correctness of the equation:
Principle:
If the equation is correct, the dimensions of the left hand side of the physical equation
should be equal to the dimensions of the right hand side. This is known as the principle of
homogeneity.
Example: 1
1. To check the equation V = u + at
Dimensions of L.H.S. = [LT-1]
Dimensions of R.H.S. = [LT-1] + [LT-2 T]
= [LT-1] + [LT-1]
[LT-1] = [LT-1] + [LT-1]
The dimensions of L.H.S. = The dimensions of R.H.S.
Both sides are dimensionally equal. Hence, the equation is correct.
Example: 2
To check the equation V2 = u2 + 2 as
Dimensions of L.H.S. = [LT-1]2 = [L2T-2]
Dimensions of R.H.S. = [LT-1]2 + [LT-2 L]
= [L2T-2] + [L2T-2]
It is found that both the sides are dimensionally equal. Hence, the equation correct.
2. To derive the equation:
The physical equation can be derived if we know the various factors on which it
depends. But the constants and numericals cannot be determined by the method of
dimensions.
Example: 1
To derive the formula for the period of a simple pendulum.
The period (t) of a simple pendulum may depend upon.
1. The mass of the bob m.
2. The length of the pendulum l and
3. The acceleration due to gravity g at a place.
The period of a pendulum executing oscillations can be given by the formula.
T = kmx lygz – (1)
Where k is the constant of proportionality.

Physical quantity Dimensions

t [T]
m [M]
l [L]
g [LT-2]

6
 Substituting the dimensions of the various quantities, we have,

T = [Mx] [Ly] [ (LT-2)z ]

Mo Lo T1 = [Mx (Ly + z) T-2z ]

According to the principal of homogeneity by comparing the dimensions on either side,

we have

x = 0, y + z = 0 and - 2z = 1

(or)

z = - ½ and y = -z = + ½

Substituting the values in the equation (1), we have

t = kmo l½ g – ½

t = k √l/g

The constant k can be found experimentally and is equal to 2.

t = 2 √𝑙/g

Limitations:

1. Using the method of dimensions the constants involved in the equations cannot be
found.

2. The trignometrical and exponential factors in the equation on cannot be determined.

CONVENTIONS TO BE FOLLOWED IN SI UNITS

1. When we write a unit in full, the first letter should not be in capital letter.

(eg) metre and not as Metre

kilogram and not as Kilogram

2. The symbols of unit in the name of scientist should be in upper case (capital) letter

(eg) N for newton, J for joule.

3. Only the singular form of the unit is to be used.

(eg) km and not as kms
Kg and not as kgs

4. There should be no full stop at the end of the abbreviations.

(eg) mm and not as mm.
kg and not as kg.

7
 5. When temperature is expressed in Kelvin, the degree sign is omitted.
(eg ) 273K not as 2730K

6. Only accepted symbols should be used.

(eg) ampere is represented as 'A' and not 'amp'
second is represented as 's' and not 'sec'

7. Use of the solidus or slash is to be avoided, but when used, not more than one solidus be
employed.
(eg) J/k.mol or Jk.mol-1 but not J/k/mol
J kg-1 K-1 but not J/kg/K

8. One letter space is always to be left between the number and the symbol of the unit.
(eg) 2.3 m and not as 2.3m
Kg m s-2 and not kg ms-2

9. The numerical value of any physical quantity should be expressed as p= a × 10m. Here 'a' is a
number between 1 and 10 and 'm' is the appropriate power of 10.
(eg) velocity of light c = 2.997 × 108 m s-1

Multiples and Submultiples of units

In SI, some units are not of convenient size to measure certain quantities. Hence multiples and sub-
multiples of the base units are used in measurements.

Multiplication factor Prefix Symbol

1 000 000 000 000 = 1012 Tera T
1 000 000 000=109 Giga G
1 000 000=106 Mega M
1 000=103 Kilo K
100=102 hecto* h
10=101 deca* da
0.1=10-1 deci d
0.01=10-2 centi* c
0.001=10-3 milli m
0.000 001=10-6 micro μ
0.000 000 001=10-9 nano n
0.000 000 000 001=10-12 pico p
0.000 000 000 000 001=10-15 femto f
0.000 000 000 000 000 001=10-18 atto a

8
 *These are not preferred ones. They are used where the other prefixes are inconvenient

Unit Conversions

1 1 HP (Horse Power) = 746 W

2 1 Calorie = 4.184 Joule

https://www.youtube.com/watch?v=Auw9MecSz2o&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj

https://www.youtube.com/watch?v=a3h7Q0JIXpI&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=2

https://www.youtube.com/watch?v=Ax-3WLaIkUU&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=3

9
 1.2 STATICS
Statics is the part of mechanics which deals with forces acting on bodies at rest. The weight of the
body, the tension of a string, the load and the reaction are the different names of forces that are used in
statics.
Any force can be represented in magnitude and direction by a straight line with an arrow head. The
beginning of the line represents the point of application of the force. The arrow head represents the
direction of the force. The magnitude of the force is given by the length of the line drawn to scale.
Scalar and Vector quantities
Physical quantities which have magnitude only are called scalar quantities.
(e.g) Mass, Volume, Speed
Physical quantities which have both magnitude and direction are called vector quantities.
(e.g) Velocity, Momentum, Force
Concurrent forces
Two or more than two forces acting at a point are called concurrent forces.
Coplanar forces
Two or more than two forces lie on same plane are called coplanar forces.
F1 F2 F3 F4

Resultant
Resultant is that single force which produces the same effect as that produced by two or more forces
acting on the body.
Equilibrant
Equilibrant is the single force which acting along with the other forces keeps the body in
equilibrium.
Resolution of a vector into two perpendicular components
The process of splitting up a vector (force) into two perpendicular component parts is known as
resolution of a vector. As the two component forces are mutually perpendicular, they are called
rectangular components.
Y

B C
R Sin θ

90-θ R
θ
X
O R Cos θ A

Let a force R act at a point O at an angle θ with x- axis. This force can be resolved into two
rectangular components along x and y axis. Two lines CA and CB are drawn perpendicular to x and y
axis as shown in the figure.
In the right angled triangle OAC,
OA
Cos  =
OC
OA = OC cosθ
= R cosθ

10
 Therefore the horizontal component of the force R is R cosθ. In the right angled triangle OBC,
OB
Cos(90 - ) =
OC
OB
Sin  =
OC
OB = OC sin θ = R sin θ
The vertical component of the force R is R sin θ.
Parallelogram law of forces
If two forces acting at a point are represented in magnitude and direction by the two adjacent
sides of a parallelogram drawn from that point, their resultant is given in magnitude and
direction by the diagonal of the parallelogram drawn through that point.
Expressions for the magnitude and direction of the resultant of two forces acting at a point,
with an acute angle between them.
B C

Q R


O P A D

Consider two forces P and Q acting at a point O with an angle θ between them as shown in the figure.
The forces P and Q are represented in magnitude and direction by the sides OA and OB of a
parallelogram OACB. The diagonal OC gives the resultant of two forces P and Q in magnitude and
direction. From C draw a perpendicular to meet OA extended at D.
In the right angled triangle OCD
OC2 = OD2 + CD2
= (OA + AD) 2 +CD2
OC2 = OA2 +AD2+2OA .AD +CD2 ...( 1)
In the right angled triangle ACD
AD AD
Cos  = AC = Q

AD = Q Cosθ
CD CD
Sin  = =
AC Q
CD = Q sinθ.
Substituting AD and CD values in eqn.(1)
R2 = P2 + Q2 cos2θ + 2PQcosθ +Q2sin2θ
R2 = P2 + Q2 (cos2θ + sin2θ) + 2PQcosθ
R2 = P2 + Q2 +2PQcosθ where cos2θ + sin2θ=1
Resultant, R = P 2 + Q 2 + 2PQ cos 
This is the expression for the magnitude of the resultant.
Let α be the angle between the resultant R and the force P.
CD CD Q sin 
tan  = = =
OD OA + AD P + Q cos 
11
  (Q sin ) 
Direction,  = tan-1  (P + Qcos ) 
 

In the right angled triangle COD,

This is the expression for the direction of the resultant.
Lami's theorem
If three forces acting at a point are in equilibrium, then each force is directly of proportional
to the sine of the angle between the other two forces.
P Q


 

Let P, Q and R be three forces acting at a point O. Under the action of three forces, the point is at
rest. Let α, β and γ be the angles opposite to the forces P, Q and R respectively.
By Lami's theorem
P
P  sin   P = K sin   =K
sin 
Q
Q  sin   Q = K sin   =K
sin 
R
R  sin   R = K sin   =K
sin 

P Q R
sin 𝛼
= sin 
= sin 
= Constant

Experimental verification of parallelogram law of forces

D

B
A 
P Q
O  O 
R
P Q
R
C

A drawing board is fixed vertically on the wall. Two frictionless pulleys are fixed at the top corners
of the drawing board. A light and inextensible string is passed over these pulleys. Another short string is
tied to the first string at O. Weight hangers P, Q and R are tied at the ends of the strings as shown in the
figure. The weights P, Q and R are adjusted suitably such that the point O is at rest.
A white paper is held just behind the string without touching them. The point O and the directions of
the strings are marked on the paper. Then the paper is removed from the board. With suitable scale the
points A, B and C are marked. OA, OB and OC represent the forces P, Q and R respectively.
The parallelogram OADB is drawn. Then the diagonal OD is drawn. The length of the diagonal OD
and COD are measured and tabulated.

12
 Sl. No P Q R OA OB O OD COD
C

The experiment is repeated for different values of P, Q and R. It is found that OC = OD and
COD = 180º. Thus the parallelogram law of forces is verified experimentally.
Experimental verification of Lami's theorem
A B

Q
P
O
O  

R
P Q
R
C

A drawing board is fixed vertically on the wall. Two frictionless pulleys are fixed at the top corners
of the drawing board. A light and inextensible string is passed over these pulleys. Another short string is
tied to the first string at O. Weight hangers P, Q and R are tied at the ends of the strings as shown in the
figure. The weights P, Q and R are adjusted suitably such that the point O is at rest.
A white paper is held just behind the string without touching them. The point O and the directions of
the forces P, Q and R are marked on the paper. Then the paper is removed from the board. Let α, β and
γ are the angles opposite to the forces P, Q and R respectively. The angles are measured with the help of
protractor and tabulated.

Sl. P Q R α β γ P Q R
No sin  sin  sin 

The experiment is repeated for different values of P, Q and R. The ratios (P/ sin α), (Q / sin β) and
(R / sin ) are calculated. For each set of readings, the ratios are found to be equal. Thus Lami's theorem
is verified experimentally.
Worked Examples:
1. A body of mass 1 kg is suspended by two light ropes 1.5 m and 2 m long from lub points in a
horizontal line 2.5 m apart. Calculate the tensions in the ropes.
̂ R =  and T1 and T2 be the tensions in RP and RQ.
PR and QR represents the two ropes. Let PQ

P S 2.5m Q
m

2m
1.2m
m
m R

13
 PQR is a right angled triangle.
 PQ2 = PR2 + RQ2
̂ Q = 90o
Then PR
̂ T = (90o + )
QR
̂ T = (180o - )
PR
By Lami’s theorem.

T1 1 T2
= =
sin QRT sin 90o sin PRT
T1 1 T2
= =
sin (90o + ) sin 90o sin (180o )

sin (90o + ) 2
T1 = = cos  cos  = = 0.8
sin(90o ) 2.5

T1 = cos  = 9.8 x 0.8 = 7.84 N

sin (180o − ) 1.5
T2 = = sin  sin  = = 0.6
sin 90o 2.5
T2 = 9.8 x 0.6 = 5.88 N
The tension in the sting PR = 7.84 N
The tension in the string QR = 5.88 N.

2. A stone of mass 5 kg is subs pended by a string 1 m long. Find the horizontal force required to
pull the stone 6 x 10-1 m to a side. Also find the tension in the string.
The points O is at rest under the action of 3 forces H, W and T.


T
1m
0.8 m kg


0 H
0.6 m
5 kg
5 kg
Applying Lami’s theorem,
T 5 H
= =
sin 90o sin (90o + ) sin (180o − )

5 x sin 90o 5 5 0.8

T = = = [∴ cos θ = ]
sin (90o + ) cos  0.8 1

T = 6.25 x 9.8 = 61.25 N

14
 5 x sin (180o − ) 5 x sin 
H = = = 5 x tan 
sin (90o + ) cos 
0.6
tan  = = 0.75
0.8
H = 5 x 0.75 = 3.75 x 9.8
H = 36.75 N
 The horizontal force H = 36.75 N
Tension T = 61.25 N
3. The forces acting on a particle keep in equilibrium. The angle between the first and the second is
90o and that between the second and the third is 120o, find the ratios of the forces. If the first
force is 3 kgf. Find the other forces.
Let F1, F2 and F3 be the forces. The angle between F1 and F2 = 90o, that between F2 and F3 =
120 and that between F3 and F1 = (360 – 90 – 120) = 150o by Lami’s theorem.
F1 F2 F3
= =
sin 120 sin 150 sin 90
F1 F2 F3
(i.e) = =
sin (180 - 120) sin (180 - 150) sin 90
F1 F2 F3
= =
sin 60 sin 30 1
F1 F2
= 1 = F3
√3/2
2

F1 3 √3
= √3 (ie) =
F2 F3 1

F2 = √3 kg f
and F3 = 2 √3 kgf.
Moment of a force
O
d

N F

Consider a body which is fixed at a point, about which it can rotate freely. Let a force F is acting on
the body. The effect of the force is to rotate the body about the fixed point, unless the line of action of
the force passes through that fixed point O. This rotating tendency or the turning effect of the force about
that point is called moment of force i.e. the turning effect of the force acting on a body about an axis or
point is called moment of force.
It is common experience that in opening or closing a door, the force we apply rotates the door about
its hinges. This rotating effect is known as the moment of force.
Moment of force is also equal to the product of the magnitude of the force and the perpendicular
distance of the line of action of the force from the axis of rotation.

Moment of force about the point O = Force × perpendicular distance of the force from the point O
= F × ON
The unit of moment of force is N m and the dimensional formula is ML2T-2.
15
Clockwise and anti-clockwise moments
If the moment of a force turns or rotates the body in clockwise direction, then it is called as clock-
wise moment.
If the moment of a force turns or rotates the body in anti-clockwise direction, then it is called
anti-clockwise moment.
d1 d2

m1 m2

Clockwise moment= m2 × d2
Anti-clockwise moment = m1 × d1
Principle of moments
The principle of moment states that if a body is in equilibrium under the action of a number of
parallel forces, the sum of the clockwise moments about any point must be equal to the sum of
anti-clockwise moments about the same point.
d1 O d4

d2 d3

m2 m3

m1 m4

According to the principle of moments,

Sum of anti-clockwise moments = sum of clockwise moments
From figure,
m1d1 + m2d2 =m3 d3 + m4 d4.
Couple:
Two equal and opposite parallel forces acting at different points in a body constitute a couple. A
body acted upon by the couple will rotate the body in clockwise direction or anti-clockwise direction.
Steering wheel and pedals of bicycles are the examples for couple, where the two forces are equal
but acting in opposite direction.
Torque acting due to couple (or) Moment of couple
Torque is calculated by the product of either of forces forming the couple and the arm of the
couple.
i.e.) Torque = one of the force × perpendicular distance between the forces.
F

O
B
A

Consider two equal forces F and F acting on the arm AB. Let O be the mid point of the arm .The
forces F and F are acting in opposite direction as shown in the figure, they constitute a couple. Then
Moment of couple or torque = F × AB.

16
 If the forces acting on the body have the same line of action then the moment becomes zero. The
torque is maximum when the forces are at right angles to the arm i.e.) θ=90º.
Determination of mass of the given body using principle of moments.

A D d2 d1
O C B

m2 m1

A meter scale AB is supported on a knife edge at its centre such that it remains in equilibrium state.
Let O be the centre of the scale. On the right side suspend the weight hanger of mass m 1 and on the left
side suspend the given body m2 whose mass is to be calculated. Adjust the positions of m1 and m2 until
the scale comes to the exact equilibrium position. Let d1 (OC) and d2 (OD) be the distances of m1 and
m2.
According to the principle of moments
m 2 × d2 = m 1 × d1
The mass of the given body m2 = (m1 × d1 ) / d2
By changing the values of m1, the experiment is repeated and the readings are tabulated.
m1 d1 d2 m2 = (m1 × d1 ) / d2
Sl.No
kg m m kg

Mean =
The average value of the last column gives the mass of the body.
WORKED EXAMPLES
4. Find the magnitude and direction of the two forces 20N and 25N acting at an angle 60º to each
other.
Given:
P = 20N Q = 25N  = 60º
Resultant R = P 2 + Q 2 + 2PQcos 

= 202 + 252 + 2  20  25cos 60

= 1525
R = 39.05N
 (Q sin ) 
Direction  = tan-1  (P + Qcos ) 
 
 25  sin 600 
 20 + 25 cos 600 
 = tan-1  

α = tan -1 (0.6661)
α = 33º40'

17
5. Find the magnitude and direction of the resultant of two forces 30N and 40N acting at right
angles to each other.
Given
P = 30N
Q = 40N  = 90°

Resultant R = P 2 + Q 2 + 2PQcos 

= 302 + 402 + 2  30  40  cos 90

= 2500
R = 50N
 (40  sin 90) 
Direction  = tan -1  (30 + 40 cos 90) 
 
= tan -1 (1.333)
α = 53º7'
6. If the resultant of two equal force is 3 times a single force, find the angle between the
forces.
Given
Two forces are equal
P=Q=P
R= 3 P
R2 = P2 + Q2 + 2 PQ cos θ
(3P)2 = P2 + P2 + 2.P.P. cos θ
3P2- 2P2 = 2P2 cos θ
P2 = 2P2 cos θ
cos θ = ½
 θ = 60º
Angle between the forces is 60º
7. If the resultant of two equal forces inclined to each other at 60º is 8 3 N, find the component
forces.
Given
Two forces are equal.
P= Q=P
R= 8 3 N
θ = 60º
R2 = P2 + Q2 + 2PQ cos θ
(8 3 )2 = P2 + P2 + 2. P.P cos 60º
192 = 2P2 + 2P2.½
192 = 3P2
64 = P2
P = 8N
Component force is 8N
18
8. The sum of two forces is 8N and the magnitude of the resultant is at right angles to the smaller
force 4N. Find the forces and the inclination between them.
Given
Sum of the two forces = 8 N
Resultant of 2 forces = 4 N
Let the smaller force be P
Larger force Q = (8–P)
C B

4N
Q=8 – P
Q=8 – P


O P A

In the right angled triangle OAB,

AB2 = OA2 + OB2
(8 – P)2 = P2 + 42
64–16P + P2 = P2 +16
16P = 48
P = 3N
Therefore the larger force Q = 8–3
Q = 5N
In triangle OCB,
BC P 3
tan  = = = = 0.75
OB 4 4
α = tan-1(0.75)
α = 36º 52'
Angle between the forces P and Q

90 + α = 90 + 36º 52'
= 126º 52’

19
 QUESTIONS
Part A and Part B
1. What are the fundamental quantities?
2. What are the derived quantities?
3. What are the supplementary quantities and their SI units?
4. What are the units for momentum , force, impulse and work?
5. What are the units for energy and power, acceleration and density?
6. What are the dimensional formula for impulse and momentum.
7. What are the dimensional formula for work or energy.
8. What are the dimensional formula for power and force
9. Define scalar quantity with example
10. Define vector quantity with example.
11. What is resultant?
12. What is equilibrant?
13. Write the parallelogram law of forces.
14. Write the Lami's theorem.
15. Write any two conventions followed in SI unit.
16. Define concurrent forces.
17. Define coplanar forces.
18. Define moment of a force.
19. Write the principle of moments.
20. Define couple.
21. Define moment of a couple.
22. Define clockwise moment and anticlockwise moment.
23. Write any three uses of dimensional formula
24. Derive dimensional formula for impulse, work and power.
25. Write any three the rules and conventions followed while writing SI.
26. Explain how a vector can be resolved into two rectangular components.
27. Find the magnitude of the two forces 3 N and 4 N acting at right angle to each other
Part- C
1. Give the rules and conventions followed while writing SI units
2. Derive expressions for the magnitude and direction of the resultant of two forces acting at a point
with an acute angle between them.
3. Describe an experiment to verify the parallelogram law of forces.
4. Describe an experiment to verify Lami's theorem.
5. Describe an experiment to determine the mass of the given body using principle of moments.
20
 Exercise Problems
1. Find the magnitude and direction of the resultant of two forces of 5N and 3N at an angle of 60º.
(Ans: R =7N, α = 21º47')
2. Two forces of magnitude 4N and 2.5N acting at a point inclined at an angle of 40º to each other. Find
their resultant.
(Ans: R= 6.149N, α =15º8')
3. Find the resultant of two forces 3N and 4N acting on a particle in direction inclined at 30º.
(Ans: R= 6.767N, α =17º19')
4. Two forces of magnitude 4N and 3N respectively, act on a particle at right angles to each other. Find
the magnitude and direction of the resultant of two forces.
(Ans: R= 5N, α =36º52')

5. If the resultant of two equal forces is 2 times the single force, find the angle between them.
(Ans: θ = 90º)
6. If the resultant of two equal forces inclined to each other at 60º is 8.660 N, find the component
force.
(Ans:5N)
7. If the resultant of two forces 6N and 8N is 12N, find the angle between them.
(Ans: 62º43')
8. The sum of two forces inclined to each other at an angle is 18N and their resultant which is
perpendicular to the smaller force is 12N. Find the forces and angle between them.
(Ans: P= 5N,Q=13N,θ =112º36')

https://www.youtube.com/watch?v=Yis3K_L4zXQ&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=4

https://www.youtube.com/watch?v=rjF7o0uxlHM&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=5

https://www.youtube.com/watch?v=Rh6Mu3r06Os&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=6

21
 SUMMARY

• Physics consists of many sub branches like acoustics, optics, magnetism, heat,
electricity, electronics etc.
• The International System of Units (SI) is the modern form of the metric system.
• Every physical quantity has unit and magnitude.
• Length, Mass and Time are the fundamental quantities.
• One hp = 746 W
• Dimensional formula is used to check the correctness of physical equations.
• Conventions of SI units shows that how to write a symbol or unit.
• Statics deals with forces acting on bodies at rest.
• Vector quantities have magnitude and direction.
• Derive the expression for magnitude and direction of a resultant.
• Explanation of Parellogram law and Lami’s theorem.
• Two equal and opposite forces acting on a body form a couple.
• The sum of clockwise moments = the sum of anticlockwise moments.
• Determine the mass of a body using principle of moments.

22
 UNIT - II

PROPERTIES OF MATTER

OBJECTIVES

▪ To differentiate elasticity and plasticity

▪ To know about stress and strain
▪ To define Hooke’s law
▪ To know about the types of Modulus.
▪ To explain the Young’s Modulus by uniform bending.
▪ To know about the viscous force
▪ To differentiate streamline and turbulent motion
▪ To understand the Reynold’s Number
▪ To compare the viscosities of two liquids
▪ To know about the applications of viscosity
▪ To define surface tension
▪ To verify the surface tension of water using experiment
▪ To know about capillarity and its applications.
▪ Can solve simple problems based on Young’s Modulus and
Surface Tension.

23
UNIT

2 PROPERTIES OF MATTER

2.1 ELASTICITY

Introduction:
When an external force is applied on a body, which is not free to move, the shape and size of the
body change. The force applied is called deforming force. When the deforming forces are removed,
the body tends to regain its original shape and size due to a force developed within the body. The force
developed within the body, which is equal and opposite to deforming force is called restoring force.
Bodies, which completely regain their original size and shape after the removal of the deforming
force, are called elastic bodies. Bodies which change the shape and size on the application of force and
which do not regain their original condition on removal of the deforming forces are said to be plastic
bodies. Bodies which do not change their shape and size on application of force are called rigid bodies.
The property by virtue of which a body tends to regain its original shape and size after removal of
the deforming force is known as elasticity.
Stress :
When an external force is acting on an elastic body, it causes deformation (change in shape or in
size or both). At the same time, due to elastic property, a force is developed within the material, which
is equal and opposite to the applied force, to bring the body to its original shape and size. This force is
‘restoring force’.
Force F
Stress = =
Area A
The stress is defined as the restoring force acting on unit area.
Since the applied force and the restoring force are equal in magnitude, the ‘stress’ is measured as the
applied force acting per unit area.
The unit for stress is newton metre-2 with symbol N m-2 or 'pascal' with symbol 'Pa'.
When the applied force tends to compress the body, the stress is compressive. When it tends to
increase the length in the direction of the force, it is tensile and when it acts parallel to the surface of a
body, the stress is tangential stress.
Hooke's law:
Within the elastic limits, the strain produced in a body is directly proportional to the stress
which causes it.
i.e., strain  stress
or stress  strain
stress
 strain = a constant
This constant is called 'modulus of elasticity'.
Types of strains:
Strain: Change in dimensions to original dimensions is known as Strain.

24
(1) Linear Strain:
When a wire or bar is subjected to two equal and opposite forces, namely pulls, at its ends, there is
an increase in the length. If the forces are tensile, the body is elongated. If the forces are compressive,
the length is shortened in the direction of the forces. This is called the 'linear strain'.
The linear strain is defined as the ratio of change in length to the original length. If the change
(increase or decrease) in length is ' l ' in a wire or bar of original length L,
As the linear strain is ratio of lengths, it has no unit.

Change in length l
linear strain = =
original length L
(2) Bulk (or) Volume Strain:
When a force is applied uniformly and normally to the entire surface of the body, there is a change
in volume of the body, without any change in its shape. This strain is called 'bulk or volume strain'.
Volume strain is defined as the ratio of change in volume to the original volume. It has also no unit.
If 'v' is the change in volume produced in a body of original volume ‘V’,

Change in volume v
bulk or volume strain = =
original volume V
(3) Shearing (or) Rigidity strain:
When a force is applied parallel to one face of a body, the opposite side being fixed, there is a change
in shape but not in size of the body. This strain is called the shearing strain.
Solids alone can have a shearing strain. It is measured by the angle of the shear 'θ' in radian.
Three modulii of elasticity
There are three types of modulii depending upon the three kinds of strain.
(a) Young's modulus (E):
It is defined as the ratio of linear stress to linear strain.
Let a wire of initial length L and cross-sectional area ' A', undergo an extension l, when a stretching
force 'F', is applied in the direction of its length.
The modulus of elasticity, in this case, is called Young's modulus and is given by
Linear stress
i.e., Young's modulus ( E ) =
Linear strain
F
Then, longitudinal or linear stress = A A

𝑙
L
and longitudinal strain =
L
F
Linear stressF / A F.L
E= = =
Linear strain l / L A.l

The unit for Young’s modulus is newton m-2 with symbol N m-2. The single term unit which is
widely used for Young's modulus is 'pascal' with symbol 'Pa'.

(b) Bulk (or) Volume modulus (k):

It is defined as the ratio of bulk stress to bulk strain.

25
 When a body is subjected to a uniform compressive force, its volume decreases and the strain
produced is a bulk or volume strain.
If 'v' is the change in volume and V is the original volume, then
If F is the total compressive force acting on a total area A, then bulk stress = F/A = P
Bulk stress
Bulk modulus (k) = Bulk strain

If P is the stress applied i.e. (force/unit area) then,

v
bulk strain =
V
P P.V
bulk modulus k = =
v/V v
The unit for bulk modulus is newton metre -2 with symbol N m-2. The single term unit which is
widely used for bulk modulus is 'pascal' with symbol 'Pa'.
(c) Shearing (or) Rigidity modulus 'n':
The ratio of the shearing stress applied to the body to the shearing strain produced is called the
rigidity modulus and denoted by the letter 'n'.
If T is the tangential force/unit area and if θ is the angle of shear measured in radian, then
T F
rigidity modulus n = =
 A
F

The unit for rigidity modulus is newton metre -2 radian -1 with symbol N m -2 rad -1. The other
unit which is widely used for rigidity modulus is pascal radian -1 with symbol Pa rad -1.
Poisson's ratio: 
When a tensile stress is applied to a wire, the wire undergoes not only an extension of length in the
direction of the force but also a contraction in its thickness. The ratio of decrease in thickness to the
original thickness in lateral direction is known as lateral contraction.
The ratio of lateral contraction to linear elongation is called Poisson's ratio.
Lateral contraction
Poisson's ratio  =
Linear elongation

Bending of beams:
A beam is a body having uniform cross section, the length of which is very large compared to its
thickness. When such a beam is fixed at one end and loaded at the other end, it is called a cantilever.
A beam is supported symmetrically on two knife edges A and B at the same level. It is loaded
symmetrically with two equal weights W, W, beyond the knife edges, at C and D, so that AC=BD=a.
Now the beam is elevated at the centre. This type of bending is called uniform bending of beam.
A beam is supported symmetrically on two knife edges A and B near its ends in a horizontal level.
A weight hanger W is suspended by means of a loop of thread from a point which is exactly midway
between the two knife edges. Now the beam is depressed at the centre. This type of bending is called
non-uniform bending of beam.
26
Experiment to determine the Young's modulus of the material of a beam by uniform
bending method:
Mg Mg
a a

C l D
A B

Mg Mg

The given bar is supported symmetrically on two knife edges in a horizontal level, so that a quarter
of the bar projects beyond each knife edge. Two weight hangers are suspended from two loops of string
tightly attached at distances about 5 cm from each end so that their distances from the knife edges are
equal. A pin is fixed vertically at the midpoint of the bar. A traveling microscope is focused on the pin
such that the horizontal crosswire coincides with the tip of the pin. The reading in the vertical scale of
the microscope is noted. Weights are added in steps of M kg and the corresponding readings of the image
of the pin in the microscope are noted. Similarly readings are taken while unloading and tabulated.
The mean elevation y is found for a load of M kg. The distance 'I' between the knife edges is
measured. The distance 'a' between the point of suspension of the weight and the nearest knife edge is
also measured. The breadth 'b' of the bar found accurately with the vernier calipers and its thickness
d is found with a screw gauge. Young's modulus is calculated using the formula.
3Mgal2
E= Nm–2
2bd 3 y

Microscope reading
Load in Elevation y
Load Load
Kg Mean For M kg
increasing decreasing
x

x+M

x+2M

x+3M

Applications of elasticity:
The elastic behavior is one such property which especially decides the structural design of the
columns and beams of a building. As far as the structural engineering is concerned, the amount of stress
that the design could withstand is a primary safety factor. A bridge has to be designed in such a way that
it should have the capacity to withstand the load of the flowing traffic, the force of winds, and even its
own weight. The elastic behavior or in other words the bending of beams is a major concern over the
stability of the buildings or bridges. For an example, to reduce the bending of a beam for a given load,
one should use the material with a higher Young’s modulus of elasticity (Y). It is obvious that the
Young’s modulus of steel is greater than aluminium or copper. Iron comes next to steel. This is the
reason why steel is mostly preferred in the design of heavy duty machines and iron rods in the
construction of buildings.

27
 Worked Examples
1. A wire of 10 m long and of area of cross section 1.25 cm 2 elongates by 1.5 cm when 4.5 kg
is suspended to it. Find the stress, strain and young’s modulus of the wire.
Given :
L = 10 m
A = 1.25 cm2 = 1.25 x 10-4 m2
F = 4.5 kg wt = 4.5 x 9.8 N
l = 1.5 cm = 1.5 x 10-2 m
F 4.5 x 9.8
Stress, =
A 1.25 x 10−4

= 35.28 x 104 Nm-2

𝑙 1.5 x 10−2
Strain , =
L 10

= 0.15 x 10-2
stress
Young’s Modulus =
strain

35.28 x 104
=
0.15 x 10−2

E = 235.2 x 106 Nm-2

2. A body of mass 5 kg suspended by means of a steel wire of length lm and diameter 1 mm.
The young’s modulus of steel 21 x 1010 Nm-2. Calculate the elongation of the wire.
Given :
F = Mg = 5 x 9.8 N
D = 1 mm  r = 0.5 mm = 0.5 x 10-3 m
E = 21 x 1010 Nm-2
FL MgL
E = =
Al πr2 l
MgL
l =
πr2E
5 x 9.8 x 1
=
3.14 x (0.5 x 10−3) 2 𝑥 21 𝑥 10 10

l = 2.97 x 10-4m (or) 0.0297 x 10-2 m

3. Calculate the Young’s modulus of an iron bar of length 2 meters and cross section
2.5 x 10-5 m2 stretches through 0.5 mm. When a mass of 112.5 kg is hung on it.

FL
E =
Al

28
 112.5 x 9.8 x 2
E =
2.5 x 10−5 x 0.5 x 10 −3

E = 1.764 x 1011 Nm-2

4. A load of 5 kg is attached to the free end of a wire of length 2 m and diameter 0.6
mm. If the extension of the wire is 0.2 mm, calculate the Young's modulus of the
material of the wire. (g = 9.8 ms-2)

Length of the wire L = 2 m

Extension of the wire I = 0.2 mm = 0.2 × 10-3 m

Force acting of the wire F = 5kgf = 5 × 9.8 N
Diameter of the wire d = 0. 6 mm = 0.6 × 10-3 m
Radius of the wire =0.3 × 10-3 m

Area of cross section of the wire a = π r2 = π (3 × 10-4)2 m2

Young's modulus = Linear stress/Linear strain

5 x 9.8
Linear stress F = = 1.734 x 108 Nm-2
A  (3 𝑥 10−4 ) 2

l 2 𝑥 10−4
Linear strain = = 1 x 10-4
L 2

1.734 𝑥 108
E = = 1.743 x 1012 Nm-2
1 𝑥 10−4

5. A copper wire of 3 m length and 1 mm diameter is subjected to a tension of 5 kg wt.

Calculate the elongation produced in the wire, if Young's modulus of elasticity of copper is
120 G Pa.

F/A
Young’s modulus E =
I/L

FL
E =
Al

FL
Elongation produced I =
AE

MgL 5 x 9.8 x 3
= = = 1.562 x 10-3 m
πr2 x E 3.14 (0.5 x 10−3) x 120 x 109

29
6. What is the force required to a steel wire to double its length when its area of cross-section
is one sq. cm and Young's modulus is 200 G Pa. As the length of the wire is doubled, the
change in length is equal to its original length.
FL
The young' s modulus E =
Al
EAl
F=
L
EAl
= L
If L is the original length, then l = L
F = EA
= 200 × 109 × 10-4
= 2 × 107 N

7. A wire of length 2m is stretched by a force of 100 N. The area of cross – section of the wire
is 0.008 m2 and the increase in length is 0.05 mm. Calculate the stress, strain and young’s
modulus.

Force = 100 N

Area of cross – section = 0.008 m2

 stress = F/A

100
= Nm-2
0.008

= 12500 Nm-2

Increase in length = 0.05 mm

length = 2 m

𝑙 0.5 x 10−3
 strain = =
L 2

= 0.000025

stress
 Young’s Modulus =
strain

12500
=
0.000025

12500 𝑥 106
=
25

= 5 x 108 Nm-2

30
8. A wire 100 cm long and 0.5 mm in diameter becomes 103 cm when stretched. If the
young’s modulus of the wire is 9.81 x 1010 Nm-2. Calculate the value of the load.
FA
y=
𝑙/L

𝑀𝑔 L
y=
πr2 𝑙

L = 1m

l = 3 x 10-2 m

y = 9.81 x 1010 Nm-2

yπr3 𝑙
M =
gL

9.81 x 1010 x  x (0.25 x 10−3 ) 2 x 3 x 10−2

=
9.81 x 1.00

= 1010 x  (0.25)2 x 10-6 x 3 x 10-2

= 102 x  x (0.25)2

= 19.625 kg

31
 2.2 VISCOSITY
Let us consider a liquid flowing over a horizontal surface. The layer in contact with the surface is at
rest. The top most layer have the maximum velocity. The intermediate layers have intermediate velocity.
To maintain this relative motion of the layers, an external force must be acting on the liquid. Otherwise
the liquid will come to rest due to internal frictional forces acting between the layers of the liquid. These
internal frictional forces that bring the liquid to rest are known as viscous force and this property is
known as viscosity.
The property by virtue of which the relative motion between the layers of a liquid is maintained is
called viscosity. We can also say viscosity is the resistance to flow.
Coefficient of viscosity of a liquid
Let F be the viscous force acting between two layers of a liquid separated by a distance dx. Let dv be
the difference in velocity between the two layers. Let A be the area of the layers. The velocity gradient
dv/dx acts perpendicular to direction of flow of the liquid.
The viscous force F is found to be directly proportional to
i. the area of the layers 'A’.
dv
ii. the velocity gradient
dx
Adv
F
dx
 Adv
F=
dx
dv
If A = 1and =1, then F = 
dx
Where  is constant, called Coefficient of viscosity of the liquid
Coefficient of viscosity of a liquid is defined as the viscous force acting between two layers of a
liquid having unit area of layers and unit velocity gradient normal to the direction of flow of the liquid.

A
V+dv
dx F
V

SI unit of η:
F
=
A dv / dx Substituting the respective units of the quantities on the R.H.S,
N
The unit of coefficient of viscosity  = m 2 .ms −1 / m

= Nsm-2
Dimensional formula of η:
F
 = A dv / dx Substituting the respective dimensional formula of the quantities on the R.H.S,

MLT−2
The dimensional formula of coefficient of viscosity η =
L2.LT −1/L

= ML-1T-1
32
Stream line flow (laminar flow) and turbulent flow:

To maintain the flow of a liquid through a tube, some pressure should be given at one end of the
tube to overcome the viscous drag. For a given external pressure, the velocity of flow depends on the
coefficient of viscosity. At low pressure the velocity is less than a certain velocity called the critical
velocity and the liquid flows in an orderly manner and the flow is called steady flow or streamline
flow.
In stream line flow,
1. The liquid particles flow in an orderly manner. i.e., the liquid particles flow along straight lines,
each line is called the line of flow.
2. The lines of flow are parallel to the axis of the tube.
3. The velocity of all particles in a line of flow is a constant.
4. The velocity of flow is always less than the critical velocity.
5. Stream line flow will happen only when the driving pressure is small.
6. The flow of water through a capillary tube at the rate of 5 to 6 drops per minute is an example for
stream line flow.
When the external pressure causing the flow of the liquid is high, the flow of the liquid is said to be
turbulent.
In turbulent flow,
1. There is no orderliness in the flow of liquid.
2. The liquid particles will not flow along straight lines, but will flow in curved paths in a zig-zag
manner.
3. The different particles of the liquid will flow in different directions with different velocities.
4. The velocity of flow is always greater than the critical velocity.
5. Turbulent flow will happen when the driving pressure is high.
6. The flow of water from a tap is an example for turbulent flow.
Distinction between streamline motion and turbulent motion
Critical velocity:
When the external pressure driving the flow of liquid increases slowly, the velocity of flow of liquid
will also increase gradually. At a particular velocity the flow of liquid will change from laminar flow to
turbulent flow. This velocity is called critical velocity. In other word the critical velocity is defined as the
velocity at which the streamline flow ceases out and the turbulent flow sets in.

Sl.
Stream line motion Turbulent motion
No.

1 Flow of liquid is orderly. Flow of liquid is zigzag and random.

2 At any point along a straight line Velocity varies along the straight
paral- lel to the axis of the tube, line parallel to the ais of the tube.
velocity is con- stant

3 Velocity is proportional to the pressure. Velocity is proportional to the

square root of pressure
4 Velocity will always be less than Velocity will always be greater
critical velocity than critical velocity.

33
Reynolds number:
Reynolds number is a pure number, which determines the type of flow of a liquid, whether a
streamline flow or a turbulent flow through a cylindrical pipe. It has no units and dimensions. It is given
by the expression
where V is velocity of the liquid, p is the density of the liquid, n is the coefficient of viscosity of the
liquid and D is the diameter of the pipe. This equation is applicable only for liquids flowing through
cylindrical pipe.

R = VρD


Experiments show that the flow of a viscous liquid is laminar or streamlined, if Reynolds number is
less than 2000. The flow of a viscous liquid is turbulent if Reynolds number is more than 2800. Between
2000 and 2800 there is a transition region in which the flow may change from streamline to turbulent.
Narrow tubes and highly viscous liquids tend to promote streamline motion, while wider tubes and
liquids of low viscosity leads to turbulence.
Experiment to determine comparison of coefficient of viscosity of two liquid - Capillary
flow method.
Aim:
To compare the coefficient of viscosity of two given liquids by capillary flow method,.
Formula:
The comparison of coefficient of viscosity

η1 / η 2 = ρ1 / ρ2 × t1 / t2

η 1 = coefficient of viscosity of first liquid

η = coefficient of viscosity of second liquid
2

ρ 1 = Density of the first liquid

ρ = Density of the second liquid
2

0
5
10
15
20
BURETTE

25
30
35
Capillary Tube

Description:
The graduated burette without stopper is mounted vertically in the stand. A rubber tube is connected
to the bottom of the burette. To the other end of the tube, a capillary tube is inserted and placed in a
perfectly horizontal position.

34
Procedure:
The burette is cleaned and filled with one of the liquids using a funnel. Adjust the tube till the
liquid through the tube comes out drop by drop. A stop clock is started when the liquid level reaches
0 c c and time taken for every 5 c c from 5,10, 15,20.. till 50 c c, is noted .The burette and the tube are
cleaned, rinsed and filled with the second liquid. The time taken to cross the every consecutive 5 ml is
noted.

The reading are tabulated .The time taken for the t1 and t2 are noted. The mean value of t / t is
calculated.

Time (S)
Burette reading t1/t2
(cc)
Liquid-I (t1) Liquid-II (t2)

0-5

5-10

-
-

-
45-50.

Mean

Terminal velocity
When a small metallic sphere is gently placed on the surface of a highly viscous liquid it descends
down with an acceleration. As its velocity increases, the viscous force opposing the motion of the sphere
also increases.
At some stage the apparent weight of the sphere becomes equal to the force due to viscous drag.
So the resultant force on the sphere is zero. Therefore the sphere moves down with a constant velocity
known as terminal velocity.
Terminal velocity is that velocity at which the apparent weight of the sphere is equal to the viscous
force acting on the sphere moving in a high viscous liquid.
Derivation of the stoke’s formula by the method of dimensions for highly viscous liquids.
Let a spherical body of radius a fall down through a viscous liquid of coefficient of viscosity 
Let V be its terminal velocity.
When terminal velocity is attained it is presumed that
1. The opposing force F is directly proportional to some power of terminal velocity of the body
(Vx).
2. F is directly proportional to some power of the radius of the body (ay).
3. F is directly proportional to some power of the coefficient of viscosity of the medium (z)
(i.e.) F  Vx

35
 F  ay
F  z
(i.e) F  Vx ay  z
(i.e) F = KVx ay z ----- (1)
Where K is a constant.
Substituting the dimensional formula for each quantity.
MLT-2 = K (LT-1)x Ly (ML-1 T-1)z
Equating the powers on L on both sides,
L = x+y–z ----- (2)
Equating the powers on M on both sides,
1 = z ----- (3)
Equating the powers on T on both sides,
-2 = -x-z ----- (4)
Substituting the value of z = 1 in (4)
-2 = -x–1
x = 1 ----- (5)
Substituting the value of x is z in (2)
1 = 1+y-1
y = 1 ----- (6)
Equation (1) becomes
P = (constant) V1 a1 1
The constant is found to be 6,
 F = 6 Va ----- (7)
Experimental determination of the coefficient of viscosity of a high viscous liquid by
Stokes' method:
Aim:
To determine the coefficient of viscosity of highly viscous liquid
(Castor oil) by Stokes' method.
Formula:
The coefficient of viscosity of the highly viscous liquid
2 ( − ) g (r 2 t)
=
9 h
where
 = the density of the material of the solid sphere (glass beads)
 = the density of the liquid (castor oil)
A
g = the acceleration due to gravity B
r = the radius of the glass beads h
C
t = the time taken by the solid sphere to travel the marked distance h, D
inside the high viscous liquid.
h = the distance between the marks B and C
36
Description:

The pure, transparent, highly viscous liquid (castor oil) is taken in the tall glass jar. On the outer
surface of the jar, two markings B and C or C and D are made at a distance of say 50 cm. The marking
B should be well below the free surface of the liquid. Then only the sphere can attain terminal velocity
when travelling from B to C or C to D.
Procedure:
First the least count, zero error and zero correction of screw gauge are determined. Then using the
screw gauge, the diameters of the given glass beads are measured. The radii of the beads are calculated.
Values of r and s are taken from standard tables.
The sphere is placed gently on the surface of the highly viscous liquid and dropped. When it crosses
the marking B, a stop clock is started and when it crosses the marking C, the clock is stopped. The time
taken (t) to travel by the sphere from B to C is recorded in the tabular column. The distance between B
and C is noted as 'h'. The same procedure is repeated for all the given spheres. Then the coefficient of
viscosity of highly viscous liquid (castor oil) is calculated, using the formula.
Tabular column:
To find r2t

Glass beads Radius (r) Radius2 (r2) Time taken (t) r

2
U m m s mm
2t
Bead 1nit m m2 s
Bead 2
Bead 3

The average of r2t = mm2s

= x 10-6 m2s
Applications of viscosity
To reduce friction between the two moving surfaces which are in contact, a drop or two of some oil
is introduced between the two contact surfaces. The oil enters into the depressions in the contact surfaces
and form a thin layer so that direct contact between the surfaces is prevented. Now, the viscous force of
the oil in between the surfaces (lubricant) reduces the friction between the two moving surfaces because
the fluid friction is always less than the friction between the solid surfaces. However a good viscous
lubricant can only reduce friction, but cannot completely eliminate the friction.
Generally, for high speeds and low pressures, highly viscous oils such as Mobil oil, grease, etc. are
used. For low speeds, oils of lower viscosity are used. It may be interesting to note that air also acts as a
sort of lubricant in reducing friction. The friction between two metallic surfaces in vacuum is found to
be greater than that in air.
Practical applications of Stoke’s law :
Since the raindrops are smaller in size and their terminal velocities are small, remain suspended in
air in the form of clouds. As they grow up in size, their terminal velocities increase and they start falling
in the form of rain.
This law explains the following:
a) Floatation of clouds
b) Larger raindrops hurt us more than the smaller ones
c) A man coming down with the help of a parachute acquires constant terminal velocity.
37
 2.3 SURFACE TENSION
Introduction
The free surface of a liquid at rest behaves like a stretched elastic membrane with a tendency to
contract in area. The following simple experiments will illustrate this property.
1. When a camel hairbrush is dipped into water, the hairs remain spread out. When the brush is
taken out, the hairs cling together on account of the films of water between them contracting.
2. A film of the soap solution is formed in a wire ring. If a wetted loop of thread is gently placed
on the film, the film is unaffected and the thread remains in any form in which it is placed on the
film. But if the film is pricked inside the loop, the film outside contracts on all sides and hence
the circular loop is formed. The area of the film is reduced to a minimum.
The above examples show that a force acts on the surface of liquid to reduce the surface area to a
minimum. This force is known as surface tension.
A water drop takes the spherical shape because for a given volume, sphere has the minimum surface
area. This is due to surface tension.
Surface tension
Surface tension is the force acting on unit length of an imaginary line drawn on the free surface of a
liquid, the force acting normal to the line and parallel to the surface. The unit of surface tension is Nm-1
and its dimension is MT-2
Angle of contact
The angle of contact is the angle by which the tangent drawn to the liquid surface at the point of
contact with the solid, makes with the surface of the solid, the angle being measured within the liquid.
For liquids like water and kerosene which spread on glass, the angle of contact θ = 0°. In the case
of mercury and clean glass, the angle of contact θ = 137°. For ordinary water and glass, the angle of
contact θ = 8°.
When a glass rod is dipped vertically in water, which wets the glass, the water is drawn up around
the glass rod at all points of contact. Here, the angle of contact is acute. When a glass rod is dipped
vertically in mercury, which does not wet the glass, the liquid is depressed around the glass rod at all
points of contact. Here, the angle of contact is obtuse.

Expression for the surface tension by capillary rise:

T Cos T Cos
B D
 
R=T R=T

T Sin T Sin
C

T T h

38
 Let a capillary tube of uniform radius (r) be dipped vertically into a liquid of density ρ taken in a
beaker. Let the liquid rises through a height (h) in the capillary tube. Let θ be the angle of contact with
glass and T be the surface tension of the liquid. The surface tension is acting tangential to the liquid
surface. The reaction R=T, offered by the walls of tube acts at an angle θ to the verticals. This reaction
T along AB can be resolved vertically as Tcosθ and horizontally as Tsinθ. Similarly the reaction T along
CD can also be resolved along the horizontal and vertical directions. The horizontal components along
the circumference are equal and opposite in direction. So they get cancelled. So the total force due to
surface tension is only due to vertical components.
The total upward force acting through the length of the meniscus 2πr
= 2 π r. T cosθ
This force balances the weight of the liquid column of height h and radius r.
The weight of the liquid column = π r2h ρ g
2πrTcosθ = πr²hρg
hrg
T = 2cos 
In the case of water θ = 0° and hence cosθ = 1
Hence for water, Surface tension
hrg
T=
2
Experimental determination of the surface tension of water by capillary rise method.

h2

h1

A capillary tube of uniform cross sectional area is first cleaned with dilute acid and washed with
distilled water and dried. It is then clamped vertically with one end inside the water taken in a beaker.
Due to surface tension, water rises to a definite height in the tube. A pointer is attached to the capillary
tube such that it just touches the surface of the water in the beaker.
Using a travelling microscope the reading in the vertical scale corresponding to the lower meniscus
of the water in the tube is taken. The beaker is removed and the microscope is brought down and the
reading corresponding to the tip of the pointer is taken. The difference between the two readings gives
the height of the capillary rise of water 'h' in the capillary tube. The diameter and hence the radius 'r' of
the capillary tube is also determined with the help of microscope. If ρ is the density of water and g, the
acceleration due to gravity, then surface tension of water,
hrg
T=
2
Applications of capillarity:
1. Lubricating oil spread easily on all parts because of their low surface tension.
2. Cotton dresses are preferred in summer because cotton dresses have fine pores which act as
capillaries for sweat.

39
 3. Dirt get removed when detergents are added while washing clothes because surface tension of
water is reduced.
4. The absorption of ink by a blotting paper is due to capillary action, as the blotting paper is
porous. When it is placed over the ink, the ink raises into the pores. Also rise of oil in the wick
of a lamp is due to capillary action.
5. If one end of a towel is dipped into a bucket of water and the other end hangs over the bucket,
the entire towel soon becomes wet due to capillary action.
6. The supply of water to the leaves at the top of even a tall tree is through capillary rise.
7. A fabric can be waterproof, by adding suitable waterproofing materials to the fabric. This
addition increases the angle of contact, thereby making the fabric waterproof.
Applications of surface tension
1. Mosquitoes lay their eggs on the surface of water. To reduce the surface tension of water, a
small amount of oil is poured. This breaks the elastic film of water surface and eggs are killed
by drowning.
2. Chemical engineers must finely adjust the surface tension of the liquid, so it forms droplets of
designed size and so it adheres to the surface without smearing. This is used in desktop
printing, to paint automobiles and decorative items.
3. Specks of dirt get removed when detergents are added to hot water while washing clothes
because surface tension is reduced.
4. A fabric can be made waterproof, by adding suitable waterproof material (wax) to the fabric.
This increases the angle of contact.
Worked Examples:
1. Calculate the surface tension of water if it rises to a height 4.2 cm in a capillary is 3.5 x 10 -4
m.
Given : r = 3.5 x 10-4 m,
H = 4.2 x 10-2 m
Known :  = 1000 kgm-3 g = 9.8 ms-2

hrg
T =
2

3.5 x 10−4 x 4.2 x 10−2 x 1000 x 9.8

=
2

T = 0.072 Nm-1

2. A liquid of density 1200 kgm-3 and a surface tension 214 x 10-3 N/m rises to a height of 2.4
cm in a capillary tube of diameter 0.5 mm. Calculate the angle of contact of the liquid with
the walls of the capillary tube.
rhg
Surface tension of liquid, T =
2 cos 
Given :
 = 1200 kgm-3, T = 214 x 10-3 Nm-1

40
 d = 0.5 mm

r = 0.25 mm

rh
Cos  =
2T

0.25 x 10−3 x 2.4 x 10−2 x 1200 x 9.8

=
21 x 214 x 10−3

= 16.49 x 10-2

= 0.1649

Angle of contact, O = cos-1 (0.1649)

= 80o 301.

3. A capillary tube of diameter 0.5 mm is dipped into a liquid vertically and the liquid rises to a
height of 6 cm. If the density of the liquid is 1000 kgm -3, calculate the value of surface
tension of the liquid. (g =9.81ms-2)

Diameter of the capillary tube 2r = 0.5 mm

Radius of the capillary tube r = 0.25 x 10-3 m

Rise of liquid in the tube h = 6 cm = 6 x 10-2 m

Density of the liquid ρ = 1000 kg m-3

Liquid being water, angle of contact θ = 0°

hgr
Surface tension of the liquid T =
2cos 

6 10−2 1000  9.81 0.25 10−3

2cos 

T = 72. 5 x 10-3 Nm-1

41
4. Calculate the diameter of a capillary tube in which a liquid rises 2.34 × 10-2 m. Surface tension
of the liquid is 25 × 10-3 Nm-1. Relative density of the liquid is 0.79. Angle of contact = 11°12'.
(g = 9.81ms-2)
Relative density of the liquid = 0.79
Density of the liquid ρ = 0.79 × 103 kgm-3
hgr
We know T =
2cos 
2T cos 
r = hg
2  25 10−3 Cos11012'
-4
= 2.34 10−2  0.79 103  9.81 = 2.705 × 10 m

Diameter of the tube = 2r = 5.41 × 10-4 m

5. A liquid of density 1000 kgm-3 is taken in a beaker. A capillary tube of diameter 1.0 mm is
dipped vertically in it. Calculate the rise of liquid in the tube. Surface tension of the liquid is
72 × 10-3Nm-1.
For water, angle of contact θ = 0°
Surface tension T =72 x 10-3 Nm-1
Diameter of the capillary tube 2r = 1 mm

Radius of the capillary tube r = 0.5 × 10-3 m

h = 2.939 × 10-2 m

2T cos 
Rise of liquid in the tube h =
rg
2  72 10−3 Cos00
h = 0.5 10−3 1000  9.8 = 2.939 × 10-2 m

6. A capillary tube of internal diameter 0.6 mm is dipped into a liquid of density 1140 kgm-3. The
liquid rises to a height of 2.6 cm in the capillary tube. If the angle of contact of liquid with glass
is 55°, calculate the surface tension of the liquid.
Diameter of the capillary tube = 0.6 mm
Radius of the capillary tube r = 0.3 x 10-3 m
Density of the liquid, ρ = 1140 kg m-3
Height through which liquid rises h = 2.6 x 10-2 m

Angle of contact, θ = 55°

hgr
Surface tension of the liquid T =
2cos 
2.6 10−2  0.3 10−3 1140  9.8
T =
2 cos 550
T = 75.96 × 10-3 Nm-1

42
7. A capillary tube of 0.5 mm bare stands vertically in a vessel containing a liquid of R.D. 0.8. If
the liquid wets the tube, find the rise of the liquid in the tube. (S.T. of liquid = 20 x 10 -3 Nm-2)
Let ‘h’ be height of the liquid.
r = 0.25 mm = 0.25 x 10-3 m
d = 0.8 x 1000 kgm-3 = 800 kgm-3
S.T. = 20 x 10-3 Nm-1
g = 9.8 ms-2
The angle of contact is zero since the liquid wets the tube.
hrdg
S.T. =
2 cos 
20 x 10−3 x 2 x 1
h =
0.25 x 10−3 x 800 x 9.8
40
=
1960
= 0.0204 m
8. A plate of area 0.1 m rests on a layer of coefficient of viscosity 1.55 Nsm -2 and 1 x 10-3m thick.
2

Calculate the force required to move the plate with a speed of 0.01 ms-1.
We know,
dv
F = 
dx
A = 0.1m2
dx = 1 x 10-3 m
 = 1.55 Nsm-2
dv = 0.01 ms-3
1.55 𝑥 0.1 𝑥 0.01
F =
1 𝑥 10−3
= 1.5 N
9. A capillary tube of 0.4 mm bore stands vertically in a wide vessel containing a liquid of surface
tension 72 x 10-3 N/m. The liquid wets the tube and has a density 1000. Calculate the rise of
the liquid in the tube.
hgr
T =
2
2𝑇
h = = ?
gr

T = 72 x 10-3 Nm-1
r = 0.2 x 10-3 m
g = 9.81 ms-2

 = 1000 Kg m-3

2 x 72 x 10−4
h =
1000 x 9.81 x 0.2 x 10−3

m = 0.0734 m

43
10. In stokes experiment the radius of the solid sphere is 0.3 mm and its density is 2500 kgm-3
which is dropped into a liquid of density 0.97 x 10 3 kgm-3 and it travels a distance of 25 cm in
43 calculate the coefficient of viscosity of the liquid.
2 x (p− )
 = x g (a2t)
9 BC

P = 2500 kgm-3, g = 9.81 ms-4

 = 0.97 x 103 = 970 kgm-3

a = 0.3 mm = 0.3 x 10-3 m

t = 43
BC = 25 cm = 0.25 m

 = ?

2 (2500−970) x 9.81 x (0.3 x 10 −3)2 x 4

 =
9 0.25

2 1530 x 9.81 x (0.3)2 x 10 −6 x 4

= x
9 0.25

= 0.004803 Nsm-2.

44
 QUESTIONS
Part -A and Part- B
1. Define stress
2. Define strain
3. Which is more elastic, rubber or steel and why?
4. What is elastic limit and plastic limit?
5. Define elastic body and plastic body.
6. What are the three modulii of elasticity?
7. Define (i) linear strain (ii) bulk strain and (iii) shearing strain.
8. Write the statement of Hooke's law.
9. Define Modulus of elasticity
10. Define (i) Young's modulus (ii) Bulk modulus and (iii) Rigidity modulus.
11. What is uniform bending of a beam?
12. What is non uniform bending of a beam?
13. Define coefficient of viscosity.
14. Derive the dimensional formula and the SI unit for the coefficient of viscosity
15. What is stream line motion?
16. What is turbulent motion?
17. What is critical velocity?
18. What is terminal velocity?
19. Write the application of viscosity.
20. Derive the S.I. unit and the dimensional formula of surface tension.
21. What is the effect of surface tension on the surface area of liquids?
22. The droplet of rain is spherical. Why?
23. How do insects run on the surface of water?
24. Define surface tension of a liquid.
25. Define angle of contact.
26. Write any two applications of capillarity.
27. What are uniform bending of beams.
28. What are non-uniform bending of beams
29. Explain young’s modulus
30. Explain Bulk modulus
31. Explain Rigidity modulus
32. Explain stream line motion
33. Explain turbulent motion

45
 Part - C
1. Explain three types of modulus
2. Describe an experiment to determine the Young's modulus of the material of a bar by
bending it uniformly.
3. Distinguish between stream line and turbulent motion.
4. Describe an experiment to determine the comparison of coefficient of viscosity of two
viscous liquid by capillary flow method.
5. Describe Stokes' method of determining the coefficient of viscosity of a transparent, high
viscous liquid.
6. Derive an expression for the surface tension of a liquid in the case of the capillary rise.
7. Describe an experiment to determine the surface tension of water by capillary rise method.
8. List the various applications of capillarity.

Exercise Problems
1. The length of a wire increases from 1.25 m to 1.2508 m when a load of 12 kg is suspended.
The radius of the wire is 0.5 mm. Find the stress, strain and young's modulus of material of the
wire.
Ans: (149.7 × 106 Nm-2, 6.4 x 10-4, 23.4 × 1010 Nm-2
2. What load in kilogram must be applied to a steel wire of length 1m and diameter 0.05 cm to
produce an extension of 1 cm. The Young's modulus of steel = 2 × 1011. Pa.
Ans: 40 kg
3. A wire 10m long of area of cross section 1.22cm2 elongates by 1.5 cm when 4.5
kg is suspended from it. Find the (i) stress (ii) strain and (iii) young's modulus of the wire.
(g = 9.8m s-2)
Ans: (i) 3.528 × 105Nm-2 (ii) 0.0015 (iii) 2.352 × 108 Nm-2
4. A body of mass 5kg is suspended by means of a steel wire of length 1 m and diameter 1mm.
The Young's modulus of steel is 21 × 1010 Nm-2, calculate the elongation of the wire.
Ans: l = 2.97 × 10-4 m
5. A uniform metal bar, one metre long, is placed symmetrically on two knife-edges separated
by a distance of 64.2 cm. When two equal weights of 0.5 kg each are suspended from
points 5cm from the two ends respectively, the mid-point of the bar is elevated by 0.76 mm.
Calculate the Young's modulus of the material of the bar if the width of the bar is 2.2 cm
and its thickness 0.62 cm.
Ans: 9.81 × 1010 Nm-2
6. Calculate the surface tension of water if it rises to a height of 4.2cm in a capillary tube dipped
vertically in it. Radius of the capillary tube is 3.5 × 10-4 m.
Ans: 72.03 × 10-3 Nm-1
7. A capillary tube of diameter 0.5 m is dipped into a liquid vertically and the liquid rises to a
height of 4.4 cm. If the density of the liquid is 1000 kg m-3, calculate the value of the surface
tension of the liquid.
8. A capillary tube of bore 0.84 mm in dipped into a liquid of density 840 kgm-3 and surface
tension 0.049 Nm-1. Find the rise of the liquid in the tube.

46
Brain teasers
1. A wire of diameter 2.5 mm is stretched by force of 980 N. If the Young's modulus of the
material of the wire is 12.5 × 1010 Nm-2. Find the percentage increase in the length of the
wire.
2. Two wires are made of same material. The length of the first wire is half of the second wire
and its diameter is double that of the second wire. If equal loads are applied on both the
wires find the ratio of increase in their lengths.
3. The diameter of a brass rod is 4 mm. Calculate the stress and strain when it is stretched by
0.25% of its length. Find the force exerted (E = 9.2 × 1010 Nm-2 for brass)

https://www.youtube.com/watch?v=tTiL42Y7TEI&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=7

https://www.youtube.com/watch?v=oRQ66FCSllY&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=8

https://www.youtube.com/watch?v=JfOiozY3GDM&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=9

https://www.youtube.com/watch?v=yc1ATWCDb3I&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=10

https://www.youtube.com/watch?v=9hsZYeXwY-Q&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=11

https://www.youtube.com/watch?v=Vjse-vR6m3Q&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=12

47
 SUMMARY

• Steel is used as shock absorbers in vehicles.

• Stress and Strain are classified by three types, i.e. Linear, Bulk and
Shearing.
• Within elastic limit stress is directly proportional to strain.
• Modulus can be classified by three types, i.e. Young’s, Bulk and Rigidity.
• Liquid consists of layers.
• Friction is the force resisting the relative motion of the solid surfaces, liquid
layers sliding against each other.
• Critical velocity, Terminal Velocity
• Stokes’s Method
• Low viscous liquids are Petrol, Diesel, Kerosine etc
• High viscous liquids are Oil, Grease, Honey etc.
• The surface of the water is like a stretched elastic membrane.
• The cohesive forces between liquid molecules are responsible for Surface
Tension.
• The surface of the water reduces its own area to minimum. e.g. Rain drops
are spherical in shape.
• Oil has low surface tension when compared to water. So oil spreads like a
thin film over the surface of the water.

48
 UNIT - III

DYNAMICS - I

OBJECTIVES

▪ To know about the types of motion

▪ To understand Newton’s laws of motion
▪ To familiarize the equations of motion
▪ To understand the Projectile motion
▪ Can derive the maximum height of a projectile
▪ To know about the maximum range of a projectile
▪ To understand circular motion
▪ To define angular velocity
▪ To relate linear velocity and angular velocity
▪ To define normal acceleration
▪ To know about the centripetal and centrifugal force
▪ Can derive the expression for normal acceleration and centripetal force
▪ To know about angle of banking
▪ Can derive the expression for angle of banking in Railway tracks
▪ To solve the simple problems based on projectile motion, circular motion
and curved path motion.

49
UNIT

3 DYNAMICS - I

3.1 STRAIGHT LINE MOTION

Introduction:
Dynamics is a branch of mechanics which deals with the motion of the bodies under the action of
forces.
A body is said to be at rest if it does not change its place or position with respect to its surroundings.
No body in this earth is at absolute rest. Because, they are sharing the motion of the earth. A body can
be at rest with respect to another body.
A body is said to be in motion if it changes its place or position with respect to its surroundings. For
example, a person in the moving train feels the motion, seeing the passing objects through the windows.
Hence rest and motion are relative terms.
The motion of the body is characterised by two quantities, Force and Momentum. If a body is in
motion with acceleration, then the body is acted upon by a force (F=ma). If a body is in motion with
uniform velocity, there is no acceleration, then the motion of the body is only under momentum (mv).
Newton's Laws of motion: These laws are very important in dynamics.

I Law: Everybody continues to be in its state of rest or of uniform motion in a straight line unless
compelled by an external force.
II Law: The rate of change of momentum of a body is directly proportional to the force acting on it
and it takes place in the direction of the force.
III Law: For every action, there is always an equal and opposite reaction.
In this unit we shall study about the Projectile motion, Circular motion, its application and Simple
Harmonic Motion.
In this chapter we shall consider the motion of the objects projected in any direction from the earth
making the following assumptions.
1) The air resistance is negligible
2) The acceleration due to gravity remains constant.
When an object is projected vertically up from the earth, we regard the upward direction as the
positive direction. The force of gravity produces acceleration downwards on the object. Acceleration is
therefore taken as negative.
Equations of motion: Let u = initial velocity, v = final velocity, a = acceleration, t = time and
s = displacement.

50
(i) For Objects in motion
(1) v = u + at
(2) s = ut +1/2 at2
(3) v2 = u2 + 2 as

(ii) For objects falling freely

Here initial velocity u = 0 and a = +g
(1) v = gt
1
(2) S = gt2
2
(3) v2 = 2 gs

(iii) For objects thrown vertically upwards

Here a = -g
(1) v = u - gt
1
(2) s = ut - gt2
2
(3) v2 = u2 - 2gs

51
 3.2 PROJECTILE MOTION

Projectile
Projectile is any object thrown obliquely into the space. The object which is given an initial velocity
and afterwards follows a path determined by the gravitational force acting on it is called projectile. Its
motion is called projectile motion.
A stone projected at an angle, a bomb released from an aeroplane, a shot fired from a gun, a shotput
or javelin thrown by the athlete are the various examples for the projectile.
Consider a body projected from a point 'O' with velocity 'u'. The point 'o' is called point of projection
and 'u' is called velocity of projection.
Y

u
A
D
I
u sin



O u cos  C B X

Velocity of Projection (u): The velocity with which the body projected is called velocity of
projection.

Angle of Projection (a): The angle between the direction of projection and the horizontal plane
passing through the point of projection is called angle of projection.

Trajectory (OAB): The path described by the projectile from the point of projection to the point
where the projectile reaches the horizontal plane passing through the point of projection is called
trajectory. The trajectory of the projectile is a parabola.

Maximum height (H): The maximum vertical displacement of the projectile from the horizontal
plane passing through the point of projection is called maximum height.

Time of flight (T): The time taken by the projectile to travel from the instant of projection to the
instant when it reaches the horizontal plane passing through the point of projection is called time of
flight (or) The time taken by the projectile to describe the trajectory.

Range (R): The distance between the point of projection and the point where the projectile reaches
the horizontal plane passing through the point of projection is called the range of the projectile.

Note: The horizontal component of velocity u cos a is constant throughout the motion and the
vertical component of velocity u sin a goes on decreasing and becomes zero at the highest point A,
because of the constant retardation '-g' acting on the projectile.

52
Expressions for maximum height, time of flight and horizontal range of a projectile.
Y

u
A
D
I

u sin

H


O u cos  C B X

Consider a body be projected upwards from the point of projection O with the velocity u. Let  be
the angle of projection. The velocity u can be resolved into two components. Let u sin  and u cos  are
the vertical and horizontal components of the velocity.
Maximum height reached by the projectile:
Let H be the maximum height reached by the projectile. At the highest point the vertical component
of velocity becomes zero.
Considering the vertical motion of the projectile,
Taking the equation of motion
v2 = u2 + 2as ...(1)
Final velocity at A, v = 0
Initial velocity at O, along vertical direction u = u sin 
Acceleration a = -g
Vertical displacement AC, s= H
Substituting these values in equation (1)

0 = u2 sin2  + 2(-g) H
0 = u2sin2  - 2 gH
2gH = u2sin2 

u 2 sin 2 
 Maximum height, H =
2g
Time of Flight of the Projectile:
The time taken by the projectile to describe the path is called time of flight. i.e., It is the time taken
by the projectile to travel from 'O' to 'B', via A.
Let T be the time of flight. Here, the vertical displacement at B is zero.
Considering the vertical motion, Taking the equation of motion
1
S = ut + 2 at2
Vertical displacement S = 0
Initial velocity at O, along vertical direction u = u sin 
Acceleration a = -g ; Time t = T; Substituting these values in equation (2)

1
53
 0 = u sin T + 2 (- g)T2
1
0 = u sin T - g T2
2
1 g T2 = u sin T
2 2u sin 
 Time of flight, T =
g

Horizontal Range of the Projectile :

The horizontal distance between the point of projection and the point where the projectile meets the
horizontal plane passing through the point of projection is called range of the projectile.
Let R be the range of the projectile.
Range R = Horizontal component of velocity × Time of flight
2u sin 
R = u cos  ×
g
u 2 sin  cos 
= g
u 2 sin 2
 Range, R =
g

Condition for maximum range (Rmax):

u 2 sin 2
The expression for Horizontal Range,R =
g
For a given velocity of projection, range will be maximum only when sin 2  is maximum. The
maximum value for sine function is 1.
i.e Sin 2  = 1 but, Sin 900 = 1
 Sin 2  = Sin 90°  2 = 90°   = 45°
For getting maximum range, the condition is the angle of projection should be 45°.
u 2 sin (2  450 )
 Maximum range, Rmax =
g
u 2 sin 900 u 2
Rmax = = ( sin 900 = 1)
g g
u2
Rmax =
g
This is the expression for the maximum range of the projectile.

54
Path of the projectile (the trajectory) is a Parabola:

Consider a body be projected from the point of projection O with the velocity u as shown in the
figure. Let  be the angle of projection of the body. The velocity u can be resolved into two components.
u sin  and u cos  are the vertical and horizontal components of velocity u. Let P (x, y)
Horizontal displacement in t second, x = u cos  x t
X
t =
u cos 
1
Taking the equation of motion S = ut + 2
at2 (1) be the position of the body after
't' second.
Vertical displacement in t second, S = y
X
Time taken to reach P, t =
u cos 

Initial velocity at O along vertical direction, u = u sin 

Acceleration a = -g
X 1 X2
y = u sin  u cos  + 2
(-g) u2 cox2 
g
y = x tan  - 𝑥2
2u2 co𝑠 2 
g 𝑥2
y = tan x - 2u2 co𝑠2 
Substituting these values in equation (1)
The above equation is similar to the equation of the parabola y = bx – cx2. Hence the path of the
projectile is a parabola.

Examples of Projectile motion:

1. A bomb released from an air plane in level flights
2. Bullet fired from a gun
3. Javelin is thrown by an athlete
4. An arrow released from the bow.

55
 Worked Examples
1. A body is thrown with a velocity of projection 49 ms-1 at an angle of projection 45°. Find
(i) Maximum height (ii) Time of flight (iii) Range (g= 9.8 ms-2).
Given
u = 49ms-1
 = 45°
u 2 sin 2 
Maximum height, H = 2g
49  49  sin 450  sin 450
=
2  9.8
= 61.25 m
2u sin 
Time of flight, T = g
249sin45
=
9.8
= 7.07 s

u 2 sin 2
Range, R = g
49  49  sin (2  45)
=
9.8
49  49  sin 90
= 9.8
= 245 m

2. Find the velocity with which a stone may be projected n order that it may have a maximum
range of 300 m on the horizontal plane through the angle of projection for maximum range
(g – 9.8 ms-2)
Rm = 300 m
g = 9.8 ms-2
u = ?
U2
Rm =
g

U 2
= Rm x g
= 300 x 9.8
U2 = 2940
U = 54.22 ms-1

56
3. If the range of the projectile in a horizontal plane is 50 m and the time of flight is 4 seconds,
find the initial velocity and the angle of projection.
R = 50 m
t = 45
U2 Sin 2 
Range =
g

U2 Sin 2 
50 = ----- (1)
9.8
2 U Sin 
t =
g

2U Sin 
4 = ----- (2)
9.8

Squaring the equation (2)

4U2 Sin 2 
16 =
(9.8) 2
----- (3)

Dividing (1) by (3)

50 U2 Sin 2 (9.8) 2
=
16 (9.8) x 4U 2 x Sin 2

50 2 Sin  cos  x 9.8

=
16 4 sin 2 
50 9.8 cos 
=
16 2 sin 
50 9.8
16
= 2 tan 
9.8 𝑥 16
tan  = 2 𝑥 50

tan  = 1.568
 = 57o201
Substituting this in equation (2)
2U Sin 570 291
4 =
9.8
4 x 9.8 19.6 19.6
U = = =
2 x Sin 570 291 Sin 570 291 0.8433

U = 23.24 ms-1

4. An aeroplane is in level flight at 180 kmh-1 at an attitude of 10000 m. How far from a given
target should it release the bomb so as to hit the target ?
Horizontal uniform velocity of the bomb = velocity of aeroplane.
= 180 kmh-1
= 50 ms-1
Height of the bomb = S = 10000 m.

57
 Substituting in equation S = ut + ½ at2
10000 = 0 x t + ½ x 9.8 x t2
20000
t2 =
9.8

t = 45.1 sec.
Distance from target (Range) = Velocity x time
= 50 x 45.1
= 2255 m.

5. An aircraft is moving horizontally with a speed of 50 m/s. At the height of 2 km, an object is
dropped from the aircraft what is time interval before the object hits the ground ?
(g = 10 m/s2)
Given data:
H = 2 km = 2000 m, g = 10 m/s2
t = ?
Solutions:
50 m/s

2000 m

s = H = ut + ½ at2
a = g, u = 0
H = ½ gt2
2000 = ½ x 10 x t2
2000
t2 =
5

t = √2000/5
= 20 seconds

6. A kicked football leaves the ground at an angle 45 o with the horizontal has an initial velocity
of 25 m/s. Determine the horizontal range. (g = 10 ms -2)
Given data:
 = 45o
u = 25 m/s
g = 10 ms-2

58
 Solution:
U2 sin 2 α
Range R =
g

(25)2 sin 2 x 45 𝑜
=
10

R = 62.5 m

7. An object is launched at the initial velocity of 20 m/s. in a direction making an angle of 25 o

upward with the horizontal. What is the maximum height reached by the object?
Given:
 = 25o
h = 20 m/s
H = ?
Solution:
U2 sin 2 α
Maximum height H =
2g

(20) 2 sin 2 250

H =
2 x 9.8
400 𝑥 0.1786
H =
19.6

H = 3.645 m

8. A ball kicked from the ground level at an initial velocity of 60 m/s and an angle  with
ground reaches a horizontal distance of 200 m. What is the angle of Projection?
Given:
u = 60 m/s
R = 200 m
 = ?
Solution:
U2 sin 2 α
R =
g

(60) 2 sin 2 
=
9.8
3600 x sin 2
200 =
9.8
200 x 9.8
= sin 2 
3600

2 = sin-1 (0.5444)
 = 16o 291

59
9. A projectile starting from the ground with an angle of projection 82o 421 hits a target on the
ground located at a distance of 1000 m after 40 seconds. At what initial velocity was the
Projectile launched ?
Given:
 = 82o 421
t = 40 sec
R = 1000 m
u = ?
Solution:
2u sin α
T =
g

2𝑢 sin 82𝑜 421

=
9.8

40 x 9.8 = 2u sin 82o 421

40 𝑥 9.8
u =
2 𝑥 sin 82𝑜 421
392
u = = 197.58 ms-1
1.984

10. A bullet fired from a gun with a velocity of 80 ms-1 strikes the ground at the same level as the
gun at a distance of 460 m. Find the angle of inclination with the horizontal at which the gun
was fired.
Given
u = 80 ms-1, R = 460 m, g = 9.8 ms-2
u 2 sin 2
Range R =
g
gR = 9.8460
Sin 2 = = (80)2 = 0.7044
u2
2 = Sin-1 (0.7044) = 44° 46'   = 22° 23’

11. The range of a projectile is equal to double the maximum height attained. Find the angle of
projection.
Given
Horizontal range = 2 × maximum height
ie., R = 2H
u 2 sin 2 u 2 sin 2 
=2
g 2g
sin 2  = sin2 
2 sin  cos  = sin  sin 
2 cos  = sin 
 tan  = 2
  = tan (2) = 63°26'
-1

60
12. A projectile is thrown at an angle  and another is thrown at (90-) from the same point
both with the velocities 78.4ms-1. The second reaches 36.4m higher than the first. Find
the individual heights.
Given
Angle of projection of first projectile = 
Angle of projection of second projectile = (90 - ) Velocity of projection u = 78.4 ms-1
and H2 = H1 + 36.4m
H2 - H1 = 36.4m

Where H1 Is the maximum height of first one and H2 is the maximum height of second one.

u 2 sin 2  u 2 sin 2 (90 − ) u 2 cos 2

H1 = and H2 = =
2g 2g
2g

u 2 cos2  u 2 sin 2 
Adding, H2 + H1 = +
2g 2g
u2 78.4  78.4
= = = 313.6
2g 2  9.8
2H2 = 36.4+313.6 = 350
H2 = 175 m
Also, H1= H2 - 36.4 = 175 - 36.4
H1 = 138.6 m
13. If the range of a projectile is and 256 √𝟑 m the maximum height reached is 64 m. Calculate
the angle of projection.
Given
H = 64m R = 256 3 m
u 2 sin 2 
Maximum height H =
2g
u 2 sin 2 
64 = (1)
2g
u 2 2 sin  cos 
Range, R =
g
u 2 2 sin  cos 
256 3 = (2)
g
Dividing eqn. (1) by (2)
64 u 2 sin 2  g tan 
=  2 =
256 3 2g u 2 sin  cos  4
1 1
tan  =  = tan-1 ( )
√3 √3

 = 30 o

61
 3.3 CIRCULAR MOTION
Introduction

In this chapter we study the motion of the bodies moving in a circular path. The motion of the body
in a circular path is called circular motion. The motion of the body in the circular path with uniform
velocity is called uniform circular motion. When it is in uniform velocity, the direction of motion at any
point is along the tangent to the circle at that point. If the body is set free, it would move in the direction
of the tangent at that point.
Angular velocity
Consider a body (or particle) moving along a circular path. If  is the angle turned by the radius
vector in t second, then

Angular velocity  =
t
The angle covered or swept or turned by the radius vector in one second is known as angular
velocity.
SI Unit of angular velocity is radian / second (rad s-1)
Radian is the angle made at the centre of the circle by an arc whose length is equal to the radius of
the circle.

Relation between angular velocity and linear velocity

Consider a particle moving along a circular path of radius r. Initially the particle is at a point A. After
t second the particle reaches the point B. Let  be the angle swept by the radius vector in t-second.

t
r v
r B
 
t
o A O r A
r

q - VERY SMALL


Angular velocity  =
t
If  is very small, the distance AB will be considered as a straight line. Let v be the linear velocity
of the particle.
s
Linear velocity, v = t
s
In  OAB, sin  =
r
If  is very small, sin  =  in radian
s
= (or) s = r 
r

Linear velocity = Radius of the circle × Angular velocity

 v = r
62
Note:
If  = 1°, 1° = 0.0175 radian; Also sin 1° = 0.0175
Then sin 0.0175 (radian) = 0.0175 (natural value)
 = 2° = 0.0349 radian; Also sin 2° = 0.0349
Then sin 0.0349 (radian) = 0.0349 (natural value) Hence sin  =  for very small values of .
It is practically valid up to = 4° = 0.0698 radian.
Period and frequency of revolution
Period (T): Period of revolution of a particle is the time taken by the particle to complete one
revolution.
Frequency (n): Frequency of revolution of a particle is the number of revolutions made in one
second.
Relation between period T and angular velocity :
For one complete revolution (in T second),
Angle turned by the radius vector in T second = 2
2π
 Angle turned by the radius vector in one second =
T
2π 2π
i.e. Angular velocity,  = or T =
T 

Relation between period T and frequency n:

Let the particle completes n rotations in one second.
This is called frequency. So, the time taken complete one rotation which gives period = 1/n.
Therefore, T = 1/n.
Relation between frequency n and angular velocity :
No. of revolutions in 1 second = n
Angle turned by the radius vector for one revolution = 2
 Angle covered by the radius vector for n revolutions in 1 second = 2n
i.e., Angular velocity, ω = 2n
Normal acceleration:
For a particle in uniform circular motion, the direction of its velocity changes continuously but the
magnitude remains constant. For this, a force must act along the radius, towards the centre of the circle,
to maintain the circular path. Hence, there is acceleration acting on the body towards the centre of the
circle.
Thus, for a particle executing uniform circular motion, there must always be acceleration directed
towards the centre of the circular path along the radius. This acceleration is called normal acceleration
and the force responsible for this acceleration is known as centripetal force.
When a particle is moving along a circular path, the particle is accelerated towards the
centre of the circular path in a direction normal to the direction of its motion. This acceleration
of the particle is known as normal acceleration.
Centripetal force and Centrifugal force:
When a particle is moving along a circular path, it is accelerated towards the centre of the circular
path. Due to this acceleration there must be force acting on the particle towards the centre and this force
is known as centripetal force.
63
 Centripetal force is defined as the force which acts towards the centre along the radius of the circular
path on which the particle is moving with uniform velocity.
Let a stone be tied to one end of a string and whirled in a horizontal plane with constant speed. The
stone is subjected to a force directed towards the centre of the circle. This force acting towards the centre
of the circle is called centripetal force.
The expression for centripetal force is given by
mv 2
or mr2
r
By Newton's third law, there must be a reaction to this force. This reactive force which is directed
away from the centre is called centrifugal force.
The centrifugal force is defined as the force acting along the radius of the circle, away from the
centre, to balance the centripetal force.
The centripetal force and centrifugal force are equal in magnitude but opposite in direction. As
centrifugal force is a reaction force, it ceases once the centripetal force stops.
Examples for centripetal force:
The centripetal force is responsible for circular motion. Without this force the body cannot move
along a circular path. This force can be produced in different ways.
1. A body tied to one end of a string is being rotated in a circle; the centripetal force is supplied by
the tension of the string.
2. The moon revolving around the earth, the centripetal force is the gravitational pull of the earth
on the moon.
3. A person riding a cycle along a circular path, the necessary centripetal force is supplied by a
sidewise pressure exerted by the road on the tyres.
Examples for centrifugal force:
1. A stone tied to one end of a string is being rotated in a circle, the stone itself exerts an equal and
opposite force on the hand. It is due to the centrifugal force.
2. A centrifugal pump is used to pump well water on the basis of centrifugal force.
Application of centripetal force and centrifugal force:
1. A centrifugal governor regulates the speed of an engine
2. Centrifugal force can be used to generate artificial gravity, as in proposed designs for rotating
space station.
3. It is used in science and industry to separate substances.
Expressions for Normal acceleration and Centripetal force
Consider a particle of mass 'm' moving along a circular path of radius r. Let v be the velocity of the
particle. Initially the particle is at a point A. After t second the particle reaches the point B. Let  be the
angle covered by the radius vector. Let B be very close to A, so that the angle covered  is small.

64
 
Angular velocity,  =
t
Initially,
The velocity component in AO direction = zero
After t second
The velocity component in BC direction = v sin 
Since  is very small, the direction BC is very close to AO direction.
 The change in velocity towards the center of the circle in
‘t’ second = v sin  - 0
= v sin 
As  is very small, sin  =  (in radian)
v sin 
The change in velocity in one second =
t
v sin 
ie., Normal acceleration, a =
t
v
 Normal acceleration, a = t


Since  = t , a = v
since v = r, a = r = r2
v v v2
since  = , a=v =
r r r
v2
Normal acceleration, a= or a = r2
r
Centripetal force
According to Newton's II Law of motion, F = ma. As 'm' is the mass of the particle moving along a
circular path of radius r,
Centripetal force = mass X normal acceleration.
v2
Normal acceleration, a = or a = r2
r
mv 2
Centripetal force, F = or F = mr2
r
SIMPLE HARMONIC MOTION
Any motion in nature is classified into two types, viz., periodic and non-periodic motion. Periodic
motion is that motion which repeats itself after equal intervals of time. The examples of periodic motion
are the motion of planets around the sun, motion of hands of a clock, motion of the balance wheel of a
watch.
If a body moves back and forth repeatedly about a mean position, it is said to possess oscillatory
motion. Vibrations of guitar strings, motion of a pendulum bob, vibrations of a tuning fork oscillations of
mass suspended in a spring, vibrations of diaphragm in telephones are few examples of oscillatory
motion. In all the cases of vibrations of bodies, the path of vibration is always directed towards the mean
or equilibrium position.
The oscillations can be expressed in terms of simple harmonic functions like sine or cosine function.
A harmonic oscillation of constant amplitude and single frequency is called simple harmonic motion
(SHM).
A particle is said to execute simple harmonic motion if its acceleration is directly proportional to the
displacement from a fixed point and is always directed towards that point.
65
Important definitions of SHM Displacement
The distance travelled by the oscillating particle from its mean position at any instant is called
displacement.
Amplitude
The maximum displacement of the oscillating particle from its mean position is called amplitude.
Oscillation
One complete to and fro motion is called one oscillation.
Period
Time taken by the oscillating particle to complete one oscillation is called period.
Frequency
The number of oscillations or vibrations made by the vibrating particle in one second is called
frequency.
Worked Examples
1. A ball weighing 0.5 kg is tied to one end of a string and is whirled at a constant speed of
10ms-1 in a horizontal plane. If the length of the string is 1m, calculate the normal
acceleration and the tension in the string.
Given
m = 0.5 kg, v = 10ms-1 l = r = 1m a=? F=?
2 2
v (10)
Normal acceleration a = = = 100 ms–2.
r 1
Tension in the string (Centripetal force)
mv 2
F = r = 0.5 × (10)2 = 50N

2. A body of mass 1 kg is tied to a string of length 1m revolves in a horizontal circle. If the

angular velocity is 3 rad s-1, calculate the centripetal force acting on the body.
Given
m = 1 kg, l = r = 1 m,  = 3 rad s-1
Centripetal force F = mr2 =1 × 1 × (3)2 =9 N
3. Find the centripetal force on a body of mass 500 g when it revolves in a circle of radius 1.5m.
The body makes 120 R.P.M.
Given
m = 500 g = 0.5 kg, r=1.5m
No. of revolutions = 120 per minute
120
i.e.,n = = 2 revolutions per second
60

Centripetal force F = m r 2
= mr (2n)2 ( since = 2n) = 42 mrn2
= 4 × 3.14 × 3.14 × 0.5 × 1.5 × 22 F = 118.4N

66
4. A mass of 5kg is moving in a circle of radius 1m. If the centripetal force acting on the mass is
20N, find the linear velocity and angular velocity.
Given
F = 20N, r = 1m, m = 5 kg
mv 2
F =
r
Fr 20 1
v2 = = =4
m 5
-1
v = 2 ms
v 2
Angular velocity, = = = 2 rad s-1
r 1
5. A rope of length 1m can withstand a maximum weight of 10 kg. Now a stone of mass 200 g is
tied to it and is whirled round in a horizontal circle. Calculate the maximum permissible speed
of the stone.
Given
r = 1m, m = 200g = 0.2kg and F = 10 × 9.8 N
mv 2
F = r
Fr Fr 10  9.8 1
v2 = m  v = =
m 0.2

v = 7.07 ms-1
6. One end of a string of length 0.64 m is attached to a bucket containing water and the bucket
is rotated about the other end in a vertical circle. Find the minimum speed with which it can
be rotated without spilling the water at the highest point. How many revolutions per minute
it is making?
Given
Length of the string (or) radius of the circle r = 0.64 m
For without spilling the water at highest point

m v2
r
mg

0.64m

mv 2 v2
mg = r (or) g =
r
 v2 = rg = 0.64 × 9.8 = 6.272
 The minimum speed v = 6.272 = 2.504 ms–1
Leet n' be the no. of revolutions per second.
 v 2.504
n= = = = 0.6231
2 r2 0.64  2
No of revolutions per minute = 0.6231 × 60
= 37.4 revolutions
67
7. An object that moves in uniform circular motion has a centripetal acceleration of 13m/s2. If
the radius of the motion is 0.02 m, what is the frequency of the motion?
Given:
r = 0.02 m
a = 13 m/s2
T = ?
Solution:
𝑉2
Normal acceleration ‘a’ =
r
2
V = ax r
= 13 x 0.02
V = 0.5 m/s
2πr
V =
T
2πr
V = V
2 x 3.14 x 0.02
=
0.51

T = 0.25 s
1
f =
T
1
=
0.25

= 4 Hz.
8. Find the centripetal acceleration for an object on the surface of the planet (at the equator) with
the following characteristics.
(radius r = 4 x 106 m and 1 day = 1000000 s)
Given:
r = 4 x 106 m , T = 1000000 s
Solution:
u2
a = r
2πr
u =
T
2πr 1
a = ( ) 2x
T 𝑟
42 r
=
T2
4 x (3.14)2 x 4 x 106
a =
(1000000)2

Normal acceleration ‘a’ = 0.016 m/s2

68
9. A car of mass 2000 kg rounds a curve of radius 250 m at 90 kmph, computation centripetal
acceleration of centrifugal force.
Given:
m = 2000kg, r = 250 m, v = 90 kmph = 25m/s
Solution:
v2
Centripetal acceleration ‘a’ =
r
(25)2
a =
250

a = 2.5 m/s2
mv2
Centrifugal force ‘F’ =
r
2000 𝑥 (25)2
=
250

F = 5000 N
10. A 200 gm ball attached to the end of a cord is revolved in a horizontal circle with an angular
velocity 5 r/s. If the cords length is 60 cm, what is the centripetal force?
Given:
m = 0.2 kg,  = 5 r/s, r = 0.6 m
F = ?
Solution:
Centripetal force ‘F’ = mr2
= 0.2 x 0.6 x 52
F = 3N
11. A motor cyclist ride in a 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 of the highest point. Also find
the angular speed.
Given:
r = 3 m, g = 9.8 m/s2
F = ? = ?
Solution:
At the highest point, the centrifugal force and height of the motorcyclist and motorcycle
are equal in magnitude.
mv2
F = r
= mg
v2 = rg
v = √rg
v = √3 x 9.8
v = 5.42 m/s
v 5.42
Angular speed  = =
r 3

 = 1.807 rad / s
69
12. An object of mass 100 gm moves around the circumference of a circle of radius 2 m with a
constant angular speed of 7.5 rad/s. Compute its Linear Velocity and force directed towards
the centre.
Given:
m = 100 gm = 0.1 kg,  = 7.5 rad/s
r = 2 m, V= ? F = ?
Solution:
V = r
= 2 x 7.5
= 15 m/s
F = mr2
F = 0.1 x 2 x (7.5)2
F = 11.25 N
Linear Velocity = 15 m/s
Centripetal force = 11.25 N radially inward.

Banking of curved paths :

Curved track is to be considered as a part of the circular path. Consider a motor car moving along
a curved road. Now two forces are acting on the car, one is the weight of the car acting downward and
second is the normal reaction acting upward. In addition to these two forces, a horizontal centripetal
force should act on the car for safe negotiation of curved motion. This force is supplied by the pressure
exerted by the road on the tyres of the car towards inner side.
According to Newton's third law of motion, the tyres exert an equal and opposite pressure on the
road towards the outer side. Hence there must be a frictional force between the road and the tyres. This
frictional force tries to balance the centrifugal force. If the road is in horizontal level, the friction may
not be sufficient to balance the centrifugal force. Then the car may skid. To avoid this the outer edge of
the road is raised a little above the level of the inner edge. This formation is called as banking of curved
paths.
In the case of train moving along a curved track, the necessary centrifugal force is supplied by the
pressure exerted by the rails on the flanges of the wheels. Also the flanges of the wheels exert an equal
and opposite pressure on the rails. Hence there must be a large amount of friction and it leads to damage
the rails also to derailment of the train.
To avoid this damage and derailment, the outer rail is suitably raised over the inner rail to eliminate
the flange pressure on the rails. This formation is called as banking of curved railway tracks.
The raising of outer edge of a curved path than the inner edge for providing the necessary centripetal
force is known as banking of curved path.
Angle of Banking
It is the angle between the normal horizontal path and the curved banked path.
The angle of banking of a curved path is defined as the angle through which the outer edge of the
road is raised over the inner edge.
The angle of banking of curved railway tracks is defined as the angle through which the outer rail is
raised over the inner rail.

70
Expression for Angle of banking of Railway tracks
Consider a train moving along a curved path of radius r. In figure, ABCD shows the vertical cross
section of a railway carriage. If the outer rail is raised over the inner rail by an angle q, the carriage is
inclined to the horizontal by the same angle . This angle is called angle of banking of the railway
tracks.
Let R1 and R2 be the reactions of the inner and outer rails.
The total vertical component of R, and R2 = R1 cos  + R2 cos
= (R1 + R2) cos 
The total horizontal component of R1 and R2 = R1 sin + R2 sin
= (R1+ R2) sin
The vertical component balances the downward weight of the carriage. The horizontal component
supplies the necessary centripetal force and it balances the centrifugal force. Let m be the mass of the
carriage and v be the speed of the train.
R2 R cos
R cos
R1
C

D 

m v2
r
R sin
B
R sin I h

A
mg

The downward weight = mg

mv2
The centrifugal force =
r
Equation the forces (R1 + R2) cos  = mg (1)
mv2
(R1 + R2) sin  = (2)
r
(2) (R1 + R2 ) sin θ mv2
=
(1) (R1 + R2 ) cos θ rmg
v2
i.e. tan  = ,
rg
V2
 = tan-1 ( rg )

 The angle of banking of railway track.

Let ‘l’ be distance (AB) between the rails.
If ‘’ small, sin  = tan 
h v2 v2 𝑙
 = , h =
𝑙 rg rg
While constructing curved railway tracks, the value of h is obtained for fixed values of v, l and r.
Therefore, on a curved rail track, every train has to move with a particular speed specified for the
track.

71
 Worked Examples
1. A cyclist is riding at a speed of 22 ms-1 takes a turn on a curved circular road of radius
44 m. What is his inclination to the vertical?
Given
V= 22 ms-1, r = 44m, g = 9.8 ms-2
v2
tan  =
rg
(22)2 484
= = 431.2
44  9.8
tan  = 1.122
 = tan-1 (1.122) = 48°181

2. A curve in the road is in the form of an arc of a circle of radius 400. At what angle should
resurface of the road be laid inclined to the horizontal. So that the resultant reaction of the
surface acting on a car running at 120 kmph is normal to the surface of the road?
Given:
v = 120 kmph = 33.33 m/s
r = 400 m, g = 9.8 m/s2
Solution:
v2
 = tan-1 ( rg )
(33.33)2
 = tan-1 ( )
400 x 9.8

 = tan-1 (0.2834)
 = 15o491
3. An electric train has to travel on a railway track with a curve of radius 120 m with a speed
of 36 kmph. Calculate the angle of banking of the rails.
Given
1000
v = 36 kmph = 36 × = 10ms-1
3600
r = 120 m g = 9.8 ms-2

v2 (10) 2
tan θ =
rg 120  9.8
tan θ = 0.0850
θ = tan-1 (0.085) = 40 511

72
4. A train of mass 105 kg rounds a curve of radius 150 m at a speed of 20 m/s. Find the
horizontal thrust on the outer rail, if the track is not banked. At what angle must the track be
banked in order that there is no thrust on the rail. (g = 9.8 m/s 2)
Given:
m = 105 kg, V = 20 m/s, r = 150 m
g = 9.8 m/s2
Horizontal Thrust F = ?
Angle of banking  = ?
Solution:
mv2
F =
r
102 𝑥 (20)2
=
150
= 2.667 N
𝑣2
 = tan-1 ( rg )

(20)2
= tan-1 ((150 x 9.8))

 = tan-1 (0.2721)
 = 15o 131

5. Find the angle of banking of the railway track of radius of curvature 3200 m, if there is no
side thrust on the rails for a train running of 144 kmph. Find the elevation of the outer rail
above the inner rail, if the distance between the rail in 1.6 m.
Given:
v = 144 kmph = 40 m/s
r = 3200 m, g = 9.8 m/s2
Distance between rail l = 1.6 m
 = ?, elevation h = ?

S
Solution:


R

 h

73
 𝑣2
 = tan-1 ( rg )

(40)2
= tan-1 ( )
(3200 x 9.8)

= tan-1 (0.0510)
= 2o 551
h
sin  =
𝑙
h = l sin 
= 1.6 x sin 2o 551
h = 81 mm

6. A vehicle enters a circular bend of radius 200 m at 72 kmph. The road surface at the bend is
banked at 10o. Is it safe?
Given:
r = 200 m, v = 72 kmph = 20 m/s
 = 10o
Vsafe = ?
r = 400 m, g = 9.8 m/s2
Solution:
𝑣2
tan  = rg

Vsafe = √rg tan θ

= √200 x 9.8 x tan 10o

Vsafe = 18.59 m/s
Now, the velocity of the vehicle is (20 m/s) greater than the safe velocity (18.59 m/s). Here it is
unsafe to drive at a speed 72 kmph.
7. A bend in level road has a radius of 18 m. Find the maximum speed with which a car
negotiates the bend without skidding. The elevation of the road is 7o.
Given
R = 18 m,  = 70 and g = 9.8ms-2
v2
tan  =
rg
v2 = rg × tan θ = 18 × 9.8 × tan 70
= 18 × 9.8 × 0.1222 =
21.65 v = 4.65ms-1

74
 QUESTIONS

PART – A and PART – B

1. What is a projectile?
2. Define velocity of projection.
3. Define angle of projection.
4. Define trajectory of a projectile.
5. Define maximum height of a projectile.
6. Define time of flight of a projectile.
7. Define range of a projectile.
8. Give the condition for maximum range of a projectile.
9. What is a circular motion?
10. Define radian.
11. Define angular velocity.
12. Define normal acceleration.
13. Define centripetal force.
14. What is centrifugal force?
15. Define angle of banking.
16. Why does the outer rail is raised over the inner rail along a curved railway track?
17. Define Newton’s Law of motion
18. Explain fundamental equation of motion
19. What are the condition for maximum range
20. Derive the relation between angular velocity and linear velocity
21. Explain centripetal force and centrifugal force
22. Explain the banking of curved path

PART C
1. Derive an expression for the time of flight of the projectile.
2. Derive an expression for the horizontal range of the projectile.
3. Show that the range is maximum when the angle of projection is 45o for a given velocity of
projection.
4. Derive the relations between angular velocity, period and frequency.
5. Obtain expressions for the normal acceleration and centripetal force of a body executing uniform
circular motion.
6. Derive an expression for the angle of banking.

75
 Exercise problems

1. A particle is thrown at an angle of 300 with a velocity of 36 kmph Calculate its (1) time of flight and
(2) horizontal range.
Ans: (i) 1.02s, (ii) 8.84m
2. The range of a projectile is equal to 3 times the maximum height attained. Find the angle of projec-
tion. Ans: 5307'
3. The maximum height of a projectile is 2 times its horizontal range reached. Find the angle of projec-
tion.
Ans: 79058'
4. A bomber flying horizontally with a constant velocity of 600 kmph at a height 480m wants to hit a
target on the ground. Find at what horizontal distance from the target should a bomb be dropped in
order to hit the target.
Ans: R = 1648 m
5. A body of mass 2 kg tied to a string of length 2m revolves in a horizontal circle. If the
angular velocity of the body is 5 rad s-1, calculate the centripetal force acting on the body.
Ans: F = 100N
6. In a circus, a man is riding a motor cycle in the form of vertical circle inside the hollow spherical
cave of diameter 6m. Find the minimum speed of the rider with which he can be riding without
falling at the highest point.
Ans: v = 7.668ms-1 or 27.6 kmph
7. A string of length 1.5 m can withstand a maximum weight of 5kg.wt. A body of mass 2kg is tied to
it and whirled round in a horizontal plane. Calculate the possible maximum number of revolutions
per minute
Ans : RPM = 38.6
8. A scooter rider negotiates a curve of 100m radius on a level road with a speed of
72 kmph. Calculate the angle of inclination to the vertical he should make to avoid falling.
Ans:θ = 22012'
9. An electric train has to travel a curved railway line of radius 50m with maximum speed of 36 kmph.
Calculate the angle through which the outer rail should be raised so that there is no lateral thrust on
the rails.
Ans: = 11031'
10. An aeroplane with a speed of 120 ms-1 negotiates a curve of radius 1200m. Calculate its angle of
banking.
Ans: = 50045'
11. The expected average speed of the traffic on a high way around the curve is 20 kmph. Calculate the
angle of banking of the high way if its radius of curvature be 12m.
Ans: = 14042'
Brain teaser
1. An electric train has to travel on a railway track with a bend of radius 120m with a speed of 45
kmph. Calculate the height through which the outer rail should be raised for safe r running of
the train, if the distance between the two rails is 1.6m.

76
 https://www.youtube.com/watch?v=UTFc1Oy-Hw4&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=13

https://www.youtube.com/watch?v=sDUpF5R51RI&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=14

https://www.youtube.com/watch?v=HS3G6ecRCnM&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=15

https://www.youtube.com/watch?v=9N1QwFQkQMI&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=16

https://www.youtube.com/watch?v=k2RDz9bYnE0&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=17

77
 SUMMARY

• Dynamics deals with the motion of bodies

• Force, F = ma and Momentum, P = mv
• According to Newton’s third law, action and reaction do not act on the
same body.
• If a body moves along a straight line, that motion is straight line motion
• The path described by the projectile is called Trajectory
• Javelin throw is an example for projectile motion
• If a body reaches the maximum height, the initial velocity will be zero.
• The condition for maximum range, α = 45o
• The number of revolutions per second is called frequency.
• Linear velocity, v = rω
• Centripetal force → acting towards the centre
• Centrifugal force → acting away from the centre
• In planetary motion, the gravitational force of Sun acts as a centripetal force.
• Banking of roads is defined as the phenomenon in which the edges
are raised for the curved roads above the inner edge.
• The angle at which the vehicle is inclined is defined as the bank angle.
• One newton(N) is how much force is required to make a mass of 1 kg
accelerate at a rate of 1 ms-2.

78
 UNIT - IV

DYNAMICS - II

OBJECTIVES

▪ To know about the rotational motion of rigid bodies.

▪ To define moment of inertia
▪ To derive the expression for kinetic energy of a rigid body
▪ To know about angular momentum
▪ To derive the expression for angular momentum of a rigid body
▪ To explain the law of conservation of momentum
▪ To familiarize Newton’s law of Gravitation
▪ To derive the value of ‘g’.
▪ To derive the expression for variation of ‘g’ with altitude.
▪ To know about natural and artificial satellites.
▪ To explain escape velocity and orbital velocity
▪ To derive the expression for escape velocity
▪ To know about the uses of artificial satellites
▪ To solve simple problems based on escape velocity.

79
UNIT

4 DYNAMICS - II

4.1 ROTATIONAL MOTION OF RIGID BODIES

Introduction:
According to Newton’s first law of motion, “Everybody continues to be in a state of rest or in a
state of uniform motion along a straight line, unless it is compelled to change that state by some external
force”.
This law consists of two parts i.e., (1) A body at rest will not move on its own unless an external
force acts on it and (2) A body which is in uniform linear motion will not change either its speed or
direction of motion on its own, without the help of an external force.
The inability of a body to change its state on its own, without the help of external force is termed as
inertia. Inertia is a fundamental property of the matter. The more is the mass of the body, the more will be
inertia. Thus in translatory (linear) motion the mass of the body is a measure of coefficient of inertia.
Rigid body
If the distance between any two points in a body is not altered by applying a force, however large the
force may be, the body is said to be a rigid body.
A rigid body may be defined as that body which does not undergo any change in its shape or size
due to the application of force.
Actually, nobody is a perfect rigid body. When the changes in the body are negligible, it can be
considered as a rigid body.
Rotational motion
Each body is made of large number of tiny particles. In the case of linear motion, all the particles
present in the body will have same linear velocity.
When the body rotates about a fixed line (axis of rotation), its motion is known as rotatory motion.
The axis of rotation may lie within the body or outside the body. If all the particles of a body move
in a circular path about the axis of rotation, the rigid body is said to have pure rotational motion.
When a body is in rotational motion about an axis, all the particles present in the body will have
same angular velocity, but different linear velocities. The values of the linear velocities of these particles
depend on the distance of the particles from the axis of rotation, since v = r 
In rotational motion also, a rigid body which is free to rotate about an axis, opposes the change to 
be produced in its state. The measure of opposition not only depends on the mass of the body but also
on the distances of the particles of the body from the axis of rotation.
Suppose a particle of mass ‘m’ is at a distance r from an axis, the product mr2 is called the moment
of inertia of the particle about that axis.
Thus in rotational motion the moment of inertia of the body about the given axis of rotation is a
measure of coefficient of inertia.
Hence, the moment of inertia plays the same role in rotational motion as that of mass in translatory
motion.
Moment of Inertia of a particle
The moment of inertia of a particle about an axis is equal to the product of the mass of the particle
and square of its distance from the axis.

80
 Consider a particle of mass ‘m’ is placed at a distance ‘r’ from the fixed axis. Then, the moment of
inertia of the particle about the axis = mr2. The S.I. unit for moment of inertia is kg m2.
X'

Moment of Inertia of a rigid body

Inertia of a body is its inability to change by itself its state of rest or of uniform motion in a straight
line.
Similarly Moment of inertia of a body is its inability to change by itself its state of rest or of uniform
rotatory motion about an axis.
y

r1
m1

r2
m2
r3
m3

An external torque (rotating effect of force) is necessary to change its state.

Let us consider the rotation of a rigid body about an axis. It consists of a large number of particles
of masses m , m , m etc., situated at distances r , r , r etc., from the axis yy’. Then m r 2 is known as
1 2 3 1 2 3 1 1

the moment of inertia of the particle of mass m1 about that fixed axis.

Then, the moment of inertia of the rigid body = Sum of moments of inertia of all the particles
present in the body.
I = m1r12 + m2r22 + m3r32 + .........................
I =  mr2
Hence the moment of inertia of a rigid body about a fixed axis is the sum of the moment of inertia
of all the particles of the rigid body.
Radius of gyration
Radius of gyration is the distance between the given axis and the centre of mass of the body. The
centre of mass of a body is point where the entire mass of the body is supposed to be concentrated.
X1

X K

81
 It is denoted by ‘ K ’.
If M is mass of the body, then moment of inertia I = MK2.
Hence, mr2 = MK2
Expression for Kinetic Energy of a Rigid body rotating about an axis:
Consider a rigid body rotating about a fixed axis XOX’. The rigid body consists of a large number of
particles. Let m1, m2, m3,….. ... etc., be the masses of the particles situated at distances. r1, r2, r3, ...... etc.,
from the fixed axis. All the particles rotate with the same angular velocity . But the linear velocities of
the particles are different.
1
Kinetic Energy = mv2
2
1
= m (r)2 (v = r)
2
1
= m r2 2
2

1
The kinetic energy of the first particle = m1 r12 2
2
1
The kinetic energy of the second particle = m2 r22 2
2
1
The kinetic energy of the third particle and so on = m3 r32 2
2

X‘

X r1
r3
r2
m1
m3

m2

The kinetic energy of the whole body is equal to the sum of the kinetic energy of all the particles
present in the body.
1 1 1
The kinetic energy of the rigid body = m1 r12 2 + m2 r22 2 + m3 r32 2 + …
2 2 2
1
= ( m1 r12 + m2 r22 + m3 r32 + …) 2
2
1
=  mr2 2
2

But,  mr2 = Moment of inertia of the rigid body about the fixed axis = 1.
1
Therefore, the kinetic energy of the rigid body = I 2
2

82
Angular momentum
The moment of linear momentum is known as angular momentum.
Consider a particle of mass m is at a distance r from the axis of rotation. When a particle is in
rotational motion about an axis, it has both linear velocity ‘v’ and angular velocity ‘’.
Then, Angular momentum of the particle = linear momentum × perpendicular distance between the
particle and the axis of rotation
=mv×r
=mr×r ( v = r)
X‘

X r

m
mv

 Angular momentum = mr2

Where  is the angular velocity of the particle.
The SI unit for angular momentum is kg m2 s–1
Expression for Angular momentum of a Rigid body rotating about an axis:
X‘

X r1
r3
r2
m1
m3

m2

Consider a rigid body rotating about a fixed axis XOX’. The rigid body consists of a large number of
particles. Let m1, m2, m3... etc., be the masses of the particles situated at distances r1, r2, r3, ... etc., from
the fixed axis. All the particles rotate with the same angular velocity, but with different linear velocities
depending on the values of ‘r’.
Angular momentum = moment of linear momentum
= linear momentum × distance
= mv × r = mr × r ( v = r)
= mr2

83
  The angular momentum of the first particle = m1r1 2 
The angular momentum of the second particle = m2r22 
The angular momentum of the third particle = m3r32 
...and so on.
The angular momentum of the whole body is equal to the sum of the angular momenta of all the
particles present in the body.
The angular momentum 
 L = m1 r1  + m 2 r2  + m 3r3  + .......
2 2 2

of the rigid body 

= (m1r12 + m2r22 + m3r32 + ..) 
= mr2 
But, mr2 = moment of inertia of the rigid body = I
 The angular momentum of the rigid body = I
Law of conservation of angular momentum
When there is no external torque acting on a rotating body, the angular momentum of
that body remains a constant. This is the statement of law of conservation of angular
momentum.
i.e., if I1 and 1 are the initial moment of inertia and angular velocity of a rotating body and if I

and 2 are new moment of inertia and angular velocity of the body, without the help of any external
torque, then according to this law,
I1 1 = I2 2

Examples
a) Consider a person standing on a turn-table with arms extended and a pair of weights, one in each
hand. The table is made to rotate by a motor and then the motor is switched off. Now, if that person
pulls his arms inwards, we can see a considerable increase in the speed of rotation. This is because,
in the new position, his moment of inertia I about the axis of rotation decrease, due to the decrease
in the value of r. Since the angular momentum is conserved, a decrease in the value
I results in an increase in the value of angular velocity . Therefore the person is found to rotate
faster.
b) A circus acrobat, a diver or skater, all take advantage of this principle. Consider a diver has just left
a diving board with his arms and legs extended, with a particular angular momentum. If he now
pulls his arms and legs in, his moment of inertia I decreases, leading to increase in angular velocity
. By increasing ω, the diver can perform more number of somersaults, before entering the
swimming pool.
Moment of Inertia
1. An object has the moment of Inertia of 1 kgm 2 rotates at a constant angular speed. 2 rad/s.
What is the rotational K.E of the object?
Given:
I = 1 kgm2,  = 2 r/s
Rot K.E = ?
Solution
1
Rotational K. E = I2
2
1
= x 1 x (2)2
2

Rotational K.E = 2 J
84
2. A 20 kg cylinder pully with radius of 0.2 m rotates at a constant angular velocity of 4 rad/sec.
𝟏
What is the rotational K.E. of the pulley? (m.I of cylinder I = mr2)
𝟐

Given:
r = 0.2 m, m = 20 kg,  = 4 r/s
Rotational K.E = ?
Solution
Moment of inertia of cylinder
1
I = mr2
2
1
= x 20 x (0.2)2
2

= 0.4 kgm2
1
Rotational K.E = I 2
2
1
= x 0.4 x (4)2
2

Rotational K.E = 3.2 J.

3. A 0.5 kg particle rotates at a constant angular velocity of 2 r/s. What is the rotational K.E. of
the particle if the radius of the circle is 10 cm.
Given:
m = 0.5 kg,  = 2 r/s, r = 10 cm
r = 0.1 m
Solution
Moment of inertia of particle I = mr2
= 0.5 x (0.1)2
I = 0.005 kgm2
1
The rotational K.E. = 2
I2
1
= x 0.005 x (2)2
2

= 0.01 J

4. A 100 gm of ball connected to one end of a cord with length of 30 cm. What is the moment of
inertia of ball about the axis of rotation? Ignore cord mass.
Given:
m = 100 gm = 0.1 kg, r = 30 cm = 0.3 m
I = ?
Solution
I = mr2
= 0.1 x (0.3)2
I = 0.009 kgm2

85
5. What is the angular momentum of a thin hoop of radius 2 m and mass 1 kg that is rotating at
a velocity of 4 r/s ?
Given:
 = 4 r/s, m = 1 kg, r = 2m
L = ?
Solution
Angular momentum L = I 
L = mr2 I = mr2
2
= 1 x (2) x 4
L = 16 kgm2/s

6. When an object with the moment of inertia of 5 kgm2 is made to rotate 1 r/s speed. Calculate
the angular momentum of the object.
Given:
I = 5 kgm2,  = 1 r/s, L=?
Solution
Angular momentum L = I
L = 5x1
L = 5 kgm2 s-1

7. The distance between a 40 kg person and a 30 kg person is 2m. What is the magnitude of the
gravitational force each exerts on the other. (G = 6.67 x 10 -11 Nm2 / kg2)
Given:
m1 = 40 kg, m2 = 30 kg, r = 2m
G = 6.67 x 10-11 Nm2 / kg2, F = ?
Solution
m1 m2
Gravitational force F = G
r2
6.67 x 10−11 x 40 x 30
F =
(2)2

F = 2 x 10-8 N

https://www.youtube.com/watch?v=EhinYBd9doA&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=18

86
 4.2 GRAVITATION
Newton’s laws of Gravitation

Law 1 : Any two particles of matter attract each other with a force
Law 2 : The force of attraction between any two objects is
i. directly proportional to the product of the masses
ii. inversely proportional to the square of the distance between them.
If m , m are masses of two particles, separated by a distance ‘d’ then the force of attraction between
1 2
the particles,
d

m1 m2

m1 m 2
F
d2
m1 m 2
F =G
d2
where G is known as “universal gravitational constant” and the value of G = 6.6733 × 10–11 N
m2kg–2
Acceleration due to gravity
Galileo was the first to make a systematic study of the motion of a body under the gravity of Earth.
He dropped various objects from the leaning tower of Pisa and made analysis of their motion under
gravity. He came to the conclusion that “in the absence of air, all bodies will fall at the same rate”. It is
the air resistance that slows down a piece of paper or a parachute falling under gravity. If a heavy stone
and a parachute are dropped where there is no air, both will fall together at the same rate.
Experiments showed that the velocity of a freely falling body under gravity increases at a constant
rate. i.e., with a constant acceleration. The acceleration produced in a body on account of the force of
gravity is called acceleration due to gravity. It is denoted by ‘g’. At a given place, the value of ‘g’ is the
same for all bodies irrespective of their masses. It differs from place to place on the surface of the Earth.
It also varies with altitude and depth.
The value of g at sea-level and at a latitude of 45° is taken as the standard (i.e) g = 9.81 m s–2.
Acceleration due to gravity at the surface of the Earth
Consider a body of mass m on the surface of the Earth as shown in the Figure. Its distance from the
centre of the Earth is R (radius of the Earth).
GMm
The gravitational force experienced by the body, F =
R2
where M is the mass of the Earth.

mg
R

Earth

87
 From Newton’s second law of motion,
Force F = mg.
GMm
Equating the above two forces, = mg
R2
GM
 g = R2

This equation shows that g is independent of the mass of the body m. But, it varies with the distance
from the centre of the Earth. If the Earth is assumed to be a sphere of radius R, the value of g on the
surface of the Earth is given by
GM
Mass and Weight g = R 2

Variation of acceleration due to gravity with altitude (height):

Let P be a point on the surface of the Earth and Q be a point at an altitude h. Let the mass of the Earth
be M and radius of the Earth be R. Consider the Earth as a spherical shaped body.
Q

The acceleration due to gravity at P on the surface is

GM
g = (1)
R2

Let the body be placed at Q at a height h from the surface of the Earth. The acceleration due to
gravity at Q is
GM
gh = (2)
(R+h)2

dividing (2) by (1)

gh R2
=
g (R+h)2

1−2h
By simplifying and expanding using binomial theorem, gh = g ( )
R

The value of acceleration due to gravity decreases with increase in height above the surface of the Earth.

https://www.youtube.com/watch? v=EZq9cTsLwOA&list=PL1b9Ht9ISqIHStwn0N EHongecMJ84S6cj&index=19

https://www.youtube.com/watch? v=6YTsmILHjSs&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=20

88
 4.3 SATELLITES
A body moving in an orbit around another bigger body is called a satellite. A body that moves around
a planet is called a satellite.
The objects that are moving in orbit by nature itself around a planet are called natural satellites.
For example, moon is the natural satellite for the earth. The earth is a satellite for the sun. Man also has
placed artificially some satellites to move in orbit around the desired planets. These satellites are called
artificial satellite.
The artificial satellites are carried by rockets to the predetermined height, a few hundred kilometers
above the surface of the earth. At this height the air is in a rarefied state and hence air friction is
negligible.
After reaching the predetermined height, the satellite is given very high horizontal velocity so that it
remains moving in a nearly circular orbit.
The artificial satellites are broadly classified as earth resources satellites, meteorological satellites
and satellites carrying microwave sensors.
Escape velocity and Orbital velocity
Escape velocity (Ve)
When a body is thrown vertically upwards, it will return to the earth’s surface after attaining certain
height. If the velocity of projection is increased, the height attained by the body becomes greater and
then the body returns to the earth. This is due to the gravitational force of attraction of the earth. If the
body is to be projected with a particular greater velocity, the body escapes from the gravitational pull so
that it never returns to the earth. This velocity of projection is called escape velocity, it is different for
different planets.
The minimum velocity required to project a body in upward direction from the surface of
the earth so that it escapes from the earth’s gravitational pull is called escape velocity.
Orbital velocity (Vo)
In order to put a satellite into the orbit around the earth, the satellite must be projected to the
particular height and then it must be turned in a direction perpendicular to the line from the centre of
the earth so that it moves in an orbit around the earth.
The velocity of the satellite along its orbit around the earth is called orbital velocity.
Expression for Escape Velocity (Ve) :
Consider a body of mass m which is to be projected with escape velocity Ve from the surface of the
earth. If the body is placed on the surface of the earth, then

V

dR

M
EARTH

The weight of the body = mg …(1)

Let M be the mass of the earth and R its radius. According to Newton’s law of gravitation,
GMm
The gravitational force, F = ... (2)
R2

89
 where G is a constant known as gravitational constant. The gravitational force on a body in the
earth’s atmosphere is also equal to the weight of the body. Equating the eqns. (1) and (2)
GMm
= mg
R2
GM = gR2
Since Ve is the escape velocity given to the body.
1
The given kinetic energy = mV02 .....(3)
2

This kinetic energy is converted into work done to escape the body from gravitational pull.
If the body moves a distance dR in vertical direction against the force F, then,
Work = Force × distance
 Work done by the body = F dR
GMm
= × dR
R2

In order to escape the body from the gravitational pull, the work done by the body to move from the
distance R to infinity (the boundary of the gravitational field) is given by

GMm
W =

R R2
dR

GMm
W = R
1 GMm
mV2 =
0
2 R
Equating the eqns. (3) and (4).
2GM
Ve =
R
Here, GM = gR2

Ve = 2gR 2
R
 The Escape velocity
Ve = 2gR

Uses of Artificial Satellites

The artificial satellites are launched for many purposes by different countries. The important uses of
artificial satellite are
i) Collection of scientific data
ii) Weather monitoring
iii) Military Spying
iv) Remote sensing
v) Communication purpose – the satellite receives microwaves and TV signals from the earth and
amplifies them and transmits them back to various stations on the earth.
90
Geo – Stationary Satellite
A geo-stationary satellite is a particular type used in television and telephone communications.
A number of communication satellites, which appear to remain in fixed position at a height of 36,000 km
above the equator, are called synchronous satellites or geostationary satellites. Some television
programmes or events occurring in other countries are often transmitted live with the help of these
satellites.
The orbit at a height of 36,000km above the equator, on which satellites are revolving, is called
geosynchronous orbit. In this orbit, the satellite takes 24 hours for revolving round the earth once.
For a satellite to appear fixed at a position above a certain place on the Earth, its orbital period
around the earth must be exactly equal to the rotational period of the earth about its axis. As the relative
angular velocity of the earth and that of satellite appears stationary to an observer on the earth.
The ground-based transmitting antenna, which is directed towards the satellite, can operate
effectively without need for expensive equipment to track the satellite’s motion. This implies enormous
savings from the satellite and are received at the destination. As the geostationary satellite can ‘see’ only
one-third of the earth’s surface, at least three such satellites are required to cover the entire globe.
Polar Satellites
The polar satellites revolve around the earth in a north-south orbit crossing over the poles as the
earth spins about its north-south axis.
The Polar satellites positioned nearly 500 to 800 km above the earth travels from pole to pole in
102 minutes. The polar orbit remains fixed in space as the earth rotates inside the orbit. As a result, most
of the earth’s surface crosses the satellite in a polar orbit. Excellent coverage of the earth is possible with
this polar orbit. The polar satellites are used for mapping and surveying.
Expression for orbital velocity of a satellite :
Consider a satellite of mass m is placed on the surface of the earth.
The weight of the satellite = mg ...... (1)
Let M be the mass of the earth, and R its radius. According to Newton's law of gravitation,
GMm
The gravitational force F = ….. (2)
R2
where G is a constant known as gravitational constant.
The gravitational force on the satellite is equal to its weight. Equating the eqns. (1) and (2).
GMm
= mg GM = gR2
R2
The satellite is to be projected vertically upto a particular height h, after it must be turned in a
perpendicular direction, then the satellite revolves around the earth in a circular orbit of radius r, with
an orbital velocity of Vo.
mV𝑜 2
The centripetal force, F = ……(3)
r

91
 The necessary centripetal force for circular motion of the satellite is given by the gravitational
force. Hence,
mVo 2 GMm
=
r r2

GM gR2
 Vo = √ r
= √ r
(∵ GM = gR2)

gR2
The orbital velocity of the satellite. Vo = √R + h
(∵ r = R + h)

Note: When the satellite is launched in near earth orbit which is 200 km to 300 km. h is small when
compared to R so that, R + h = R. Then
Vo = √gR
Worked Examples

1. If the radius of the earth is 6400 km and the acceleration due to gravity is 9.8ms-2.
Calculate the escape velocity
Given: R = 6400 km and g = 9.8 ms-2

Escape velocity Ve = 2gR = 2  9.8  6400 103

Ve = 11.2 km / s

2. The distance between a 40 kg person and a 30 kg person is 2m. What is the magnitude of the
gravitational force each exerts on the other (G = 6.67 x 10-11 Nm2 / kg2)
Given:
m1 = 40 kg, m2 = 30 kg, r = 2m
G = 6.67 x 10-11 Nm2 / kg2, F = ?
Solution:
m1 m2
Gravitational force F = G r2
6.67 x 10−11 x 40 x 30
F =
(2)2

F = 2 x 10-8 N
3. Determine the distance between two bodies. If the mass of two bodies are 80 kg and 100 kg
and they excerts gravitational force of 148.28 x 10-10 N each on the other. (G = 6.673 x 10-11
Nm2 kg-2)
Given:
M1 = 80 kg
M2 = 100 kg
F = 148.28 x 10-10 N
G = 6.673 x 10-10 Nm2 kg-2,
r = ?
92
 Solution :
m1 m2
Gravitational force F = G r2
m1 m2
r2 = G
F
6.673 x 10−11 x 80 x 100
= 148.28 x 10−10

6.673 x 10−11 x 80 x 100

r = √
148.28 x 10−10

r = 6m

4. Calculate an escape velocity of moon. The acceleration due to gravity and radius of the moon
are 1.6 m/s2 and 1750 km.
Given:
g = 1.6 m/s2, R = 1.75 x 106 m
escape velocity of moon Ve = ?
Solution :

Ve = √2gR = √2 x 1.6 x 1.75 x 106

Ve = 2.367 km / s
5. The acceleration due to gravity at the Moon’s surface is 1.67 m/s 2. If the radius of the Moon
is 1.74 x 106m, Calculate the mass of the Moon.
Given:
g = 1.67 m/s2, R = 1.74 x 106 m,
G = 6.67 x 10-11 Nm2 / kg2, M = ?
Solution :
GM = gR2
gR2 1.67 x (1.74 x 106 ) 2
M = =
G 6.67 x 10 −11

M = 7.58 x 1022 kg

93
 QUESTIONS
Part – A and B
1. Define rigid body
2. Define moment of inertia of a rigid body
3. Define radius of gyration
5. Define angular momentum
6. State the law of conservation of angular momentum
7. State Newton’s I law of gravitation
8. State Newton’s II Law of gravitation
9. What is a satellite?
10. Define escape velocity
11. Define orbital velocity
12. Give any two uses of artificial satellites
13. Derive an expression for the moment of inertia of a rigid body about an axis.
14. Explain Newton’s Law of Gravitation
15. Explain escape velocity and orbital velocity
16. Write the uses of artificial satellites
Part – C
1. Derive an expression for the kinetic energy of a rigid body rotating about an axis.
2. Derive an expression for the angular momentum of a rigid body rotating about an axis.
3. Obtain an expression for the acceleration due to gravity on the surface of the earth.
4. Obtain an expression for the variation of acceleration due to gravity with altitude.
5. Derive an expression for the escape velocity from the surface of the earth
Exercise Problems
1. Find the escape velocity at the surface of the moon, given that the radius of moon is 2 x 106 m
and acceleration due to gravity on the surface of moon is 1.69 ms-2.
Brain teaser
1. The escape velocity from the surface of the earth is 11.2 kms-1. Find the escape velocity from the
surface of another planet where mass and diameter are twice that of earth.
2. What is the acceleration due to gravity at a distance from the centre of the Earth, which is equal
to diameter of the Earth ?

https://www.youtube.com/watch?v=CtKPpNeAkP0&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=21

94
 SUMMARY

• A rigid body is a solid body in which deformation is zero or so small it can be

neglected.
• The linear velocity of a particle in a rigid body is different
• The moment of inertia is a physical quantity which describes how easily a
body can be rotated about a given axis.
• Moment of inertia, I = mr2
• The K.E of a rigid body = ½ I ω2
• The moment of linear momentum is called angular momentum.
• The angular momentum of a rigid body = I ω
• The angular momentum of a rigid body is always constant.
• In vaccum, the bodies of different masses when dropped from a same point
reached the ground at the same time.
• Sir Issac Newton proposed the universal law of Gravitation.
𝑮𝑴𝒎
• Newton’s law of gravitation, F = 𝒓𝟐

• If the height increases, the value of ‘g’ decreases.

• There are 8 planets revolve around the Sun.
• Earth has 1 moon, Saturn has 82 moons and Jupiter has 79 moons.
• The moons revolve around the planets are called natural satellites.
• Manmade objects that revolve around the planets are called artificial satellites
• Escape velocity refers to the minimum velocity which is needed to leave a
planet or moon.
• Escape velocity, ve = √𝟐𝒈𝑹
• If a satellite revolves around a planet, it needs orbital velocity.
• India's first satellite, the Aryabhata, was launched by the Soviet Union on
19 April 1975.
• IRNSS - The Indian Regional Navigation Satellite System.

95
 UNIT - V

SOUND AND MAGNETISM

OBJECTIVES

▪ To know about the wave motion

▪ To explain progressive waves
▪ To define period, frequency and wavelength.
▪ To explain stationary waves
▪ To know the types of vibrations
▪ To explain the frequency of a tuning fork using Sonometer.
▪ To define echo and reverberation.
▪ To know about the Architectural acoustics.
▪ To define retentivity and coercivity
▪ To draw Hysteresis loop.
▪ To know about the uses of hysteresis loop.
▪ Can solve simple problems based on intensity of magnetization.

96
 UNIT

5 SOUND AND MAGNETISM

5.1 SOUND
Introduction:
Sound is produced due to the vibrations of the body. These vibrations are transferred to the air
medium and propagated in all directions in the form of waves. When the vibrational sound waves reach
our ear, the diaphragm of the ear vibrates with equal vibrations produced by the body. Hence we are able
to hear the sound by the sensation perceived by the nerves of the ear.
The number of vibrations made in one second is known as frequency of the sound. It is expressed
in hertz (Hz). The range of the frequency between 20 Hz and 20,000 Hz is the audible range, Human
ears cannot respond to the sound below and above this range. In this sense, all vibrating bodies cannot
produce audible sound. The vibrations of frequency below 20 Hz are called infrasonic and above 20,000
Hz are called ultrasonic.
Sound is a mechanical wave and hence it requires a medium to propagate. So we can't hear the sound
in vacuum. It travels with the velocity of 330 ms-1 in air. Sound waves are reflected and refracted like
light waves.
ULTRASONIC WAVES AND USES:
The sound waves above the frequency of 20000 Hz (20 KHz) are known as ultrasonic waves.
Ultrasonic waves are inaudible to humans, because, their high frequencies are unable to vibrate the
human ear drum. The ultrasonic waves are high frequency sound waves smaller wave length. Its wave
length at room temperature is 0.0175m. Since they are sound waves, they require a material medium to
travel and have most of the properties of sound waves. In addition to this, the ultrasonic waves are having
high energy and it helps in large number of applications in engineering and medical fields.
The ultrasonic waves are usually generated by the principle of piezo electric effect.
Applications of ultrasonic waves:
Ultrasonic waves have a large number of practical applications. Some important applications are
given below.
1. The ultrasonic waves are used to find the depth of the sea, distance and direction of submarine
and depth of the rocks in the sea.
2. They are used in non-destructive testing to detect cracks in the materials, welding, castings etc.
SONAR:
SONAR stands for SOund Navigation And Ranging. Sonar is used in navigation, forecasting
weather and for tracking aircraft, ships, submarines and missiles.
Wave motion:
When a stone is thrown on a pond of water, ripples spread out in all directions on the surface of
water. The stone disturbs the water medium at one place but the disturbance is transferred in all
directions continuously. This continuous movement of the disturbance is called a wave.
If a wave passes in a medium, the particles of the medium vibrate about their mean position. The
particles do not move along with the wave, only the vibrations are transferred from one particle to
adjacent particle of the medium in the form of energy.
97
 There are two types of wave motion. They are
1) Longitudinal wave motion and
2) Transverse wave motion
1) Longitudinal wave motion:
If the particles of the medium vibrate parallel to the direction of propagation of the wave, the wave
is known as longitudinal wave.
Examples:
1. The propagation of sound in air
2. The propagation of sound in gas
3. The propagation of sound inside the liquid
a)
C

b)
R C

c)
C R C

d)
R C R C R C

The longitudinal waves travel in a medium in the form of compressions and rarefactions. The place
where the particles of the medium crowded together are called compressions and the places where the
particles spread out are called rarefactions.
The compressions and rarefactions produced by a vibrating tuning fork are as shown in the figure.
When the prong (arm) of the fork moves to the right, it compresses the medium in front of it to form
compression. Meanwhile the prong returns to the left, a temporarily vacuum is created there. To fill it,
the particles of the medium spread out in that place to form a rarefaction.
Thus as the prong of the fork vibrates to and fro, alternate compressions and rarefactions are trans-
mitted in the medium. As a result, the particles of the medium simply move back and forth about their
mean position parallel to the direction of the propagation of the wave.
2) Transverse wave motion:
If the particles of the medium vibrate perpendicular to the direction of propagation of the wave, the
wave is known as transverse wave.
Examples :
1. Ripples travelling on the water surfaces.
2. Waves travelling along a rope.
3. Other waves like light waves, heat radiations, radio waves etc.
The transverse waves travel in a medium in the form of crests and troughs. The points where the
particles of the medium displaced maximum in the upward direction are called crests. The points where
the particles displaced maximum in the downward direction are called. troughs

Crests

Troughs

98
 The crests and troughs produced by the transverse wave motion are as shown in the figure. In
transverse wave, alternate crests and troughs are transmitted in the medium. As a result, the particles of
the medium move up and down about their mean position perpendicular to the direction of propagation
of the wave.
Progressive Waves:
If a wave travels continuously in a medium without any disturbance, then the wave is said to be
progressive wave. Longitudinal waves and Transverse waves are two types of progressive waves and
they can travel continuously in any medium if there is no obstruction.
Amplitude:
When sound wave propagates in a medium, the maximum displacement of the vibrating particles of
the medium from their mean position is called amplitude.
Wavelength () :
The wavelength is the distance between two consecutive particles of the medium which are in the
same state of vibration.
It is also defined as the distance travelled by the wave during the time the vibrating particle
completes one vibration.
In longitudinal waves, the wavelength is the distance between two successive compressions or
rarefactions. In transverse waves, the wavelength is the distance between two successive crests or
troughs.
Period (T):
The time taken by the vibrating particle to make one vibration is called period.
Frequency (n):
The frequency is the number of vibrations made by the vibrating particle in one second.
Velocity (v):
The distance travelled by the sound wave in one second is known as velocity of sound.
Relation between Wavelength, Frequency and Velocity of a Wave :
Let n be the number of vibrations made by the vibrating particle in one second. It is also known as
its frequency.
Time taken for one vibration = period (T) = 1/n.
Let  be the wavelength of the wave produced
Velocity of the wave is the distance through which the wave advances in the medium in one
second
distance travelled
 Velocity of wave motion, V =
time taken

V = /T =/(1/n) =  n
V = n
Stationary Waves:

If a progressive wave travelling in a medium meets the surface of an obstacle, it is reflected.

The reflected wave is superimposed on the incident wave to form a new type of wave called
stationary wave.

Also, when two identical waves having equal wavelength and amplitude travel in opposite directions
they superimpose on each other forming stationary wave.

99
 A stationary wave formed by a vibrating string is as shown in the figure. Consider a string P Q fixed
at the end Q and it vibrates up and down at the free end P. Then a transverse wave is setup and it proceeds
towards the fixed point Q and gets reflected back to the end P. Now the stationary wave is formed in the
string.

At certain points of the medium, the displacement due to the two waves cancel each other and
those points remain at rest. Such points are called nodes (N). At certain other points there is maximum
displacement. Such points are called antinodes (A). The distance between two successive nodes
or antinodes is  / 2.
λ
A A A A A

P Q
N N N N N N
A A A A A

λ/2

The distance between a node and the next antinode is  / 4

The longitudinal waves also produce the stationary waves.
VIBRATIONS

Free Vibrations :
The vibrations of anybody with its natural frequency are called free vibrations.
When a body is set in vibration and left free, it executes vibrations. The frequency depends upon
the dimensions and elastic constants of the body. Such vibrations are called free vibrations and the
frequency of vibration is known as the natural frequency.
If a tuning fork is set in vibration, it vibrates with its own frequency. Such vibrations are called free
vibrations or natural frequency.
Forced Vibrations :
The vibrations of a body with a frequency induces vibrations on another vibrating agent are called
forced vibrations.
Suppose a vibrating tuning fork is placed with its stem on a table, the vibrations of the fork are
impressed on the table and the table is forced to vibrate. The vibrations set up on the table are called
forced vibrations.
Resonance :

When the forced vibrations given on the body is equal to its natural frequency of vibrations, the
body vibrates with maximum amplitude. This phenomenon is called resonance.

When a vibrating tuning fork is kept on a table, the table is forced to vibrate with the frequency of
the tuning fork. If the natural frequency of the table is equal to the frequency of the tuning fork, the table
vibrates with its natural frequency and hence resonance occurs.
Laws of vibration in stretched strings:
I Law : The frequency of vibrating string is inversely proportional to its length, when the tension
and linear density of the string are kept constant.
1
n 
𝑙

100
 II Law : The frequency of vibration is directly proportional to the square root of the tension,
when the length and linear density of the string are kept constant.
n  √T
III Law: The frequency of vibration is inversely proportional to the square root of the linear
density of the string, when the tension and length of the string are kept constant.
1
n 
√m

According to above laws,

1 T 1 T
n  √ (or) n=k √
𝑙 m 𝑙 m

Where k - is a constant and its value is equal to ½.

1 T
n  √
2𝑙 m

1 Mg
n = √ (∵ T = Mg)
2𝑙 m

1 M g
n = 2 √(𝑙2)
m

Note: The linear density,

mass Volume x density r2𝑙p
m = length
= length
= 𝑙
= r 2 p

Sonometer :
The phenomenon of resonance is used in sonometer. In sonometer, the frequency of a tuning fork
is equal to the frequency of the vibrating string. Here resonance takes place and the string vibrates with
maximum amplitude.
Experimental determination of frequency of a tuning fork using sonometer:
L
A C B

The sonometer consists of a hollow wooden box. A nail is fixed at one end and a smooth pulley is
fixed at the other end of the box. One end of a sonometer string is tied to the nail and other end of the
string passes over the smooth pulley. The free end is attached to the weight hanger. A, B and C are three
knife edges placed on the box under the string. A, B are fixed and C is movable.
A suitable tension (T = Mg) is applied to the string. A small paper rider is placed on the string in
between A and C. The tuning fork of frequency ‘n’ is excited with a rubber hammer and its stem is kept
on the sonometer box. Now the string is made to vibrate. The movable knife-edge C is adjusted such
that the string vibrates with the same frequency of the fork. At that time, the paper rider placed on the
string between A and C, is violently thrown off from the string. Now the vibrating length of the string
AC= l is measured.
Let r be the radius of the string measured by using screw gauge and '' be the density of the material
of the wire, then

101
 The linear density, m =  r2 
The experiment is repeated by changing the value of tension (load). The readings are tabulated.

Vibrating
length
Sl. No. Load (M) l2 M/l2
AC
=l

Mean :
1 M g
2  l2  m
The frequency of the tuning fork, n =

M
The mean value of 2 is substituted in the above formula and the frequency of the tuning fork ‘n’
l
is calculated.

DOPPLER EFFECT

Whenever there is a relative motion between a source and a listener, the frequency of the
sound heard by the listener is different from the original frequency of sound emitted by the source.
This is known as “Doppler Effect”.

The whistle of a fast moving train appears to increase in pitch as it approaches a stationary
listener and it appears to decrease as the train moves away from the listener. This apparent change in
frequency was first observed and explained by Christian Doppler, an Austrian Mathematician and
Physicist. He observed that the frequency of the sound as received by a listener is different from the
original frequency produced by the source whenever there is a relative motion between the source and
the listener. This is known as Doppler Effect.
Application of Doppler Effect
1. It is used to measure the speed of a motorist by the police.
2. It is used to measure the temperature of a gas in a plasma state.
3. It is used in medical imaging and blood flow measurement.
4. Used in satellite communication
5. Used in underwater acoustics in submarine

102
ACOUSTICS OF BUILDINGS:
Echo : The first reflected sound is known as echo. The sound produced by a source is propagated
continuously in a medium if there is no disturbance. But if it meets the hard surface of an obstacle, it is
reflected. The clear echo depends upon the following factors.
1. Good reflector of sound
2. Maximum surface area of the reflector and
3. The distance of the reflector from the source of sound.
Reverberation :
The sound produced in a hall suffers multiple reflections before it becomes inaudible. As a
result of these reflections, the listener continues to receive sound, even if the source of sound is cut off.
This prolonged reflection of sound in a room even after the sound source has been stopped is called
reverberation. It is the persistence of sound due to multiple reflections from the walls, floor and ceiling
of a hall. The reverberation is also called multiple echoes. In a room, the walls, floor and other flat
surfaces reflect sound with a small loss of energy.
Reverberation time:
If a building is to be acoustically correct, its reverberation time must be in optimum level. It
should not be too long or too short. if it is too short, then the room becomes dead in sound aspect. If it is
too long, then the reverberation will be there inside the building for long duration. The reverberation
produces continuous sound with decreasing intensity upto a particular time after that it disappears.
This time is known as reverberation time.
The reverberation time is defined as the time taken by the sound to fall from its original intensity to
one millionth of its original intensity.
Sabine formula :
Sabine derived an equation for the reverberation time.
Where V is the volume of the hall,  is the coefficient of absorption of each reflecting surface present
in the hall and A is the area of the each
0.16 V
T= second
A
sound absorbing surface present in the hall.
Coefficient of absorption of sound energy:
The co-efficient of absorption of sound energy of any surface is defined as the ratio of the sound
energy absorbed by the surface to the total sound energy incident on the surface.
Let α be the coefficient of absorption of sound energy of a surface, then
The sound energy absorbed by the surface
=
The total sound energy incident on the surface
Acoustics of Buildings (Architectural acoustics):
The term 'Acoustics of building' describe the production, transmission and reception of sound
inside the building. The following points should be taken into consideration for the acoustic design of
the buildings, auditorium, cinema theatres etc.
1. The sound heard by the audience should be sufficiently loud in any part of the hall.
2. The quality of the speech and music should not be changed anywhere inside the hall.
3. There should not be focusing of sound due to walls and ceiling, in any part of the hall.
4. There should not be any vibrations due to resonance.
5. There should not be any other noise from other sources, both from outside the hall and from
with in the hall.
Two more important factors should also be considered for good acoustic design of the buildings.
1. Echo and 2. Reverberation
103
1. Echo :
The direct sound from the source and the reflected sound (echo) from the walls produce confusion
in certain buildings. A hall with large number of open windows is free from the defect. Echoes can be
eliminated by making the walls rough. But in the case of musical hall, however echoes are desirable, to
a certain extent.
2. Reverberation :
The Reverberation is the prolonged reflections of sound from the walls, floor and ceiling of a hall.
The sound produced in the hall undergoes multiple reflections before it becomes inaudible. As a result
the clarity of the successive speech is affected. Due to this reason, the reverberation time should be
optimum to hear the clear sound in the hall.
To minimize this defect, Sabine derived an equation for the reverberation time,
Where V is the volume of the hall, α is the coefficient of absorption of each reflecting surface present
in the hall and A is the area of the each sound absorbing surface present in the hall.
0.16V
T= second
A
In the above Sabine's equation, T will be minimized by increasing the value of α. So the
reverberation is minimized by covering the inner surfaces of the hall using absorption materials like
carpets, wall screen, porous tiles etc.
The reverberation time value depends on the use for which the building is designed. The acceptable
limit for reverberation time is
For speech, 0.5 s
For music, between 1.0 and 2.0 s
For small theatres, between 1.0 and 1.5 s
For larger theatres, up to 2.3 s
Noise Pollution :
The unwanted sounds which are dumped into the atmosphere, thus producing a nuisance and
adverse effect on the health of human beings, are known as noise pollution. Noise pollution leads to
severe health problems by creating either physical or mental problems. As a result, it affects the working
efficiency, personal comfort, and in some cases it may lead to industrial accidents. There are varieties
of noises such as domestic noise, traffic noise, aircraft noise, etc. Even though the sources of noise are
numerous, the types of sources of noise are generally classified into two types.
(i) Industrial noises and (ii) Non- industrial noises
The noises which are produced due to the movements of car, motor truck, motor cycle, aircraft,
machines, explosions, etc. are known as industrial noises. The noises which are produced due to the
loudspeaker, construction work, radio, etc. are known as non-industrial noises. The industrial and
non-industrial noises are generally measured by using the sound pressure level in decibel (dB).

Some important sound pressure level for domestic and industrial noise is listed below.

Source of noise. Sound pressure level (dB)

Garden 20
Library 30
Ringing alarm clock 80
Heavy traffic 90
Electric generator 76
Concrete mixer 85
Power saw 90
Noisy factory 100
Threshold of pain 140
104
 Worked Examples

1. A wire 50 cm long and of mass 6.5 × 10-3 kg is stretched so that it makes 80 vibrations per
second. Find the stretching tension.
mass 6.510−3
m= = = 13 10−3 kgm −1
length 0.5

1 T 1 T
n= or n 2 = 2
2l m 4l m
 Tension T = 4 l2 m n2 = 4 × 0.5 × 0.5 × 13 × 10-3 × 80 × 80
= 83.2 N
2. The density of a sonometer wire of radius 0.3mm is 7800 kgm-3. Find its linear density.
Linear density m = r2 
m = 3.14 × (0.3 × 10-3)2 × 7800
= 2204.28 × 10-6
= 2.204 × 10-3 kg m-1

3. A sonometer wire 0.5m long gives a vibration of 256Hz when stretched with a load of 5kg.
Find the linear density of the material of the wire.
Length of the vibrating wire l = 0.5 m
Frequency of sonometer is n = 256Hz
Load M = 5 kg
Linear density m= ?
1 Mg
Frequency of sonometer is n = 2l √ m
1 Mg
Squaring both sides, n2 = 4l2
m
1 Mg
Linear density, m =
4l2 n2
1 5 × 9.8
=
4 ×0.5 ×0.5 x 256 × 256
1 49
=1 ×
65536

= 1.4768×10-4 kgm-1.
Linear density = 1.4768×10-4 kgm-1

4. Find the frequency of sound produced by a string 25 cm long stretched by load of 5 kg and
the linear density of the wire is 4.9 × 10-3 kgm-1.
Length of the vibrating wire l = 25 cm = 25 ×10-2 m
Load M = 5 kg
Linear density m = 4.9 × 10-3 kgm-1
Frequency of sound produced, n = ?
1 𝑀𝑔
Frequency n = √
2𝑙 𝑚

105
 1 5 × 9.8
= × √
2 × 25 × 10 −2 4.9 × 10−3

49
= 2 x √4.9 × 10−3 = 2 x √10000 = 2 x 100

= 200 Hz
Frequency of sound produced= 200 Hz

5. When the tension in the string is increased by 44% the frequency is increased by 10 Hz.
Find the frequency of the string. Length of the string and linear density are kept constant.
nα√T
n1α√T1 and n2α√T2
n T
n1 = √T1
2 2

If n1 = n and T1 = T
then n2 = n + 10
44
and T2 = T + 44% of T= T + 100 × T
n n T
n1 = n+10 = √ 44
2 T+ ×T
100

T 1
=√ = √
1.44 T 1.44

n 1
=
n+10 1.2
1.2 n = n + 10
0.2 n = 10
10
n= = 50
0.2

Frequency of the string = 50 Hz

6. If the tension and diameter of a sonometer wire of fundamental frequency 'n' is doubled
and density is halved then find its fundamental frequency.
The fundamental frequency, n of stretched string of length l having mass per unit length m and
density  is given by the following relation
1 𝑇
n = 2𝑙 √𝑚 where T represents tension of the string.

Now, linear density, m =π r2 where r represents the radius of the string
1 𝑇
 n = 2𝑙 √𝜋𝑟2 

1 𝑇
n=
2𝑙 𝑟
√ 𝜋 ------- (1)

106
 Let n’ be the frequency when tension = 2T, radius = 2r and density = /2
1 2𝑇
So, n’ = √𝜋/2
2𝑙 ×2𝑟

1 4𝑇 1 𝑇
= 4 𝑙 𝑟 √ 𝜋 = 2𝑙 𝑟 √𝜋 ------ (2)

Comparing equations (1) and (2), n’ = n

 Frequency remains unchanged

https://www.youtube.com/watch?v=YIuEampSCSc&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=22

https://www.youtube.com/watch?v=TO0Cm8NOuqM&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=23

https://www.youtube.com/watch?v=R7WAs_ln-7k&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=24

107
 5.2 MAGNETISM
The word 'magnetism' is derived from the name of a dark brown coloured ore called magnetite found
in Magnesia. The ore possesses the property of attracting small pieces of iron. The ore magnetite
available in nature is called as natural magnet. Artificial magnets also can be prepared from iron or
steel materials. Iron materials are mostly used to prepare electromagnets and temporary magnets
whereas the steel materials are used to prepare permanent magnets.

Magnetic field:
The space around the magnet in which the magnetic lines of force act is called the magnetic field.
The direction of magnetic field at a point is given by the direction of the force acting on the north pole
of a magnetic needle placed at the point.
If the magnetic lines of force formed are parallel, the field is called uniform magnetic field and
have equal strength at all the points.

N S
N S

Pole Strength (P) :

The pole strength of a magnet is defined as the force acting on the pole when it is placed in a uniform
magnetic field of unit intensity.
Let P be the pole strength placed in a uniform magnetic field of intensity H, Force acting on the pole,
F = HP. If H = 1, then F = P The unit of pole strength is ampere metre (Am.)
Magnetic Induction (B) :
The magnetic induction or magnetic flux density is defined as the total number of magnetic lines of
force passing normally, through unit area of cross section.
The unit of magnetic induction is weber/metre².
Intensity of magnetic field (H) :
The intensity of the magnetic field at a point in a magnetic field is defined as the force experienced
by a unit north pole placed at that point.
The unit of Intensity of magnetic field is ampere/metre (Am-1)
Magnetic moment (m) :
Magnetic moment of a magnet is defined as the moment of the couple acting on a magnet when it is
placed perpendicular to a uniform magnetic field of unit intensity.
Consider a magnet of length 2l placed perpendicular to the uniform magnetic field of intensity H. If
P is the pole strength of the magnet,
Force experienced by the poles
F = HP
If H = 1, unit intensity of magnetic field, Force F = P

108
 Two equal forces P and P acting on the magnet in opposite directions to form a couple.
Moment of the couple = P × 2l

i.e. Magnetic moment M = 2lP

Hence, the magnetic moment of a magnet can also be defined as the product of pole strength and the
distance between the poles. But, the distance between the poles is approximately equal to the length of
the magnet. So, the moment can be taken as the product of pole strength and length of the magnet.
The unit of magnetic moment is ampere metre² (Am²)

Magnetic permeability (m):

The magnetic permeability of the medium (material) is the ratio of the magnetic induction in the
medium to the intensity of magnetic field
magnetic induction B
Magnetic permeability = Intensity of Magnetic field
i.e.  = H
producing that induction

The unit of magnetic permeability is weber/ampere-metre (WbA m )or henry/metre.

Intensity of Magnetisation (M) :
Intensity of magnetisation of a magnet is defined as the magnetic moment per unit volume of the
magnet.
Let m be the magnetic moment of the magnet having volume v, then
Magnetic moment
Intensity of magnetisation =
Volume of the magnet
m
i.e. M = v

The unit of intensity of magnetisation is ampere/metre (Am-1)

Hysteresis :
In electrical method, a magnetic material can be magnetised by keeping it inside an insulated coil
(solenoid) through which a current is passed. Before passing the current, the magnetising field around
the coil is zero. Then there is no magnetism induced in the material. If the magnetising field H is
gradually increased by increasing the current ‘I’ the intensity of magnetisation induced in the material
also increases upto a particular value. After that the intensity of magnetisation is not increased even
though the magnetising field is increased. Now the material is said to be saturated.
If the magnetising field is gradually decreased to zero value, the intensity of magnetisation in the
material also decreases but not to zero value. There is a tendency in the material to retain some amount
of magnetism. Then, if the field is increased in the reverse direction from zero value, for a particular
value the retained magnetism in the material is removed completely. Further increasing the field, the
material attained the saturated value of magnetism in the reverse direction and so on.

109
Definition :
The lagging in intensity of magnetisation behind the magnetising field is known as hysteresis.
A graphical curve which represents the variation in the intensity of magnetisation (M) with magnetic
field intensity (H) is known as hysteresis loop (M-H curve).
Retentivity, Coercivity and magnetic saturation :
1. Retentivity or residual magnetism :
Retentivity or residual magnetism is the amount of intensity of magnetisation retained in the
material after removing the magnetising field.
2. Coercivity or coercive force :
Coercivity or coercive force is the amount of magnetising field applied in the reverse direction to
remove the residual magnetism completely from the material.
3. Magnetic saturation :
When the magnetising field is increased, the intensity of magnetisation induced in the material also
increases up to a particular value. After that the intensity of magnetisation is not increased even though
the magnetising field is increased. Now the material attains the saturation value of intensity of
magnetisation.
Experiment to draw the Hysteresis loop:
The experimental arrangement is shown in the figure. A long solenoid is connected in a circuit.
The direction of the current can be reversed by using a commutator C. The given specimen in the form
of a rod is placed inside the solenoid. A deflection magnetometer is placed in Tan-A position at a short
distance along the axis of the rod. The deflection magnetometer is adjusted zero position.
The intensity of magnetic field H can be calculated by using the formula H = (NI) / L, where N is
the number of turns in the solenoid, 'I' the strength of the current passing through the solenoid and L, the
length of the solenoid. The value of I can be measured by using ammeter. The intensity of magnetisation
M induced in the specimen can be calculated by using the formula M = k tan  / V where k is
constant,  is the deflection in the magnetometer and V is the volume of the magnet.

I TAN-A N

+ - -
A +
Ba k Rh
The strength of the current I is gradually increased in steps of 0.2A up to a required value. Every time
the deflection in the magnetometer is noted. Then the values of H and the corresponding values of M are
calculated. The current is decreased in steps up to zero value and again the corresponding values of H
and M are calculated. The experiment is also repeated by passing the current in the reverse direction.
A graph is drawn by taking the values H along X-axis and M along Y-axis. A loop OABCDEFA is
obtained as shown in the figure.
In the figure, when H is increased from zero to G, the value of M is increased from zero to A
represented by OA. At the point A, the specimen reached the saturated value.
If H is decreased from G to zero, the value of M decreases from A to B, but not to zero. Now the
specimen retains some amount of magnetism known as retentivity (OB).

110
 When H is increased in the reverse direction from zero to C, the value M reaches zero value at a
point C. The amount of H applied in the reverse direction to remove the residual magnetism is known
as coercive force or coercivity (OC). Further repeating the process the remaining portion CDEFA is
obtained. The closed loop ABCDEFA is called one cycle of hysteresis loop or M-H curve.
When the material is taken through a cycle of magnetisation, there is some expenditure of
energy. This loss of energy dissipates in the form of heat in the specimen. The loss of energy is
directly proportional to the area of the hysteresis loop.
Y
M
A
B

–X C F X
K O G H

E
D

-Y

Selection of magnetic materials for permanent and temporary magnets (Uses of

Hysteresis loops):
The hysteresis loops are very useful in the selection of magnetic materials for permanent and
temporary magnets and other industrial purposes. The hysteresis loops for different materials are
compared to one another and then the materials are selected for suitable purposes.
For example, let ABCDEFA be the hysteresis loop for soft iron and A' B' C' D' E' F' A' the hysteresis
loop for steel. Here the area of the hysteresis loop for steel is more than that of soft iron. Therefore the
loss of energy is more in steel than that in soft iron. Hence the quantity of heat emitted by steel is more
than that by soft iron. For this reason soft iron is selected as a core materials for transformers, chokes,
AC motors, dynamos, etc.
The retentivity of soft iron (OB) is more than that for steel (OB'). So, soft iron retains more
magnetism than steel. But the coercivity for soft iron (OC) is less than the coercivity for steel (OC'). So
the strength of the magnetic field needed to remove the residual magnetism in soft iron is less than in
steel. That means it is easy to remove the residual magnetism in soft iron than steel. Hence steel retains
the magnetism during long period. For the above reasons steel is used for permanent magnets and soft
iron is used for temporary magnets and electromagnets.

M Y
A
B
A'
B'
-X C F X
C' O E' F' H

E
D'
D
–Y

111
MAGNETISM:
Types of magnetic materials and their applications:
1. Ferro magnetic materials are strongly attracted by the magnets. (eg. Iron, steel)
Para magnetic materials are weekly attracted by the magnets. (Aluminium, Platinum)
Dia magnetic materials are repelled by the magnets. (Cadmium, Silver)
2. Electromagnets: An electromagnet consists of a U shaped soft iron core called yoke having
large permeability. Coils of thick copper wire are wound over the vertical limbs. Two pole pieces
P1 and P2 are kept as shown in the figure and intense magnetic field is created in the space
between P1 and P2. One pole becomes positive and the other one negative. Intensity of the
magnetic field developed is found to depend in the product of the number of turns of the coil and
the current through the coil. The ends of the pole pieces facing one another are made conical so
as to concentrate the field in the air gap between the poles.

Uses of Electromagnets:
1. Electromagnets are used in relays, remote control switches etc.
2. In nuclear physics giant electromagnets are used in huge accelerators like cyclotron etc.
3. In industrial establishment where ferrous metals have to be lifted and moved, electromagnets
are used. The electromagnet is energized when the current is switched in and the ferrous
metals are held. When the current is switched off the load is released.
Various Application and Uses of magnets:
1. Magnets are used in speakers and are important in reproduction of sound in TVs and radios.
The small coil of wire and a magnet inside a speaker transforms the electronic signal to
sound vibrations.
2. Magnets are used in a generator to transform mechanical energy to electrical energy where
there are other kinds of motors which use magnets to change electrical energy to mechanical
energy.
3. Electrically charged magnets can help cranes to move large metal pieces.

112
 Worked Examples
1. The length ,breath and thickness of a bar magnet are 30 cm, 2 cm and 1 cm respectively. if
its magnetic moments is 6×10-6 Am2,calculate the intensity of magnetisation
l =30 cm = 0.30 m, b = 2 cm = 0.02 m and h = 1 cm = 0.01 m
m = 6×10-6 Am2
Volume V = lbh = 0.30 × 0.02 × 0.01 =0.00006 m2
Intensity of magnetization (M) = m/V = 6×10-6 / 0.00006
M = 0.1 Am-1

2. The moment of a bar magnet is 0.6 Am 2 and it’s volume is 3 x 10-5 m3. Calculate the intensity
of magnetisation of the magnet.
Given Data
Moment of the magnet, m = 0.6 Am2
Volume of the magnet, V = 3 x 10-5 m3
𝑚 0.6
Intensity of magnetisation, M = 𝑣
= 3 𝑥 10−5

M = 0.2 x 10-5 m1

https://www.youtube.com/watch?v=TpYjNkJ92kA&list=PL1b9Ht9ISqIHStwn0NEHongecMJ84S6cj&index=25

113
 QUESTIONS
Part - A and Part – B

1. Define wave motion.

2. Define transverse wave motion.
3. Define longitudinal wave motion.
4. Define progressive wave.
5. Define amplitude of a wave.
6. Define wavelength of a wave.
7. Define period of wave motion.
8. Define frequency of a wave.
9. Define velocity of sound wave.
10. Define stationary wave.
11. Define free vibrations.
12. Define forced vibrations.
13. Define Resonance.
14. What is the use of a sonometer?
15. What is an echo?
16. What is reverberation?
17. What is reverberation time?
18. Write sabine's formula.
19. Define co-efficient of absorption of sound energy.
20. Define Pole Strength.
21. Define magnetic induction.
22. Define intensity of magnetic field.
23. Define magnetic moment of a magnet.
24. Define intensity of magnetisation.
25. Define Hysteresis.
26. Define Retentivity
27. Define coercivity
28. Define magnetic saturation.
29. Explain transverse wave motion.
30. Explain longitudinal wave motion.
31. Explain stationary wave
32. What are the important factor of good acoustics
33. Explain noise pollution
114
34. Explain resonance
35. Explain Hysterisis
36. Write the uses of hysteresis loop.

Part – C

1. Explain transverse wave motion and longitudinal wave motion.

2. Distinguish between transverse and longitudinal wave motion.
3. Describe how the frequency of a tuning fork is determined using a sonometer.
4. Write a note on acoustics of buildings.
5. Explain noise pollution and the methods of controlling industrial noise.
6. Describe the method of drawing hysteresis loop of a specimen using a solenoid.
7. Explain and uses of hysteresis loop.

Exercise Problems
1. A sonometer wire of 0.5 m long gives vibrations of 256 Hz when stretched with a load of 5 kg. Find
the linear density of the material of the wire.
2. Find the frequency of sound produced by a string 25 cm long stretched by load of 5 kg. The linear
density of the wire is 4.9 × 10-3 kgm-1
3. A string 75 cm long and weighing 15 g produces a note of frequency 100 cycles per second on
plucking. What is the tension in the string?
Ans: 450 N
4. A wire of 50 cm long and of mass 6 x 10 –3 kg is stretched so that it makes 60 vibrations per second.
Find the tension in the wire.
Ans: 43.20 N
5. The vibrating length of 0.75 m of a sonometer wire is unison with a tuning fork when stretched by
weight of 5 kg. The linear density of the wire is 0.5 x 10-3 kg m–1. Calculate the frequency of the
tuning fork.
Ans: 209 Hz
6. 6.A wire 50 cm long and of mass 0.6 x 10-3 kg is stretched by a tension of 4 kg.wt. When
sounded, it is found to vibrate in 2 loops. Calculate the frequency of the note emitted by the wire.
Ans: 361.Hz
7. 7.A wire 0.5 m long vibrates 100 times a second. If the length of the wire is shortened to 0.4
m and the stretching force is increased to 4 times its original value, what will be the new
frequency?
Ans: 250 Hz
Brain Teaser
1. Thunder was heard 6 second after a flash of lighting was seen. If the velocity of sound is 345ms-1,
calculate the distance at which flash occurred.
2. Two magnetic poles, one of which is twice stronger than the other repel one another with a force of
2 × 10-5 N when kept separated at a distance of 20 cm in air. Calculate the strength of each pole
115
 SUMMARY

• The study of sound is called Acoustics.

• The unit of sound is decibel (dB).
• 20 Hz to 20 kHz is the audible range of Humans.
• Ultrasonic or Ultrasound is above 20 kHz, used for medical diagnostics such
as sonograms.
• Infrasonic or infrasound is below 20 Hz, used to detect volcanic eruptions.
• Whales, elephants and other animals can detect infrasound and use it to
communicate.
• The continuous movement of disturbance is called wave.
• Sound does not travel through vacuum.
1 𝑀 𝑔
• The frequency of a tuning fork, n = √ 2 ( )
2 𝑙 𝑚

• The first reflected sound is called Echo.

0.16 𝑉
• Sabine equation, T = s
𝛼𝐴

• The first magnets were not invented, but rather were found from a naturally
occurring mineral called magnetite.
• A magnet is a material or object that produces a magnetic field.
• Artificial magnets are prepared from iron and steel materials.
𝑚
• Intensity of magnetization, M = 𝑣

• A graphical curve which represents the variation of M with H is known as

hysteresis loop.
• Hysteresis loop is used to select the magnetic materials for permanent and
temporary magnets.

116
 Model Question Paper

TIME: 3hrs ENGINEERING PHYSICS - I MARKS: 100

PART A (5 X 1 = 5)
Note: Answer ALL questions. All questions carry equal marks.

1. Mention any two fundamental physical quantities.

2. Why rain drops fall slowly?
3. What is the condition for maximum range of a projectile?
4. Define moment of inertia of a rigid body.
5. Define magnetic moment.

PART B (10 X 2 = 20)

Note: Answer any TEN questions. All questions carry equal marks.

6. What are the uses of dimensional formula?

7. What is meant by coplanar force?
8. State Lami’s theorem.
9. What are three moduli of elasticity?
10. Explain turbulent flow.
11. Define angle of contact.
12. State Newton’s laws of motion.
13. Define normal acceleration.
14. Define angle of banking.
15. State law of conservation of angular momentum.
16. State Newton’s law of gravitation.
17. Write any two uses of artificial satellites.
18. Define resonance.
19. Define Doppler effect.
20. Explain hysteresis.
PART C (5 X 15 = 75)
Note: Answer ALL questions by choosing either a & b (OR) c & d

21. a) State the conventions to be followed in the SI units. (8 Marks)

b) Explain how to resolve a vector quantity in to two rectangular components. (7 Marks)
OR
c) Describe an experiment to verify the parallelogram law of forces. (8 Marks)
d) If the resultant of two equal forces is √3times each force. Find the angle (7 Marks)
between the forces.

22. a) Explain uniform and non uniform bending of beams. (8 Marks)

b) Describe an experiment to determine the Young’s modulus of the material of (7 Marks)
the beam by uniform bending method.
OR
c) Describe an experiment to compare the coefficient of viscosities of two (8 Marks)
liquids.
d) Calculate the surface tension of water if it rises to a height of 4.2 cm in a (7 Marks)
capillary tube dipped vertically in it. Radius of the capillary tube is 3.5 X 10 -4 m
and density of water is 1000 kgm-3

117
23. a) Derive an expression for the maximum height and time of flight reached by (8 Marks)
the projectile.
b) Derive expression for normal acceleration and centripetal force acting on a (7 Marks)
body executing uniform circular motion.
OR
c) Derive an expression for the angle of banking. (8 Marks)
d) A ball weighing 0.5 kg tied to one end of a string of length 2 m is rotated at a (7 Marks)
constant speed of 10 ms-1 in a horizontal plane. Calculate the centripetal force
on the ball.

24. a) Derive an expression for angular momentum of rotation of a rigid body (8 Marks)
rotating about an axis.
b) Derive an expression for variation of acceleration due to gravity with altitude. (7 Marks)
OR
c) Write short notes on polar and geostationary satellites. (8 Marks)
d) If the radius of the earth is 6400 km, and acceleration due to gravity is 9.8 ms - (7 Marks)
2
. Calculate the escape velocity on the surface of the earth.

25. a) Distinguish between longitudinal and transverse wave. (8 Marks)

b) Write a note on acoustics of buildings. (7 Marks)
OR
c) Explain the method of drawing hysteresis loop of a given specimen. (8 Marks)
d) The vibrating length of 0.24 m of a sonometer wire is unison with a tuning fork (7 Marks)
when stretched by a weight of 4.5 kg. The linear density of the wire is 0.65 X 10 -3
kgm-1. Calculate the frequency of the fork.

118
 SEMESTER – I
40006 ENGINEERING PHYSICS PRACTICAL
1. MICROMETER (SCREW GAUGE)

To measure the thickness of the given irregular glass plate, using a micrometer. To measure area of
the glass plate using a graph sheet and to calculate the volume of the glass plate.
Apparatus required:
Screw gauge, graph sheet and given glass plate.
Formula:
Volume of the glass plate = Thickness × area of the glass plate m3
Thickness of glass plate = Observed Reading + Zero Correction
= {PSR+(HSC × LC)}+ZC
Where PSR = Pitch scale reading
HSC = Head scale coincidence
LC = Least Count
ZC = Zero Correction
Head Scale Reading (HSR)= Head Scale coincidence × Least Count
Observed Reading (OR) = Pitch Scale Reading + Head Scale Reading
Correct Reading (CR) = Observed Reading + zero Correction
Description:
The screw gauge consists of a screw working in a hollow cylindrical nut attached to one end of a
frame G. On the hollow cylinder a scale is marked in mm. This scale is called the pitch scale (PS). There
is a fixed stud A at one end of the frame and exactly opposite to A, there is a movable stud B of the screw.
The other end of this screw has a head H. To the head, a hollow sleeve is attached. On the tapering of the
sleeve 50 or 100 divisions are marked. This scale is called the Head Scale (HS). The Screw gauge head
is having a ratchet arrangement, to avoid over- working of the screw.
Procedure:
i) To find the least count (LC):
Least count is the minimum measurement that can be made with the given instrument.
First, the pitch of the screw is determined; by giving a known number of rotations to head scale and
noting the distance advanced in pitch scale in mm. Then the number of divisions in the head scale is
also noted.
The pitch and least count are determined using the formulae.
ii) To find Zero Error (ZE) and Zero Correction (ZC):
The Screw gauge is checked to find whether there is any initial (zero) error in the instrument. If there
is any initial error, suitable correction is to be made.
When the studs A and B touch each other, if the zero of the head scale lies on the same line as that
of the pitch scale index line(I.L), the instrument has no error.
If the zero of the head scale is above the index line, it has negative error. So the zero correction is
positive.

119
 If the zero of the head scale is below the index line, it has positive error. So the zero correction is
negative.
The type of error (ZE) and the suitable zero correction (ZC) for the given micrometer is determined
with the help of figures and formulae given.
iii) To find thickness of the glass plate:
The given glass plate is placed gently in between the two studs A and B and the ratchet is rotated
till the glass plate is firmly but gently gripped. Note the number of completed divisions in mm on the
pitch scale as Pitch scale reading (PSR) and the divisions on the head scale, which coincides with the
index line as head scale coincidence (HSC). The PSR and HSC are entered in the tabular column. Then
the head scale reading (HSR), observed reading (OR) and correct reading (CR) are calculated. The
procedure is repeated for different positions of the glass plate and the average thickness of the glass plate
is calculated.
iv) To measure the area of the glass plate:
The given irregular glass plate is placed on a graph sheet. The trace of the glass plate is drawn on the
graph sheet. The number of squares inside the trace in mm2 is counted. This gives the area of the glass
plate in mm2.
By multiplying the thickness of the glass plate and the area of the glass plate, the volume of the glass
plate is determined.
Result:
The thickness of the given glass plate = ×10-3 m
The area of the glass plate = × 10-6 mm2
The volume of the glass plate = × 10-9 mm3

Pitch Scale Ratchet

20
0
A B 15
10
5

Head Scale

G Frame
Observations:
(i) Least Count:
Pitch of the screw
Least count =
No.of head scale division
Dis tan ce moved in the pitch scale
Pitch of the screw =
No. of rotations given to HS
= mm
Number of head scale divisions(HSD) = div
pitch
Therefore, Least Count (LC) = No.of HSD = mm
LC = mm
Three possible initial error and initial correction:
HS HS HS
5 10 0
I.L 0 I.L 5 I.L 95

95 0 90

120
 No error Positive error Negative error
No correction Negative correction Positive correction
HSC = 5 div HSC = 95 div

ZE = + (HSC × LC) ZE = - (100 – HSC) × LC

= + (5 × 0.01 mm) = - (100) × 0.01 mm
= + 0.05 mm = - 5 × 0.01 mm
ZE = + 0.05 mm ZE = -0.05 mm
ZC = - 0.05 mm ZC = + 0.05 mm
(ii) Zero error and zero correction for the given screw gauge:

Tabular Column:
To find the thickness of the given glass plate:
LC = mm ZE = mm ZC = mm

Thickness
Sl. No. PSR HSC HSR = HSC × LC OR = PSR + HSR CR = OR + ZC
Unit mm div Mm mm mm
1.
2.
3.
4.
5.
The average thickness of the given glass plate = mm
= × 10-3 m
Area of the glass plate using graph sheet = mm2
= × 10-6 m2

Volume of the glass plate = Thickness × Area

= m3

121
 2. VERNIER CALIPERS
Aim:
To measure the length and diameter of the given solid cylinder using vernier calipers and to calculate
the volume of the solid cylinder.
Apparatus required:
Vernier calipers and solid cylinder.
Formula:
Volume of the solid cylinder = π r2 l m3
Where
l = length of the cylinder
r = radius of the cylinder
Length / diameter of the solid cylinder = Observed reading + Zero correction
= {MSR + (VSC × LC)} + ZC
Where MSR = Main Scale Reading
VSC = Vernier Scale Coincidence
LC = Least Count
ZC = Zero Correction
Vernier Scale Reading (VSR) = Vernier Scale Coincidence (VSC) × Least Count (LC)
Observed Reading (OR) = Main Scale Reading (MSR) + Vernier Scale Reading (VSR)
Correct Reading(CR) = Observed Reading (OR) + Zero Correction (ZC)
Description:
The vernier calipers consists of main scale (MS) graduated in cm along the lower edge of the frame
and inches along the upper edge of the frame. To one end of the frame, two fixed jaws are provided.
There are two movable jaws, which slide over the main scale, and it can be fixed at any point using a
screw S. The movable jaw attachment carries a vernier scale (VS).
Procedure:
i) To find the Least Count :
Least Count is the minimum measurement that can be made with the given instrument.
The value of 1 main scale division (MSD) and number of divisions (n) on the vernier scale are
determined. The least count of the instrument is calculated using the formula
1
Least Count (LC) = × value of 1MSD
n
ii) To find the Zero Error (Z.E) and Zero Correction(Z.C) :
The instrument is checked to find whether there is any initial error. If there is any initial error,
suitable correction is to be made.
When the two jaws are in contact, the graduations are made such that the zero of the MS coincides
with the zero of the VS. Now the instrument has no error.
If the zero of the VS is on the right side of the zero of the MS, it has positive error. Then the zero
correction is negative.
If the zero of the VS is on the left side of the zero of the MS, it has negative error. Then zero
correction is positive.
In each case, note the division of the vernier scale which is coinciding with any one of the main scale
divisions. This is vernier scale coincidence (VSC). Applying this in the given formula, the value of zero
error is determined. Hence zero correction (ZC) is also determined.

122
iii) To find the length and diameter of the solid cylinder :
The given cylinder is gently placed in between the two lower jaws of the vernier calipers such that
the length of the cylinder is parallel to the scale.
The completed main scale reading (MSR) is taken by noting the position of zero of the vernier scale.
The vernier scale coincidence (VSC) is then noted. VSC is that particular division which coincides with
any one of the main scale divisions, making a straight line.
The MSR and VSC are noted in the tabular column I for different settings of the cylinder. Then
the Vernier Scale Reading (VSR), Observed Reading (OR) and Correct Reading (CR) are calculated
using the formula given. The average value of CR is determined, which gives the length of the given
cylinder.
Similarly the readings for diameter of the solid cylinder are taken by placing the cylinder suitably in
between the lower fixed jaws and the readings are noted in the tabular column II.
Result:
The length of the solid cylinder l = × 10-2 m
The diameter of the solid cylinder d = × 10-2 m
The volume of the solid cylinder v = × 10-6 m3

Observations :
(i) Least count :
Value of 10 MSD = cm
Value of 1 MSD = cm
No. of division on the vernier scale (n) =
(ii) The three possible initial error and initial correction :
1
Least Count (LC) = × value of 1 MSD = = cm
n
No error Positive error Negative error
No correction Negative correction Positive correction
M.S M.S M.S

10 5 10
0 5 10 0 5

0 V.S 10 0 10 0 10
V.S V.S

VSC = 3 div VSC = 7 div

ZE = + (VSC) × LC ZE = -(10-VSC) x LC
= + 3 × 0.01 cm = - (10-7) × 0.01cm
= + 0.03 cm = - 0.03 cm
ZC = - 0.03 cm ZC = + 0.03 cm
Zero error and zero correction of the given vernier calipers:

123
Tabular column I :
To find the length of the solid cylinder
LC = cm ZE = cm ZC = cm

Length
Sl. No MSR VSC VSR = VSC x LC OR = MSR + VSR CR = OR + ZC

Unit cm div cm cm Cm
1
2
3
4
5
The average length of the solid cylinder (L) = cm
= × 10-2 m
Tabular column II :
To find the diameter of the solid cylinder
LC = cm ZE = cm ZC = cm

Diameter
Sl. No MSR VSC VSR = VSC × LC OR = MSR + VSR CR = OR + ZC

Unit cm div cm cm Cm
1
2
3
4
5

The average diameter of the solid cylinder (D) = cm

= × 10-2 m
The average radius of the solid cylinder ‘r’ = × 10-2 m
Volume of the solid cylinder = π r2 l
= m3

124
 3. PARALLELOGRAM LAW OF FORCES
Aim:
To verify Parallelogram law of forces.
Apparatus required:
Vertical drawing board, two pulleys with clamps, three sets of slotted weights, thread, white paper,
scale, protractor, compass and pencil.
Formula:
To verify the parallelogram law of forces, it is to be shown that Resultant = Equilibrant and they
are opposite in directions. i.e. OD = OC and COD = 180º where OD and OC are lines drawn
representing the resultant and equilibrant of P and Q.
where P, Q and R are the forces  and  are the angles opposite to P, Q and R respectively.
Description:
A drawing board is supported vertically on a wall. The two smooth pulleys are fixed at the two top
corners of the board, so that they can rotate freely. The two weight hangers P and Q are attached to the
two ends of a long thread, after passing through the pulleys. The third weight hanger R is attached to the
middle of the thread as shown in the figure.
Procedure:
The weights in the hangers P, Q and R are adjusted such that the point O is at rest. A small
displacement is given to the system so as to overcome the friction in the pulleys. A drawing paper
is fixed on the board, behind the strings. The trace of the three strings is taken on the sheet of paper,
using an adjustable lamp arrangement in front of the threads. The trace sheet is removed and the lines
are joined to get the image of the threads. These lines meet at a point O.
To verify parallelogram law of forces:

Taking a suitable scale (0.050 kg = 1 cm) OA, OB and OC are marked to represent the forces
P, Q and R respectively both in magnitude and in direction. With OA and OB as adjacent sides, a
parallelogram OADB is completed. The length of the diagonal OD is measured and entered in the
tabular column. Also the COD is measured and entered in the tabular column I.
According to parallelogram law, the diagonal OD gives the resultant of P and Q, where OC is the
equilibrant of P and Q.
The procedure is repeated thrice with different combinations of P, Q and R.
Result:
In the tabular column, in all cases, it is seen that Resultant = Equilibrant i.e. OD = OC and also
COD = 180°, which verifies parallelogram law of forces.
D

B
A 
P Q
O  O 
R
P Q
R
C

125
Tabular Column :
To verify parallelogram law of forces
Scale : 0.05 kg.wt = 1 cm

Equili
Sl. Force Line Force Line Force Resultant
- brant COD
No. P OA Q OB R OD
OC

kg. kg. kg.

Unit cm cm cm cm degree
wt wt wt
1.
2.
3.
4.
5.

126
 4. LAMI'S THEOREM
Aim:
To verify Lami's theorem.
Apparatus required:
Vertical drawing board, two pulleys with clamps, three sets of slotted weights, thread, white paper,
scale, protractor, compass and pencil.
Formula:
To verify Lami's theorem, it is to be shown that
P Q R
= =
Sin  Sin  Sin 

where P, Q and R are the forces  and  are the angles opposite to P, Q and R respectively.
Description:
A drawing board is supported vertically on a wall. The two smooth pulleys are fixed at the two top
corners of the board, so that they can rotate freely. The two weight hangers P and Q are attached to the
two ends of a long thread, after passing through the pulleys. The third weight hanger R is attached to the
middle of the thread as shown in the figure.
Procedure:
The weights in the hangers P, Q and R are adjusted such that the point O is at rest. A small
displacement is given to the system so as to overcome the friction in the pulleys. A drawing paper
is fixed on the board, behind the strings. The trace of the three strings is taken on the sheet of paper,
using an adjustable lamp arrangement in front of the threads. The trace sheet is removed and the lines
are joined to get the image of the threads. These lines meet at a point O.
To verify Lami's theorem:
In all the three figures drawn in the trace sheet, the angles between the forces are marked. The angle
between the forces Q and R is marked as . The angle between the forces R and P is marked as . The
angle between the forces P and Q is marked as  as shown in the figure. The angles are measured and
entered in the corresponding tabular column II. To check the accuracy of angle measurement, for each
case, the sum of the three angles measured should be equal to 360°. i.e.(  +  +  ) = 360°.
Then the calculations are done using the given formula and the values are entered in the tabular
column.
Result:
In the tabular column II, in all cases, it is seen that the values in the last three columns are found to
be equal.
which verifies Lami's theorem.
P Q R
i.e = =
Sin  Sin  Sin 

A B

Q
P
O
O  

R
P Q
R
C

127
Tabular Column :
To verify Lami's Theorem
Forc Angl
es es
Sl. P Q R
No. P Q R    Sin  Sin  Sin 

kg. kg. kg. kg. kg. kg.

Unit degree degree degree
wt wt wt wt wt wt
1
.2
.3
.
Calculation:

1. P = kg. wt, Q= kg. wt, R= kg. wt

= ° = ° = °
P
Sin  =
Q
=
Sin 

R
=
Sin 

2 P= kg. wt, Q= kg. wt, R= kg. wt

= ° = ° = °

P
Sin  =
Q
=
Sin 

R
=
Sin 

3. P = kg. wt, Q= kg. wt, R= kg. wt

= ° = ° = °
P
Sin  =
Q
=
Sin 

R
=
Sin 
128
 5. Comparison of Co-efficients of viscosity of two
liquids by capillary flow method
Aim:
To compare the coefficient of viscosity of two given liquids by capillary flow method.

Apparatus required:
Burette stand, graduated burette without stopper, Capillary tube, rubber tube, beaker, stop watch,
water, funnel, wooden bench, etc.

Formula:
The comparion of coefficient of viscosity

(η1 / η2) = (ρ1 / ρ2) × (t1 / t2 )

η1 = coefficient of viscosity of first liquid

η2 = coefficient of viscosity of second liquid
ρ1 = Density of the first liquid
ρ2 = Density of the second liquid

0
5
10
15
20
BURETTE

25
30
Capillary
35 Tube

Description:
The graduated burette without stopper is mounted vertically in the stand. A rubber tube is connected
to the bottom of the burette. To the other end of the tube, a capillary tube is inserted and placed in a
perfectly horizontal position.

Procedure:
The burette is cleaned and filled with one of the liquids using a funnel. Adjust the tube till the liquid
through the tube comes out drop by drop. A stop clock is started when the liquid level reaches 0 cc and
time taken for every 5 cc from 5,10, 15,20..till 50cc, is noted .The time taken by the liquid (t1) to flow
from 0 to 15, 5 to 20, 10 to 25… are found out The burette and the tube are cleaned, rinsed and filled
with the second liquid The time taken to cross the every consecutive 5 ml is noted. The reading are
tabulated.

129
The time taken for the t1 and t2 are noted. The mean value of t1 / t2 is calculated

Result:
The ratio of coefficients of viscosities of the two liquids =

Tabular column

Sl. Burette Time of flow of liquid (s)

No
Reading CC
Liquid I Liquid II
1
2
3
4
5
6
7
8
9
10

Sl. Time of flow (second) 1

Range
No. Liquid I Liquid II t / t2
CC
(11) (12)

Mean

Observations:
Density of the first liquid (water) = 1000 kgm-3
Density of the second liquid (kerosene) = 800 kgm-3
Averge ratio of the time of flow (t1 / t2 ) =

Calculation
The comparison of coefficient of viscosity (1/2) = (1 / 2) x (t1 / t2)

130
 6. COEFFICIENT OF VISCOSITY OF HIGH VISCOUS
LIQUID – STOKES' METHOD

Aim:
To determine the coefficient of viscosity of high viscous liquid (Castor oil) by Stokes' method.

Apparatus Required
A tall glass jar, castor oil, glass beads of different radii, stopwatch, screw gauge, etc.

Formula:
2 ( −) g (r 2 t )
The coefficient of viscosity of the highly viscous liquid h = Nsm –2
9h
Where
 = the density of the material of the solid sphere (glass beads)
 = the density of the liquid (castor oil)
g = the acceleration due to gravity
r = the radius of the glass beads
t = the time taken by the solid sphere to travel the marked distance h, inside the high viscous
liquid.
h = the distance between the marks B and C

Description:
The pure, transparent, high viscous liquid (castor oil) is taken in the tall glass jar. On the outer
surface of the jar, two markings B and C are made at a distance of say 50 cm. The marking B should be
well below the free surface of the liquid. Then only the sphere can attain terminal velocity when
travelling from B to C.

Procedure:
First the least count, zero error and zero correction of screw gauge are determined. Then using the
screw gauge, the diameters of the given glass beads are measured. The radii of the beads are calculated.
Values of r and s are taken from standard tables.
The sphere is placed gently on the surface of the high viscous liquid and dropped. When it crosses
the marking B, a stop clock is started and when it crosses the marking C, the clock is stopped. The
time taken (t) to travel by the sphere from B to C is recorded in the tabular column. The distance
between B and C is noted as 'h'. The same procedure is repeated for all the given spheres. Then the
coefficient of viscosity of highly viscous liquid (castor oil) is calculated, using the formula.

Result:
The coefficient of viscosity of the high viscous
liquid (castor oil) = N s m-2

131
 Glass jar

High viscous liquid

B

Glass bcads
Observations:
The density of the material solid sphere (glass beads)r = kg m-3
The density of the viscous liquid (castor oil) s = kg m-3
The acceleration due to gravity g = m s-2
The distance between B and C h = × 10-2 m
Distance moved in pitch scale
Pitch of the screw = No. of rotations given to HS
(I) Least Count: Least Count = Pitch / No. of Head Scale Divisions
= mm = mm
pitch
Therefore, Least Count (LC) = = - - - - - - - mm
No. of HSD
Number of head scale divisions(HSD) = Div
LC = Mm
(ii) Zero Error and Zero Correction:
Tabular column I:
To find the radii of the solid spheres (glass beads)
LC = mm; ZE = mm; ZC = mm

HSR = OR = Mean Radius

Sl. CR = OR +
Solid PSR HSC (HSC × PSR
No. ZC diameter r = d/2
sphere LC) + HSR
Unit mm div mm mm Mm mm mm
1.
2.
Bead 1 3.
4.
5.
1.
2.
Bead 2 3.
4.
5.

132
 Bead 3 1.
2.
3.
4.
5.

Tabular column II:

To find r2t

Radius Radius2
Glass beads Time taken t r2t
R r2

Unit mm mm2 mm2

Bead 1
Bead 2
Bead 3

The average of r2t = mm2 m2

= × 10-6
Calculation:
The coefficient of viscosity of the highly viscous liquid

2 ( − ) g (r2 t)
(castor oil)  =
9 h
=

= Nsm-2

133
 7. SONOMETER

Aim:
To find the frequency of the given tuning fork using a sonometer.
Apparatus Required:
1 M g
The frequency of the tuning fork n =  l2  m Hz
2  

Sonometer, screw gauge, tuning fork, hammer, weight hanger set and paper rider.
Formula:
Where
M = total mass at the end of the sonometer wire.
i.e. Mass of the weight hanger + Mass placed on it
g = acceleration due to gravity.
m = linear density of the wire
i.e., mass per unit length of the wire.
ℓ = first resonating length.
The linear density of the wire, m = r2
Where
 = density of the material of the wire.
r = radius of the sonometer wire.
Description :
A sonometer consists of a hollow wooden box about one metre with bridges A and B fixed at two
ends on the top surface. A uniform wire is stretched over the two bridges, one end of the wire being fixed
to the nail at one end of the box. The other end passes over a smooth pulley and carries a weight hanger
on which weights can be placed. M and N are two movable bridges in between the fixed bridges A and
B. By changing their position, the length of the vibrating segment of the wire can be altered.
Procedure:
A suitable load say 0.5 kg is placed on the hanger. The two movable bridges M and N are placed
close to each other. A light paper rider is placed at the centre of the segment MN of the sonometer wire.
The given tuning fork is excited and its stem is pressed on the sonometer box. The length of the vibrating
segment MN is gradually increased by moving the bridge M or N, till the paper rider flutters violently
and fall down. The first resonating length ℓ between M and N is measured using the metre scale fixed
on the sonometer box. The load is changed to 1 kg, 1.5 kg, 2 kg, 2.5 kg, 3 kg etc and in each case the
resonating length is determined and M/ ℓ2 is calculated.
The diameter d of the wire is found using a screw gauge. The radius of the wire r is calculated. The
mass per unit length of the wire is m = r2. Here  is the density of the material of the wire.
The frequency of the given turning fork is calculated by substituting the values of M/ℓ2 and 'm' in
the formula.
Result:
The frequency of the given tuning fork (n) = Hz.

134
 A M N B

Observations:
Acceleration due to gravity g = ms-2
Density of the material of the wire (steel)  = kg m-3
Distance moved in pitch scale
Pitch of the screw = =
No. of rotations given to HS
(i) Least Count: Least Count = Pitch / No. of Head Scale Divisions
pitch
Therefore, Least Count (LC) = No. of HSD = mm

= ------------ mm = mm
Number of head scale divisions (HSD) = div
LC = mm
(ii) Zero Error and Zero Correction:
Tabular column I:
M
To find
l2

Mass placed at the end of

the sonometer wire [mass First resonating M
Sl. No. ℓ
of the hanger + mass length ℓ l
placed on the hanger] M
Unit Kg cm m m kg m-2
1. Hanger + 0.5 1.0
2. Hanger + 1.0 1.5
3. Hanger + 1.5 2.0
4. Hanger + 2.0 2.5
5. Hanger + 2.5 3.0

Tabular Column II:

To find the diameter of the wire
LC = mm ZE = mm ZC = mm
Sl. No. PSR HSC HSR = HSC × LC OR = PSR + HSR CR = OR + ZC
Unit mm div mm mm mm
1.
2.
3.
4.
5.
Mean

135
 The average diameter of the wire = mm
Radius of the wire r = d/2 = mm
R = × 10-3 m
Calculation:
The Mass per unit length of the wire m =  r2
=
= kg m-1
1 M g
The frequency of the tuning fork n =  2 
2 l m
=
= Hz

136
 8. DEFLECTION MAGNETOMETER
(TAN A-POSITION) COMPARISON OF MAGNETIC MOMENTS

Aim:
To compare the magnetic moments of the given two bar magnets using a deflection magnetometer
in Tan A position, by equal distance method.
Apparatus Required:
m1 tan 1
= (No unit)
m11 tan 11
Deflection magnetometer, two bar magnets, metre scale, etc.
Formula:
The ratio of the magnetic moments of two given short bar magnets where
mI = the magnetic moment of first magnet
mII = the magnetic moment of second magnet
I = mean deflection for first magnet
II = mean deflection for second magnet
Description:
The magnetometer is arranged in tan A position. For arranging tan A position the following
adjustments are to be made.
i) The magnetometer is placed on a horizontal table and the arms of the magnetometer are kept along
the east-west direction and it is adjusted so that the arms are parallel to the length of the aluminium
pointer.
ii) Without disturbing the adjustment, the compass box is slowly rotated until the pointer reads 0°
– 0°.
Procedure:
Equal distance method:
The magnetometer is arranged in tan A position. One of the two given bar magnets of magnetic
moment mI is placed on one side (Eastern arm) of the two arms with its axis along the east-west
direction in the magnetometer. The distance of the magnet is suitably adjusted so that the deflection lies
between 30° and 60°. The distance between the centre of the magnet and the centre of the
magnetometer is measured as'd'. The readings of both the ends of the pointer are taken as 1 and 2. The
magnet is reversed from pole to pole, in the same place and two more readings 3 and 4 are taken.
The magnet is then kept at the same distance on the other side of the arm (Western arm) and four
deflections  5, 6, 7 and 8 are noted, as mentioned above. The average of the eight deflections is found
and taken as I. The same procedure is repeated with the second magnet of magnetic moment m II,
keeping at the same distance and the average deflection 1I is found. The experiment is repeated for
different values of d and the readings are tabulated. The ratio of magnetic moments (mI/ mII) is
calculated using the formula.
Result:
i) The magnetic moments of the given two bar magnets are compared.
ii) The ratio of the magnetic moments of two given bar magnets by equal distance method =

137
Tan A position:

138
 40023 - ENGINEERING PHYSICS – II

UNIT I

HEAT

OBJECTIVES

▪ To understand the difference between the concepts of Heat and

Temperature

▪ To know the Measurement of Temperature

▪ To understand the Conduction, Convection and Radiation - Definitions and

Explanations - Properties of Thermal Radiation.

▪ To familiarise the properties of Thermal Radiation.

▪ To know the definition of Root Mean Square (RMS) velocity

▪ Can derive the expression for Pressure of an ideal gas

▪ Able to obtain the relation between Pressure and Kinetic Energy

▪ To solve simple problems based on the expression for pressure of a gas

▪ To explain specific heat capacity of a solids and liquids.

▪ To define Cp and Cv

▪ To derive the Mayer’s relation

▪ To calculate the value of Universal gas constant R

▪ Can solve problems based on Mayer’s relation.

139
 UNIT

1 HEAT

1.1 TRANSFER OF HEAT

Introduction:

Heat is a form of energy, which is responsible for the hot condition or the cold condition of a body.
Since it is a form of energy, the quantity of heat is measured in 'joule' with symbol 'J'.

We can realize the hotness or coldness of a body by our sense of touch. But there should be a method
to determine the degree of hotness or coldness.

Temperature is a measure of the relative degree of hotness or coldness of the body. For quantitative
measurement of temperature some standard temperatures are required, which can be easily reproduced
in any laboratory. Generally, we take the melting point of ice as lower fixed point and the boiling point
of water (steam point) as upper fixed point.

To measure the temperature, different scales are in use.

1. Centigrade scale :

In this scale, the temperature is measured in 'degree Celsius' with symbol °C. In this scale, the lower
fixed point is taken as 0 °C for ice point and the upper fixed point is taken a 100° C for steam point.

2. Fahrenheit scale :

In this scale, the temperature is measured in 'degree Fahrenheit' with symbol °F. Here, the lower
fixed point (ice point) is taken as 32 °F and the upper fixed point (steam point) is taken as 212 °F.

3. Absolute scale or Kelvin scale:

In this scale, the temperature is measured by the unit 'kelvin' with symbol 'K'. Usually, the
temperatures are measured in centigrade scale in laboratories. To convert the temperature from
centigrade scale to Kelvin scale, we have to add 273.15, because 0 K corresponds to -273.15 °C.

Ex: The room temperature = 32° C

= (32 + 273.15) K = 305.15 K

The relation between Celsius 'C', Kelvin 'K' and Fahrenheit 'F' is

C F−32 K−273
= =
100 180 100

140
Good and poor thermal conductors:

Substance  (Wm-1 K-1) Substance  (Wm-1 K-1)

Silver 420 Water 0.6

Copper 390 Red brick 0.6

Aluminium 240 Rubber 0.5

Steel 14 Wood 0.04 to 0.1

Lead 35 Glass 0.8

Concrete 1.1 Ebonite 0.2

We require both good and poor thermal conductors according to the need. Materials having
higher values of coefficient of thermal conductivity are termed as good thermal conductors and the
materials having lower values of coefficient of thermal conductivity are called as poor thermal
conductors or insulators. The presence of free electrons in metals helps the easy transfer of heat from one
part of the metal to the other part. So, all metals are good thermal conductors of heat. Some important
materials and their values of coefficient of thermal conductivity are listed above.
Coefficient of thermal conductivity:
Let us consider a metallic rod of cross sectional area A. Let the two ends be separated by a
distance d, maintained at temperatures 1, and 2, Let 1 be greater than 2. Heat flows from the end at the
higher temperature to the end at the lower temperature.

When the steady state is reached the quantity of heat Q conducted is

i) directly proportional to the area of cross section A
ii) directly proportional to the difference in temperature between the ends (1 - 2)
iii) directly proportional to the time for which the heat is conducted (t)
iv) and inversely proportional to the distance between the two ends.

141
 (θ1 − θ2 )t
Hence Q  A
d
or
(θ1 − θ2 )t
Q= A d

where  is a constant known as the coefficient of thermal conductivity of the material and
(1 - 2) / d is known as temperature gradient.
(θ1 − θ2 )
when A=1, = 1 and t = 1, then Q = 
d

Hence, Coefficient of thermal conductivity () of the material of a conductor is defined as the
quantity of heat conducted per second per unit area per unit temperature gradient at the steady state. unit
of  :
(θ1 − θ2 )t
Q = A or
d
Qd
 =
A (θ1− θ2 )t

Substituting the units,

Jm
Unit of  = m2Ks
= Js-1 m-1 K-1 = Wm-1 K-1 (as Js-1 is W)
Heat conversion:

The average body temperature of a material is 101.5 °F. What is this temperature in Celsius?
Solution:
The formula needed to convert Fahrenheit to Celsius is:

°C = 5⁄9(°F – 32)

where °F and °C are the temperatures in Fahrenheit and Celsius respectively.

Plug 101.5 °F into the formula

°C = 5⁄9(101.5 – 32)
°C = 5⁄9(69.5)
°C = 38.6 °C

Answer:
The average body temperature of the material in Celsius is 38.6 °C.
The boiling temperature of liquid oxygen at normal pressure at -182.96 °C. What is the value in
Kelvin Scale ?
K = ºC + 273.15
ºC= -182.96

Put the value of ºC in the formula : K = -182.96 + 273.15

Value in Kelvin Scale : K = 90.19 K
The temperature at which liquid oxygen boils under normal pressure is 90.19 K

142
Heat Transfer :
There are three methods of transfer of heat from one place to another. They are Conduction,
Convection and Radiation. Conduction takes place both in solids and fluids. Convection cannot occur in
solids. Radiation does not require the presence of any medium.
Conduction:
In Conduction, the molecules of the body are responsible for the heat transfer. Here there is no
actual movement of molecules from one place to another place. When a rod is heated at one end, the
molecules at the hot end vibrate about their mean position and transfer the heat energy to the neighbouring
molecules and thus the heat energy reaches the other end of the rod.
Conduction is the process by which heat is transmitted from one point to another without the actual
movement of the particles from their equilibrium positions. Conduction takes place in solids, liquids
and gases.
Convection:
In the case of heat transfer by convection, there is actual movement or motion of molecules of the
fluid ( liquid or gas ).The portions of the fluid that get warmed up by contact with the heat source, expand
and so move up through the body of the fluid due to the decrease in density. There is an inflow of cooler
molecules to take the place of heated mass of the fluid which has moved up. This circulatory motion of
the fluid mass by which heat is transferred from place to place is called Convection.
Convection is the process in which heat is transmitted from one place to the other by the actual
movement of heated particles.
Convection takes place only in liquids and gases. It cannot take place in solids.
Applications of Convection
1. The wind flow is due to the convection currents in the atmosphere. During day time, parts of
earth get heated by the Sun. As the air expands, it rises up and its place is taken by the flow of
air from colder areas.
2. The land breeze and sea breeze are due to the convection in the atmosphere. During day time,
land mass is heated to a higher level than the sea. So, the warm air over the land rises giving
place to the cool air flow from the ocean. This gives the sea breeze. During night time, the land
mass cools quickly than the water in the sea. So, cool air flows from land mass towards sea
which gives the land breeze.
Radiation:
For heat transfer by radiation, no medium is required. The heat from the Sun is received by earth
only by radiation. Both heat and light travel through space with same velocity and they are similar in
nature. Both are generated and transmitted in the same way. They are transverse waves and
electromagnetic in character, but they differ in wavelength and frequency. The thermal radiations are
detected in the 'infra-red'' region of the electro-magnetic spectrum.
Radiation is the process by which heat energy is transmitted from one place to another without the
aid of an intervening material medium.
Properties of thermal radiation :
The nature of thermal radiation is similar to that of light. Following are some of the properties of
thermal radiation.
1. Thermal radiation travels with the velocity of light, which is 3 × 108 ms-1.
2. Thermal radiation obeys the same laws of reflection, refraction etc., as light.
3. Thermal radiation travels through vacuum.
4. It obeys the law of inverse square, as light.

143
 5. It travels in straight lines.
6. When thermal radiation falls on anybody, which can absorb it, then converted into ordinary
heat, which raises its temperature.
7. It is absorbed by dark rough surfaces and reflected by light smooth surface.

Applications of Radiation

1. White coloured dresses are used in hot countries to keep the inside cool.
2. In some countries, shining aluminium sheets are used to cover the roof of the house to reflect
back the radiant heat and to keep the inside cool.
3. A cooking vessel is painted black at the bottom for greater absorption of heat, but polished at the
top to minimize radiation losses.
4. In cold countries, hot air or water runs through the pipes along the walls inside a building and
the radiant heat energy keeps the occupants warm by 'Central Heating'.
Applications of Heat
Heat and energy are quite related and are known to man before advancements in science. It has a
significant role in human lives and also the environment. Heat energy helps to
1. Water cycle
The water cycle is a phenomenon responsible for the rain and life on earth. When the heat from
the sun contacts water on the earth, it evaporates by forming water vapor. This evaporation mostly
happens from the oceans to form the clouds. These vaporous clouds, in turn, reach the land through the
sky, get cooled, and cause rains.
2. Cooking
Cooking is a daily routine at home, followed by the man to make his food. Cooking helps keep
the food tasty, easy to digest, and also kill any microorganisms which might be harmful to health.
3. Sterilization
This is a process to kill any microbes in drugs and other healthcare material. This aims to keep
the preparation sterile until use. So heat sterilization is the most common and highly effective method of
doing it.

144
 1.2 KINETIC THEORY OF GASES
A gas consists of a large number of particles called molecules. The molecules are in the state of
continuous motion. The pressure of a gas is due to the collision of individual molecules on the walls
of the container. The behaviour of the gas can be explained on the basis of kinetic theory of gases The
kinetic theory of gases has found better applications to explain various properties of gases. This theory
was developed by Clausius, Boltzmann and Maxwell
1. Postulates of Kinetic Theory of gases:-
1. A gas consists of large number of tiny, rigid particles called molecules.
2. The gas molecules are identical in all respects, like mass, size, etc.
3. The molecules are perfectly elastic spheres.
4. The molecules are in random motion and travel with all possible velocities in all possible
directions.
5. The molecules collide with each other and also with the walls of the container.
6. Between two successive collisions, the molecules travel in the straight line.
7. The average distance between two successive collisions is called the mean free path.
8. The time of collision is negligible when compared to the time taken by the molecule to travel
its mean free path.
9. The volume of the molecules is negligible when compared to the volume of the gas.
10. The force of attraction between the molecules is negligible.
11. As the molecules of a gas are always in random motion, the energy of the gas is in the form of
kinetic energy and it is directly proportional to the absolute temperature of the gas. When the
temperature of the gas remains constant, the molecular velocities will not change.
Mean square velocity and Root Mean Square velocity (R.M.S. velocity) :-
Let C1, C2, C3, C4 be the velocities of 1st, 2nd, 3rd …..nth molecule present in the gas, then the mean
square velocity is,
C12 + C2 2 + …. Cn 2
𝐶2 =
n

The mean square velocity is defined as the average of the square of velocities of all the molecules.
Now R.M.S. velocity is obtained by taking square root of the mean square velocity of the
molecules. R.M.S velocity is

2
√ 𝐶 2 = √ C1 + C2 2 + …. Cn 2
n

The R.M.S. velocity is defined as the square root of the average of squares of velocities of all
molecules
Expression for the pressure of a gas:
Consider a cubical vessel of side one metre containing a perfect gas. The gas contains n-number
of molecules and m be the mass of each molecule. The molecules are in continuous motion and they
collide with the wall of the container. The force experienced by the wall per unit area, due to the
molecular collision gives the pressure of the gas.
Force
Pressure = Area

145
 Consider a single molecule moving in the X-direction with velocity v. It collides with the wall
BCGF of the container and then it rebounds with velocity -v.
The momentum of the molecule before collision = mv
The momentum of the molecule after collision = - mv
The change in momentum for single molecule = mv - (-mv)
= 2mv.
After recoil from the wall BCGF, the molecule travels towards the wall AEHD and collide then
rebound back to the wall BCGF. So, the molecule travels 2 metre (1m + 1m) for every next collision on
the same wall.

Time required to travel 2m for next collision t = distance / velocity = 2/v second

The rate of change in momentum = change in momentum/ time taken

2mv
= = mv2
2/v
According to Newton's second law of motion,
Force = Rate of change in momentum.
 Force acting on the wall due to 1 molecule= mv2
An average n/3 molecules can be considered to move in each of the three directions
X, Y and Z.

The force acting on the wall due to n/3 molecules = n/3 ( mv2)
Pressure on the wall = Force / Area
(n / 3) mv2
=
1
{Since the area of any side A = 1 × 1 = 1 m2}
Since the pressure will be the same on all sides,
Pressure of the gas P = (⅓) nmv2

But all the molecules are not moving with the same velocity v.
Let C1, C2, C3 …………… Cn be the velocities of the molecules.

The mean square velocity C2 = (C12 + C22 + C32 ……. + Cn2) / n

Substituting this average square of the velocity C2 for v2

146
 Then, Pressure of the gas P = 1 nm 2
3 C

As we have taken m as the mass of each molecule and there are n molecules in the gas,
The total mass of the gas present in the vessel of 1 m3 volume = nm.
But, mass per unit volume of the gas = nm =  the density of the gas.
1
 Pressure of the gas P =  C2
3
Relation between Pressure and kinetic Energy of a Gas:
Consider a perfect gas having unit volume with density  and C 2 be the mean square velocity of
gas molecules, then

 pressure of the gas P = 3  C

2
1

1
 The Pressure , P = M C2
3

M
The density  = = M {Since volume V=1 m3}
V
Dividing and multiplying by 2,

2 1 2 1
P=  MC 2 =  MC2
2 3 3 2
The mean Kinetic energy per unit volume of the gas
1 2
K.E. = MC
2
2
The Pressure P = K.E
3

The pressure of a gas is equal to two third of the mean kinetic energy per unit volume of the gas.
But, the mean square velocity of the molecules is C2 is directly proportional to the absolute
temperature of the gas.
Hence, the pressure of a gas is directly proportional to the absolute temperature of the gas.
Relation between kinetic energy and absolute temperature of a gas.
The pressure of a gas is P = 1/3 c2
P = 1/3 c2
M M
P = 1/3 V c2 [ Density  = ]
V

3PV = Mc2
3 PV = nmc2 [ M = nm where n is the number gas molecule]

147
 The universal gas equation is PV = RT
3RT = nmc2
(multiply by ½ on both sides)
3 R
T = ½ mc2 [  N – Avogadro number ]
2 N

3
KT = K.E R
2 [ is the Boltzmann’s constant ]
N

Thus kinetic energy is 3/2 of Temperature.

WORKED PROBLEMS
1. Find the r.m.s. velocity of hydrogen molecule at S.T.P. if density of hydrogen is 0.0899 kgm -3
at S.T.P.
1
P = 3 𝑐 2
3𝑃
c =√


Pressure at S.T.P = 0.76 m of Hg

= 0.76 x 13.6 x103 x 9.8

= 101292.8 N/s2
 = 0.0899 kgm-3

3 ∗ 101292.8 303878.4
c =√ 0.0899
= √ 0.0899

= √3380182.425

c = 1838.527 ms-1

2. The root mean square velocity of hydrogen molecules is 1838 ms-1. Calculate the density of
hydrogen at N.T.P.

1
P = 𝑐 2
3

3𝑃
 =
𝑐2

3 ∗101292.8
=
18382

 = 0.0899 kgm-3

148
3. Find the temperature at which the root mean square velocity of hydrogen will double its
value at S.T.P. pressure remaining constant.

Let t-be the temperature at which the r.m.s. velocity of hydrogen will double its value at S.T.P.
Let C0 and Ct be the R.M.S Velocity of molecules at S.T.P and at temperature ‘t’ . i.e. ct = 2c0
Known, c  T C1  (273 + t) and Co  273

C1 = 2C 0 = (273 + t)

2 273 = 273 + t

Squaring both sides

4 × 273 = 273 + t
 t = 1092 - 273 = 819 K or 546° C

149
 1.3 SPECIFIC HEAT CAPACITY

When a substance is heated, it gains heat and its temperature increases. The quantity of heat gained
by different substances is specified in terms of the value known as specific heat capacity.
Specific heat of solid and liquid :
To increase the temperature of a substance, heat energy should be supplied to it. It is found that the
quantity of heat energy supplied to the substance (or) taken by the substance (Q) is directly proportional.
to
i) the mass of the substance (m) and
ii) increase in temperature of the substance (T2 – T1 )
where T1 initial temperature of the substance and T2 final temperature of the substance.
Q  m (T2 – T1)
(or) Q = S m (T2 – T1) (1)
Where S is the constant of proportionality and is called as the specific heat capacity of the
substance.
In equation (1) when m = 1 kg and (T2 – T1 ) = 1 K, then,
Q=S
 Specific heat capacity of solid or liquid (or substance) is defined as the quantity of heat energy
required to increase the temperature of 1 kg of the substance through 1 K.
Unit for specific heat capacity :
We know Q =Sm (T2 – T1)
Q
 S = m (T − T )
2 1
joule
Substituting the respective units in R.H.S, the unit for specific heat capacity is
kilogram-kelvin
with symbol J kg-1 K-1.
For example, the specific heat capacity of water is 4190 J kg-1 K-1. It means, to increase the
temperature of 1 kg of water through 1 K, 4190 J of heat energy is required.
Similarly, the specific heat capacity of copper is 385 J kg-1 K-1. It means, to increase the temperature
of 1 kg of copper through 1 K, 385 J of heat energy is required.
Specific heat capacities of a gas :
When a solid or liquid is heated, its temperature alone is increased, neglecting the little increase in its
volume. But, when a gas is heated, its pressure and volume change in addition to its temperature. So, to
study the effect of heating on a gas, we have to keep either the pressure or volume of the gas as constant.
Accordingly, for each and every gas, there are two specific heat capacities.
1. Specific heat capacity at constant pressure (Cp) :
The specific heat capacity of a gas at constant pressure is the quantity of heat energy required to
increase the temperature of one mole of the gas through 1K, when the pressure is kept constant.
2. Specific heat capacity at constant volume
v
(Cv) :
The specific heat capacity of a gas at constant volume is the quantity of heat energy required to
increase the temperature of one mole of the gas through 1K, when the volume is kept constant.
The unit for Cp and Cv is J K-1 mol –1

150
 Ratio of specific heat capacities () :
For a gas, the ratio of the specific heat capacity at constant pressure to the specific heat capacity at
constant volume is a constant. i.e.
Specific heat capacity at cons tan t pressure (C p )
=  (constant)
Specific heat capacity at cons tan t volume (C v )
Since Cp is always greater than Cv, the value of  is always greater than one for any gas.
Explanation for Cp greater than Cv :
The Specific heat capacity of a gas at constant pressure (Cp) is the heat energy required to increase
the temperature of 1 mole of a gas through 1K, at constant pressure.
The Specific heat capacity of a gas at constant volume (C v) is the heat energy required to increase
the temperature of 1 mole of a gas through 1K, at constant volume.
When a gas is heated, keeping the volume of the gas as constant, the entire heat energy supplied is
used only to increase its internal energy i.e. temperature
When a gas is heated, keeping the pressure of the gas as constant, the supplied heat energy is
simultaneously used for two purposes. A part of heat energy is used to increase the temperature of the
gas and rest of the energy is used to do the external work of pushing the piston up, to maintain the
constant pressure. Hence, in this case more heat energy is to be supplied to the gas. Thus Cp is greater
than Cv.
3. Ratio between Cp and Cv
The ratio of the specific heat capacity of gas at constant pressure to the specific heat capacity of gas
at constant volume is a constant and is denoted by the letter ()
Cp
Then,  =
Cv
When a gas is heated at constant volume the heat supplied to the gas is wholly used to raise the
temperature of the gas whereas when the gas is heated at constant pressure, a part is used to external
work to keep the pressure constant.
For monoatomic gases  = 5/3 for diatomic gases,  = 7/5 and for triatomic gases  = 4/3
Mayer's relation Cp - Cv = R (Relation between Cp and Cv)
Consider 1 mole of a perfect gas inside a cylinder in which there is a tight piston moving without
any friction.The internal energy of the gas is directly proportional to its absolute temperature. Let dQ1
be the heat energy supplied to the gas at constant volume. Consequently, the temperature increases
by dT.
We know, quantity of heat supplied =
mass × specific heat capacity × increase in temperature
So,dQ1 = 1 Cv dT (1)

dx

P P
GAS GAS

dQ 1 dQ2

Here, the entire heat is used to increase the temperature of the gas i.e. to increase the internal energy
of the gas and no external work is done.
151
 Let us now consider one mole of a perfect gas enclosed in a cylinder as before. Let the pressure and
volume of the gas be P and V1 respectively. Let A be the area of the piston. A quantity of heat dQ2 be
given to the gas at constant pressure, such that the increase in temperature is same dT.
Now d Q2 = 1 CP dT (2)
As the pressure of the gas is constant, the volume increases to V2.
So, some external work is done by the gas on the piston. Here the heat supplied is used for two
purposes.
A portion of the heat dQ2 is used to increase the internal energy and the rest is used to do external
work. i.e. dQ2 is used to
(i) to increase the temperature of the gas by dT and (ii) to push the piston up.
 dQ2 = dQ1 + work done in pushing the piston up.
dQ2=dQ1 + {Force x displacement of the piston } (3)
We know that, Pressure = Force / Area
 Force acting on the piston = Pressure × Area= P A
Let dx be the distance moved by the piston, as shown in fig (ii). Therefore (3) becomes,
dQ2= dQ1 + P A dx. But A dx = dV
where dV = V2 – V1 is the increase in volume of the gas.
dQ2 = dQ1 + P dV (4)
Using (2) & (1), (4) becomes,
1 CP dT=1 Cv dT+P dV
CP dT= Cv dT+P dV
(CP - Cv ) dT = P dV
The equation for a perfect gas is PV=RT
On differentiating, P dV = R dT
So, the above expression becomes
dT dT
(Cp – Cv). dT = R.dT ; (Cp – Cv) = R. ; But, =1
dT dT
 (Cp – Cv) = R
Note: Since value of the Universal gas constant R is positive, the value of CP is always greater than
Cv for any gas.

Value of Universal Gas Constant: (R)

Calculate the value of R from the gas equation PV = RT (OR) Calculate the universal gas constant
of given gas, given one gram molecule of at gas at N.T.P. occupies a volume of 22.4 litres.
AT N.T.P. Pressure P = 0.76m of Hg
= 0.76 × 13.6 × 103 × 9.8
Volume V = 22.4 litres
= 22.4 × 10-3 m3
Temperature T = 273 K
Gas equation PV = RT
R = PV/T
0.76  13.6  103  9.8  22.4  10−3
=
273
Universal gas constant R = 8.31 JK-1 per gram mole (or) R = 8310 JK-1 per kg mole

152
Worked Examples:
1. A gas at 1.5 atmospheres occupies a volume of 2m3 at 25o C. What will be the volume of
gas at 40o C and at 3 atmospheres?
PV
Gas equation, PV = RT (or) =R
T
P2 V2
P1 V1 =
Also 𝑇2
𝑇1

P1 = 1.5 Atm.pre., T 1 = 25°C + 273 = 298K

P2 = 3 Atm.pre., T2 = 40°C + 273 = 313K
And V1 = 2 m3
V2 = (P1 V1 T2 )/P2 T1 = (1.5 × 2 × 313) / (3 × 298)
= 1.05 m3
2. The ratio between two specific heats of a gas is 1.36. The specific heat capacity of the
gas at constant volume is 23094 JK-1 per kg mole. Find the value of Universal gas constant.
Cp / Cv = 1.36 and Cv = 23094 JK-1 per kgmole
CP = 1.36 × Cv = 1.36 × 23094
= 31407.84 JK-1 per kg mole.
By Mayers relation CP - CV = R
i.e., R = 31407.84 - 23094
R = 8313.84 JK-1 per kg mole
3. Find the specific heat capacity at constant pressure of helium if Cv = 12525 JK-1 per kg mole.
One gram molecule of the gas occupies 22.42 litres at NTP Given g = 9.8ms-2 and density of
mercury = 13600 kgm-3

Given: CV = 12525 JK–1 per kg mole

V = 22.42 litres 22.42 × 10-3 m3
At. N.T.P., P = 0.76 × 136000 × 9.8 and T = 273 K
Gas equation PV = RT
PV 0.76  136000  9.8  22.42  10−3
R = =
T 273

= 8.313JK-1 per gm mole

= 8313 JK-1 kg mole-1

By Mayer's relation, Cp - Cv = R

CP = R + CV
= 8313 + 12525
i.e., CP = 20838 JK–1 per kg mole

153
4. The ratio of the specific heats of methane is 1.313. The specific heat capacity at constant
pressure is 35200 JK-1 per kg mole. Calculate the value of specific heat at constant volume
and hence to calculate the value of gas constant.

𝐶𝑃
Given: =  = 1.313 CP = 35200 JK-1
𝐶𝑉

Calculate CV and R:

𝐶𝑃
CV = 

35200
= 1.313

CV = 26808.834 JK-1 per kg mole

CP – CV = R = 35200 – 26808.834

R = 8391.165 JK-1 per kg mole

5. The ratio of two specific heats of Ammonia is 1.31 and universal gas constant is 8314 JK-1
per kg mole. Calculate the specific heat of the gas at constant pressure and at constant
volume.

The ratio, Cp / Cv = , CP =  CV
By Mayer's Relation, CP - CV = R, CP = R + CV
Then R + CV =  CV
R =  CV - CV = CV ( - 1)
R
CV = = 8314 / (1.31- 1)
( – 1)
i.e.,CV = 26819 JK-1 per kg mole
CP = R + CV = 8314 + 26819
i.e., CP = 35133 JK-1 per kg mole

154
 QUESTIONS
Part -A and Part – B
1. What is Heat?
2. What is temperature?
3. What are the three important scales of temperature?
4. Define Conduction.
5. Define Convection.
6. Define Radiation.
7. Give the relation between pressure and kinetic energy of a gas.
8. Define mean square velocity of molecules.
9. Define Root Mean Square velocity of molecules.
10. Give the relation between two specific heats of a gas with universal gas constant.
11. State any two postulates of kinetic theory of gases.
12. Define mean free path.
13. Define specific heat capacity of a solid.
14. Define specific heat capacity of a liquid
15. Define specific heat capacity of a gas at constant pressure.
16. Define specific heat capacity of a gas at constant volume
17. Define ratio of specific heat capacities of gases.

Part –C
1. Explain the process of conduction, convection and radiation.
2. List the properties and uses of thermal radiation.
3. List the postulates of kinetic theory of gases.
4. Derive an expression for the pressure of a gas on the basis of Kinetic theory of gases.
5. Derive the relation between pressure and kinetic energy.
6. Derive Mayer's relation between two specific heat capacities Cp and Cv of a perfect gas (or)
Derive Cp – Cv = R, for a perfect gas.
7. Calculate the value of universal gas constant from the equation PV = RT.
Exercise problems:
1. Find the RMS velocity of hydrogen molecules at S.T.P., if the density of hydrogen is 0.0899
kg m-3 at S.T.P. (Ans: 1838.52 m s -1)

(Hint: - S.T.P. means standard temperature and pressure Standard pressure = 1 atmospheric
pressure = 76 cm of mercury P = hρg = 0.76 × 13,600 × 9.81 = 101396.16 N m-2)

2. At atmospheric pressure, the density of Nitrogen gas is 1.25 kg m-3. Find the R.M.S. velocity of
Nitrogen molecules. (Ans: 493.05 m s-1)

155
 3. The density of carbon dioxide gas at S.T.P. is 1.977 kg m-3.Find the R.M.S. velocity of carbon
dioxide molecules. (Ans: 392.05 m s-1)
4. Calculate R.M.S. velocity of air molecules at N.T.P., if the density of air is 1.29 kg m-3.
(Ans: 485.35 m s-1)
5. The density of Helium gas at S.T.P. is 0.1786 kg m-3. Calculate the RMS velocity of Helium
molecules. (Ans: 1304.4 m s-1)
6. The molecular weight of carbon di-oxide is 44.Calculate the RMS velocity of CO2
molecules at N.T.P. (Ans: 393.32 m s-1)

7. The specific heat capacity of a gas at constant volume is 23,094 J K-1 kg mol-1.The ratio of
specific heat capacities of the gas is 1.36.Calculate the specific heat capacity of the gas at constant
pressure and also the value of the universal gas constant R for that gas.
( Ans: 8313.84 J K-1 kg mol-1)

8. The specific heat capacity of Helium gas at constant volume is 12,525 J K-1 kg mol-1. Find the
specific heat capacity of the gas at constant pressure. 1 gram molecule of the gas occupies
22.4 litres at N.T.P. ( Ans: 20836.2 J K-1 kg mol-1)

9. The ratio of specific heat capacities of Helium is 1.667.Calculate its specific heat capacity at
constant volume and at constant pressure.Given R = 8.3151 J K-1 gm mol-1.
(Ans: C = 12.430 J K-1 gm mol-1 & C = 20.72 J K-1 gm mol-1)

10. Calculate the value of gas constant for nitrogen if its density is 1.251 kgm-3 at N.T.P. (Molecular
mass of nitrogen is 28g.) (Ans: R = 8.3048 JK -1 gm mol-1)
Brain Teasers:
1) Calculate the value of gas constant for hydrogen, given density of hydrogen at NTP is 0.09 kgm-3
and molecular mass is 2 g (Ans: R = 8.140 JK-1 gm mol-1)
2) The ratio of the specific heats of methane is 1.313. The specific heat capacity at constant
pressure is 35200 JK-1 per kg mole. Calculate the value of specific heat at constant volume and hence
calculate the value of gas constant. (Ans: C =26808.8 JK-1 kgmol-1 R= 8911JK-1 kg mol-1)

https://www.youtube.com/watch?v=xzECamQRUVE&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=1

https://www.youtube.com/watch?v=qi6BPJmE1CU&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=2

https://www.youtube.com/watch?v=qb8DyQuDurg&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=3

https://www.youtube.com/watch?v=1R8vrwibfNo&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=4

156
 SUMMARY

• Heat is a form of energy

• Temperature is a measurement

• Relation between different scales of temperature is

𝐶 𝐹 − 32 𝐾 − 273
= =
100 180 100
• Three ways of heat transfer are Conduction, Convection and Radiation

• Properties of Thermal Radiation

• Average distance between two successive collision is mean free path

• Root Mean Square (RMS) velocity

1
• Expression for the pressure of a gas P = 3 c2

• Specific heat capacity of a solid and liquid

• Specific heat capacity of air, Cp and Cv

• Mayer’s Relation (Cp – Cv = R)

• Problems based on Pressure of a gas and Mayer’s Relation.

157
 UNIT II

THERMODYNAMICS, LIQUEFACTION OF GASES

AND NON CONVENTIONAL ENERGY

OBJECTIVES

▪ To define the First law of thermodynamics

▪ To understand the difference between Isothermal and Adiabatic changes

▪ To know the equations for isothermal and adiabatic changes

▪ To define the Clausius and Kelvin’s statements of Second law of

thermodynamics.

▪ To know the definitions of critical temperature, critical pressure and critical

volume.

▪ Can explain the process of liquefaction of oxygen by Cascade method.

▪ To define Joule Thomson effect and Temperature of inversion.

▪ Can explain the liquefaction of air by Linde’s process.

▪ To familiarise the Non-renewable and Renewable energy sources.

▪ To understand the working of solar energy, wind energy systems.

▪ To know the advantages and disadvantages of renewable energy

▪ To solve simple problems based on equations P1V1 = P2V2 and

P1V1= P2V2.

158
UNIT
THERMODYNAMICS, LIQUEFACTION OF
2 GASES & NON-CONVENTIONAL ENERGY

2.1 THERMODYNAMICS

Introduction:
Energy exists in various forms in nature. They are mechanical energy (kinetic and potential), heat
energy, electrical energy, radiant energy, chemical energy, etc.
In a hydroelectric power station, mechanical energy is converted into electrical energy. In a steam
engine, heat energy is converted into mechanical energy. When a metal piece is hammered mechanical
energy is converted into heat energy.
Thermodynamics deals with the conversion of heat energy into mechanical energy and vice versa.
But in a mere general sense, it includes the relation of heat to other forms of energy like electrical energy,
chemical energy, etc. In our discussions we will be concerned mainly about the conversion of heat energy
into mechanical energy.
The science of thermodynamics is guided by the three laws of thermodynamics which will be
discussed here.
First Law of Thermodynamics:-
When a certain quantity of heat is supplied to a system, the whole of the heat energy may not be
converted into work. Part of the heat may be used in doing external work and the rest of the heat might
be used in increasing the internal energy of the molecules. Let the quantity of heat supplied to the system
be Q, W is the work done by the system and U is the change in internal energy of the system.
Q = W + U
Hence the first law of thermodynamics states that the amount of heat energy supplied to a
system is equal to the sum of the change in internal energy of the system and the work done
by the system.
According to First Law of Thermodynamics, heat energy can be converted into mechanical work
and vice versa.
ISOTHERMAL AND ADIABATIC CHANGES:
In general, when a quantity of heat is given to the gas a part of the heat is used to increase the internal
energy of the gas and the rest is used to do external work by the gas. Let dQ be the quantity of heat given
to the gas, dE the increase in internal energy and W the external work done by the gas. Then
dQ = dE + W
This equation is called Gas - energy equation.
Isothermal Change
Any change in the pressure and volume of a gas at constant temperature is called as isothermal
change.
Consider a gas inside the cylinder with a piston moving in it without friction. When the gas is
compressed slowly, the temperature of the gas increases due to the work done on the gas. If the cylinder
and the piston are made up of conducting materials then the gas gives out the excess heat to the
surroundings. Hence there is no rise in temperature.

159
 GAS

When the piston moves upwards slowly, the gas expands and the temperature of gas decreases due
to the work done by the gas. Then the gas gained shortage of heat from the surroundings. Hence the
temperature and therefore there is no change in internal energy. The gas- energy equation for Isothermal
change becomes,
dQ = change in internal energy + External work for isothermal change, dQ = 0 + W
For a perfect gas, isothermal change can be represented by the equation,
PV = a constant Also P V = P V
1 1 2 2

2. Adiabatic Change :

When the pressure and volume of a gas change, but no heat is allowed to enter or leave the gas,
this change is called as adiabatic change.
Consider a gas inside the cylinder with a piston moving in it without friction. When the gas is
compressed suddenly, the temperature of the gas is increased and the work is done on the gas.
Similarly when the piston moves upwards suddenly, the temperature of the gas is decreased and work
is done by the gas. Due to the sudden compression or expansion, the gas neither gives out heat to the
surroundings nor receives heat from the surroundings, when the cylinder and the piston are made up of
non conducting materials.
During the adiabatic change, no heat leaves or enters the gas. That is dQ = 0. The work is done at the
cost of internal energy. The gas energy equation, for the adiabatic change becomes,

GAS

dQ = Change in internal energy + External work

For a perfect gas, adiabatic change can be represented by the equation,
PV =a constant Also P1V1 = P 2V2
 −1
The equation relating the temperature and volume of the gas is T1 V1 = T2 V2  −1

The equation relating the temperature and pressure of the gas is T  P 1−  = T  P 1− 

1 1 2 2

Where  = CP / CV is the ratio of the two specific heat capacities of the gas.

160
 WORKED PROBLEMS
1. A gas at 2 ATP is compressed to half of its original volume. Calculate the final pressure, if the
compression is (i) isothermal and (ii) adiabatic ( = 1.4)
Let V be the original volume

Given; P1 = 2 atmosphere, V1 = V and V2 = ½ V

i) For Isothermal change, the equation is P1V1 = P2V2

𝑃 1 𝑉1 𝑃 1𝑉1 2xVx2
P2 = = 1 = = 4 atmospheric pressure
𝑉2 𝑉 𝑉
2

ii) Adiabatic equation is P1V1 = P2V2


𝑃1 𝑉1 𝑉
P2 =  =2x 𝑉 = 2 × 2 = 2 × 21.4
𝑉2
2
= 5.278 Atmospheric Pressure.

2. Air at a pressure of 0.75 m of mercury and of volume 1 litre is compressed to a pressure of

1.5 m of mercury under isothermal process. Calculate the resulting volume.
The equation for Isothermal change is P1V1 = P2V2

Given; P1 = 0.75m of Hg

P2 = 1.5m of Hg
V1 = 1 litre
 V2 = ( P1V1) / P2= ( 0.75 × 1 ) / 1.5 = 0.5 litre
3. A certain mass of gas at 3 atmosphere is compressed adiabatically to half of its volume.
Calculate the resulting pressure if  = 1.4
Given; V1 = V, V2 = ½ V and P1 = 3 atmosphere pressure

P2 =P1 V1 / V2

P2 = P1V  / (½V ) 
P2 = (3 × V1.4) / ( ½ V)1.4 = 3 × 21.4 = 7.92 atmosphere pressure

4. The ratio of the specific heats of a gas is 1.5. It is initially at one atmospheric pressure. The gas
is compressed to half of its original volume. Calculate the final pressure, if the compression is
i) isothermal and ii) adiabatic.
Let V be the initial volume. Final volume is V/2.
Initial pressure P1 = 1 atp
𝐶
Ratio of specific heats, 𝐶𝑃 =  = 1.5
𝑉
i) Isothermal:
P1V1 = P2V2
𝑃1 𝑉1 1 ×𝑉
P2 = = 1 = 2
𝑉2 𝑉
2

final pressure, P2 = 2 atp.

161
 ii) Adiabatic:
  𝐶𝑃
P1V1 = P2V2 where  =
𝐶𝑉
𝑃1 𝑉1 𝑉1 𝛾
 P2 = = 𝑃1 ( )
𝑉2 𝑉2


𝑉
= 𝑃1 (1 ) = 𝑃1 x 21.5 = 1 x 2.828 = 2.828
𝑉
2

final pressure, P2 = 2.828 atp.

5. The ratio of the specific heats of a gas is 1.4. One atmosphere of the gas is compressed
adiabatically to one-fifth of its original volume. Find its final pressure.

Let V be the initial volume. Final volume is V/5.

Initial pressure P1 = 1 atp

Ratio of specific heats, 𝐶𝑃 =  = 1.4

𝐶
𝑉

 
where  =
𝐶𝑃
P1V1 = P2V2
𝐶𝑉

𝑃1 𝑉1 𝑉 𝛾
 P2 =  = 𝑃1 (𝑉1)
𝑉2 2


𝑉
= 𝑃1 (1 ) = 𝑃1 x 51.4 = 1 x 9.518 = 9.518
𝑉
5

final pressure, P2 = 9.518 atp.

6. Air at 5 atmospheres is suddenly compressed to half of its volume. Find the resulting pressure
of the air. ( = 1.4)

Let V be the initial volume. Final volume is V/2.

Initial pressure P1 = 5 atp

𝐶
Ratio of specific heats, 𝐶𝑃 =  = 1.4
𝑉

  𝐶𝑃
P1V1 = P2V2 where  =
𝐶𝑉
𝑃1 𝑉1 𝑉1 𝛾
 P2 = = 𝑃1 (𝑉 )
𝑉2 2


𝑉
= 𝑃1 (1 ) = 𝑃1 x 21.4 = 5 x 2.639 = 13.195
𝑉
2

final pressure, P2 = 13.195 atp.

162
7. A certain mass of gas at one atmospheric pressure and 27C is suddenly compressed to one-
fourth of its original volume. Calculate the resulting temperature and pressure ( = 1.4).

Let V be the initial volume. Final volume is V/4.

Initial pressure P1 = 1 atp and T1 = 27C = 27+273 = 300 K

𝐶𝑃
Ratio of specific heats, =  = 1.4
𝐶𝑉

𝑃1 𝑉1 𝑃2 𝑉2
PV = RT hence =
𝑇1 𝑇2

 
We know, P1V1 = P2V2

𝑃1 𝑉1 𝑉1  𝑉
P2 = = 𝑃1 (𝑉 ) = 𝑃1 (1 ) = 𝑃1 x 41.4 = 1 x 6.964 = 6.964
𝑉2 𝛾 2 𝑉
4

final pressure, P2 = 6.964 atp.

Also we have a relation, T1V1-1 = T2V2-1 ( for adiabatic process, TV -1 = a constant)
𝑇2 𝑉 −1
= (𝑉1)
𝑇1 2

Here, V1 = V , V2 = V/4 and T1 = 300 K

−1
𝑇2 𝑉
We get, 300
= (1 ) = 4 (1.4-1) = 1.741
𝑉
4

T2 = 1.741 x 300 = 522.3

final temperature T2 = 522.3 K = 249.3 C

Second Law of Thermodynamics:-
A heat engine is mainly used for the conversion of heat energy into mechanical work. A refrigerator
is a device used to cool a certain space below the temperature of its surroundings. The first law does not
contradict the existence of 100% efficient heat engine or a refrigerator, which are not attainable in
practice. These phenomena led to the formulation of second law of thermodynamics. There are two
versions of this second law of thermodynamics.
a. Clausius Statement :
It is impossible for a self acting mechanism working in a cyclic process unaided by an
external agency to transfer heat from a body at a lower temperature to the body at a higher
temperature.
This part of the law is applicable in the case of ice plants and refrigerators. i.e., Heat itself cannot
flow from a body at a lower temperature to a body at a higher temperature, on its own.
b. Kelvin's Statement:
It is impossible to derive a continuous supply of work from a body by cooling it to a
temperature lower than that of its surroundings.

163
 This part of the law is applicable in the case of Heat engines. In a heat engine the working
substance does some work and rejects the remaining heat to the sink. The temperature of the source
must be higher than the surroundings. The engine will not work when the temperatures of the source
and the sink are same.
Carnot's Engine:
The Carnot Engine is an ideal engine, operating between a source and a sink is free from friction
and the other imperfections of actual engines. Which absorbs all its heat at a constant high temperature,
works in a cycle of operations which is perfectly reversible and rejects remaining heat at constant low
temperature
A Carnot's Engine consists of the following parts.
1. Source:
A source is a hot body at a constant temperature T1K. The heat engine can draw heat from the
source.
2. Sink:
This sink is a cold body at constant temperature T2 K in which any amount of heat can be rejected.
3. Working substance:
The working substance is an ideal gas enclosed in a cylinder piston arrangement. The cylinder has
perfectly non-conducting walls but a perfectly conducting bottom fitted with a perfectly non conducting
and frictionless piston.

A perfectly non-conducting stand is provided so that the working substance can undergo adiabatic
processes.
Working of Carnot Engine (Carnot's Cycle):
The Carnot Engine is based on a reversible cycle of four operations.

Pressure Volume Indicator diagram

1. The working substance is placed on the source (T1). Now the gas expands isothermally. In this
processes gas does work on the piston whence its temperature fall which is immediately taken from
the source. The pressure of gas decreased from P 1 to P2 and volume of gas increased from V1 to V2.
This isothermal expansion of gas at constant temperature T1 is represented by AB on the indicator
diagram.

164
2. The cylinder is removed from the source and placed over the insulating stand. Now, the gas is thermally
isolated and allowed to undergo an adiabatic expansion at the expense of its internal energy till
temperature fall to T2. In this adiabatic expansion pressure and volume of gas changes from P 2, V2 to
P3, V3 which is represented by BC of the indicator diagram
3. The working substance at C is unable to perform work (P 3 very low) hence it is brought back to the
initial state A. For this cylinder is kept on the sink and the piston is moved down so that the gas is
compressed from P3, V3 to P4, V4. Heat produced due to compression passes to the sink. This isothermal
processes is represented by CD of the PV diagram. Here, work done is equal to the Heat rejected.
4. At this stage D, the cylinder is again kept over the insulating stand and compressed adiabatically so
that the initial state is attained.
Efficiency:
The efficiency of the Carnot's engine depends on the temperature of the source (T1) and the
temperature of the sink (T2).
Output Heat Converted into Work
Efficiency  = =
Input Heat received from the Source

Application of thermodynamics
1. Used in all refrigerators, deep freezers and industrial refrigeration systems.
2. Used in all types of air-conditioning systems.
3. Used in heat pumps and all types of air and gas compressors, blowers and fans.

165
 2.2. LIQUEFACTION OF GASES
The work on the liquefaction of gases started in 1823 by Faraday. Gases like chlorine, hydrogen
chloride, hydrogen sulphide, sulphur dioxide, ammonia and carbon dioxide were liquefied under high
pressure with the aid of simple freezing mixtures. But this method cannot be used to liquefy oxygen,
nitrogen, hydrogen and helium.
In 1863, Andrews had discovered the phenomenon of critical state of the gas. According to
Andrew's discovery, a gas cannot be liquefied when it is at above a particular temperature. That particular
temperature is called critical temperature. The pressure and volume of the gas corresponding to that
temperature are called the critical pressure and critical volume. After Andrew's discovery, the
liquefaction was done easily by cooling the gas below its critical temperature, even though the
process is difficult to liquefy oxygen, nitrogen, hydrogen and helium since they have very low critical
temperatures. For example, the critical temperature for oxygen is - 119°C and for hydrogen is - 240°C.

Then in 1877, Pictet liquefied the oxygen by cascade (step by step) process in which low
temperatures are obtained in step by step by the evaporation of suitable liquids. In 1896, Linde performed
a process to liquefy air by using Joule - Thomson effect.

Critical constants:
The critical temperature, critical pressure and critical volume are called critical constants Tc.
1. Critical Temperature (Tc) : The critical temperature of gas is defined as that temperature above
which the gas cannot be liquefied however great the pressure applied may be.
2. Critical pressure (Pc) : The critical pressure is the minimum pressure required to liquefy a gas
when it is at its critical temperature.
3. Critical volume (Vc) : The critical volume is the volume of unit mass of a gas when it is at its
critical temperature and critical pressure.
Note: Below the critical temperature the gas is termed as vapour and above the critical
temperature it is termed as gas. The gas can be liquefied easily when it is in the state of vapour. Gases like
oxygen, nitrogen, hydrogen and helium are termed as permanent gases since they have very low critical
temperatures.
Cascade Process - Liquefaction of oxygen:
The critical temperatures for oxygen -119°C and critical pressure is 49.7 atm.
Principle: When a liquid is allowed to evaporate under reduced pressure, it produces high cooling.
The apparatus arrangement used in this process is shown in the figure. It consists of three narrow
tubes. A,B and C enclosed by three outer jackets P, Q and R respectively. The narrow tubes and the outer
jackets are linked with the compression pumps P1, P2 and P3 as shown in the figure. The methyl chloride

gas of critical temperature 145° C is compressed by the pump P1 through the tube A. It is cooled by the
cold water circulating in the jacket P. Here the methyl chloride reaches the temperature lower than its
critical temperature. Then it is liquefied under high pressure. The liquid methyl chloride is collected in
the jacket Q and evaporates under reduced pressure lowering the temperature to -90° C.
The ethylene gas of critical temperature 10° C is compressed by the pump P through the tube B.
2
It is cooled to -90° C by liquid methyl chloride. Then it is liquefied under high pressure. The liquid
ethylene is collected in the jacket R and evaporates under reduced pressure lowering the temperature to
-160° C.

166
 The oxygen gas of critical temperature - 119° C is compressed to 50 atmospheric pressure by the
pump P3 and passed through the tube C. It is cooled to - 160° C by liquid ethylene in R. Then it is
liquefied and the liquid oxygen is collected in the Dewar flask D.
Disadvantage: The lowest temperature obtained in this process is -160° C. Therefore the cascade
process cannot be used to liquefy hydrogen and helium whose critical temperatures are lower than
- 160° C.
3. Joule - Thompson Effect:
Using the principle of Joule - Thompson effect, air, hydrogen and helium were also liquefied. Joule
performed an experiment to study the effect of change in the temperature of a gas when it is allowed to
pass through a porous plug from the high pressure side to the low pressure side.

Joule - Thompson effect : When a gas is allowed to expand from a high pressure side to a low
pressure side through a porous or nozzle or jet then it produces intense cooling, if the temperature
of the gas is less than the temperature of inversion. This effect is called Joule - Thompson effect.

Temperature of inversion : If a gas is at a particular temperature no cooling or heating is observed

when it is passed through the porous plug. If the temperature of the gas on the high pressure side is more
than that temperature, heating will takes place instead of cooling. That particular temperature is called
the temperature of inversion.
Temperature of inversion may be defined as the temperature at which there neither cooling nor
heating due to Joule Thomson effect.
4. Linde's Process - Liquefaction of Air:
The principle of Joule - Thompson effect is used in this process. The apparatus arrangement used in
this process is shown in the figure. It consists of two compressors C1 and C2, a cold water jacket, freezing
mixture jacket, narrow tube having nozzle N1 and a wide tube having nozzle N2 .

Air is compressed to 25 atmospheres by the compressor C1 . The compressed air is passed through
a tube surrounded by a cold water jacket. Again this air is compressed to 200 atmospheres by the
compressor C2. This compressed air is passed through a spiral tube surrounded by a freezing mixture
jacket. Now the air reaches to a temperature of -20° C. This cooled air at high pressure is allowed to
expand through a nozzle N1 and it is further cooled by Joule - Thompson effect. This cooled air is
circulated back to the compressor C2 and is compressed again.

167
 As the process continues, after a few cycles, the air gets cooled to a sufficiently low temperature.
Finally the air expands from high pressure side to the low pressure side through the nozzle N2 .
Again the temperature is reduced by Joule - Thompson effect and the air gets liquefied. The liquid air
is collected in the Dewar flask. The unliquefied air is again circulated back to the compressor C 1 and the
process is repeated. The whole apparatus arrangement is packed in cotton or thermal wool to avoid any
conduction and radiation.

168
 2.3 NON CONVENTIONAL
ENERGY
Energy is an important input for all sectors of a country's economical development. For India,
the energy needs is increasing rapidly day by day. It is a well-known fact that the currently available
conventional energy resources like coal, oil, natural gas, etc., are depleting very fast.
54% of our energy need is provided by coal, which may be available for another 80 years at the
current rate of usage. 32% of energy need is fulfilled by oil, which may last up to 22 years and the rest
14% is met by hydro, nuclear, natural gas, etc. So, we will have to increasingly depend on other non
conventional energy resources like solar, wind, tidal, geothermal, hot dry rock, bio mass, etc.
The sources of energy utilised by human beings fall into two categories:
1. Non-renewable energy resources are sources of power that cannot be replaced once they are used,
because the energy source has taken millions of years to form (ex. coal, oil).These energy resources are
also called as finite energy resources.
2. Renewable energy resources are sources of power that quickly replenish (fill again)
themselves and can be used again and again repeatedly (ex. Wind, tidal). These energy resources are also
called as infinite energy resources.
Conventional or Non – renewable sources of energy:
The conventional or non – renewable energy sources are classified as primary and secondary energy
sources. The primary energy sources are which provide a net supply of energy, like coal, oil, natural gas,
uranium, etc. These fuels are called primary fuels, from which the energy yield ratio is very high. i.e.,
the energy required to obtain these fuels is much less than what they can produce by combustion or
nuclear reaction. The secondary fuels like wood, animal wastes, agricultural wastes, etc., produce no net
energy.
Non – conventional or renewable sources of energy:
To meet our future energy requirements, we have to tap other available non conventional or
renewable energy sources. The renewable sources of energy are also called as non conventional (or) al-
ternate sources of energy. They are Solar, Wind, Tidal, Geothermal, Bio mass, Hot dry rock, Chemical,
etc. Various technologies are designed and developed to tap these earth's natural energy resources which
are renewable. They are called renewable because, they are available in plenty and freely.
Solar energy:
The sun is a large sphere of very hot gases, the energy being generated by various kinds of nuclear
fusion reactions.
Sun gives us 1000 times more power than we need. Therefore, solar energy can be a major source of
power. Even if we could tap one thousandth of solar power, that will meet our requirements. But so far
it could not be developed on a large scale.
Energy produced and radiated by the sun, which reaches the earth, is termed as “Solar energy”.
Energy is radiated by the sun, as electromagnetic waves. Solar energy, received in the form of electro
magnetic radiation, can be converted directly or indirectly into other forms of energy, such as heat and
electricity, which can be utilised by human being. Since the sun can radiate energy at constant rate for a
few billion years, it may be regarded as an inexhaustible source of useful energy.
As India is situated near equator, where sunlight is abundant for a major part of the year, utilising the
solar energy to a greater extent is the best option available to us.

169
 Some important applications of solar energy are
(i) Solar water heating, (ii) Heating and cooling of residential buildings, (iii) salt production by
evaporation of sea water, (iv) Solar cookers, (v) Solar furnaces, (vi) Solar pumps for water pumping and
(vii) Solar electric power generation.
From solar energy electricity can be produced with the help of photovoltaic or solar cells. The
solar cells directly generate electricity by photovoltaic process. When the photons from the sun's light is
absorbed by special type of semi conducting material they create free electrons.
These free electrons induce higher energy electrons to flow out of the semiconductor. The flow of
higher energy electrons in the circuit constitutes the electric current.

BOILER
REFLECTOR
PARABOLIC

HELIOSTAT
MIRROR

Solar water heating system

This system is used to drive turbo-generators to produce electricity. A plant of this type is operating
in France and produces 1000 KW.
2. Wind energy :

Wind is basically caused by the solar energy irradiating the earth. Energy of wind can be
economically used for the generation of electrical energy. Wind is caused due to two main factors:
(a) Due to air in motion. Air motion arises from a pressure gradient. The surface winds flow from the
poles toward the equator. Solar radiation heats the air near the equator, and this low density heated air
buoyed up. At the surface, it is displaced by cooler, more dense higher pressure air flowing from the
poles. In the upper atmosphere near the equator the air tend to flow towards the poles and away from
the equator. The net result is a flow of surface winds from north to south in the northern hemisphere.
Similarly the flow of surface winds is from south to north in the southern hemisphere of the
earth.
(b) Due to the rotation of the earth with respect to atmosphere, and its motion around the sun.

Wind energy which is an indirect source of solar energy conversion can be utilized to run windmill,
which in turn drives a generator to produce electricity. Wind can also be used to provide mechanical
power, such as for water pumping.

170
 Some characteristics of wind energy are stated below:
i) It is a renewable source of energy.
ii) Wind-power systems are non-polluting, so it does not affect the environment.
iii) Wind energy systems avoid fuel provision and transport.
iv) Large areas are needed to install wind forms for electrical power generation and wind energy
needs storage means because of its irregularity in the speed of flow.
v) Manufacture and implementation of wind farms can be costly.

As our country has lot of wind throughout the year, the energy in the wind can be used to generate
power that can be used for domestic and industrial uses. The wind mill is the form in which we can
convert the wind energy to electricity. The potential of wind energy is of the order of 1.6 × 107 MW.
The wind mill generally consists of a turbine (a dynamo which can produce electricity when rotated)
connected to large rotating blades. The blades rotate by using the wind.

canopy
containing blade
generator

Tower
Foundations

The setting up cost of these windmills is high but the running cost is very low. Use of this would help
in producing cheap power. Tamilnadu has set up a large number of wind mills in the state especially in
places like Tiruppur, Muppandal in Kanyakumari, Coimbatore and their surroundings.
Tidal Energy:
Tide is a periodic rise and fall of the water level of sea due to gravitational effect of the sun and
moon on the water of the earth. Due to fluidity of water mass, there is a periodic rise and fall in levels of
sea water which is in rhythms with the daily cycle of rising and setting of sun and moon. As the earth
rotates, the position of a given area relative to the moon changes, and thus a periodic succession of high
and low tides occur. This periodic rising and falling of the water level of sea is known as tidal power.

171
Advantages of renewable energy:

1. Use of renewable energy resources is less pollutant.

2. Energy from renewable resources can be produced quickly.
3. Several renewable options are financially and economically competitive for certain
applications in remote locations, where the costs of transmitting electrical power or transporting
conventional fuels are high.
4. The plants which use renewable sources of energy can be built nearer to the site, where energy
is needed, and reducing transportation cost of conventional fuels.
5. They can be produced in large numbers and introduced quickly, unlike large power stations
which take often 10 years or more to become operational.
6. They can be built on, or close to the site where the energy is required. This minimises
the transmission costs.
7. The important advantage is the availability in plenty, almost inexhaustible.
Disadvantages of renewable energy :
1. Even a bad weather can interrupt the supply of electricity produced by wind, tidal, solar energy,
etc.
2. The electrical energy produced by these renewable sources has to be stored suitably for
continuous reliability.
3. Non existence of a proper planning, policy to finance, nodal agency for implementing renewable
energy systems, etc. at the government level.
4. Lack of qualified trained persons to design, manufacture, market, operate and maintain
renewable energy sources plants.
5. Inadequate documentation and evaluation of past experience and a lack of clear priorities for
future work.

172
 QUESTION
Part – A and Part – B
1. Write First law of thermo dynamics.
2. What is an isothermal change?
3. What is an adiabatic change?
4. Write the Clausius statement of second law of thermo dynamics.
5. Write the Kelvin's statement of second law of thermo dynamics.
6. Define Critical temperature.
7. Define Critical pressure.
8. Define Critical volume.
9. What is the difference between a gas and a vapour ?
10. What is the principle used in cascade process?
11. What is the disadvantage of cascade process of liquefaction of gases?
12. What is temperature of inversion?
13. What is Joule–Thompson effect?
14. What are the renewable sources of energy?.
15. What are the non-renewable sources of energy?
16. Define conventional energy with examples.
17. Define non-conventional energy with examples.
18. Write any two advantages of Non-conventional sources of energy.
19. Write any two disadvantages of Non–conventional sources of energy.

Part – C
1. Explain isothermal and adiabatic change.
2. Describe the cascade process for the liquefaction of oxygen.
3. Describe Linde's process for the liquefaction of air.
4. What are the advantages and disadvantages of non- conventional energy over conventional
energy?
5. Write a note on solar energy and wind energy.
Exercise Problems:
1. Air at 2 atmospheric pressure is compressed isothermally to one fourth of its original volume.
Find the final pressure. [Ans: P2 = 8 At.pr]

2. A gas at a pressure of 0.75 m of mercury and volume one litre is compressed slowly to a pressure
of 1.5 m of mercury. Find the resulting volume (Ans: 0.5 litre)
3. Air at a pressure of 5 atmosphere is compressed slowly to half of its original volume. Find the
resulting pressure. (Ans: 10 atm)
4. A certain mass of gas at 1 atm pressure is suddenly compressed to one-fourth of its original
volume. Find the resulting pressure. Given  = 1.4 ( Ans: 6.964 atm)

173
 5. A certain mass of gas at 50 atm is suddenly expanded to 10 times of its initial volume. Find the
resulting pressure. Given  = 1.4 (Ans: 1.9905 atm)
6. A gas at 2 atm pressure is compressed to half of its original volume. Calculate the final pressure,
if the compression takes place, (i) slowly and (ii) suddenly. Given  = 1.4.
(Ans: (i) 4 atm (ii) 5.2780 atm)
7. Ten litres of gas at 50 atmospheric pressure is suddenly expanded to 5 atmospheres.Find the
resulting volume. Given  = 1.4 (Ans: 51.7129 litre)
8. Five litres of a gas at 5 atmospheric pressure is suddenly compresses to a pressure of 5 times the
original pressure. Find the resulting volume. Given  = 1.4. (Ans: 1.5837litre)

Brain Teasers:
1. The ratio of the specific heat capacities of an ideal gas is 1.4. One atmosphere of the gas is
compressed adiabatically to one-fifth of its original volume. Find its final pressure.
[Ans: 9.52 At.pr]

2. Air at 5 atmospheres is suddenly compressed to half of its volume. Find the resulting
2 pressure of
the air (γ =1.4) [Ans: P= 13.19 At.pr]

https://www.youtube.com/watch?v=IIvGUE9z5oU&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=5

https://www.youtube.com/watch?v=YtgEZkcPW5E&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=6

https://www.youtube.com/watch?v=Ddkw3Hbblk4&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=7

https://www.youtube.com/watch?v=f94y-vsjSE8&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=8

https://www.youtube.com/watch? v=3ICesiyGdBw&list=PL1b9Ht9IS qIFUWvOqFTEdUPmtZadrXNk4&index=9

174
 SUMMARY

• First law of thermodynamics Q = dE + W

• Explanation of isothermal and adiabatic process

• Equations PV = a constant and P1V1= P2V2.

• Clausius and Kelvin statements of second law of thermodynamics

𝑄1 − 𝑄2
• Efficiency of Carnot’s reversible engine =
𝑄1

• Definitions of the terms:

▪ critical temperature

▪ critical pressure

▪ critical volume

• Liquefaction of oxygen by Cascade process

• Joule Thomson effect and temperature of inversion

• Liquefaction of air by Linde’s process.

• Classification of energy sources

• Different types of energy sources

175
 UNIT III

LIGHT AND REMOTE SENSING

OBJECTIVES

▪ To know the laws of refraction

▪ To define refractive index of a medium

▪ To understand the working of spectrometer.

▪ Can determine the refractive index of glass using spectrometer

▪ To know the phenomenon of total internal reflection

▪ To familiarise the fibre optics and the advantages of OFCs.

▪ To know the definition of LASER

▪ To understand the principle and characteristics of LASER

▪ To understand the difference between Spontaneous emission and

Stimulated emission

▪ To explain the process of population inversion.

▪ Able to explain the construction and working of Ruby LASER

▪ To know the uses of LASER

▪ To understand the concept of remote sensing

▪ To know the difference between active and passive remote sensing

▪ To know about the various components of remote sensing

▪ Can explain the principle and working of RADAR

▪ To solve simple problems based on refractive index.

176
UNIT

3 LIGHT AND REMOTE SENSING

LIGHT
Light is a form of energy radiated by luminous bodies. When it enters our eyes, it stimulates the
retina to a sense of vision. An object is visible only by the scattering of light from the object reaches the
eyes.
The objects which emit light rays (energy) are called light sources. For example, Sun, Stars, lamp
etc. are the light sources. Light can pass through the transparent medium like glass, air, water etc. Light
rays undergo reflection, refraction interference, diffraction, polarization, etc. Light is propagated with a
finite velocity 3x108 ms-1. Light plays dual role. In some places it behaves like particles and in some
places it behaves like waves, traveling in a straight line.

3.1 OPTICS
Refraction
When a ray of light travels from one transparent medium into another medium, it bends while
crossing the interface, separating the media. This phenomenon is called refraction.
Laws of Refraction
I. Law : The incident ray, the refracted ray and the normal to the refracting surface at the point
of incidence are in the same plane.
II. Law : The ratio of the sine of the angle of incidence to the sine of the angle of refraction is
a constant for any two given medium for one particular colour of light. This is known as Snell's law.
Refractive index of a medium
When a ray of light passes from air or vacuum to the given medium, the ratio between the
sine of the angle of incidence and the sine of the angle of refraction is called the refractive index
of the given medium.
If 'i' is the angle of incidence and 'r' is the angle of refraction,
Sin i
Refractive index of the medium  =
Sin r
According to wave theory of light, if ‘c’ is the velocity of light in air or vacuum and ‘v’ is the velocity
of light in the medium, then
Refractive Index of the medium
Velocity of light in air (or) vacuum (c)
=
Velocity of light in the medium (v)

177
Expression for the refractive index of the material of the prism

ABC is a glass prism. AB and AC are the refracting faces. BC is the base of the prism. A ray of
light PQ is incident on the face AB at angle of incidence i1. QR is the refracted ray. The angle of
refraction is r1. The refracted ray QR is incident on the face AC at an angle of incidence r2, RS is the
emergent ray. The angle of emergence is i2, N1 T1 and N2 T2 are the normals drawn to the faces AB and
AC at Q and R respectively. By producing PQ and SR the angle of derviation d is obtained.
In Δ ΑQR,
AQR = 90°-r1, and ARQ = 90° - r2
Again in Δ AQR.
A + AQR + ARQ = 180°
ie A + (90°-r1) + (90° - r2) = 180°
A - (r1 + r2) + 180° = 180°
A - (r1 + r2) - 180° - 180o = 0
or A = r1 + r2 ………….. (1)
Here A is the angle of the prism.
In  MQR, MQR = i1 – r1 and MRQ = i2 –r2
The exterior angle = sum of interior angles
i.e d = MQR + MRQ
d = (i1 – r1) + (i2 – r2) = i1 + i2 - (r1 + r2)
d = i 1 + i2 - A
i.e. A + d = i1 + i2 ................. (2)
As the angle of incidence increases, the angle of deviation decreases to a minimum value and
then increases. The minimum value of the angle of deviation is known as the angle of minimum deviation
‘D’.
When the deviation is minimum
i1 = i2 = i and r1 = r2 = r
Therefore from eqn. (1)
A = r + r = 2r
A
r=2
From equation (2), A + D = i + i = 2i
A+D
i=
2

178
Refractive index of the glass prism is
sin i
=
sin r
A+D
sin ( )
 = 2
A
sin( )
2

SPECTROMETER
Spectrometer is used to determine the refractive index of the material of the prism and to study
spectra.
The three main parts of the spectrometer are Collimator, Prism table and Telescope.

Prism table
Collimator
Telescope

Collimator
The collimator consists of two co-axial brass tubes, one is capable of sliding into the other. An
adjustable slit is provided at one end and a convex lens is fixed at the other end of the collimator. The
distance between the slit and the convex lens can be changed using the screw provided. The width of the
slit can also be changed. The collimator is rigidly fixed to the instrument. It is used to get parallel rays
of light.
Prism table
It consists of two circular metal discs with three levelling screws. The prism table can be rotated
about an axis passing through its centre. The prism table can be raised or lowered and can be fixed at
any desired height. The prism table can be made horizontal by adjusting the levelling screws with the
help of a spirit level.
Telescope
It consists of two co-axial brass tubes, one is capable of sliding into the other. An eye-piece is
provided at one end and an objective lens is fixed at the other end of the telescope. The telescope can
be rotated about the central axis of the instrument. A circular scale graduated in half degree is
attached to the telescope. Therefore when the telescope rotates, the circular graduated scale also
rotates. The telescope can be fixed in any position using the clamping screw. For fine adjustment of
the telescope, the tangential screw is used.
Initial adjustments of a spectrometer
1. The eye-piece in the telescope is gently pulled out or pushed in, till the cross wires are seen
clearly.
2. The telescope is turned towards a distant object. Then the distance between the eye-piece and
the objective lens of the telescope is adjusted, till the clear image of the object is seen in the field
of view of the telescope.

179
 3. The slit is illuminated by a monochromatic source of light. The width of the slit is adjusted
suitably by viewing through the convex lens in the collimator. Then the telescope is brought in
line with the collimator. Clear image of the slit is obtained by using the screws in the collimator
and telescope. Using the tangential screws, the vertical cross wire is made to coincide with
the left edge of the image of the slit.
4. The prism table is made horizontal by adjusting the levelling screws in it using a spirit level.
Refraction
When a ray of light passes from one medium to the second its direction changes at the point
where it enters the second medium. This phenomenon is known as refraction.
Experiment to determine the refractive index of the material of the prism:
The refractive index of the material of prism is determined by measuring the angle of prism and the
angle of minimum deviated ray by using Spectrometer.
Before, starting the experiment the following initial adjustment is made for the spectrometer
1. The telescope is kept in focus by capturing the clear image of distance object.
2. The prism table is kept horizontal by levelling screws in it using sprite level.
3. The slit is illuminated by monochromatic source of light. The collimator screw is adjusted
to produce narrow slit. Then, the clear image of the slit is viewed through the telescope and
coincide with vertical cross wire of the telescope. Now, the vernier table is adjusted to
0o –180o.
Determination of angle of prism A:

After making initial adjustments, the least count of the spectrometer is determined (as in
Vernier Calipers).

Having a very fine slit, the telescope is adjusted for direct ray. A prism is placed on the prism table as
shown in the figure. The telescope is turned to catch the reflected image from one polished face AB of
the prism. Fixing in that position, the tangential screw is adjusted till the vertical cross wire is made to
coincide with the fixed edge of the image of the slit. The readings in the scales I and II are noted.

The telescope is then turned to the other polished face AC of the prism. The readings in the scales I
and II are noted when the vertical cross wire coincides with the fixed edge of the image of the slit. The
difference between both the scale I readings is 2A. Similarly the difference between both the scale II
readings is 2A. Then the average of the angle of the prism A is calculated.

180
Tabular column I:
To determine the angle of the prism A LC =
Spectrometer readings in degree
Scale I Scale II
Ray
VSR = OR = VSR = OR =
MSR VSC MSR VSC
VSC  LC MSR + VSR VSC  LC MSR + VSR
Reading
of
the X1 = X3 =
reflected
ray (Left)
Reading
of
the
X2 = X4 =
reflected
ray
(Right)

Determination of angle of minimum deviation:

Tabular column II:

To determine the angle of the minimum deviation D LC =
Spectrometer readings in degree
Scale I Scale II
Ray
VSR = OR = VSR = OR =
MSR VSC MSR VSC
VSC  LC MSR + VSR VSC  LC MSR + VSR
Reading
of
Minimum d= d=
deviated
ray
Reading
of d= d=
Direct ray

The prism is placed on the prism table as shown in the figure. The ray of light from the collimator
incident on one polished face of the prism gets refracted and emerges out of the other polished face.
Viewing the image of the slit through the telescope, the prism table and telescope are rotated simultane-
ously through small angle such that the image of the slit moves towards the direct ray. At one stage, it is
found that the image of the slit is stationary for a moment and on rotating the prism table further in the
same direction the image begins to retrace its path. The telescope is fixed at that position, the cross wire
is made to coincide with the fixed edge of the image of the slit. At this minimum deviation position, the
readings of scales I and II are noted.
181
 The prism is removed from the prism table. The telescope is brought in line with the collimator and
the image of the slit is seen through the telescope. The vertical cross wire is made to coincide with the
fixed edge of the image of the slit. The readings in the scales I and II are noted.
The difference between the two readings of scale I or scale II gives the angle of minimum deviation
D. Hence the refractive index of the material of the prism is calculated using the formula.
The refractive index of the prism is determined by using the formula,

Total Internal Reflection

When a ray of light passes from a denser to a rarer medium, it is bent away from the normal. When
the angle of incidence is gradually increased, the corresponding angle of refraction also increases. At
a particular angle of incidence, the angle of refraction reaches to 90°. That particular value of angle of
incidence is called the critical angle of the medium.
Critical angle is defined as a particular angle of incidence in the denser medium for which
the corresponding angle of refraction in the rarer medium is 90°.

r
Rarer Rarer r=90o Rarer
Denser
i c
i>c
Denser Denser
When the angle of incidence is further increased above the critical angle, the ray is totally reflected
back in the same denser medium itself. This phenomenon is called total internal reflection.
Two conditions are to be satisfied for total internal reflection.
1) Light should travel from denser medium to rarer medium.
2) The angle of incidence should be greater than the critical angle of the medium.
According to Snell's law,
Sin i
rd =
Sin r
The refractive index of the rarer medium with respect to denser medium,
Sin r
d r =
Sin i
Also, the refractive index of the denser medium with respect to rare medium
In critical angle position, i = c and r = 90°
Sinr Sin90 0
 d r = =
Sin i Sin c
1
i.e.,  d = (Sin900 = 1)
r

Sin C
The sparkling of the diamond, the phenomenon of mirage and the propagation of light waves along
optical fibre are due to the total internal reflection.

182
 Sl.No. Medium Refractive index  w.r.t air Critical angle : C

1 Water 1.33 48°45'

2 Glass 1.5 41°48'

3 Diamond 2.42 24°24'

Applications of Total internal reflection:

Glittering of diamond:
Diamond appears dazzling because ‘total internal reflection of light’ happens inside the diamond.
The refractive index of only diamond is about 2.417. It is much larger than that for ordinary glass which
is about only 1.5. The critical angle of diamond is about 24.4°. It is much less than that of glass. A
diamond cutter makes use of this larger range of angle of incidence, to ensure that light entering the
diamond is total internally reflected from the many cut faces before getting out. This gives a sparkling
effect for diamond.
Mirage and looming:
The refractive index of air increases with its density. In hot places, air near the ground is hotter
than air at a height. Hot air is less dense. Hence, in still air the refractive index of air increases with
height. Because of this, light from tall objects like a tree, passes through a medium whose refractive
index decreases towards the ground. Hence, a ray of light successively deviates away from the normal at
different layers of air and undergoes total internal reflection when the angle of incidence near the ground
exceeds the critical angle. This gives an illusion as if the light comes from somewhere below the ground.
For of the shaky nature of the layers of air, the observer feels as if the object is getting reflected by a pool
of water or wet surface beneath the object. This phenomenon is called mirage.
In the cold places the refractive index increases towards the ground because the temperature of
air close to the ground is lesser than the temperature above the surface of earth. Thus, the density and
refractive index of air near the ground is greater than at a height. In the cold regions like glacier sand
frozen lakes and seas, the reverse effect of mirage will happen. Hence, an inverted image is formed little
above the surface. This phenomenon is called looming.
FIBRE OPTICS
Fibre optics deals with the light propagation through thin glass fibres. Fibre optics plays an
important role in the modern communication systems. In this system, television voice and digital data
signals are transmitted from one place to another place in the form of light waves. The light waves are
transmitted along a thin cylindrical glass fibre by using the phenomenon of total internal reflection. It
was first demonstrated by John Tyndall in 1870. But in the field of communication it was tried only from
1927. After the discovery of laser beam in 1960, fibre optical communication further developed
and extended to medical field.
A bundle of fine fibres is used in the fibre optical communication. The bundle has the advantage of
flexibility. A beam entering at one end can be transmitted at the other end through the fibre even though
the fibre is curved. The study of such properties of a bundle of fibres is known as fibre optics.
Optical fibre cable as a wave guide:
The optical fibre cables are used as wave guides for transmitting electromagnetic waves at
optical frequencies. These frequencies are in the order of 1015 Hz higher than that of radio wave
frequency of 1010Hz. The number of information transmitted is directly proportional to the frequency of

183
the transmitted signals. Hence 15,000 and more simultaneous telephone conversations can be transmitted
in the form of electric current.
An optical fibre cable is a cylindrical core made of silica glass of radius 0.0002cm to 0.001cm.
Around this core cladding is formed by the addition of small amount of boron, germanium or phosphorus
to form the rarer medium. The cladding is necessary to retain the light waves within the core and also to
provide mechanical strength. Further the cladding is coated with buffer jacket made of plastic to protect
the fibre from moisture and scratches.

The buffer jacket is also surrounded by a thick tougher like Kevlar that provides toughness and
tensile strength to the fibre. Finally the cable is covered by black polyurethane outer jacket that provides
flexibility to the fibre. When the optical signal enters at one end of the fibre, it is incident on the layer
separating the core and cladding.
The angle of incidence is greater than the critical angle. Since the core is a denser medium and
the cladding is a rarer medium, total internal reflections take place continuously throughout the core
region as shown in the figure. Finally it comes out from the other end of the fibre. Hence the optical fibre
functions as a wave guide.

Advantages of Optical Fibre Cables (O.F.C)

1. The fibre material is made of silica glass which is available in surplus in the world
compared to copper or aluminium.
2. Silica is very much cheaper than copper.
3. There is no electrical danger, since no electric current flows through the fibre. It carries
only optical signal.
4. Optical fibre can be handled more easily because of their light weight, smaller in size and
flexibility.
5. The optical signal transmitted through the fibre cannot be disturbed by the radio frequency
interference.
6. Tapping (overhear secrets) of information is impossible in fibre optic systems.
7. A large number of signals can be simultaneously transmitted along the optical fibre cables
compared to the metallic cables.
8. The loss of intensity of energy is less in optical fibre cables compared to the metallic
cables. In the optical fibre cables the loss is 12.5 dB per km, whereas in the metallic cables
the loss is 49 dB per km.

184
 9. The optical fibre cables are having additional advantages of longer life, easy maintenance,
low cost transmission, high quality signal transmission and efficient transmission.
10. The optical fibre cables are electrical insulators. They are not affected by any other
electromagnetic wave and lightning.
Problem:
Light travelling through transparent oil enters in to glass of refractive index 1.5. If the refractive
index of glass with respect to the oil is 1.25, what is the refractive index of the oil?
Solution:
Given, ngo = 1.25 and ng = 1.5
ng
Refractive index of glass with respect to oil, ngo = no
ng 1.5
Rewriting for refractive index of oil, no = ngo
= 1.25 = 1.2

The refractive index of oil is, no = 1.2

WORKED PROBLEMS
1. Calculate the angle of minimum deviation for an equilateral glass prism of refractive index 1.4.
Refractive index, μ = 1.5
For an equilateral glass prism, A = 60o
Angle of minimum deviation, D = ?
 A+D 
sin
2
The refractive index, =
 A
sin
2
 A+ D  A  60o 
sin =   sin = 1.5  sin 
2 2  2 
 A+D 
i.e. sin = 1.5  sin 30o = 0.75
2
A+D
 = sin –1 (0.75) = 48o 35
2
A + D = 2 × 48o 35 - 60o = 37o 10'
2. The angle of prism is 60o and a refractive index of its material is 1.512. Calculate the angle of
minimum deviation.
Refractive index μ = 1.512
Angle of the prism, A = 60o
Angle of minimum deviation, D = ?
 A+D 
sin
2
The refractive index, =
 A
sin
2
 A+ D  A  60o 
sin =   sin = 1.512  sin 
2 2  2 

185
  A+D 
i.e. sin = 1.512  sin 30o = 0.756
2
A+D
 = sin –1 (0.756) = 49o 6
2
A + D = 2 × 49o6' = 98o12'
D = 98o12' – A = 98o12' – 60o = 38o12'
i.e The angle of minimum deviation D = 38o12'

3. Calculate the angle of the prism if the refractive index is 1.5 and angle of minimum deviation
is 38.
Refractive index µ = 1.5
Angle of minimum deviation D = 38 Angle of the prism A = ?
(A+D)
Sin
2
The refractive index µ = A ----- (1)
Sin
2

We know that Sin (A+B) = Sin A . Cos B + Cos A . Sin B

A+D A D A D A D
Sin ( ) = Sin ( + ) = Sin Cos + Cos Sin
2 2 2 2 2 2 2

Substituting in equation (1), we get

A D A D A D A D
Sin Cos + Cos Sin Sin Cos Cos Sin
2 2 2 2 2 2 2 2
µ= A = A + A
Sin Sin Sin
2 2 2

A
D Cos D
2
Hence µ = Cos 2
+ A Sin 2
Sin
2

D A D
= Cos 2
+ Cot 2
. Sin 2
---- (2)

Substituting the values for µ and D in equation (2)

38 A 38 A
1.5 =Cos + Cot . Sin = 0.9455 + Cot x 0.3255
2 2 2 2

A
Cot 2 x 0.3255 = 1.5 - 0.9455 = 0.5545
Hint for using calculator:
A 0.5545 𝐴 1
Hence, Cot 2 = = 1.7035 Tan =
0.3255 2 1.7035

A
 = 30.4 and A = 60.8  = 0.587
2
𝐴
 = Tan-1(0.587)
2
Angle of the Prism = 60.8 

186
4. Calculate the refractive index of the material of an equilateral prism. The angle of minimum
deviation is 40.

For an equilateral prism, the angle A = 60

Angle of minimum deviation D = 40 Refractive index µ= ?
(A+D)
Sin
2
The refractive index µ = A
Sin
2

(60+40)
Sin
2
µ = 60
Sin
2

Sin 50 0.766
= = = 1.532
Sin 30 0.5

The refractive index µ = 1.532

5. Calculate the refractive index of a prism, if the angle of the prism is 59o and the angle of
minimum deviation is 40o.

Angle of the prism, A = 59o

Angle of minimum deviation, D = 40o
Refractive index, μ = ?

6. In an experiment the angle of the prism and the angle of minimum deviation were found to
be 59o 52' and 40o 30'. Find the refractive index of the material of the prism.

Angle of the prism, A = 59o 52'

Angle of minimum deviation, D = 40o 30'
Refractive index, μ = ?

Refractive index of the material of the prism is 1.539.

187
7. The minimum deviation produced by a glass prism having an angle of 60o is 30o. If the
velocity of light in vacuum is 3 × 108 ms–1, calculate the velocity of light in glass.

A = 60o, D = 30o
Velocity of light in vacuum, C = 3 × 108 m s–1
Refractive index of glass, μ = ?

Refractive index of glass is

Velocity of light in air or vacuum (C)

=
Velocity of light in the medium (V)

C 3 108
Velocity of light glass (V) = =
 1.414 = 2.122 108 ms –1

188
 3.2 LASER

An atom consists of an electron cloud revolving around the central nucleus having protons and
neutrons. The atoms are constantly in motion, they vibrate, move and rotate. Also they are in different
states of excitation and hence have different energies.

If some external energy in the form of light, heat, or electricity is supplied to an atom, the electrons
from low energy level called ground state may jump to higher energy level called excited state. The
electrons in the higher energy orbit cannot stay there for a long time. Therefore they return to the
ground state spontaneously releasing the absorbed energy in the form of radiations.

LASER
The term LASER stands for Light Amplification by Stimulated Emission of Radiation. The
LASER is a device which amplifies light waves similar to the transistor which amplifies electric current.
The laser is a powerful source of light.

Characteristics of LASER
Laser radiation is different from normal light radiation. The characteristics of laser are given
below:
1. Laser radiation is monochromatic. It contains only one particular wavelength (colour) of light.
The wavelength is determined by the amount of energy released.
2. Laser radiation is coherent. In emitted radiation all the photons are having the same phase and
amplitude.
3. Laser radiation is highly directional. Laser beam of light can travel very long distance, without
much divergence.
4. Laser radiation is very intense. The intensity and hence the brightness of the laser radiation is
high.
The radiation having single wavelength, same amplitude and same phase is called ‘coherent
radiation’.

Principle of LASER
Laser works on the principle of quantum theory of radiation. When an atom in the lower energy E1
absorbs the photon energy of incident radiation, it is excited to the higher energy level of energy E2.
This process is called absorption.

The number of absorption transition take place per unit time per unit volume is directly
proportional to the number of atoms in the lower energy level and the number of photons in the
incident radiation.

The energy difference E2 - E1 is called excitation energy and it is equal to the energy of the photon
absorbed by the atom. In higher energy level (E2 ), the atom cannot remain long time, since it has a
tendency to drop energy. So the atom jumps to the lower energy level (E1 ). At that time, the energy
difference E2 - E1 is emitted in the form of radiation as photon.

The energy of the photon released, E2 - E1=h, where  is the frequency of the radiation and h, the
Planck's constant. (6.623 × 10–34 Js)

189
 Spontaneous emission
Spontaneous emission is the process of photon emission takes place immediately without any
inducement during the transition of atoms from higher energy levels to lower energy levels. This is a
random process in which the photons are emitted at different time. So, in the emitted radiation each
photon has different phase and are incoherent.

The number of spontaneous transition (Nsp) taking place per unit time per unit volume is directly
proportional to the number of atoms in the higher energy level (N2).

is Nsp  N2 (or) Nsp = AN2

where A is a constant known as the probability of spontaneous transition per unit time. Also it is
called Einstein's co-efficient of spontaneous emission.

Stimulated emission
Einstein found a new process called stimulated emission to increase the number of transition of
atoms from higher energy levels to lower energy levels.

Stimulated emission is the process of photon emissions takes place by an inducement given by
another photon incident on the atoms in higher energy levels. The energy of the photon emitted is equal
to the energy of the photon incident.

Consider an atom in the higher energy level (E2 ). When an external radiation of photon energy E2 - E1
is incident on the excited atom, the photon stimulates the atom to make transition from higher to lower
energy level. As a result the same photon energy E2 - E1 is emitted in the form of radiation. During this
process, the stimulating photon and the photon emitted by the excited atom are emitted simultaneously
in the same direction.

Hence they are identical in phase, direction and frequency and are coherent. This process of
stimulated emission is used to produce laser beam.

The number of stimulated transition (Nst) taking place per unit time per unit volume is directly
proportional to

1) the number of atoms in the higher energy level (N )

2

2) The number of photons in the incident radiation (Q)

i.e., Nst  N2 Q (or) Nst = BN2 Q

Where B is a constant known as the probability of stimulated transition per unit time. Also it is called
Einstein's co-efficient of stimulated emission.
The stimulated emission can be multiplied through a chain reaction as shown in the figure. When a
single photon hits an atom in the higher energy level, two photons are emitted by stimulated emission.

190
 E2

E1

Then these two photons hit on two atoms of higher level, four photons are emitted. This process
is continued as a chain reaction and the photons are getting multiplied. Finally it leads to the emission
of the powerful, coherent, monochromatic and highly directional beam of laser light. This is called
Light amplification by stimulated emission of radiation. This amplification takes place only if there are
more number of atoms in the excited state (higher energy level) than in the ground state (lower energy
level).
Population inversion
Population inversion is the establishment of the situation to make more number of atoms
in the excited state required for the chain reaction of stimulated emission.

For achieving the population inversion, the number of atoms must be raised continuously to the
excited state. When a sufficient energy is given to the atoms in the ground state, they are excited to the
higher energy levels. This process is called pumping. If the atoms in the ground state are brought to the
excited state by means of light energy, the process is called optical pumping.
Construction and working Ruby Laser
Ruby rod

Highly
silvered
end Laser beam

Partially
silvered end
Flash tube

Ruby is a crystal of Aluminium oxide (Al2 O3) with an addition of 0.05% chromium oxide (Cr2O3 ).
ink ruby is machined into a rod of about 10 cm in length and 0.5 cm in diameter. One end of the ruby
rod is highly silvered and the other end is partially silvered.
Energy Level diagram for Ruby Laser
E2
Optical
pumping
5600 Ao

E1

h  (6943 Ao)

Ground state
E0

191
 The flash light from the Xenon lamp makes the chromium ions (Cr3+) in the ground state (E0) to go
to the meta stable state (E1 ) by giving out energy in the form of heat. From the E1 level the atoms jump
to the E0 level, giving out light photons of energy (E1 – E0) = h υ.

Uses of Laser :

1. Laser is used to find the long distances by direct pulse reflection method. Using this method, the
distance between the moon and earth was measured accurately.

2. It is used to produce high temperature to melt and vapourise the metals in a very short period.

3. It is used to bore holes in hard substances like tungsten and diamond.

4. It can be modulated to transmit hundred messages at a time on radio, television and telephone.

5. The big size objects like aeroplanes, missiles etc can be destroyed in a few seconds by passing a
powerful laser beam on to them. For this reason, laser is called a “Death ray” instrument.

6. In surgery, a laser beam is used in spot-welding of a detached retina in the eye ball. It is also used
for painless drilling and welding the surface of a tooth to prevent its decay.

7. The laser beam is not easily absorbed by water. So it can be used in under-water communication
between submarines.

192
 3.3 REMOTE SENSING
Remote sensing is the process of sensing and measuring the objects in a long distance without
physical contact with them. A device used for the above purpose is called remote sensor.
Remote sensors are mounted on the aircraft or satellite platforms. The sensors receive the
reflected or emitted radiations from a particular place on the earth's surface and measure the energy. These
measurements are being made at a large number of points from the image of the earth's surfaces. The
sensed images are interpreted in terms of geographical variations. The active power of the remote
sensors depends upon the detection of energy emitted or reflected by the object.
Types of remote sensing (Active and Passive) :

The system of remote sensing is classified into two types. They are
1) Active remote sensing
2) Passive remote sensing

(1) Active remote sensing is the process of sensing the distant target in which the sensors direct
their own generated radiations towards the target and then it detect the amount of energy
reflected by the target.

Examples: Radar, Radar Altimeter, Wind scatterometer, Synthetic Aperture Radar (SAR) and
Camera with flash.

(2) Passive remote sensing is the process of sensing the distant object in which the sensors detect
only the amount of energy radiated by the object.

Examples: Aerial photography, Camera without flash and Microwave radiometer.

Components of remote sensing:
The electromagnetic remote sensing process is divided into two components namely
(1) Date acquisition and
(2) Date analysis

193
 Date acquisition : The schematic diagram of data acquisition is shown in the figure. In this process,
five distinct elements are involved.
They are
1. Energy Sources.
2. Propagation of energy through the atmosphere.
3. Energy interaction on earth surface.
4. Space borne sensors record the reflected energy.
5. Conversion of sensors data in the form of pictures or digital information.
The electromagnetic radiations emitted by the energy source are propagated through the atmosphere
and incident on the surface of the earth. Then the radiations are reflected back towards the space borne
remote sensors. The sensors generate the receiving data into picture or digital forms.

(1) Visual
Interpretation
Information Users
(2) Products
Digital
analysis

Data analysis: The schematic diagram of data analysis is shown in the figure. The data analysis is
divided into two separate processes.
They are
1) visual interpretation technique
2) Digital analysis
In visual interpretation techniques, various viewing instruments are used for examining and
analysing the pictorial data. In digital analysis the computers are used to analyse the digital data. All
analyzing data are viewed in the information products by users. The data analysis can be carried only
with the help of reference data.
The process that involves examination and analysis of pictorial data is called visual image
interpretation techniques. This process involves the use of fundamental picture elements like tone,
texture, pattern, shape and size in order to identify the various objects. Stereoscopic instruments
are used for this purpose.
Use of computers to analyse digital data is known as digital image processing techniques. If the data
is available in digital form, it can be analysed on interactive computer systems for getting statistical data.
This statistical data is analysed using computers by comparing with the actual “signatures” of the objects
collected through field visits. This system of classification of objects is quite accurate.
Reference data:
Whether it is pictorial data analysis or digital data analysis, reference data should be generated as
supporting data. Reference data is called ground truth or field check. The reference data is used to
analyse and interpret remotely sensed data and also to verify information obtained from remote sensing
data. The acquisition of reference data is through actual field visits.
The collection of reference data may depends upon time-critical or time-stable measurements.
The time-critical measurements are those vegetation conditions, water pollutants and climatic conditions.
The time- stable are other geological measurements.

194
RADAR

Synchronizer Transmitter Duplexer or

T/R Switch

Antenna
Display Receiver

The RADAR is an acronym derived from Radio Detection and Ranging. Radar is an electromagnetic
device for detecting the presence and location of objects. The presence of the object, their range and
direction are determined by the transmission and return of electromagnetic energy. The radar antenna
transmits short burst pulses of energy to the target and echoes from this target carry information about
the position and quality of the illuminated object. The block diagram of an elementary pulsed radar set
is given above.
Radar consists of a transmitter and a receiver each connected to a directional antenna through a
Duplexer or Transmitter/Receiver switch. The transmitter sends out microwave power through the
antenna. The receiver collects as much energy as possible from the echoes reflected in its direction by
the target. In most of the radar systems the receiving antenna is same as the transmitting antenna.
The transmitter generates a short pulse and sends it to the antenna via duplexer. As soon as a small
fraction of the pulse is fed to the duplexer, it disconnects the transmitter from the antenna and connects
the receiver to it. The transmitted pulse is reflected by the target and a portion of it reachers the antenna.
Now the duplexer connects the receiver to the antenna so that the returning echoes can be processed. The
processed pulses are fed to the necessary display devices which gives the nature and position (range)
of the target. In most radars, the antenna moves in a predetermined pattern and this process is known as
scanning.

195
 QUESTION
Part –A & B
1. What is Refraction?
2. Write the laws of refraction.
3. Define refractive index.
4. What is total internal reflection?
5. Define critical angle.
6. What is spectrometer?
7. What are the three major part of spectrometer?
8. Define angle of minimum deviation.
9. Give the formula for refractive index of the material of the prism.
10. Expand the acronym LASER.
11. Write any two of the characteristics of LASER.
12. What is the principle of LASER?
13. What is spontaneous emission?
14. What is the stimulated emission?
15. What is population inversion?
16. Write any two uses of LASER.
17. What is remote sensing?
18. What is active remote sensing?
19. What is passive remote sensing?
20. What are the important components of remote sensing?
21. Expand the acronym RADAR.

Part-C
1. Explain the phenomenon of total internal reflection with the help of ray diagram.
2. Describe an experiment of determine the refractive index of the material of prism using angle
of minimum deviation method.
3. Write a note on spontaneous emission, stimulated emission and population inversion.
4. Describe the construction and working of RUBY LASER.
5. List the uses of Laser
6. Describe the data acquisition and data analysis of remote sensing.
7. Explain the working of RADAR with block diagram.

196
 PROBLEMS:

1. The minimum deviation for a glass prism is 38o and the refractive index of glass is 1.509. Find
the angle of the prism.
Note: use sin (A + B) = sin A cos B + cos A sin B (Ans: A=60o8’)

2. The refractive index of the material of a prism is 1.5. When the prism is placed in
the minimum deviation position, the angle of incidence is 51o. Calculate the angle of prism and
angle of minimum deviation.
(Ans: 62o24’,39o36’)

3. Calculate the refractive index of the material of an equilateral prism. The angle of
minimum deviation is 38o.
(Ans: 1.509)

4. The angle of glass prism is 59o. Calculate the angle of minimum deviation, if the refractive
index of the material of the prism is 1.642.
(Ans: 48o52’)

https://www.youtube.com/watch?v=anolwfe_ZsE&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=10

https://www.youtube.com/watch?v=mb1eWsr2BQs&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=11

https://www.youtube.com/watch?v=aJSgOWI1_CQ&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=12

https://www.youtube.com/watch?v=HTZVdh0MRYk&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=13

https://www.youtube.com/watch?v=1ND59chGkHU&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=14

197
 SUMMARY
𝑆𝑖𝑛 𝑖
• Laws of refraction (Snell’s law) = µ = , whereµ is refractive index
𝑆𝑖𝑛 𝑟

• Definition of minimum deviation

(𝐴+𝐷)
𝑆𝑖𝑛
• Refractive index of prism µ = 2
𝐴
𝑆𝑖𝑛
2

• Main parts of spectrometer: Collimator, Vernier disc, Prism table,

Telescope and Eye-piece

• Determination of refractive index of glass prism using spectrometer

• Conditions for total internal reflection:

1) Light must travel from denser to rarer medium

2) The angle of incidence must be greater than critical angle

• The advantages of OFCs

• LASER – Light Amplification by Stimulated Emission of Radiation

• List ofcharacteristics of LASER and it’s uses.

• Definitions of Spontaneous emission and Stimulated emission

• A state at which the number of atoms in the excited state is more than that
in the ground state is known as population inversion

• Construction and working of Ruby LASER

• Definition examples ofactive and passive remote sensing

• Data acquisition and Data analysis are the components of remote sensing

• Working of RADAR using block diagram

• Simple problems on refractive index

198
 UNIT IV

ELECTRICITY

OBJECTIVES

▪ To know Ohm’s law

▪ To understand laws of resistances

▪ To define Resistivity, Conductivity, Super conductivity and Meissner effect

▪ To state Kirchhoff’s current and voltage laws

▪ To define capacitance of a capacitor and farad

▪ To derive expressions for effective capacitance when capacitors are

connected in series and in parallel

▪ To know Joule’s law of heating

▪ Experimentally determine specific heat capacity of a liquid using Joule’s

calorimeter

▪ To familiarise Faraday’s laws on electrolysis.

▪ To define electro chemical equivalent (e.c.e) of an element

▪ Able to determine of e.c.e. of copper

▪ Can derive expression for the force acting on a current carrying straight
conductor placed in a uniform magnetic field

▪ To understand Fleming’s Left Hand rule

▪ To derive expression for the torque experienced by a rectangular current

carrying coil placed inside a uniform magnetic field

▪ Understanding the working of a moving coil galvanometer and its merits

▪ Can convert Galvanometer into an Ammeter and Voltmeter

▪ To solve simple problems based on resistors, capacitors, e.c.e of materials.

▪ Able to do the calculations for converting Galvanometer into an Ammeter

and Voltmeter

199
UNIT

4 ELECTRICITY

Introduction
The flow of free electrons in a conductor constitutes electric current. The current is defined as the
rate of flow of charges across any cross sectional area of a conductor. The current is expressed in ampere
and it is a scalar quantity. The direction of current is the direction of flow of positive charges or opposite
to the direction of flow of electrons. The electric current is the flow of positive charge from a point at a
higher potential to a point at a lower potential. Current flows only if there are charge carriers present in
a conductor which can move freely when an electric field source is connected between the ends of the
conductor. When charges are in motion, they produce magnetic effects. The current flowing through a
conductor is one ampere if a charge of one coulomb flows through any section of it per second.

4.1 ELECTRICAL CIRCUITS

In an electrical circuit there must be a source of energy which provides energy to electrons so that
they can move and constitute an electric current. For easy flow of current, the connecting wires must
be made of conductor material. The energy source creates a potential difference across the circuit and
therefore it forces electric charges to move along the conductor. When a voltage (V) is applied across a
conductor, a current (I) flows in the conductor. The magnitude of current in a conductor depends on its
resistance and the resistance plays an important part in the conduction.
Ohm's law
George Simon Ohm in 1826 established the relationship between potential difference and current in
a conductor which is known as Ohm's law.
It states that the current (I) flowing through a conductor is directly proportional to the
potential difference (V) across the ends of the conductor, if its temperature remains the same.
ie I  V
or V  I
or V = R I
Where R is called the resistance of the conductor. The unit of resistance is ohm with symbol ''
There are several materials which do not obey Ohm's law. Conductor which do not obey Ohm's law
are called non-ohmic conductors. This is due to increase in the resistance of the conductor when the
current increases. Transistors, crystal rectifiers etc., do not obey Ohm's law.
Laws of Resistances
When more than one resistor is to be connected in an electric circuit they can be connected either in
series or in parallel. The equivalent resistance offered in the circuit by a group of resistor depends on the
way of their combination.
Resistances in Series
The resistances are connected in series combination will increase the effective resistance in the
circuit. When a number of conductors are connected end to end so that the same current flows through
all of them, then they are said to be in the series combination.

200
 Consider AB, BC and CD are three conductors of resistances R1, R2 and R3 respectively connected
in series as shown in the figure. Let I be the current flowing through them and V 1, V2 and V3 be the
potential differences across R1, R2 and R3 respectively.

A I B C D
v1 v2 v3

Rs
I A D

Let AD be the equivalent resistance (RS) carrying the same current I and having a potential
difference V across the ends A and D such that
V = V1 + V2 + V3
Applying Ohm's law we get,
IRs = IR1 + IR2 + IR3
i.e. RS = R1 + R2 + R3

Thus, when a number of conductors are connected in series, the equivalent resistance is equal to the
sum of the individual resistances.
Resistances in Parallel
The resistances are connected in parallel combination to decrease the effective resistance in the
circuit. When a number of resistors are connected to common terminals so that the potential difference
across each conductor is the same, then they are said to be in parallel combination.
R1
I1
R2
I I2
A B
I3 R3

RP
A B

Consider R1, R2 and R3 are the resistances of the conductors connected between the point A and
B and V is the common potential difference across AB. Let I1, I2 and I3 be the current through the
individual branches.
Let AB be the equivalent resistance (Rp) having a potential difference V and carrying a current I
such that,
I = I1 + I2 + I3
Applying Ohm's law we get,
V V V V
= + +
R p R1 R 2 R 3

1 1 1 1
i.e., = + +
R p R1 R 2 R 3

201
 Thus, when a number of conductors are connected in parallel, the reciprocal of the equivalent
resistance is equal to the sum of the reciprocals of the individual resistances.
Resistivity
The resistance of a conductor is directly proportional to the length of the conductor and inversely
proportional to the area of cross section of the conductor.
Rl
1
R A

l
i.e., R 
A
l
R = ()
A

RA
 =
l
ρ is called the resistivity or specific resistance of the material of the conductor.
The resistivity or specific resistance of a material is defined as the resistance offered by a conductor
of unit length having unit area of cross section. The unit of resistivity is ohm metre ( Ω m ).
Conductivity
The reciprocal of resistivity of the material of a conductor is called as its conductivity. It is denoted
by σ. Based on the above expression of conductivity, the unit of conductivity is mho/m.
1
=

l
=
RA
Superconductivity
At very low temperature the resistance of the most of the conductors becomes very low essentially
zero. And so, the current passes freely through them, this phenomenon is called Superconductivity.
The temperature below which a conductor becomes superconductor are called transition temperature.
The material which goes to super conducting state when called below transition temperature are called
superconductors.
Meissner Effect
When a specimen is placed in a magnetic field and then cooled below its transition temperature
for superconductivity the magnetic flux originally present inside the specimen is ejected from the
specimen.

This process of ejection of magnetic flux from the super conducting material as the
material goes transition from conducting state into super conducting state is known as Meissnor
effect.

Kirchoff's current and voltage laws

In electrical circuits which consist of simple arrangement of resistors in series or in parallel with
a given source of emf, all calculations can be carried out using Ohm's law. But in more complicated
circuits where there are junctions between more conductors and where more than one source of emf are
active, then Kirchoff's laws can be useful for calculation purposes. There are two laws called Kirchoff's
current law and Kirchoff's voltage law.
202
 First law – Kirchoff's Current law or junction law
“In any network of conductors, the algebraic sum of the currents meeting at any junction is zero”.
The convention applied in the junction rule is that the current flowing towards a junction is taken as
positive and the current flowing away from the junction is taken as negative.
Let i1 , i2 , i3 , i4 and i5 be the current passing through the five conductors and meeting at the junction
'O'. Then according to Kirchoff's first law,

i4 i5
i1
i2
i3

i1 - i2 + i3 - i4 + i5 = 0
or i1 + i3 + i5 = i2 + i4
That is, the sum of the currents flowing towards the junction is equal to the sum of the currents
flowing away from the junction.
Second law – Kirchoff's Voltage law or mesh law
“In any closed electrical network, the algebraic sum of the products of current and resistance in each
branch of a network is equal to the algebraic sum of the emfs in that network.”
The current direction may be assumed either clockwise or anti- clockwise. The convention applied
is, if the current flows in the same direction in which we are considering the network or mesh, it is taken
as positive and the current in opposite direction is taken as negative. The emf which sends the current
in the same direction in which the mesh is considered, it is taken to be positive and emf which sends the
current in the opposite direction is taken to be negative.

For the network BAGFB we have

I1R1 + I1R2 + I1R3 + I3R6 = E

For the network FBCDF we have

I3R6 – (I1 – I3 ) R5 – (I1 – I3) R4 = 0

Condition for balancing the Wheatstone's Bridge

Wheatstone bridge is a very accurate and precise method of determining an unknown resistance
and it is an important application of Kirchoff's laws. Since it is a null method, the measurement of
unknown resistance is not affected by the internal resistance of the battery used.
Wheatstone bridge or network consists of four resistances P.Q.R and S connected in the form of
a quadrilateral ABCD. A galvanometer of resistance G is connected between the opposite end points B
and D. A cell of emf E and a key are connected between the opposite end points A and C.

203
 When the circuit is closed different currents flow through various branches of the network as
shown in the diagram. According to Kirchoff's current law Ig is the current passing through the
galvanometer. Considering the network ABDA, applying Kirchoff's second law,

I1 P + 1g G - (I -I1) R = 0
or I1 P + 1g G = (I -I1) R (1)
Considering the network BCDB. applying Kirchoff's second law
(I1 - Ig) Q – (I - I1 + Ig) S - Ig G = 0
or (I1 - Ig) Q = (I - I1 + Ig) S - Ig G (2)
The four resistances P, Q, R and S are adjusted so that there is no deflection in the galvanometer.
i.e. Ig = 0. Now the bridge is said to be balanced.
Now, equation (1) becomes
I1 P = (I - I1) R (3)
and equation (2) becomes
I1 Q = (I - I1) S (4)
dividing (3) by (4) we get
𝐼1 𝑃 (𝐼− 𝐼1)𝑅
𝐼1 𝑄
= (𝐼− 𝐼1)𝑆
𝑃 𝑅
=
𝑄 𝑆

This is the balancing condition of Wheatstone's bridge network.

CAPACITORS
Apart from resistors and insulators, a capacitor is also the other basic passive component commonly
used in electrical circuits. Capacitor is a device which has the ability to store charges in the form of
electrical energy, which neither a resistor nor an insulator can do.
An ideal capacitor may retain the charges stored in it for a long time, after power supply is removed
from the circuit. That is, it can store charges for ever. This stored charge can discharge through a
conductor (or) wire. This discharge can cause shock or damage to the connected equipment, during the
misuse of capacitors in a circuit.

204
Construction of a capacitor :
Two similar metallic conductors separated by an insulator (or) dielectric medium forms a capacitor.
Conductor – insulator – conductor sequence is the basic idea behind a capacitor.
The basic difference between insulator and dielectric medium is :
The materials which do not have free electrons in them are unable to conduct electricity and they are
termed as insulators.
Some material like ceramic behaves as a dielectric medium. Charges are induced in the opposite
faces of it, because of the external applied electric field. Such material is called dielectric.
All dielectric materials are insulators, but not all insulators make good dielectric materials.
The dielectric medium could be air, mica, ceramic, paper, polyester, Teflon, etc.
In the figure shown below two metallic conductors of equal area “A” separated by a dielectric
medium, forms a two terminal device, named as Capacitor.

The plates are kept perfectly parallel to each other. So it is called as parallel plate capacitor. The
conducting plates of a capacitor can be either square or circular or rectangular or cylindrical or spherical
in shape. Shape selection depends on its application and voltage rating.
Capacitance:
An ideal capacitor is characterized by a single constant value named as capacitance. The capacitance
of a capacitor is defined as amount of charge required to create a unit potential difference between its
plates or capacitance can also be defined as the ratio of charge to potential difference.
Q
C = Charge / p.d = farad
V
If Q = 1 coulomb and V = 1 volt
Then C = 1 farad (F)
Capacitance of a capacitor depends upon the area of the plate, distance of separation between the
plates “d” or thickness of the dielectric medium used.
The unit of capacitance is' farad' with symbol 'F' named after British Physicist Michael Faraday.
Definition of ‘farad’:-
One farad is defined as the capacitance of a capacitor which requires a charge of one coulomb to
establish a p.d of one volt between the plates.
How many electrons are present in one coulomb of charge?
Let there be “n” electrons in one coulomb. But charge q = ne, where ‘e’ is the charge of one
electron.
q 1
Therefore n= e = −19 = 6.25 × 1018 electrons
1.6  10
205
 So to establish one volt p.d. between the two plates, large number of electrons have to move from
one plate to another plate. So one farad is too large for practical purposes.
In most of the cases we have smaller unit like
one milli farad = 1 mF = 1 × 10-3 F = 0.001 F
one micro farad = 1 µF = 1 × 10-6 F = 0.000 001 F
one nano farad = 1 nF = 1 × 10-9 F = 0.000 000 001 F
The symbol for capacitors is not unique. In the modern electronic world, the symbol for capacitors
are given below.

PRECAUTION:
Capacitors in series:-
Let us consider three capacitors of capacitances C1, C2 and C3 respectively connected in series as
shown in the figure.

Due to the inducement of charges on the plates by the applied potential difference
1) Charge on each capacitor is same irrespective of its capacitance.
2) Potential difference across each capacitor is different
3) Sum of the voltage across the capacitors is equal to the applied battery
Voltage “V”.

So, V = V1 + V2 + V3 -----------------→ (1)

𝑄 𝑄
We know C = 𝑉 (or) V = 𝐶
𝑄 𝑄
So the potential difference between the first capacitor is C 1 = 𝑉 or V1 = 𝐶
1 1
Similarly for the second and third capacitors
𝑄 𝑄
V2 = 𝐶 and V3 = 𝐶
2 3

substituting all in equation (1)

Q Q Q Q
= + +
C C1 C2 C3
If Cs is the effective capacitance of the series connection, then

Q 1 1 1
= Q ( + + )
Cs C1 C2 C3

206
 Cancelling Q on both sides

1 1 1 1
= ( + + )
Cs C1 C2 C3

When capacitors are connected in series the reciprocal of the effective capacitance is equal to the
sum of the reciprocals of the individual capacitances.
Hence connecting capacitors in series is equivalent to increasing the thickness of the electric
medium. When we increase the thickness, distance increases. So capacitance decreases.
Capacitors connected in series block DC and allow AC.
Capacitors Connected in Parallel :-
Let us consider three capacitors of capacitances C1, C2 and C3 respectively connected in parallel as
shown in the diagram.
The charge Q supplied by the battery is divided among the three capacitors as Q1 , Q2 and Q3.
Therefore,
1) Charges across each capacitor are different.
2) p.d. across all capacitors is same
3) Sum of the charges stored in the capacitors is Q

 Q = Q1 + Q2 + Q3 (1)
Q (or) Q = CV
We know C =
V
The charge stored on the first capacitor is
Q1 = C1 V.
Similarly for other two capacitors
Q2 = C2 V and Q3 = C3 V.
Substituting this in equation (1) above
CV = C1 V + C2 V + C3 V
If Cp is the effective capacitance of the parallel connection,

then Cp V = V (C1 + C2 + C3)

207
 Cancelling the “V” on both sides
Cp = (C1 + C2 + C3)
When capacitors are connected in parallel, the effective capacitance is equal to the sum of
the individual capacitances
Connecting capacitors in parallel is equivalent to adding their plate areas. Hence, combined
capacitance equals to sum of the individual capacitances.
Application of Capacitors:
1. Capacitors are used to accurately measure the fuel level in airplanes fuel tank. As the fuel
level between the plates varies, the Capacitance also varies.
2. In Radio, the variable capacitors are used with inductance to tune the receiver to a different
resonant frequency for receiving the programmes of different stations transmitting different
frequency waves.
3. A camera flash unit contains a capacitor. The light energy for each “flash” from the camera
is from electrical energy stored in the capacitor. Here electrical energy is converted as light
energy.
WORKED PROBLEMS
1. Three capacitors each of capacitance 9 pF are connected in series. Calculate the
effective capacitance of the capacitors and potential difference across each capacitor,
when they are powered by 120V DC supply.
Given values:-
C1 = 9 pF, C2 = 9 pF and C3 = 9 pF
In series connection with p.d. of 120V.
C1 = C2 = C3 = 9pF

For series connection,

1 1 1 1 1
= +𝐶 +𝐶 =3x𝐶
𝐶𝑠 𝐶

1 1 1
=3 −12
=
Cs 9 10 3 10−12
Therefore Cs = 3 × 10 –12 F = 3 pF
In series connection, sum of the p.d across the capacitor is equal to supply voltage 120 V
V = V +V +V
1 2 3
q q q q q q
120 = + + = + +
C1 C2 C3 C C C
q 3q q
120 = 3 = 120 = 3  10−12
C 9  10−12
q = 3 × 120 × 10-12 = 360 pC
Therefore, potential difference across each capacitor
q 360  10−12
V= = = 40 V
Capacitance 9 10−12

208
2. Calculate the resistance of a wire of length 0.75m and diameter 0.3 mm. Specific resistance of
the material of the wire is 48 x 10-4 ohm m.
Length of the wire l =0.75 m
Diameter of the wire, d = 0.3 mm Radius of the wire = 0.15 mm = 0.1510-3 m
Specific resistance of the wire, = 48 x 10-4 ohm m Resistance of the wire = ?
𝑙
R = 𝐴 Where A = area of cross section of the wire

0.75
 R = 48 x 10-4 x
3.14 × (0.15 × 10−3)2

3.6 × 10−4
=
0.07065 × 10−6

Resistance of the wire = 5095.5 ohm

3. A thin wire of radius 0.2 mm has a resistance of 7 ohm. If the length of the wire is 2 m
calculate the resistivity of the material of the wire.

Radius of the wire = 0.2 mm = 0.210-3 m

Length of the wire l =2 m
Resistance of the wire = 7 ohm Specific resistance of the wire = ?
𝐴
=R Where A = area of cross section of the wire
𝑙
2
3.14 × (0.2 × 10−3 )
  = 7x
2
0.8792 × 10−6
 = = 0.4396  10-6m
2

The resistivity of the wire,  = 0.4396  10-6m

4. Calculate the effective capacitance of three capacitors of capacitance 1 mF, 2 mF and 4

mF respectively, when they are connected in series and parallel configuration in a circuit.

Given values:-
C1 = 1 mF, C2 = 2 mF and C3 = 4 mF
(a) When they are connected in series
1 1 1 1
= (C + + )
Cs 1 C2 C3

209
 Cs =0.571 × 10–3 = 0.571 mF
(b) When they are connected in parallel
Cp = C1 + C2 + C3 = (1 + 2 + 4) × 10–3 F
Cp = 7 × 10–3 F C = 7 mF
p

5. Calculate the resistance of a wire of length 1.5 m and diameter 0.8 mm.
The specific resistance of the material of the wire is 45 × 10-8 ohm m
Given
l = 1.5 m
d = 0.8 mm r = 0.4 mm
-3
= 0.4 × 10 m
 = 45 x 10-8 ohm m
RA
We know specific resistance  =
𝑙
l
Resistance R = 𝐴
𝑙
i.e, R = 𝜋𝑟2
45 𝑥 10−8 𝑥 1.5
= 1.343 ohm.
3.14 𝑥 (0.4 𝑥 10−3 ) 2

6. Calculate the resistivity of a material of a wire whose resistance is 1.5 ohm with a
diameter of 0.32 cm and length 80 cm.
Given
R= 1.5 ohm
Diameter = 0.32 cm r = 0.16 cm
-2
r = 0.16 × 10 m
l = 80 cm
= 80 × 10-2 m.

RA
We know specific resistance  =
l

Rr 2
=
l

210
 1.5  3.14  (0.16  10−2 )2
=
80 ×10-2
= 15.07 × 10-6 ohm m

7. Three capacitors of capacitances 5, 10 and 20 micro farads are connected in series. Calculate
the equivalent capacitance.

C1 = 5 µF C2 = 10 µF C3 = 20 µF CS= ?
1 1 1 1
𝐶𝑆
= 𝐶1
+ 𝐶2
+ 𝐶3

1 1 1
= + + = 0.2 + 0.1 + 0.05
5 10 20

1
= 0.35
𝐶𝑆

1
Cs = 0.35 = 2.857

Equivalent capacitance = 2.857 µF

8. The capacitors of values 20 µF, 30µF and 60µFare connected (i) in series and (ii) in parallel.
Calculate the effective capacitance in each case.

C1 = 5 µF C2 = 10 µF C3 = 20 µF

(i) In series:

1 1 1 1
= + +
𝐶𝑆 𝐶1 𝐶2 𝐶3

1 1 1
= + + 60=0.05 + 0.0333 + 0.0166
20 30

1
= 0.0999
𝐶𝑆

1
Cs = 0.0999 = 10

Equivalent capacitance when connected in series = 10 µF

(ii) In parallel:

CP = C1 + C2 + C3

= 20 + 30 + 60 = 110

Cs = 110 µF

Equivalent capacitance when connected in parallel = 110 µF

211
9. Show when four equal capacitors are connected in parallel the system has sixteen times the
capacitance of that obtained when they are joined in series.

Let us consider four equal capacitors, C1 = C2 = C3 = C4 = C

When connected in parallel, CP = C1 + C2 + C3 + C4 = 4 C

CP = C + C + C + C = 4 C - - - - - - - - (1)
1 1 1 1 1
When connected in series, 𝐶 = 𝐶1
+ 𝐶2
+ 𝐶3
+𝐶
𝑆 4

1 1 1 1 1 4
𝐶 = 𝐶 + 𝐶 +𝐶 + 𝐶 = 𝐶
𝑆

i.e., C = 4 Cs - - - - - - - (2)

Substituting equation (2) in (1), we get Cp = 4  4 Cs

CP = 16 Cs, hence proved.

212
 4.2 EFFECTS OF CURRENT
If a positive charge is moved from a point at a higher potential to a point at a lower potential, energy
is liberated. When current flows in a conductor, the electrons collide with the ions because of the
acceleration of electrons and energy is transferred to the ions. Therefore, there is increase in the energy
of the ions and the temperature of the conductor rises and thereby it is heated. This is called heating
effect of electric current. This heating effect was studied by Joule.
Joule’s law of heating
The quantity of heat developed ( H ) in a conductor is
(1) directly proportional to the square of the current passing through the conductor ie H α I 2
(1) directly proportional to the resistance of the conductor ie H α R
(iii) directly proportional to the time of flow of current ie H α t
ie H α I2 Rt
H = I2 Rt
where I is in ampere ,R is in ohm and t is in second
But R = E/I
E
∵ H = I2 t
I
H = EI t joule

The heating effect of electric current is applied in the working principle of electric fuse wire, electric
heater, electric welding and electric furnace.
Application of heating effect of electric current:
The incandescent electric lamp:
The heating effect of electric current is used for the production of artificial light. If a current is
passed through a filament of high resistance and high melting point, it is rendered incandescent and the
filament emits bright light. The higher the melting point of the material, the higher the temperature it can
withstand and higher will be the light emitted.
Edison first discovered the carbon filament lamp, the filament being enclosed inside an
evacuated glass bulb. In modern lamps tungsten filaments are used. The filament is enclosed inside a
glass bulb containing an inert gas.
Electrical fuses:
Electrical fuses are included in electric circuits as a safety device. An electric fuse consists of a
wire made of tin or an alloy of tin of low melting point.
If the current in the circuit exceeds the safe limit due to accidental short circuiting, the fuse wire
melts and the circuit gets broke, thus preventing damage.
The heating effect of electric currents is used in the electric heater, electric iron, electric
immersion heater electric solder etc. All this consists of a suitable heating element made of nichrome.
Experimental determination of specific heat capacity of a liquid using Joule’s
calorimeter
The specific heat capacity of a liquid can be determined experimentally using Joule’s calorimeter.

The Joule’s calorimeter is a polished copper or aluminium vessel containing the given liquid.

213
 It is covered by a non-conducting lid and a suitable resistance coil is immersed in the liquid and its
ends are connected to connecting terminals provided in the lid. There are holes in the lid for a stirrer and
inserting a thermometer. The calorimeter is placed inside a wooden box and it is filled with cotton or
wool to minimize the radiation losses.

The mass of the clean, empty calorimeter with stirrer is determined and it is taken as m1 kg. Now
the calorimeter is filled with the given liquid and its mass is determined with stirrer and it is taken as
m2 kg. The heating coil should be immersed well in the liquid .The calorimeter is now placed inside the
wooden box.
A battery, an ammeter, a key and a rheostat are connected in series with the coil and a voltmeter is
connected in parallel across the coil. The current is adjusted for about one ampere. The circuit is kept
open. The ammeter reading and the voltmeter reading are noted as I and E. Now, a thermometer is
inserted through the hole in the lid. The initial temperature of the liquid is noted (1 C). When the
circuit is closed, simultaneously a stop clock is started. The liquid is stirred well and when the
temperature rises about 50 C, the circuit is kept open and the stop clock is stopped and the final
temperature  0 C is noted.
The time is noted as t second.
The heat produced by the heating coil = E I t joule.
Heat is absorbed by both the calorimeter and the liquid.
Heat absorbed by the calorimeter = m1 s( -  ) joule
Heat produced = Heat absorbed
i.e. EIt = m1s(2 - 1) + (m2 – m1 ) S (2 - 1 ) J kg-1 K-1
The specific heat capacity of the liquid S is given by
EIt m1s
− J kg-1 K-1
(m2 − m1 ) (2 − 1 ) (m2 − m1)

CHEMICAL EFFECT OF CURRENT

The phenomenon of conduction of electricity through liquid was first studied by Faraday in 1833.
Pure liquid does not conduct electricity, but the conductivity of a liquid increases if the solution contains
an acid. The molecules of a salt in solution are supposed to break up into two parts, positive and negative
ions and this process is called dissociation.
When an electric field is applied between two electrodes dipping in the solution, the
positively charged ions move towards the negative electrode (cathode) and they are called as
Cations. The negatively charged ions move towards the positive electrode (anode) and they are called
as Anions.
The process of chemical decomposition of an electrolyte by the passage of electric current through
it is called electrolysis.

214
 The liquid containing the compound is called an electrolyte and the vessel containing the electrolyte
is called a voltameter.
Faraday’s laws of electrolysis
The two laws of electrolysis are stated as follows.
First law
The mass of an element liberated at cathode from an electrolyte by passing of electric current is
directly proportional to
(1) The strength of the current (I)
(ii)The time for which the current is passed through the electrolyte (t)
ie m  I
and m  t
ie m  I t
or m = ZIt
Here, the constant Z is known as the electrochemical equivalent (e.c.e) of the element.
Second law
If the same current passes through different electrolytes for the same time then the masses of the
elements liberated from electrolytes are directly proportional to their respective chemical equivalent
weights.
Let m1, m2 and m3 be the masses of the elements liberated by the passage of the same current through
the electrolytes for the same time whose chemical equivalent weights are E1, E2 and E3, then according
to second law,
m1  E1; m2  E2 and m3  E3
m1 m2 m3
or = =
E1 E 2 E3
Electro-chemical equivalent (e.c.e) of an element
According to Faraday’s First law of electrolysis
m α It
or m = ZIt
Where Z is a constant called electrochemical equivalent of the given element
When I = 1 ampere and t = 1 second, then Z = m
Therefore the electro-chemical equivalent of an element is defined as the mass of the element
liberated from an electrolyte when one ampere current is passed through it for one second. Since
Z = m/It the unit of e.c.e is kilogram per ampere second or kg C-1 (ampere second = coulomb).
Experimental determination of e.c.e of copper

215
 Copper voltameter is used for the determination of electro-chemical equivalent of copper. It
consists of a glass vessel containing copper sulphate solution and this is the electrolyte. There are three
electrodes of copper plates dipped in the solution. The two outer plates are connected to the positive
terminal of the battery through the ammeter and they act as anodes. The central copper plate is connected
to the negative terminal of the battery and this plate is the cathode plate.
A battery, a plug key, a rheostat and an ammeter are connected in series with the copper voltameter.
The circuit is closed and the rheostat is adjusted to send suitable current for a while and then the circuit
is kept open.
The cathode plate is taken out, polished well with sand paper, washed and dried. The mass of the
cathode is determined carefully correct to a milligram, using a digital balance. The cathode is then kept
in its position and when the circuit is closed, simultaneously a stop clock is started. The current value is
maintained the same by adjusting the rheostat, if necessary, and it is allowed to pass for thirty minutes.
After the time the current is stopped and the cathode plate is taken out, rinsed with water, dried and then
weighed correct to a milligram.
The difference between the final and the initial mass gives the mass of copper deposited on the
cathode plate and this is taken as m. Let I be the current that is passed through the electrolyte and t the
time for which current is passed. Then electro-chemical equivalent
Z = (m / It) Kg C–1
Thus, the e.c.e of copper is determined using the above equation.

WORKED PROBLEMS
1. Calculate the electro-chemical equivalent of copper when a current of one ampere passing
through a copper voltmeter for one hour liberates 1.2 g of copper.
Given,
m = 1.2 g
= 1.2 x 10-3 kg
t = 1 hour
= 3600 s
I = 1 ampere
m
ECE of copper Z =
It
1.2  10−3
=
1  3600
= 3.33 × 10-7 kg C-1
2. Calculate the mass of copper deposited when a current of one ampere flows for 30 minutes
in a copper voltameter. e.c.e of copper = 3.33 × 10-7 kg C-1.
Given,
I = 1 ampere
t = 30 minutes
= 30 × 60
= 1800 s
Z = 3.33 × 10-7 kg C-1

216
 m
We know Z =
It
m=ZIt
m = 3.33 × 10-7 × 1 × 1800
m = 5.994 × 10-4 kg.

3. A current of one ampere passing through a copper voltameter for one hour liberates 1.2 g of
copper. Calculate the electro chemical equivalent of copper metal.

Current = 1 A

Time of current flow t = 1 hr = 1  60  60 = 3600 s

Mass of the copper liberated m = 1.2 g = 1.2  10-3 kg

𝑚
e.c.e of copper, e =
𝐼𝑡

1.2 × 10−3
= = 3.33 10-7
1 ×3600

e.c.e of copper, e = 3.33  10-7 kg C-1

4. Calculate the strength of the current which will deposit 10 -3 kg of gold in 30 minutes. Electro
chemical equivalent of gold is 0.678 10-6 kg C-1.

Mass of the gold deposited m = 10-3 kg

Time of current flow t = 30 minutes = 3060 =1800 s

e.c.e of gold, e = 0.678  10-6 kg C-1 Current = ?

𝑚
We know that, e.c.e of gold, e =𝐼𝑡

𝑚
 Current I = 𝑒 𝑡
10−3
= = 0.819
0.678 ×10−6 ×1800

Strength of the current, I = 0.819 A

5. How long will it take a current of 0.5 ampere to deposit 0.5  10-3 kg of copper? ECE of
copper is 3.33  10-7 kg/C.
Current = 0.5 A
Mass of the copper deposited m = 0.5 10-3 kg

217
 e.c.e of copper, e = 3.33  10-7 kg C-1
Time of current flow t = ?
𝑚
We know that, e.c.e of copper, e =𝐼𝑡
𝑚
 Time, t = 𝐼 𝑒
0.5 × 10−3
= 0.5 ×3.33 × 10−7
= 3030

Time of current flow t = 3030 s

6. A thin metal plate having total surface area 1  10-2 m2 is silver coated electrically. If the
current of 2 A is passed for 2 hour, find the mean thickness of deposits. Density of silver is
10.4  103 kgm-3 and e.c.e. of silver is 1.118  10-6 kgC-1.

Surface area of the plate, A = 1  10-2 m2

Current = 2 A

Time of current flow t = 2hr = 2 60  60 = 7200 s

Density of silver,  = 10.4  103 kgm-3

e.c.e of silver, e = 1.118  10-6 kg C-1

𝑚
We know that, e.c.e of silver, e =
𝐼𝑡

mass of silver deposited, m = e I t -------(1)

𝑚𝑎𝑠𝑠
Density,  = 𝑉𝑜𝑙𝑢𝑚𝑒

mass, m = volume  = area  thickness 

= A  d  ------ (2)

Equating equations (1) and (2), we get,

A d  = e I t
𝑒𝐼𝑡
d = 𝐴 

1.118 × 10 −6 × 2 ×7200 0.016099

= = = 1.548  10-4
1 × 10−2 × 10.4 × 103 104

Mean thickness of the deposit, d = 1.548  10-4 m

218
7. Calculate the current required to deposit 3.3 g of silver in 45 minutes. e.c.e of silver = 1.118 ×
10-6 kg C-1
Given,
m = 3.3g
= 3.3 × 10-3 kg
t = 45 minutes
= 45 × 60
= 2700 s
Z = 1.118 × 10-6 kg C-1

m
We know that Z = It

𝑚
I = 𝑍𝑡
3.3  10−3
=
1.118  10−6  2700
I = 1.093 ampere

8. 2 × 10-3 kg of gold is to be deposited electrolytically by passing a current of 1.2 ampere. What

will be the time taken ? e.c.e of gold is 68 × 10–6 kg C–1
Given,
m = 2 × 10-3 kg
I = 1.2 ampere
Z = 0.68 × 10-6 kg C-1
We know,
m
Z =
It
m
t=
ZI
2  10−3
t=
0.68  10−6  1.2
t = 2451 s

219
 4.3 MEASURING INSTRUMENTS
When charges are in motion they produce magnetic effect. An electric current produces
magnetic field. Therefore moving charged particles in magnetic field experience force.
Expression for the force acting on a current carrying straight conductor placed in a
uniform magnetic field.
Let us consider a small element PQ of a straight conductor of length l, which carries a current I .
This is placed in a uniform magnetic field of magnetic induction B. The current passing through the
conductor is due to the flow of electric charges. Let θ be the angle between the direction of current flow
and the magnetic induction B. Let there be ‘n’ number of free charges each of charge ‘q’ moving with a
velocity v through the conductor.

We know force on each charge F= B q v sin θ

Since there are n numbers of charges moving in the conductor of length l,
The total force experienced by the conductor is given by
F = B n q v sin θ
In a conductor current is the rate of flow of charges
Charge
i.e. Current = time
nq
I =
t
Here t is the time taken by a charge to travel a distance l of the conductor.
From the above relation n q = I t
The total force experienced by the conductor is given by
F = B I t v sin θ
The velocity of charges is the distance travelled by them in time t
l
i.e., v =
t
l
F = BIt sin 
t
F = B I l sin 
This expression gives the force on a straight conductor placed in a uniform magnetic field.
Suppose if the straight conductor is placed at right angles to the magnetic field then θ = 900
The force acting on the conductor is given by
F = BI l sin 900
ie F = BI l (sin 900 = 1)
The direction of the force is given by Fleming's left hand rule.

220
Fleming's Left Hand rule

The direction of motion of current carrying conductor placed in a uniform magnetic field is given
by Fleming's Left Hand rule.
According to this rule, when the forefinger, middle finger and thumb of the left hand are
stretched mutually perpendicular to each other and if the forefinger represents the direction of
magnetic field, the middle finger represents the direction of current in the conductor, then the
thumb gives the direction of motion of the conductor.
Expression for the torque experienced by a rectangular current carrying coil placed in a
uniform magnetic field
Consider a rectangular coil ABCD placed in a uniform magnetic field of induction B. Let l be
the length of the coil and ‘b’ its breadth. The magnetic field acts parallel to the plane of the coil and
perpendicular to the sides AB and CD. Let I be the current passing through the coil.
Now the side AB of length l experiences a force B I l ,since the angle between this side and the
magnetic field is 90o. The direction of this force is given by Fleming's left hand rule. The side CD also
experiences an equal force of B I l but in the opposite direction, since the current on this side flows in
opposite direction.
But the sides BC and AD are parallel to the direction of the magnetic field and hence there is no force
acting on these two sides.

Therefore, the two equal unlike and parallel forces acting on the sides AB and CD constitute a
couple. Torque is the moment of the couple in the coil.
Moment of the couple = One of the forces × the perpendicular
distance between the forces.
This perpendicular distance is the distance between the two sides AB and CD, measured along the
direction of the field. It should be noted that this distance is not a constant and varies from position to
position of the coil. The breadth BC of the coil (b) gives the maximum distance.
The maximum torque on the coil = B I l × b = B I A
(where l × b = A, The area of cross section of the rectangular coil )
If there is 'n' number of turns in the coil then the maximum torque is given by n B I A.

221
Working of a Moving coil galvanometer
The principle of the moving coil galvanometer is that a coil of conductor carrying current placed in
a magnetic field experiences a torque.
The moving coil galvanometer consists of a rectangular coil of wire of many turns wound on a
nonmagnetic metallic frame and is suspended between two strong permanent magnetic poles N and S.
The coil is suspended by means of a fine phosphor bronze wire from a movable torsion head. The
coil is held tight by a hair spring S. A small light mirror M is attached to the suspension fibre and the
angle of deflection of the coil can be measured with a lamp and scale arrangement. A small soft iron
cylinder is fixed at the centre of the coil for producing a radial magnetic field.

Now, when a current of I ampere passes through the vertical limb AD of the rectangular coil ABCD,
it experiences a force perpendicular to it and directed outwards according to Fleming's left hand rule.
Similarly the limb BC experiences an equal and opposite force to that of AD and directed inwards.
It should be noted that in all the position of the coil, the forces on the sides AD and BC will always
perpendicular to the magnetic field, since the plane of the coil remains parallel to the magnetic field. No
force acts on the sides AB and CD, since they are parallel to the magnetic field. The coil will come to
rest in a deflected position.
According to Fleming's left hand rule the forces experienced by the sides AD and BC are B I l and
B I l where, l is the length of the sides and B is the magnetic induction. These forces constitute a couple
and the controlling couple is provided by the torsion in the phosphor bronze suspension fibre.
For one turn of the coil, the deflection couple produced = B I l × b where b is the breadth of the
coil.
For 'n' turns of the coil, the deflection couple produced = n BI l × b
= nB IA
where l × b = A, the area of coil.
If C is the couple per unit twist of the suspension wire then for a twist θ, the restoring couple = C θ
(The deflection θ can be measured from the lamp and scale arrangement)
For equilibrium position,

Deflecting Couple = Restoring couple;

i.e; nBIA = cθ
C
I= .
nBA

222
 The current required (I / ) to produce unit deflection is called the figure of merit of the Galvano
C
meter, K =
nBA
Since C, n , B and A are all constants I = K
i.e, I

The current is directly proportional to the deflection produced. For a small value of I, if the
deflection θ is large, then the galvanometer is said to be sensitive.
Merits of moving coil galvanometer
(i) The external magnetic fields do not alter the deflection of the coil since there is a strong
magnetic field between the pole pieces, and the galvanometer can be placed in any position.
(ii) Since the deflection of the coil is directly proportional to the current, a uniformly divided scale
can be used to measure the current.

Conversion of a galvanometer in to an ammeter

A galvanometer is a very sensitive device used to detect the strength of flow of current and large
current cannot be passed through it, because large current flow can easily damage the coil. Therefore a
low resistance called shunt is connected in parallel to the galvanometer to avoid the damage of the coil
and only a fraction of the current will pass through the galvanometer and most of the current passes
through the shunt.
A galvanometer can be converted into an ammeter by connecting a suitable low resistance called
shunt in parallel to it.

Consider a galvanometer of resistance G ohm. If the galvanometer is to be converted into an

ammeter, a low resistance called shunt S should be connected in parallel with the galvanometer.
Let Ig be the maximum current that can be passed through the galvanometer and this current will
give the full scale deflection in the galvanometer. Therefore out of the total current of I ampere in the
circuit, the current passing through the shunt is
= (I - Ig)
Since the galvanometer and the shunt are parallel.
Ig G = (I – Ig) S
𝐼𝑔 𝐺
S = 𝐼− 𝐼
𝑔

The current Ig is a fraction of I and the scale is marked in ampere. In an electrical circuit, ammeter
is always to be connected in series.

223
Conversion of a galvanometer into a voltmeter
The voltmeter should have high resistance, then only it will not draw any appreciable current from
the circuit. It is used to measure the potential difference across two points of an electrical circuit.
A galvanometer can be converted into a voltmeter by connecting a high resistance in series with the
galvanometer.

Consider a galvanometer of resistance G ohm. If it is to be converted into a voltmeter, a high

resistance R should be connected in series with it to read E volt.
Let Ig be the current required to produce full scale deflection in the galvanometer. A and B are two
terminals of the voltmeter. Since the high resistance R is connected in series with the galvanometer, the
current Ig passes through both the galvanometer and the high resistance.
E = Ig (G + R)
E
i.e. Ig = G + R
E
and R = Ig – G

The scale is marked in volt. In an electrical circuit, a voltmeter is always connected in parallel
between the two terminals across which the p.d. is to be measured.
WORKED PROBLEMS
1. A galvanometer has a resistance of 5 ohm and full scale deflection is produced by 15
mA. Calculate what resistance must be used to enable it to read (i) 1.5 ampere (ii) 1.5 volt.

Given,
Full scale deflection Ig = 15 mA = 15 × 10-3
A Galvanometer resistance G = 5 ohm
I = 1.5 A
Ig G
We know, S = I – Ig

1.5  10−3  5
S = = 0.05 ohm
1.5 − (15  10−3 )
A shunt of 0.05 ohm is to be connected in parallel with the galvanometer so as to measure
upto 1.5 ampere.

Potential difference E = 1.5 volt

224
 We know, R = E -G
Ig
1.5
= −5
15  10−3
= 95 ohm
A resistance of 95 ohm is to be connected in series with the galvanometer so as to measure up to
1.5 V.

2. A galvanometer of resistance 20 ohms shows maximum deflection for a current of 5 mA. How
will you convert it to read 5 A.

G = 20  Ig = 5 mA = 5  10-3 A I=5A S=?

𝐼𝑔 𝐺
S =𝐼− 𝐼
𝑔

5 × 10−3 ×20 100 × 10−3

= = = 0.02
5−5 × 10−3 4.995

S = 0.02 

by connecting a 0.02 ohm resistance in parallel with galvanometer, we can convert.

3. A galvanometer of 30 ohms gives full scale deflection for a current of 15 mA. How will you
convert it into a voltmeter to read 100 V.

G = 30  Ig = 15 mA = 15  10-3 A V = 100 V R=?

𝑉
R=𝐼 −𝐺
𝑔

100
= - 30 =6666.66 – 30
15 × 10−3

R = 6636.66 

by connecting a 6636.66 ohm resistance in series with galvanometer, we can convert.

4. A galvanometer has a resistance of 200 ohm and when a current of 1mA passes
through it, produces a deflection of 1 division. How this can be converted into a voltmeter to
read 10 volt per division ?
Given,
G = 200 ohm
E = 10 volt
Ig = 1 mA = 1 × 10-3 A

225
 E
We know, R = -G
Ig
10
R = − 200
10−3
R = 10000 - 200 = 9800 ohm

A high resistance of 9800 ohm is to be connected in series with the galvanometer so as to measure
up to 10 volt / division

5. The resistance of a galvanometer is 100 ohm and it can withstand a maximum current of 100
mA. How this can be converted into an ammeter to read 10A.
Given,
G = 100 ohm

Ig = 100 mA = 100 × 10-3 A

I = 10 A

Ig G
We know, S = I – Ig

100 x 10−3 x 100

S=
10−100 x 10−3

S = 1.01 ohm

A shunt of 1.01 ohm should be connected in parallel with the galvanometer to measure up to 10 A.

226
 QUESTIONS
PART A & B
1. Write the law of resistances when the resistances are connected in series.
2. Write the law of resistances when the resistances are connected in parallel.
3. Define the term Resistivity.
4. Define the term Conductivity
5. What is super conductivity?
6. What is Meissner effect?
7. Write Kirchoff's current law.
8. Write Kirchoff's voltage law.
9. What is meant by heating effect of electric current?
10. Write Joule's law of heating.
11. Define electrolysis.
12. Write Faraday's first law of electrolysis.
13. Write Faraday's second law of electrolysis.
14. What is a Voltameter?
15. Define Electro-chemical equivalent (e.c.e) of an element.
16. Define capacitance of a capacitor.
17. Define farad.
18. Write the expressions for effective capacitance when two capacitors are connected in Series and
in parallel
19. Find the effective capacitance when two equal capacitors are connected in series.
20. Write Fleming's left hand rule.
21. Write the principle of a moving coil galvanometer
22. What is meant by the figure of merit of a galvanometer?
23. What are the merits of a moving coil galvanometer?

Part–C
1. Describe an experiment to determine the specific heat capacity of a liquid using Joule's
calorimeter.
2. Describe an experiment to determine the electrochemical equivalent of copper using copper
voltameter.
3. Derive the expression for effective capacitance when three capacitors are connected in series.
4. Derive the expression for effective capacitance when three capacitors are connected in
parallel.
5. Derive an expression for the force acting on a current carrying straight conductor placed in a
uniform magnetic field
6. Derive an expression for the torque experienced by a rectangular current carrying coil placed
inside a uniform magnetic field.
7. Describe the construction and working of a moving coil galvanometer with a diagram
8. Describe how a galvanometer can be converted into i) an ammeter and ii) a voltmeter.

227
 Exercise problems
1. Calculate the resistance of the wire of length 1.5 m and diameter 0.5mm if the resistivity of the
material of the wire is 49 × 10-8 ohm metre. [Ans 3.745 ohm]

2. Calculate the resistivity of the material of a wire of length 25.6 m and diameter 0.84 mm whose
resistance is 0.72 ohm [Ans 1.557 ×10-8 ohm metre]

3. Calculate the resistance of a wire of 1.5m long and area of cross - section 0.4mm2 whose
resistivity is 46 × 10-8 ohm metre [Ans 4.312 ohm]

4. Calculate the electro chemical equivalent of a metal when a current of one ampere liberates in one
hour 1.2 g of metal from its electrolyte [Ans 3.33 × 10-7 kg C-1]

5. What will be the time taken for 2.2 × 10-3 kg of gold is to be deposited electrolytically by passing
a current of one ampere ? Given, e.c.e of gold = 0.68 × 10-6 kg C-1 [Ans 3235 s]

6. Calculate the amount of current that should be passed through a copper voltameter to deposit
3.12g of copper in one hour (e.c.e of copper = 3.33 ×10-7 kg C-1) [Ans 2.6A]

7. A galvanometer of resistance 30 ohm shows maximum deflection for a current of 2 mA. How this
can be converted into a voltmeter to read 10V?
[Ans 4970 ohm connected in series with the galvanometer.]

8. An ammeter of resistance 0.60 ohm can measure current up to 1A. What must be the shunt
resistance to enable the ammeter to measure current upto 5A ? [Ans 0.15 ohm]

9. Show that when two equal capacitors are connected in parallel, the system has four times
capacitance of that obtained, when they are connected in series (C p =4Cs )

10. Find the resultant capacitance of three capacitors of capacitance 20mF, 30mF and 50mF,
when they are connected in series and in parallel.

11. The capacitors of capacitance 3mF and 5mF are connected in series. Calculate the effective
capacitance of the capacitors and potential difference across each Capacitor, when they are
powered by 60 V DC supply.
Brain Teaser
1. In the electrification of a house how the various lamps are connected? Explain Why?
2. Why does a silver voltameter give accurate results than a copper voltameter?
3. Two capacitors of capacitance 3mF and 6mF are connected in series and charged by
connecting a battery of 10V. Next replace the battery from the circuit, connect a wire
between the terminals (show circuiting process). What is the final charge stored on the
capacitors?

228
https://www.youtube.com/watch?v=PPUaEMPLwUc&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=15

https://www.youtube.com/watch?v=5B3vw6mCJjg&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=16

https://www.youtube.com/watch?v=9qd4rDY7ZB0&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=17

https://www.youtube.com/watch?v=keGuki4tDOs&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=18

https://www.youtube.com/watch?v=qx528dZ1jhc&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=20

229
 SUMMARY
• At constant temperature, the strength of the current passing through a
conductor is directly proportional to the potential difference between the
ends. I α R (or) V = IR
• Equivalent resistance when three resistors connected in series,
RS = R1 + R2 + R3
• Equivalent resistance when three resistors connected in parallel,
1 1 1 1
𝑅𝑃
= 𝑃1
+ 𝑃2
+ 𝑃3

• Resistivity is defined as the resistance of the conductor having unit length

and unit area of cross section.
• Conductance is inversely proportional to resistance
• The complete ejection of magnetic field lines from the interior of the
superconductor as it transitions into the superconducting state is known as
Meissener effect.
• The algebraic sum of the currents meeting at any junction is equal to zero.
• In any closed circuit, the algebraic sum of the products of current and
resistance is equal to the algebraic sum of the e.m.f’s in the circuit.
• Equivalent capacitance when three capacitors connected in series,
1 1 1 1
𝐶𝑆
= 𝐶1
+ 𝐶2
+ 𝐶3
. Simple problems.
• Equivalent capacitance when three capacitors connected in parallel,
CP = C1 + C2 + C3 . Simple problems.
• Joule’s law of heating: H = I2RT
𝑚
• Electro-chemical equivalent of a substance e = . Simple problems.
𝐼𝑡
• Force acting on a current carrying conductor, F = BIl
• Torque on a rectangular coil carrying current placed in a uniform magnetic
field, Torque = B I A.
• Construction of moving coil galvanometer.
• A galvanometer can be converted into an ammeter by connecting a low
resistance in parallel to the galvanometer.
• A galvanometer can be converted into a voltmeter by connecting a high
resistance in series to the galvanometer. Simple problems.

230
 UNIT V

ELECTRONICS

OBJECTIVES

▪ To know about different types of semiconductors

▪ To understand about Intrinsic semiconductors

▪ Familiar with the concept of doping

▪ To understand about the extrinsic semiconductors.

▪ Concepts behind P and N type semiconductors.

▪ To know the formation of PN junction diode

▪ Able to connect the diode in forward and reverse bias.

▪ To understand the working of a diode as rectifier.

▪ Can explain the working of full wave bridge rectifier using P N junction
diodes

▪ To know the PNP and NPN transistors and different configurations.

▪ To understand the working of NPN and PNP transistor in common base

configuration

▪ To know the concepts of digital electronics.

▪ To familiar the OR, AND, NOT, NAND and NOR gates.

▪ To know the Boolean expression for the gates.

▪ To draw the Truth table for logic gates.

▪ To know about Integrated circuits

▪ To define different levels of formation of Integrated circuits and its

advantages.

231
 UNIT

5 ELECTRONICS

5.1 SEMICONDUCTORS
Introduction
The branch of science and engineering dealing with current conduction through vacuum tube or
semiconductor is known as electronics.
Initially, vacuum tubes were used in the process of electrical signal generation, amplification and
transmission. With the advent of semiconductor devices like diode, transistor and other solid state
electronic components, the vacuum tubes were replaced completely from all fields of applications. All
modern gadgets like television, computer, CD player, automatic washing machine, etc., use
microprocessors having integrated chips which consist a large number of logic gates, diodes,
transistors, resistors, etc.,
Materials are classified as conductors, insulators and semiconductors according to their electric
conductivity. This classification is purely based on the available number of free electrons within the
materials, apart from the bonded orbital electrons.
So the word conductivity is used to describe a material’s ability to transport electricity. The metals
like copper and gold are good conductors. Glass and plastics, on the other hand, are very bad
conductors. Semi conducting materials like germanium and silicon has conductivity somewhere
between good conductors and insulators.
In semiconductors, charges movement can be manipulated according to our need to make electronic
devices. So silicon and germanium are used as a base material for making electronic components like
diode, transistors, etc.
New emerging organic semiconductors with similar properties resembling conventional semi-
conductors are available now. In general, now a day's semiconductors are classified as inorganic and
organic semiconductors. One can fabricate diode, transistors and other devices using organic
semiconductors which are bio degradable electronic devices.
Devices fabricated using inorganic semiconductors like silicon, Germanium, Gallium Arsenide,
Silicon carbide etc., are not bio degradable which poses threat to our green environment. So the new
organic semiconductors will give us biodegradable materials, to lead green electronic world.

Energy band diagram for Conductors, insulators and semiconductors:

Our interest is focused on conducting property of a material. In general, physical and chemical
properties of an element are decided by the valence orbital electrons. For example, if valence orbit is
filled, with element is inert like He, Ne, etc. On the other hand the materials with unfilled valence orbit
exhibit electrical and magnetic properties like metals. So, only valence band is considered for further
studies. The other low level bands are not contributing any significant changes in the conducting
properties.
The energy band occupied by the valence electrons is called valence band (V.B.)
Next, the conducting band (C.B) lies above the valence band. Conduction band is due to the free
electrons from the atoms of the crystal or solid. These free electrons possess kinetic energy and acts as a
carrier within the crystal.
The energy band diagrams for good conductors, insulators and semi conductor are shown in the
figure below.

232
 The above energy band diagram has a gap between V.B. and C.B, named as forbidden energy
gap. So based on the energy gap between the valence band and conduction band, the materials are
classified as good conductor, bad conductor and semiconductor.
(a) For good conductors, Eg = 0. Here valence band and conduction band is overlapping each
other:
Zinc is a good example for conductor which has overlap of valence band and conduction band. In
Zinc, valence band is completely filled and the conduction band is empty, but they overlap each other.
Because of overlap. it is considered as single band. Hence conduction band is assumed as partially filled.
Therefore, on applying a small electric field, the metals conduct electricity
(b) For Insulator, Eg value is large.
In insulators like diamond, the forbidden energy gap is quite large. The forbidden energy gap
value is 6 e V, so minimum of 6 eV energy is required for electron to move from valence band to
conduction band. Normally conduction band is empty and valence band is full for insulators. Since there
is no electron for conduction, these type of materials behave as insulators.
(c) For Semiconducting materials, the Eg value is very small.
The energy band structure of the semi conductor is similar to the insulator, but the forbidden gap
is very narrow. The forbidden energy gap is 1.1 eV for Silicon. The semi conductor acts as an insulator at
0 K and at room temperature the thermal energy absorbed from the surrounding is enough for the electron
to jump from the valence band to the conduction band. So a semiconductor starts conducting at room
temperature.
Intrinsic Semiconductors:-
In general, semiconductors are classified as intrinsic and extrinsic semiconductors. Semiconductors
in pure form are called intrinsic semiconductor. Example:- A crystal formed by silicon atoms alone. The
silicon atoms are arranging themselves by sharing an electron between the neighbouring atoms. Such
bond is covalent bond.

233
 In intrinsic semiconductor like silicon forbidden energy gap value is 1.1 eV. This energy
is available for semiconductors placed at room temperature. Due to that thermal energy some covalent
bond within the crystal breaks (or) some electrons are pumped from valence band to conduction band.
In the bond from which electron is freed, a vacancy is created there. This absence of electron is named
as hole. Electron and hole pair is created. Thus at room temperature, a pure semiconductor will have
both electrons and holes wandering in random directions. This electron and holes are called intrinsic
carriers and such a semiconductor is called intrinsic semiconductor.
As the crystal is electrically neutral, the number of electron is equal to number of holes, in intrinsic
semiconductor.
 Fermi level lies at the middle of the forbidden gap.
Fermi level:
At 0 K, the electrons start filling the energy levels in Valence band, from bottom to top level.
Few electrons fill the bottom level and other electrons above this level. So filling of energy starts from
bottom of the band to top of the band.
The highest energy level, which an electron can occupy in the valence band at 0 K, is named as
Fermi level.
Fermi level is also defined as the energy level corresponding to the average energy of electrons
and holes present in the crystal.
Concept of holes:
The vacant place created in the valence band due to the jumping of electron from the valence band
to conduction band is called 'hole', which is having positive charge.
Doping:-
Pure semiconductor at room temperature possesses free electrons and holes but their number is so
small that conductivity offered by the pure semiconductors cannot be used for any practical purpose like
device making.
By the addition of certain selected impurities to the pure semiconductor in a very small ratio (1:106),
the conductivity of a silicon or germanium crystal can be remarkably improved.
The process of adding impurity to a pure semiconductor crystal to improve its conductivity is
named as doping.
Extrinsic Semiconductor :-
The impurity added semiconductors are named as extrinsic semiconductors. The extrinsic
semiconductors are classified as P-type and N-type semiconductors, based on the type of impurity
atoms added to the semiconductors.
N-Type (Extrinsic) Semiconductor:-
Pentavalent element like antimony (Sb) or arsenic (As) is added to pure silicon crystals. These
impurity atoms replace some of the silicon atoms, here and there in the crystal. The added arsenic
(As) atom shares it four electrons with the neighbour atoms and release it fifth electron to the crystal
for conduction. So these pentavalent elements are called donor impurities, as they donate electrons
without creating holes.
In silicon, electron needs 1.1 eV to move from valence band to conduction band. This energy
becomes available to the semiconductor even at room temperature. So at room temperature few covalent
bonds within the material are broken by the thermal energy from the surrounding and some electrons
from the valence band are pumped to conduction band. This process leaves some absence of electrons

234
in the valence band. Electron and hole pairs are created. At the same time, the number of electrons in
the conduction band are increasing further by the addition of pentavalent impurities without any
addition of holes which already exist within the crystal. So the number of electrons are donor electrons
plus thermal electrons at room temperature.
This thermal excitation produces very less pair of electrons and holes, whereas the added impurity
donates more electrons. The majority charge carriers electrons are of the order of 1024, whereas the
minority charge carriers holes are of the order of 108 at 300 K.

Si Si Si
Conduction Band Fermi
Level
EF
Si As Si
ED
Donor
HOLE
Level
Si Si Si
Valence Band
FREE
ELECTRON

Hence, the majority charge carriers are electrons in this material. The electron carries negative
charge, so it is named as N-type semiconductor and conduction is due to large number of electrons.
As the number of electrons in the conduction band is more than the number of holes in the valance
band in N type semiconductor, the Fermi level lies nearer to the conduction band
P-type Semiconductor:-
When trivalent element like Indium, Aluminium, Boron is doped with pure silicon, the added
impurity atoms replace some of the silicon atoms, here and there in the crystal and establish covalent
band with the neighbouring atoms. Indium has three electrons but that Indium is covered by four
silicon atoms as shown in the figure below. So, one of the covalent bonds is not completed by
sharing of electrons between them. There is an absence of electron which creates a hole.
Indium needs one more electron to complete its covalent band. So indium is an acceptor of electrons.
Now, this extrinsic semiconductor gains thermal energy from the surrounding at room temperature. So
some electron absorbs this thermal energy and jumps to the conduction band. This creates electrons and
holes pair due to thermal excitation. So the total number of holes in the valance band is due to donor's
atom plus thermally generated holes.
The holes are majority and electrons are minority. Hence conduction is due to majority charge
carriers which are holes. Here the holes are behaving like positive charge carriers. So this material is
named as P-type.
As the number of holes in the valence band is more than the number of electrons in the conduction
band, the Fermi level lies nearer to the valence band in P type semiconductor.

235
 5.2. DIODES AND TRANSISTORS
Introduction :-
To understand electronic devices and circuits, brief idea about semiconductor diodes is must. The
semiconductor diode is a fundamental two terminal electronic device, similar to a resistor. The
volt–ampere (V-I) relationship of a resistor is linear. However, the V-I characteristic of a diode is not
only nonlinear but also depends on the operating condition. That is resistor allow the charge carriers
at any condition and behaves like passive element. A diode allows current to pass through it in one
direction and acts as a switch in electronic circuits.
P-N junction diode:-
At the room temperature, a piece of P-type material has majority of holes and N-type material has a
majority of electrons. When a part of intrinsic semiconductor piece is doped with pentavalent impurities
and the remaining part is doped with trivalent impurities, a P-N junction diode is formed.

P-region -ve ion +ve ion N-region

Holes Majority Electrons Majority

Symbol

p n

Potential barrier

Potential barrier and depletion region:

In the PN junction diode as shown in the above figure, the P region has circles with a negative sign
indicating immobile ions and the mobile holes are represented by small circle.
In the N region, the circles with a positive sign inside represents immobile ions and the mobile
free electrons are represented by small dots.
During the junction formation, the free electrons and holes on both sides of the junction migrate
across the junction by the process of diffusion. The electrons passing through the junction from N
region into the P region recombines with holes in the P region very close to the junction.
Similarly the holes crossing the junction from the P region into the N region recombine with the
electrons in the N region very close to the junction. This recombination of free or mobile charges
produces a narrow region on either sides of the junction of width about 10–4 cm to 10–6 cm.
This region is called the depletion region, where there are no mobile charges available.
In the depletion region, the atoms on the left side of the junction become negative ions (-) and the
atoms on the right side of the junction become positive ions (+).Thus an internal potential difference
(p.d.) is produced across the junction. This p.d. is called the internal potential barrier. This prevents the
further flow of charge carrier from P-region to N-region. The potential barrier for germanium is 0.3
V and that for silicon is 0.7 V. The potential barrier of a PN junction diode can be decreased or
increased by applying external voltage.

236
 This barrier acts like a battery. However the value of potential barrier depends upon the number of
diffused impurity atoms within the silicon crystal (or) depends upon the dopant concentration.
The arrow mark or arrow-head represents holes current flow direction (Conventional Current flow
direction), when they form a circuit with an external source.
Forward and Reverse biasing of P-N junction diode:
Biasing:
Applying a suitable d.c. voltage to a diode is known as biasing. It can be done in two ways.
Forward Biasing:
When the positive terminal of the battery is connected to the P-type semiconductor and the
negative terminal to the N-type semiconductor of the P-N junction diode, the junction is said to be
forward biased.
When the applied voltage is increased from zero, the holes and the electrons move towards the
junction. Therefore the depletion layer is decreased and disappeared i.e., the potential barrier is
disappeared.

While the holes and the electrons move across the junction, they combine together and get
neutralized. Electrons from the battery move into the N-type semiconductor. Further they move across
the junction and come out of the P-type semiconductor. Hence there is a flow of current. The current
is due to flow of electrons. Thus the P-N junction diode conducts electricity. As the battery voltage
increases, the current also increases. The current is of the order of milli ampere (mA).
Reverse biasing:
When the negative terminal of the battery is connected to the P-type semiconductor and the positive
terminal to the N-type semiconductor of the P-N junction diode, the P-N junction diode is said to be
reverse biased. The negative potential of the battery attracts the holes. Similarly, the positive potential
of the battery attracts the electrons. Therefore the holes and electrons move away from the junction.
Hence the width of the depletion layer increases and there is no current flow through the junction diode
during reverse bias. However, as the reverse bias voltage increases, the minority charge carriers move
across the junction. Therefore there is a very feeble current, which is of the order of microampere
(A).

237
 Diode acts as a switch. In forward biased state diode is in “ON” position and in the reverse
biased condition the same diode is in “OFF” position. So diode is an electronic device which allows the
conventional current flow in one direction alone.
Diodes can be used in a number of ways. For example, a device that uses batteries often contains a
diode that protects the device if you insert the batteries backward. This diode simply blocks any
current from leaving the battery if it is reversed. It protects the sensitive electronic devices.
P-N junction diode –Rectification:
A P-N junction diode conducts electricity when it is forward biased and it does not conduct
electricity when it is reverse biased. Hence it is used to rectify alternating voltage (A.C).
The process in which an AC voltage is converted into a unidirectional (D.C) voltage is known as
rectification and the circuit used for the conversion is called a rectifier.
An electronic device which converts a.c. signal into d.c. signal is named as rectifier.
Principle:
It is based on the principle that a junction diode offers low resistance path for input signal, when
forward biased and high resistance under reverse biased condition. When a.c. input is applied to a
junction diode, it gets forward biased during the positive half cycle and reverse biased for negative half
cycle.
The Diode Bridge Rectifier
The two diodes labelled D3 and D1 are arranged in series pair with respect to each other.
Similarly other two diodes D2 and D4 are in series pair with respect to each other. Here the diodes act
as a switch, during the forward bias. During the reverse biased state it acts as non conductor. So two
diodes are in “ON” and other two diodes are in “OFF” condition for each half of the input signal. Only
two diodes are conducting current during each half cycle. During the positive half cycle of the supply,
diodes D1 and D2 conduct in series while diodes D3 and D4 are reverse biased and the current flows
through the load resistance as shown below.

During the negative half cycle of the supply, diodes D3 and D4 conduct in series, but diodes D1
and D2 switched off as they are now reverse biased. The current flowing through the load resistance is
in the same direction as before. That is current flow direction in the load resistance is from top to
bottom. As the current flowing through the load is unidirectional, the voltage developed across the load
is also unidirectional and thus we get d.c. voltage as output across the load resistor.

238
 Transistor :
A junction diode cannot be used for amplifying a signal. For amplification another type of
semiconductor device called 'transistor' is used. Transistor is a three sectioned semiconductor. Transistor
is a solid state device. Two P-N junction diode placed back to back form a three layer transistors.
base base

P N P N P N

emitter collector emitter collector

The three sections of the transistors are called emitter [E], base [B] and collector [C]. In a transistor
the emitter is heavily doped, since emitter has to supply majority carriers. The base is lightly doped. Two
type of transistors are available, namely N-P-N and P-N-P transistor.
Symbol for transistors:-
C C
collector collector

B B
PNP NPN
base base

E E
emitter emitter

In the symbolic representation for a transistor, the arrow mark is placed on the emitter in the
direction of conventional current flow, i.e., from P to N direction.
Applications of Transistors:
1. Transistors:
(i) Used as switches
(ii) Used in amplification
(iii) Used in high frequency amplifiers.
Three different Configurations :-
In an electronic circuit a transistor can be connected in three different ways. They are,
(I) Common base (CB)
(ii) Common emitter (CE)
(iii) Common Collector (CC)

The term common is used to denote the lead that is common to the input and output circuits. The
three different modes are shown above for NPN transistor .
For proper working of a transistor, the input junction should be forward biased and the output
junction should be reverse biased.

239
Working of PNP transistor in the common Base Configuration
In a PNP transistor the central N region is called base. The left hand P region is called emitter and
the right hand P region is called collector. The junction between the emitter and the base is called emitter
junction or input junction. Similarly the junction between the base and the collector is called collector
junction or output junction.

For proper functioning of the transistor the emitter junction is forward biased with the emitter
base voltage VEB and the collector junction is reverse biased with the base collector voltage VBC. The
positive potential of the VEB battery repels the holes in the emitter region towards the junction. Similarly,
the negative potential of the VEB battery repels the electrons in the base region towards the junction.

Hence the potential barriers is decreased and disappeared. Therefore the hole and electrons move
across the junction. Now about 5% of the holes combine with electrons and neutralized. The remaining
about 95% of the holes reaches the base region. Since the base is very thin and is connected to the
positive of the voltage VBC the 95% of the holes are repelled towards the collector junction. As soon as
they reach the collector region, they are attracted by the negative of the voltage VBC. Hence there is
collector current. To compensate the loss of electrons in the base region, the electrons from the negative
of the voltage VEB move to the base region. This constitutes the base current. In PNP transistor the
current is due to the flow of holes. If Ic is the emitter current, Ib is the base current and Ic is the collector
current according to Kirchhoff’s law,
Ie = Ib + Ic

Working of a NPN transistor in the Common Base Configuration

In NPN transistor the central P region is called base. The left hand N region is called emitter and
the right hand N region is called collector. The junction between the emitter and the base is called emitter
junction or input junction. Similarly the junction between the base and collector is called collector
junction or output junction.
For proper functioning of the transistor the emitter junction is forward biased with an emitter
base voltage VEB and the collector junction is reverse biased with a base collector voltage VBC. The
negative potential of the VEB battery repels the electrons in the emitter region towards of the emitter
junction similarly, the positive potential of the battery VEB repels the holes in the base region towards the
emitter junction. Hence the potential barrier is decreased and disappeared. Therefore the electrons and
holes move across emitter junction. Now about 5% of the electrons combine with holes and get
neutralized. The remaining about 95% of the electrons reaches the base region.

240
 Since the base is very thin and is connected to the negative of the voltage V BC the 95% of the
electrons are repelled towards the collector junction. As soon as they reach the collector region, they are
attracted by the positive of the voltage VBC. Hence there is collector current. The loss of holes in the base
region is compensated by the positive of the voltage VEB. This constitutes the base current.

In NPN transistor the current is due to flow of electrons. If Ie is the emitter current, Ib is the base
current and Ic is the collector current, according to Kirchhoff’s law.

Ie = Ib + Ic

241
 5.3 DIGITAL ELECTRONICS
Introduction
With the invention of the transistor in 1948 by W.H. Brattain and J. Bardeen, the electronic circuits
became considerably reduced in size. It was due to the fact that a transistor is cheaper, easily available,
smaller compare to vacuum tube, less power consuming. It is widely used in many electronic circuits.
With the development of printed circuit board (PCB) which further reduced the size of
electronic equipments in the early 1960's a new field of “micro electrics” was born primarily to meet the
requirements of the military, which wanted to reduce the size of its electronic equipment. The drive for
extremely small size of microelectronic circuits ends up with an invention of integrated circuits (ICs).
Integrated circuits can contain anything from one to millions of logic gates, flip-flops, multiplexers, and
other circuits in a few square millimetre. The small size of these circuits allows high speed, low power
dissipation and reduced manufacturing cost compared with board-level integration (PCB).
This increased capacity per unit area can be used to decrease cost and increase functionality.
In electronics, an integrated circuit (also known as IC chip, or microchip) is a miniaturized
electronic circuit (consisting mainly of semiconductor devices, as well as passive components) that has
been manufactured on the surface of a thin substrate of semiconductor material. Integrated circuits are
used in almost all electronic equipment in use today and have revolutionized the world of electronics.
The integrated circuit was invented by Jack Kilby and Robert Noyce. This invention is a boon for
digital technologies like computer, mobile phones, MP3 and DVD's. This list could be almost infinite.
Digital electronics :
In the modern electronic world, signal processing plays vital role. Different types of signals with
different shapes are generated by different devices. Two shapes of signals or waves are considered
here.
Analog signal and Digital signal :
The voltage signals which vary continuously with time are called continuous or analog voltage
signals. The Fig (a) below shows a typical voltage signal, varying as a sinusoidal wave of 0 to 5 v.
In general, symmetrical square wave forms a digital signal. Fig (b) below shows a digital voltage
signal, which does not vary continually with time. The values will be equal to 0V or 5V. By representing
these two voltage levels as a binary numbers, 0 and 1 can be formed.
Digital signal processing is familiar and formed a digital world. The counters, computers, etc are
outcome of digital electronics.
+5 -
5

t
+5 - V0
t

(a) (b)

Positive and Negative Logic:

In computing system, the binary number symbols “0” and “1” represent two possible states of a
circuit or an electronic device.
In positive logic, the “1” represents
1. an “ON” circuit 2. a “CLOSED” switch 3.a “HIGH” voltage
242
 4. a ”PLUS” sign 5. a ”TRUE” statement.
Consequently, the “0” represents
1. an “OFF” circuit 2. an “OPEN” switch 3. a “LOW” voltage
4 a “MINUS” sign 5. a “FALSE” statement.
In Negative logic, just opposite conditions prevail.
Suppose, a digital system has two voltage levels like 0V and 5V.
If we say that value 1 stands for 5V and value 0 for 0V, then we have positive logic system.
If on the other hand, we decide that”1” should represent 0V (low voltage) and 0 should represent 5V
(high voltage), then we have negative logic system.

Logic gates :

The logic gates are building blocks of digital electronics. They are used in digital electronics
to change one voltage level (input voltage) into another (output voltage) according to some logical
statement relating them.
Thus, logic gate is a digital circuit, which works according to some logical relationship between
input and output voltage.
The logic gate may have one or more inputs, but only one output.
Truth table of a logic gate is a table that shows all possible input combinations and the corresponding
output for the logic gate.
The logical statements that logic gates follow are called “Boolean expressions”.
1. In Boolean algebra, the addition sign (+) is referred as OR. The Boolean expression is
y =A+B
This Boolean expression is read as y is equal to A 'OR' B.
2. The multiplication sign (.) is referred as AND in Boolean algebra. The Boolean expression is
y = A.B
This Boolean expression is read as y is equal to A 'AND' B.
3. The bar sign (-) is referred to as NOT in Boolean algebra. The Boolean expression is
y= A
This expression is read as y is equal to 'NOT' A.

BASIC LOGIC GATES:

The following three gates viz., OR gate, AND gate and NOT gate are called as basic logic gates.

OR-GATE
In general, the simple OR gate is a two input and one output logic gate. It combines the inputs A
and B to give the output y, following the Boolean expression y = A + B. The symbol, truth table and
equivalent electrical circuit are shown below.

243
 The function of the OR gate is, the output is “TRUE” if any one of its inputs is in 'TRUE'
condition.
A B y
0 0 0
1 0 1
0 1 1
1 1 1

Two parallel Switches and Lamp is connected as shown above.

Lamp “ON” is equal to 1
Lamp “OFF” is equal to 0.
Case 1 :-
If both the Switches A and B are open. No current flow through the external wire. So Lamp is “OFF”
i.e. equal to “0”.
Case 2 :-
If the Switch A is closed and B is open. The conventional current pass through “A” Lamp is “ON”
i.e.. equal to 1
1+0=1
Case 3 :-
If Switch A is open and Switch B is closed. Current pass through “B”, So Lamp is “ON” i.e.. equal
to 1.
0+1=1
Case 4 :-

If both the Switches are closed, Lamp is “ON”, i.e. equal to 1.

1+1=1
So, for all possible input and the obtained output values are tabulated. It is called truth table.
AND gate :-
In general, the simple AND gate is also a two inputs and one output logic gate. It combines the inputs
A and B to give the output Y, following the Boolean expression
y = A.B
i.e. y is equal to A “AND” B. The symbol, equivalent circuit and truth table are shown below.
The function of the AND gate is, the output is 'TRUE' if and only if all the inputs are in 'TRUE'
condition.
A B y
0 0 0
1 0 0
0 1 0
1 1 1

244
 In the above equivalent diagram the switches and the lamp is in series connection mode.
Case 1 :-
If both the switches A and B are OPEN (i.e., A = 0, B = 0) then the lamp will not glow (i.e.y = 0).
The current will not pass through the lamp.
Case 2 :-
If the switch “A” is closed and “B” is open, current will not pass through the lamp. The lamp is OFF,
so y = 0.
Case 3 :-
If the switch “B” is closed and “A” is open, current will not pass through the lamp. The lamp is OFF,
so y = 0.
Case 4 :-
If both switches A and B are closed, current will pass through the circuit. Now the lamp is “ON” and
glowing. So y = 1.
NOT gate:-
The NOT gate is a one input and one output logic gate. It inverts or complements the input A to give
output y following the Boolean expression.
y= A
This gate is also called as 'inverter'. The symbol, equivalent circuit and truth table are shown below.
The equivalent diagram is shown below.
The function of the NOT gate is to invert or complement its input. i.e., the output is 'ON' if input is
'OFF' and vice versa.
A y
+
0 1
Ba - A L
A y 1 0
OFF

Case 1:-

If the switch “A” is open, the current will pass through the lamp and lamp glows.
So, y = 1 when A = 0
Case 2:-
If the switch “A” is closed, the current will take shortest path. That is current will pass through the
switch above. Hence the lamp is “OFF”, so y = 0 when A = 1.
UNIVERSAL LOGIC GATES:
Any logic gate can be made from combinations of NAND gate or NOR gates. Hence, the NOR gate
and NAND gate are called as Universal logic gates.
NOR gate
It is a logic gate in which OR gate is followed by a NOT gate. The symbol, equivalent circuit and
truth table are shown below.
The function of this gate is 'inverting' the output of the OR gate.

245
 A y’ A
y v
B
B

It combines the inputs A and B to give the output y, by the following Boolean expression
y = A+B

A B y’ y A B y
A
0 0 0 1 0 0 1
1 0 1 0 1 0 0
B
0 1 1 0 0 1 0
L 1 1 0
i 1 1 1 0
+ Ba -

Case 1 :-
If both the switches A and B are open, the current flow through the lamp, Lamp is “ON”. So y = 1.
Case 2 :-
If the switch A is closed and B is open, the current will pass through the shortest path or low
resistance path. Hence conventional current flow through the switch “A” and reaches the cathode of the
battery. The lamp is “OFF”, so y = 0.
Case 3 :-
If the switch A is open and B is closed, the conventional current will pass through the shortest path.
The current will pass through the switch B and reaches the cathode of the battery. Current will not pass
through the lamp, so lamp is OFF, i.e. y=0.
Case 4 :-
If the switch A and B are closed, the current from the battery will pass through two parallel switches
and reach the cathode of the battery. Current will flow through shortest path (or) low resistance path.
The lamp is “OFF”, so y = 0.
NAND gate :-
It is logic gate in which AND gate is followed by NOT gate. The symbol, equivalent circuit and truth
table are shown below.
The function of this gate is “inverting” the output of the AND gate.
It combines the inputs A and B to give the output y, by the following Boolean expression
y = A.B
y = A.B
A y’ A
y y
B B

A B y’ y A B y
0 0 0 1 0 0 1
A B L
1 0 0 1 1 0 1
0 1 0 1 0 1 1
i 1 1 1 0 1 1 0
Ba
+ -
246
 Case 1:-

If both switches A and B are open, the current will flow through the lamp. The lamp is ON, so
y=1
Case 2:-
If the switch A is closed and B is open, the current will again flow through the lamp. The lamp is
ON, so y = 1.
Case 3:-
If the switch A is open and B is closed, the current flows through the lamp. The open switch A act as
a inter circuit breaker. The lamp is ON, so y = 1.
Case 4:-
If the switch A and B are also in closed position, the current will pass through the switches and reach
the cathode of the battery. Current will flow through shortest path (or) low resistance path. Hence, the
lamp is OFF, So, y = 0
Uses of Gates and IC’s:
(i) Gates are used in computers, calculators, washing machines etc
(ii) Used for switching purposes
(iii) Used in control systems
(iv) Used in alarms.
IC’s:
(i) Used in computers, calculators and many electronic devices.
Integrated Circuits :-
An Integrated Circuit (IC) is just a compact electronic circuit. It has both active and passive
components.
A discrete circuit is one that is built by connecting separate components. In this case, each
component is produced separately and then all are assembled together to make the electronic circuit. For
example, consider house wiring. In that, basically switches, lamp and wires are used. They are different
components made by different manufactures. We assemble them to form a circuit to energise the lamp
and other equipments.
In IC, all components are manufactured on the same silicon wafer. Diodes, transistors, resistors,
capacitors and their inter connection is also established well to perform a specific task. Hence all active
and passive components are formed on the same chip, using microelectronic techniques.
Scale of Integration :-
In scale of integration, the circuit density and component density is playing important role. For
example, 1 lakh components can form 100 circuits. Another circuit designer can form 1000 circuits
using the same 1 lakh components. Hence comparing, circuit density and component density is not
unique. It all depends upon the application and circuit designer's idea.
An approximate method of classifying the amount of circuit or components density is as follows:-
1. SSI – Small Scale Integration :
In this case, the number of circuits packed in one IC is less than 12 (or) number of components is
less than 50.
247
 2. MSI – Medium Scale Integration :
Here, number of circuits per package is between 13 and 99 or number of components is between 50
and 500.
3. LSI – Large Scale Integration :-
In this case, circuit density is between 100 and 9999 or components density is between 5000 and
100 000 (hundred thousand).
4. VLSI – Very Large Scale Integration :-
Here the number of circuits per package is between 10,000 to 99,999 or number of components is
between 100 000 and 1 000 000.
In Summary;
Circuits per Chip

SSI < 12
MSI 12-99
LSI 100 – 9 999
VLSI 10 000 – 99 999

Advantages of ICs:-
ICs have the following advantages:-
1. Extremely small physical size. Often the size is thousands of times smaller than discrete
circuit.

2. Weightless when compared with a discrete circuit. Since many circuit functions can be packed
into a small area, in many applications where weight and space are critical, such as in air craft,
space shuttle, satellite, etc., the IC invention and application brought a very big change in the
satellite communication and satellite launching.

3. Life time of ICs is long which is most important from both military and consumer application
point of view. Most significant factor is the absence of soldered connections.
Extremely high reliability. IC damages are less under proper usage.

4. In signal processing, the signal transfer from one circuit to another circuit or from one device
to another device is fast when we use ICs. Moreover, because of short distance between the
internal circuits, the stray signal pick is less and nil in many ICs. Response time and speed are
more, on the other hand.

5. Low power consumption.

6. Easy replacement
7. Cost effective due to mass production of IC and less failure rate in mass production.

248
1. A locker has been rented in the bank. Express the process of opening the locker in terms of
digital operation.
The locker gate (Y) can be opened by using one key (A) which is with the client and the other
key which is with the bank (B). When both the keys are used the locker door opens.
Locker door is opened i.e. Y=1 when both keys are applied A=B=1. So the process can be
expressed as an AND operation.

Y = A.B

2. Calculate the value of the output Y, in the following circuit.

From the truth table for OR,

if A is 1 then A + A = 1 + 0 = 1,

while if A is 0 then A + A = 0 + 1 = 1.

This can be summarised in the rule that A+A=1

So, the output of the above circuit is 1.

The output, Y = 1

3. Draw the truth table for the following circuit:

249
We know that the outputs of first two NOT gates are 𝐴 𝑎𝑛𝑑 𝐵

And Boolean expression for AND gate is Y = X .Y , where X and Y are two inputs.

Here A and B are inputs to the circuit.

Y = 𝐴 . 𝐵

A B 𝐴 𝐵 Y = 𝐴 .𝐵
0 0 1 1 1
0 1 1 0 0
1 0 0 1 0
1 1 0 0 0

The output Y = 𝐴 . 𝐵

250
 QUESTIONS
Part – A and Part – B

1. What is Doping?
2. What is the purpose of doping?
3. Name the two types of extrinsic semiconductor and write the majority and minority carriers in
each.
4. What is a hole in P type semiconductors?
5. What is potential barrier in P-N Junction diodes?
6. Draw the symbol for PNP and NPN transistor.
7. How will you connect the battery for forward bias? For this bias, does the diode act as a
conductor or non-conductor?.
8. How will you connect the battery for reverse bias? For this bias, does the diode act as
conductor or non-conductor?
9. Write the three different configurations for a transistor use in electronic circuits.
10. What is the difference between analog and digital waves?
11. What is positive logic?
12. What is negative logic?
1 3 . What are the basic logic gates?
14. Draw the symbol for OR gate and write the Boolean Expression for the same.
15. Draw the symbol for AND gate and write the Boolean Expression for the same
16. Draw the symbol for NOT gate and write the Boolean Expression for the same
17. What are the universal logic gates?
18. Draw the symbol for NAND gate and write the Boolean Expression for the same
19. Draw the symbol for NOR gate and write the Boolean Expression for the same
20. What is an Integrated Circuit?
21. Scaling of ICs are based on what factors?
22. Give any three advantages of ICs.

Part – C
1. Write short notes about intrinsic semiconductor.
2. Explain the Formation of Extrinsic P-type semiconductors.
3. Explain the Formation of Extrinsic N-type semiconductors.
4. Explain the P-N junction diode formation, depletion region and barrier potential.
5. Explain the forward bias action in a P-N junction diode, with a neat diagram.
6. Explain the reverse bias action in a P-N junction diode, with a neat diagram.
7. Explain the working of common base configuration in PNP transistor.
8. Explain the working of common base configuration in NPN transistor.

251
9. Explain all the three basic logic gates with symbols, Boolean expressions and Truth tables.
10. Explain the NAND and NOR gates with symbols, Boolean expressions and truth tables.
11. Explain level of integration in ICs based on circuit densities and component densities.
12. Discuss the advantages of ICs.

Brain teasers:
1. Analyze the circuit given below and explain the current flow direction through the circuit. When
will the lamp glows connected in series with the capacitor?
M N

Bulb

A.C input

2. In the following circuits, which of the diodes is forward –biased and which is reversed biased ?
+5V R
-10V
R
R

-10V

3. How the P.N.P transistor shown below is biased?

+1V
C
B
0V

E
-2V

4. Prove that the following diagram's truth table is equal to NAND gate truth table.
y1
A

B
y2

252
https://www.youtube.com/watch?v=jErh36lBH4s&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=19

https://www.youtube.com/watch?v=hzt48azWcrU&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=21

https://www.youtube.com/watch?v=gbWDa7dPszQ&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=22

https://www.youtube.com/watch?v=YMUBsSxlrUo&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=23

https://www.youtube.com/watch?v=hlbXBIK3C_g&list=PL1b9Ht9ISqIFUWvOqFTEdUPmtZadrXNk4&index=2

253
 SUMMARY

• Semiconductors are the materials having electrical conductivity lying

between good conductors and insulators.

• Pure form of semiconductors is known as the intrinsic semiconductors.

• The process of addition of impurities to a semiconductor is known as

doping.

• After doping the semiconductors are called as extrinsic semiconductors.

• When a trivalent element is added to the pure semiconductors it

becomes P-type semiconductor.

• When a pentavalent element is added to the pure semiconductors it

becomes N-type semiconductor.

• The PN junction diode is a combination of P-type and N-type

semiconductors of the same material.

• PN junction diode conducts electric current in forward bias

• PN junction diode does not conduct electric current in reverse bias

• Diode can be used to rectify the AC signal to DC signal. This is called

rectification.

• Working of full wave bridge rectifier using diodes

• Working of common base configuration of NPN and PNP transistors.

• Symbols, Boolean expression and Truth tables of gates.

• SSI, MSI, LSI and VLSI – different levels of integrated circuits.

254
 Model Question Paper

TIME: 3hr ENGINEERING PHYSICS - II MARKS: 100

PART A (5 X 1 = 5)
Note: Answer ALL questions. All questions carry equal marks.

1. Define mean free path.

2. State first law of thermodynamics:-
3. Expand the acronym of RADAR?
4. State Ohm’s law.
5. What is rectifier?

PART B (10 X 2 = 20)

Note: Answer any TEN questions. All questions carry equal marks.

6. Define conduction:-
7. Define root mean square velocity of gas molecules.
8. Define specific heat capacity of a solid.
9. State Kelvin statement of second law of thermodynamics.
10. What is the disadvantage of casecade process?
11. What is meant by temperature of inversion?
12. Give the condition for total internal reflection.
13. Give any two properties of laser.
14. What is active remote sensing?
15. State any one law of the laws of resistance.
16. State Joule’s law of heating.
17. State the Fleming’s left hand rule.
18. What are intrinsic semiconductors?
19. What is dopping?
20. Give any two advantages of an integrated circuit.

PART B (5 X 15 = 75)
Note: Answer ALL questions by choosing either a & b (OR) c & d

21. a) What are the properties of thermal radiation? (8 Marks)

b) Derive the expression for the pressure of a gas on the basis of kinetic theory (7 Marks)
of gases.
OR
c) Derive Mayer’s relation for a perfect gas. (8 Marks)
d) Calculate the value of universal gas constant R from the gas equation PV = RT. (7 Marks)

22. a) Explain isothermal and adiabatic change. (8 Marks)

b) A certain mass of gas at 1 atm pressure is suddenly compressed to one-fourth (7 Marks)
of its original volume. Find the resulting pressure (=1.4).
OR
c) Describe the Linde’s process for the liquefaction of air. (8 Marks)
d) Write an essay on wind energy. (7 Marks)

255
23. a) Describe an experiment to determine the refractive index of glass prism using (8 Marks)
spectrometer.
b) Describe the construction and working of the Ruby LASER. (7 Marks)
OR
c) Describe the components of remote sensing. (8 Marks)
d) Explain the working of RADAR with block diagram. (7 Marks)

24. a) State and explain Kirchoff’s current and voltage law. (8 Marks)
b) The capacitors of values 20µf, 30µf and 60µf are connected (i) in series and (ii) (7 Marks)
in parallel. Calculate the effective capacitance in each case.
OR
c) Describe an experiment to determine the specific heat capacity of a liquid (8 Marks)
using Joule’s calorimeter.
d) Derive an expression for the torque on a rectangular coil placed on a uniform (7 Marks)
magnetic field.

25. a) Explain N type and P type semiconductors. (8 Marks)

b) Explain the working of full wave bridge rectifier using PN junction diode. (7 Marks)
OR
c) Explain AND, OR and NOT gates with the help of truth tables. (8 Marks)
d) Explain the various levels of integration circuit. (7 Marks)

256
 Model Question Paper

TIME: 3hr ENGINEERING PHYSICS - II MARKS: 100

PART A (5 X 1 = 5)
Note: Answer ALL questions. All questions carry equal marks.

1. Define ratio of specific heat capacities of gases.

2. State Kelvin statement of second law of thermodynamics.
3. Write any two uses of LASER?
4. Write Fleming’s left hand rule.
5. What is Doping?

PART B (10 X 2 = 20)

Note: Answer any TEN questions. All questions carry equal marks.

6. State any four postulates of kinetic theory of gases.

7. Define Conduction.
8. List the properties of thermal radiation.
9. What is an adiabatic change?
10. What are the advantages of renewable energy?
11. What is joule Thompson effect?
12. What are the important parts of spectrometer?
13. Explain stimulated emission and population inversion.
14. Write the principle of LASER?
15. Write Kirchhoff’s current law and voltage law.
16. Write Faraday I law and Faraday II law of electrolysis.
17. Describe how Galvanometer converts to ammeter.
18. Explain p type semiconductor.
19. What are the Universal logic gates?
20. What is an integrated circuit?

PART B (5 X 15 = 75)
Note: Answer ALL questions by choosing either a & b (OR) c & d

21. a) State the postulates of kinetic theory of gases. (8 Marks)

b) Derive the relation between pressure and kinetic energy of the gas. (7 Marks)
OR
c) Derive Cp – Cv = R, for a perfect gas. (8 Marks)
d) The ratio of specific heat capacities of a gas is 1.66. Assuming the value of (7 Marks)
universal gas constant R as 8.12JK-1 per gm mole. Find the values of specific heat
capacity at constant pressure and specific heat capacity at constant volume of
the gas.

22. a) Describe the cascade process for liquefaction of oxygen. (8 Marks)

b) A certain mass of gas at 2 atm pressure is suddenly compressed to one half of (7 Marks)
its original volume. Find the resulting pressure (=1.4).
OR

257
 c) Write a note on solar energy and wind energy. (8 Marks)
d) Air at 5 atmosphere is compressed to half its original volume under isothermal (7 Marks)
process. Calculate the resulting pressure.

23. a) Explain Optical fibre cable as a wave guide. (8 Marks)

b) Explain construction and working of Ruby LASER. (7 Marks)
OR
c) Explain the working principle of RADAR with a block diagram. (8 Marks)
d) Explain components of remote sensing. (7 Marks)

24. a) Describe an experiment to determine the specific heat capacity of a liquid (8 Marks)
using Joule’s calorimeter.
b) Calculate the electrochemical equivalent of a metal when a current of one (7 Marks)
ampere liberates in one hour deposit 1.2g of metal from its electrolyte.
OR
c) Derive an expression for the torque experienced by rectangular coil carrying (8 Marks)
current placed inside a uniform magnetic field.
d) Describe an experiment to determine the electrochemical equivalent of (7 Marks)
copper using copper voltmeter.

25. a) Explain the working of NPN transistor in common emitter configuration. (8 Marks)
b) Explain the full wave rectifier with a neat circuit diagram. (7 Marks)
OR
c) Explain all the three basic logic gates with symbols, Boolean expressions and (8 Marks)
truth tables.
d) Explain various level of integration. (7 Marks)

258
 SEMESTER - II
9. REFRACTIVE INDEX
Aim :
To determine the refractive index of a transparent liquid (water) using travelling microscope.
Apparatus required :
Given liquid, glass beaker, travelling microscope, lycopodium powder, pin, etc.
Formula :

The refractive index of liquid (water)

Real depth of the liquid
 =
Apparent depth of the liquid

Reading 3−Reading 1
= Reading 3−Reading 2

Description :

A travelling microscope is an ordinary microscope fixed on a stand in such a way that it may be
made to travel in vertical as well as horizontal direction without disturbing its adjusted focus. The
readings are recorded by means of main scale and vernier scale of high accuracy (0.001 cm) attached to
the instrument.
Procedure :
The least count of the microscope is determined as in the case of vernier calipers.
A small pin is fixed horizontally with wax or cello tape at the bottom of the empty beaker. The tip of
pin is focussed clearly on the microscope and the corresponding main scale reading (MSR) and vernier
scale coincidence (VSC) in the vertical scale are noted in the tabular column. (Reading 1)
The given liquid is taken in the beaker. Now the pin is apparently raised through a height. Therefore,
it will be no longer in focus. The microscope is moved up without changing the adjusted focus, so that
the image of the pin is clearly seen through the microscope. The corresponding main scale reading (MSR)
and vernier scale coincidence (VSC) in the vertical scale are noted in the tabular column. (Reading 2)
Finally, a little lycopodium powder or saw dust which can float on liquid is scattered on the
surface of the liquid. The microscope is further moved up without changing the adjusted focus, so that
the clear image of lycopodium powder is seen through the microscope. The corresponding main scale
reading (MSR) and vernier scale coincidence (VSC) in the vertical scale are noted in the tabular column.
(Reading 3)
The difference between this Reading 3 and Reading 1 gives the real depth of the liquid, whereas the
difference between Reading 3 and Reading 2 gives the apparent depth of the liquid.
By substituting the readings in the formula, the refractive index of the liquid (water) is determined.
Result :
The refractive index of the liquid (water) =

259
 E
M

Pin at the Pin viewed Lycopodium powder

bottom of the through the scattered on the
beaker liquid surface of the liquid

To find Least count of microscope :

Least count: Value of 20 MSD = cm
Value of 1 MSD = cm
No. of divisi ns on the vernier scale(n) =
=

1
Least Count (LC) = × value of 1 MSD
n
= cm
Tabular column:
Least count = cm

Microscope Readings
Position
MSR VSC VSR = VSC × LC OR = MSR + VSR
Unit cm div cm cm
Clear image of tip of the pin fixed
at the bottom of the beaker. (Reading 1)

Clear image of tip of the pin, after

(Reading 2)
the liquid is taken in the beaker.
Clear image of lycopodi-
um powder scattered on the (Reading 3)
surface of liquid.
Calculation:
The refractive index of liquid (water)
Re al depth of the liquid
 =
Apparent depth of the liquid
Real depth of the Liquid = Reading 3 – Reading 1
= cm - cm
= cm
Apparent depth of the liquid = Reading 3 – Reading 2
= cm - cm
= cm
Refractive index of the liquid  =
260
 10. SOLAR CELL
Aim :

To draw the V-I characteristics curve for the given solar cell.
Apparatus required:
Solar cell, milliammeter (0 – 100 mA), voltmeter (0 – 2 V), variable resistor (0 – 1 kΩ), 60 W bulb,
connecting wires, plug key.
Theory :
The solar cell is a semiconductor device that converts solar energy (electromagnetic radiation) into
electrical energy by photovoltaic effect. Solar cells are widely used to produce electrical energy from
sun's radiation.
A solar cell consists of a p-n junction diode, in which electrons and holes are generated by the
incident photons (i.e. electromagnetic radiation). When an external circuit is connected through the p-n
junction device, a current passes through the circuit. Therefore, the device generates power when the
electromagnetic radiation is incident on it.
Procedure :
A voltmeter, milliammeter and a variable resistor of 0 to 1000 ohm and a plug key are connected as
shown in the circuit diagram.
The solar cell is illuminated by a filament bulb of 60 W. Now the two pins of the solar cell are
plugged in to the two terminals (A and A) of the milliammeter and the short circuit current Isc is noted,
keeping the plug key in open condition.

Then the two pins of the solar cell are plugged into the two terminals (V and V) of the voltmeter and
the open circuit voltage Voc is noted, keeping the plug key in open condition.

Now the plug key is closed to include the variable resistor in the circuit. By slowly changing the
value of resistance included in the circuit, the milliammeter readings are noted for various values of
voltmeter readings in steps of 0.1 V or 0.2 V.
The readings of both the milliammeter and voltmeter are entered in the tabular column. With these
readings, the V – I characteristics graph of the solar cell is drawn.
Result :

Thus the V-I characteristics of the solar cell is drawn.

Note:

i) The Voc and Isc values depend on the type of solar cell and number of solar cells used in the solar
panel.

ii) These values also depend on the source of electromagnetic radiation, which we use to irradiate the
solar cell.
iii) Care should be taken so that the bulb and solar cell do not come into direct contact. (I.e. touch each
other). In such case the readings of the meters may go beyond the range and the solar cell may get
damaged.

261
Circuit Diagram
A + -

Variable resis for

+ AA Isc
mA
SOLAR V + (key open)
PANEL
- VV Voc
- V (key open)
E.M
Radiation key

Tabular Column:

Short circuit current Isc = mA

Open circuit voltage Voc = volt

Voltage generated by the Current delivered by the

S.No
solar cell (V) solar cell (I)
Unit volt milliampere

Model Graph:

Isc
Current delivered by
the solar cell (I)
in mA

O X
Vcc
Voltage generated by the
Solar cell (v) in volt

262
 11. SPECTROMETER
Aim:

To determine the angle of the prism using spectrometer.

Apparatus required :
Spectrometer, given glass prism, Sodium vapour lamp, Reading lens, black cloth, etc.
Formula:
Spectrometer is an optical instrument having three parts
(i) Collimator (ii) Telescope and (iii) Prism table. The following initial adjustments should be carried
out before starting any experiment with a spectrometer.
a) The telescope is turned towards a white wall of background and the eyepiece is gently pushed in
or pulled out so that a clear image of the cross wires is seen without parallax.

b) To adjust the telescope for parallel rays, the telescope is directed towards a distant object like
distant tree or building and the distance between the objective lens and the eyepiece is altered till
the clear image of the distant object is seen in the field of view of telescope.
c) To adjust the collimator for parallel rays, the slit of the collimator is illuminated with a
monochromatic source of light (sodium vapour lamp). The telescope is brought in line with the
collimator in order to catch the image of the slit in the telescope. The distance between the convex
lens in the collimator and the slit is altered to get the clear image of the slit through the telescope,
so that the width of the slit is minimized.

d) The prism table is made perfectly horizontal by adjusting the screws provided with the help of the
spirit level.
Procedure :
Angle of the prism A :
After making initial adjustments, the least count of the spectrometer is determined (as in Vernier
Calipers).
Having a very fine slit, the telescope is adjusted for direct ray. A prism is placed on the prism table
as shown in the figure. The telescope is turned to catch the reflected image from one polished face AB of
the prism. Fixing in that position, the tangential screw is adjusted till the vertical cross wire is made to
coincide with the fixed edge of the image of the slit. The readings in the scales I and II are noted.
The telescope is then turned to the other polished face AC of the prism. The readings in the scales I
and II are noted when the vertical cross wire coincides with the fixed edge of the image of the slit. The
difference between both the scale I readings is 2A. Similarly the difference between both the scale II
readings is 2A. Then the average of the angle of the prism A is calculated.

263
Tabular column:
To determine the angle of the prism A LC =

Spectrometer readings in degree

Scale I Scale II

Ray
VSR = OR = VSR = OR =
MSR VSC MSR VSC
VSC × LC MSR + SR VSC × LC MSR + VSR

Reading of the
reflected ray
(left) x 1= x3 =

Reading of
the reflected
ray(right) x2 = x4 =

2A = (x1 ~x2) 2A = (x1 ~ x2)

A= = A=
2 2

+
The average angle of the prism A = A =
2

Observations :

Least count of Spectrometer :

Value of 20 MSD =

Value of 1 MSD =

No. divisions on the Vernier scale (n) =

1
Least Count = x value of 1 MSD
𝑛

Result :
The angle of the prism  =

264
 12. OHM'S LAW
Verification of Laws of Resistances
Aim :
To verify the laws of resistances when the two given standard resistances are connected
i) in series and ii) in parallel, using Ohm's law.
Apparatus required :
Two known standard resistances, battery, key, rheostat, ammeter, voltmeter, connecting wires,
etc.
Formulae:
i) When two resistances R1 & R2 are connected in series, their effective resistance is given by the
formula RS = R1 + R2
E
Rs = 
I
The formula for determining the effective resistance by experiment is
ii) When two resistances R1 & R2 are connected in parallel, their effective resistance is given by the
formula

R1 R 2 
RP =
R1 + R 2
The formula for determining the effective resistance by experiment is
E
RP = 
I
where
R1 = the value of the one standard resistance ()
R2 = the value of the other standard resistance ()
E = Potential difference between the ends of the resistances (V)
(Voltmeter reading)
I = Current flowing through the resistances (Ammeter reading) (A)

Circuit diagram for series connection

265
Procedure :
Resistances in series :
The given two resistances R1 and R2 are connected in series. This combination is again connected in
series with a battery eliminator, key, rheostat and an ammeter. The voltmeter is connected in parallel to
the combination of resistances to measure the potential difference across the resistances.
Now the circuit is closed. By adjusting the rheostat for different values of current, the corresponding
values of potential difference are noted in tabular column I. The effective resistance Rs is calculated in
each case using the formula. Then the average value of effective resistance Rs is calculated.

Resistances in parallel :

The given two resistances R1 and R2 are connected in parallel. This combination is again connected
in series with a battery eliminator, key, rheostat and an ammeter. The voltmeter is connected in parallel
to the combination of resistances, to measure the potential difference across the resistances.

Now the circuit is closed. By adjusting the rheostat for different values of current, the corresponding
values of potential difference are noted in tabular column II. The equivalent resistance Rs is calculated
in each case using the formula. Then the average value of effective resistance Rs is calculated.

Result :

1) When the resistances are connected in series:

i. The effective resistance by calculation Rs = ohm

ii. The effective resistance by experiment Rs = ohm

2) When the resistances are connected in parallel :

i. The effective resistance by calculation RP = ohm

ii. The effective resistance by experiment RP = ohm

The laws of resistances are verified, using Ohm's law.

266
 Tabular column I:
To find the value of effective resistance when R1 and R2 are in series

Current flowing Potential difference between

through the the ends of resistances E
Sl. resistances (Ammeter (Voltmeter reading) Rs =
I
No reading) E
Unit d I ampere d v ohm
1 i i ol
.2 v v t
.3
.4
.5
.
Average value of effective resistance when R1 and R 2 
 =
ohm
are connected in series R S 
Calculation:
E
1) E = volt ; I= ampere; RS =
I

Circuit diagram for parallel connection

Resistance R1 = ohm; Resistance R2 = ohm
R 1 R 2
Effective resistance R P = =
R1 + R 2

Tabular column II : = ohm

Current flowing Potential difference between

through the the ends of resistances E
Rs =
Sl. I
resistances (Ammeter (Voltmeter reading)
No reading) E
Unit d I ampere ohm v ohm
1 i ol
.2 v t

.
3
.4
.
5
.
267
 Average value of effective resistance when R1 and R 2 
 = ohm
are connected in series RS 

Calculation:

E
1) E = volt ; I = ampere; Rs =
I

268
 13. JOULE'S CALORIMETER
Aim:
To determine the specific heat capacity of water (liquid), using Joule's calorimeter.
Apparatus required :
Joule's calorimeter, battery eliminator, ammeter, voltmeter, rheostat, plug key, stop watch,
thermometer, digital balance, liquid (water), connecting wires, etc.
Formula :
EIt m1Sc -1 -1
S= − J kg K
(m 2 − m1 ) (T2 − T1 ) (m 2 − m1 )

The specific heat capacity of the liquid

where

E = potential difference across the heating coil (Voltmeter reading) (V)

I = current flowing through the heating coil (Ammeter reading) (A)

t = time for which the current flows to increase the temperature of water (S)

m1 = mass of empty and dry calorimeter with stirrer (kg)

m2 = mass of calorimeter with stirrer and water (kg)

S = specific heat capacity of the material of the calorimeter (J kg-1 K-1 )
c

T1 = initial temperature of the liquid and (K)

T2 = final temperature of the liquid (K)
Description :

The Joule's calorimeter consists of a brightly polished copper or aluminium vessel closed by an
ebonite lid with two holes. A heating coil, connected to two binding screws provided on the ebonite
lid is kept within the calorimeter. Through holes in the lid a sensitive thermometer and a stirrer can be
inserted. The calorimeter is placed inside a wooden box stuffed with cotton or wool.
Procedure :

The mass of the clean, dry calorimeter with stirrer (m1) is determined using a digital balance. The
calorimeter is filled to two- third of its volume with the given liquid (water) and the mass (m2) is once
again determined.
The calorimeter is kept in the wooden box and the heating coil is completely immersed in the liquid.
Connections are given as shown in the circuit diagram.
The circuit is closed for a moment and the rheostat is adjusted for a suitable current of 0.8 A or 1.0
A. The ammeter and voltmeter readings are noted as I and E respectively. The circuit is kept open. Now,
the liquid is stirred well and the initial temperature (T1) of the liquid is determined using a sensitive
thermometer.

269
 The circuit is again closed and simultaneously a stopwatch is started. The liquid is stirred slowly
and uniformly. The temperature of the liquid is noted for every minute in the tabular column. When
the temperature of the liquid is increased by 5°c, the key is removed and the stopwatch is stopped
simultaneously.
The time (t second) for which the current was flowing through the circuit is noted from the
stopwatch. The specific heat capacity of the material of the calorimeter (Sc) is taken from tables.
Then, the specific heat capacity of the water (S) is calculated using the given formula.
Result :
The specific heat capacity of water is found to be S = J kg-1 K-1
Battery Key
+ -
Rheostat
Voltmeter

+ -
Ammeter V

+ -
A

Thermometer

Stirrer
Liquid
Heating coil
Joule’s Calorimeter
Insulating materiel
Wooden box

Observation :
Mass of the empty calorimeter with stirrer (m1) = g

= kg
Mass of calorimeter with stirrer + water (m2) = g
= kg
Current flowing through the heating coil 
 (I) = A
(Ammeter reading) 
Potential difference across the coil
 (E) = V
(voltmeter reading) 
Initial temperature of water (T1 ) = °C
Final temperature of water (T2 ) = °C
Time for which current is passed to (t) = minutes
increase the temperature of water = second

Specific heat capacity of the 

 = J kg-1 K-1
material of the calorimeter (copper)  (SC )

270
Tabular column :

To record the time taken for 5 C raise in temperature of water

Time in minutes Temp. of water in °C

Time taken for 5°c raise in temp. of water (t) = s

Calculation :

The specific heat capacity of water

EIt m1Sc
S = −
(m 2 − m1 ) (T2 − T1 ) (m2 − m1 )
= J kg-1 K-1

271
 14. COPPER VOLTAMETER
Aim:

To determine the electro chemical equivalent (e.c.e) of copper, using copper voltameter.
Apparatus required :
Copper voltameter, battery eliminator, plug key, rheostat, ammeter, stop watch, digital balance,
connecting wires, emery sheet, etc.
Formula :
The electro chemical equivalent of copper
m m2 − m1
Z = = kg C-1
It It

where
m = mass of copper deposited on the cathode (kg)
m1 = mass of the cathode before passing current(kg)
m2 = mass of the cathode after passing current (kg)
I = current flowing through the electrolyte (Ammeter reading)(A)
t = time of passage of current (S)
Description :
The copper voltameter consists of a glass vessel, containing a strong solution of copper sulphate.
Three plates of copper are dipped in the solution. The two outer plates are connected together and they
form the anode. The middle plate serves as the cathode.
Procedure :
The copper voltameter is connected in series with the battery, plug key, rheostat and ammeter. It
should be noted that the negative of the battery is connected to the central cathode plate. After making
all the connections, the circuit is closed for a moment and the rheostat is adjusted for a suitable current,
say 1 ampere, in the ammeter. The circuit is kept open, the cathode plate is removed and cleaned with
emery sheet. It is then washed in water, dried and its mass m1 is determined accurately using a digital

balance.
The cathode plate is then fixed in its position and the circuit is closed, simultaneously starting a
stopwatch. The current is passed for 30 minutes and during the passage of current the ammeter value
should be maintained as the same.
After 30 minutes the cathode is taken out, rinsed with water, dried and its mass m2 is found

accurately. The time of passage of current is noted. The mass of copper deposited is determined and
hence e.c.e. of copper is calculated using the given formula.

272
Result :
The electro chemical equivalent of copper Z = kg C-1

+ -
Ammeter

Anode Anode
Cathode

Copper sulphate
Solution (CuSo4)

Observation : = g
= kg
Mass of the cathode before passing current m1
= g
Mass of the cathode after passing current m2
= kg
Mass of the copper deposited at cathodem m = (m2 – m1)

m = kg
Current flowing through the electrolyte

 I = A
(Ammeter reading) 

Time of passage of current t = minutes

= second
Recording of time of passage of current (in minutes) :

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Calculation :
The electro chemical equivalent of copper

m m2 − m1
Z = =
It It

= kg C-1

273
 15. CHARACTERISTICS OF P-N JUNCTION DIODE
Aim :
To draw the voltage - current characteristics of a P-N Junction diode in forward bias and to find the
'dynamic forward resistance' and 'knee voltage' from the graph
Apparatus required :
P-N junction diode (Germanium or Silicon), D.C. power supply (0 - 10 V), D.C. milliammeter
(0 – 15 mA), D.C. Voltmeter (0 – 1.5 V), resistance (500 ), etc.
Formula :
∆𝑉
The dynamic forward resistance r =
∆𝑖
where
 V = a small change in potential difference (p.d) applied to the diode

 i = the corresponding change in the current flowing through the diode

Knee voltage (Vk): It is the approximate voltage above which the diode current increases sharply.
Theory :

When the positive terminal of the battery is connected to P-type and negative terminal to N-type of
a semiconductor diode, current flows easily through the diode. Now it is said to be in forward bias.
When the P-type material is connected to negative and N-type to positive of the battery, the diode
offers a high impedance path and little current flows through the diode. Now it is said to be in reverse
bias.
This property of P-N Junction diode, allowing the flow of current in one direction easily is made use
of in many electronic circuits, especially in solid state half-wave and full-wave rectifiers.
Procedure :

The anode and cathode of a P-N junction diode can be identified by seeing a colour band or spot on
the cathode. Also the cathode may be having pointed edge in some type.
For forward biasing, the connections are given neatly as shown in the diagram. Suitable ranges of
meters, resistance and power supply can be selected.
The current flowing through the diode is noted for various values of p.d. applied across the diode
and the readings are noted in the tabular column. With these readings, a neat graph is drawn by taking
forward voltage applied across the diode in X - axis and forward current flowing through the diode
(in mA) along Y - axis.
This curve is called as the forward characteristics of the diode. From this graph, the approximate
voltage above which the diode current sharply increases, can be determined, which gives the "Knee
Voltage" of the diode.
For germanium diode, the knee voltage is 0.3 volt and for silicon diode the knee voltage is 0.7 volt,
which is almost equal to the barrier potential.
In the upper region of the curve, two perpendicular lines OA and AB are drawn to find the dynamic
forward resistance, as shown in the model graph. In this OA gives the value of small change in p.d.
applied across the diode (v) and AB gives the corresponding change in the current flowing through the
diode (i). These values are determined.
Then the dynamic forward resistance of the diode is determined by using the given formula.

274
Result :
The forward characteristics of a P-N Junction diode is drawn.
From the graph,
(i) The dynamic forward resistance r = ohm

(ii) The knee voltage Vk = volt

milli
Ammeter
P Diode
N
500W

DC power supply
mA

Forward bias
(0-15mA)

(0-10v)
Voltmeter
(0-1.5v)

Circuit diagram for drawing the forward characteristics of P-N junction diode.
Tabular column :
Readings for drawing the forward characteristics of P-N junction diode

P.d applied across Current flowing through

Sl. No. the diode (v) the
diode
Unit d v d (i) milliampere (mA)
1 i ol i
. v t v
2
.
3
.
4
.
5
.
6
.
7
.
8
.
9
.

Calculation :
From graph,
V = div ; i = div
For calculating V in volt
In X – axis 10 div = volt ; 1 div = volt
V = div = x = volt

275
 B

Di
O
Dv A

o X
VK
Forward p.d. applied across the
diode v (volt)

For calculating i in mA
In Y – axis 10 div = mA ; 1 div = mA
i = div = x = mA
= × 10-3 A

V
(i) The dynamic forward resistance is r = 
i

= ohm
(ii) The knee voltage is Vk = div
= x
= volt

276
 16. LOGIC GATES
Aim :

To find the output condition for different combinations of the inputs for NOT gate and two inputs
AND, OR, NAND and NOR logic gates, using I C Chips.
Apparatus required :
Logic gate apparatus kit with chip mounting base, integrated circuit chips (IC chips) for the above
logic gates.
Description :
The logic gate kit consists of required DC power supply, light emitting diodes (LEDs), to indicate
the input and the output signal, toggle switches to provide various combinations of input and breadboard
to mount the desired IC chip.
In the kit when the LED glows, the corresponding condition is “ON”,
and the binary is 1. If the LED does not glow the corresponding condition is “OFF”, and the binary
is 0.
Procedure :
(a) NOT gate :
It is the simplest form of the logic gates. It has only one input and one output. The function of this
gate is to produce one logic level from the opposite logic level. i.e. If the input condition is “ON”, the
output condition will be “OFF” and if the input condition is “OFF”, the output condition will be “ON”.
The NOT gate is available in IC chip number 7404. The pin configuration for the power supply, ground,
input and output of the IC chip is given in the figure and the Boolean expression is also given.
The 7404 IC is mounted on the base provided in the kit. Suitable input level is given and the
corresponding output level is noted in the truth table of NOT gate.
The function of this gate is inverting or complementing the input signal. So, this gate is also called
as a 'Complementer or Inverter gate'.
(b) AND gate :
This gate has two or more inputs and only one output. The function of this gate is to produce a true
condition on its output, if and only if all its input conditions are true. i.e., the output of the gate is in
“ON” condition if and only if all its inputs are in “ON” condition.
The two inputs AND gate is available in IC chip number 7408. The pin configuration for the power
supply, ground, input and output of the IC chip is given in the figure and the Boolean expression is also
given.
The 7408 IC is mounted on the base provided in the kit. Suitable combinations of input levels are
given and the corresponding output levels are noted in the truth table of AND gate.
This gate is also called as a 'Multiplier gate'.
(c) OR gate :
This gate has two or more inputs and only one output. The function of this gate is to produce the true
condition on its output, when one or more of its input conditions are true. i.e., The output of the gate is
in “ON” condition if any one of its inputs is in “ON” condition.
The two inputs OR gate is available in IC chip number 7432. The pin configuration for the power
supply, ground, input and output of the IC chip is given in the figure and the Boolean expression is also
given.
277
 The 7432 IC is mounted on the base provided in the kit. Suitable combinations of input levels are
given and the corresponding output levels are noted in the truth table of OR gate.
This gate is also called as a 'Adder gate'.
These above three gates are called 'Basic Gates'.
(d) NAND gate :
It is a combination of AND and NOT gates. The output of NAND gate is complementation of that
of an AND gate.
The two inputs NAND gate is available in IC chip number 7400. The pin configuration for the power
supply, ground, input and output of the IC chip is given in the figure and the Boolean expression is also
given.
The 7400 IC is mounted on the base provided in the kit. Suitable combinations of input levels are
given and the corresponding output levels are noted in the truth table of NAND gate.
(e) NOR gate :
This gate is a combination of NOT and OR gates. The output of the NOR gate is Complementation
of that of an OR gate.
The two inputs NOR gate is available in IC chip number 7402. The pin configuration for the power
supply, ground, input and output of the IC chip is given in the figure and the Boolean expression is also
given.
The 7402 IC is mounted on the base provided in the kit. Suitable combinations of input levels are
given and the corresponding output levels are noted in the truth table of NOR gate.
These above two gates are known as 'Universal gates'.
Result :
Thus the output condition for different combinations of the inputs for NOT gate and two inputs
AND, OR, NAND and NOR logic gates are noted in the truth table.
a) NOT gate :
A y=A

Symbolic representation
+vce
14 13 12 11 10 9 8

I C : 7404

1 2 3 4 5 6 7
GND
NOT gate IC chip
IC No: 7404

NOT gate IC chip

IC No: 7404
Pin configuration :

Input pins : 1,3,5,9,11,13

Output pins : 2,4,6,8,10,12
Ground pin : 7
Power pin (+V)cc : 14

278
Truth table for NOT gate

INPUT OUTPUT
A Y= A
0
1

b) AND gate :
A Y = A.B

Symbolic representation

14 13 12 11 10 9 8

I C : 7408

1 2 3 4 5 6 7
GND
Two input AND gate IC chip
IC No: 7408
 Two input AND gate IC chip
IC No: 7408

Pin configuration:

Input pins : 1,2,4,5,9,10,12,13

Output pins : 3,6,8,11
Ground pin :7
Power pin (+ Vcc) : 14

Truth table for AND gate

INPU OUTPUT
A TS B Y = A.B
0 0
0 1
1 0
1 1
c) OR gate :
A y=A+B
B

Symbolic representation
+Vcc
14 13 12 11 10 9 8

I C : 7402

1 2 3 4 5 6 7
GND

Pin configuration:
Input pins: 1, 2, 4, 5, 9, 10, 12, 13
Output pins: 3, 6, 8, 11 Two input NOR gate IC chip
IC No: 7402
Ground pin : 7
Two input OR gate IC chip
IC No: 7432
Power pin (+ Vcc): 14

Truth table for OR gate

INPU OUTPUT
A TS B Y=A+ B
0 0
0 1
1 0
1 1

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d) NAND gate:

A y=A.B
B

Symbolic representation
+Vce
14 13 12 11 10 9 8

I C : 7400

1 2 3 4 5 6 7
GND

Two input NAND gate IC chip

IC No: 7400
Pin configuration:
Inputpins:1,2,4,5,9,10,12,13
Output pins: 3,6,8,11
Ground pin : 7

Power pin (+ Vcc): 14

Truth Table for NAND gate

INPUTS OUTPUT
A B Y = 𝐴. 𝐵
0 0
0 1
1 0
1 1
e) NOR gate :
A y=A+B
B

Symbolic representation
+V cc
14 13 12 11 10 9 8

I C : 7402

1 2 3 4 5 6 7
GND

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 Two input NOR gate IC chip
IC No: 7402
Pin configuration:
Input pins: 2, 3, 5, 6, 8, 9, 11, 12
Output pins: 1, 4, 10, 13
Ground pin : 7
Power pin (+ Vcc) : 14

Truth table for NOR gate

INPU OUTPUT
A TS B Y= 𝐴 + 𝐵
0 0
0 1
1 0
1 1

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 40006 ENGINEERING PHYSICS PRACTICAL

MODEL QUESTION PAPER

1. Measure the thickness of the given irregular glass plate using micrometer. Determine the
area of the glass plate using a graph sheet and calculate the volume of the glass plate.

2. Measure the length and diameter of the given solid cylinder using Vernier calipers and
then calculates the volume of the solid cylinder.

3. Verify the parallelogram law of forces using concurrent forces.

4. Verify the Lami’s theorem using concurrent forces.

5. Compare the coefficient of viscosity of two Liquids by capillary flow method, using
graduated burette.

6. Determine the coefficient of viscosity of a highly viscous liquid by Stokes’ method.

7. Determine the frequency of the given tuning fork using Sonometer.

8. Compare the magnetic moments of the two bar magnets using deflection magnetometer
in Tan-A position, by equal distance method.

9. Determine the refractive index of the given transparent liquid using travelling Microscope.

10. Measure the angle of the prism using Spectrometer.

11. Draw the V – I characteristics of the solar cell.

12. Verify the laws of resistances by connecting the two given standard resistances

(i) in series and (ii) in parallel, using Ohm’s law.

13. Determine the specific heat capacity of water, using Joule’s calorimeter.

14. Determine the electro chemical equivalent (e.c.e.) of copper using Copper Voltameter.

15. Draw the voltage – current characteristics of a P-N junction diode in forward bias and
then find the ‘dynamic forward resistance’ & ‘knee voltage’ from the graph.

16. Find the output conditions for different combinations of the input for NOT gate and 2
inputs AND, OR, NAND & NOR logic gates using IC chips.

The college authority should ensure the safety of all the students during the lab practical

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Notes

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Notes

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Notes

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Notes

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