HC Verma-1.pdf

CONCEPTS OF PHYSICS [VOLUME 1] H C VERMA, PhD Retired Professor Department of Physics IIT, Kanpur  Dedicated to Indian Philosophy & Way of Life of which my parents were an integral part ...

CONCEPTS OF PHYSICS [VOLUME 1] H C VERMA, PhD Retired Professor Department of Physics IIT, Kanpur  Dedicated to Indian Philosophy & Way of Life of which my parents were an integral part  FOREWORD A few years ago I had an occasion to go through the book Calculus by L V Terasov. It unravels intricacies of the subject through a dialogue between Teacher and Student. I thoroughly enjoyed reading it. For me this seemed to be one of the few books which teach a difficult subject through inquisition, and using programmed concept for learning. After that book, Dr Harish Chandra Verma’s book on physics, CONCEPTS OF PHYSICS is another such attempt, even though it is not directly in the dialogue form. I have thoroughly appreciated it. It is clear that Dr Verma has spent considerable time in formulating the structure of the book, besides its contents. I think he has been successful in this attempt. Dr Verma’s book has been divided into two parts because of the size of the total manuscript. There have been several books on this subject, each one having its own flavour. However, the present book is a totally different attempt to teach physics, and I am sure it will be extremely useful to the undergraduate students. The exposition of each concept is extremely lucid. In carefully formatted chapters, besides problems and short questions, a number of objective questions have also been included. This book can certainly be extremely useful not only as a textbook, but also for preparation of various competitive examinations. Those who have followed Dr Verma’s scientific work always enjoyed the outstanding contributions he has made in various research areas. He was an outstanding student of Physics Department of IIT Kanpur during his academic career. An extremely methodical, sincere person as a student, he has devoted himself to the task of educating young minds and inculcating scientific temper amongst them. The present venture in the form of these two volumes is another attempt in that direction. I am sure that young minds who would like to learn physics in an appropriate manner will find these volumes extremely useful. I must heartily congratulate Dr Harish Chandra Verma for the magnificent job he has done. Y R Waghmare Professor of Physics IIT Kanpur.  PREFACE Why a new book ? Excellent books exist on physics at an introductory college level so why a new one ? Why so many books exist at the same level, in the first place, and why each of them is highly appreciated ? It is because each of these books has the privilege of having an author or authors who have experienced physics and have their own method of communicating with the students. During my years as a physics teacher, I have developed a somewhat different methodology of presenting physics to the students. Concepts of Physics is a translation of this methodology into a textbook. Prerequisites The book presents a calculus-based physics course which makes free use of algebra, trigonometry and co-ordinate geometry. The level of the latter three topics is quite simple and high school mathematics is sufficient. Calculus is generally done at the introductory college level and I have assumed that the student is enrolled in a concurrent first calculus course. The relevant portions of calculus have been discussed in Chapter 2 so that the student may start using it from the beginning. Almost no knowledge of physics is a prerequisite. I have attempted to start each topic from the zero level. A receptive mind is all that is needed to use this book. Basic philosophy of the book The motto underlying the book is physics is enjoyable. Being a description of the nature around us, physics is our best friend from the day of our existence. I have extensively used this aspect of physics to introduce the physical principles starting with common day occurrences and examples. The subject then appears to be friendly and enjoyable. I have taken care that numerical values of different quantities used in problems correspond to real situations to further strengthen this approach. Teaching and training The basic aim of physics teaching has been to let the student know and understand the principles and equations of physics and their applications in real life. However, to be able to use these principles and equations correctly in a given physical situation, one needs further training. A large number of questions and solved and unsolved problems are given for this purpose. Each question or problem has a specific purpose. It may be there to bring out a subtle point which might have passed unnoticed while doing the text portion. It may be a further elaboration of a concept developed in the text. It may be there to make the student react when several concepts introduced in different chapters combine and show up as a physical situation and so on. Such tools have been used to develop a culture: analyse the situation, make a strategy to invoke correct principles and work it out. Conventions I have tried to use symbols, names, etc., which are popular nowadays. SI units have been consistently used throughout the book. SI prefixes such as micro, milli, mega, etc., are used whenever they make the presentation −6 more readable. Thus, 20 µF is preferred over 20 × 10 F. Co-ordinate sign convention is used in geometrical optics. Special emphasis has been given to dimensions of physical quantities. Numerical values of physical quantities have been mentioned with the units even in equations to maintain dimensional consistency. I have tried my best to keep errors out of this book. I shall be grateful to the readers who point out any errors and/or make other constructive suggestions. H C Verma  ACKNOWLEDGEMENTS The work on this book started in 1984. Since then, a large number of teachers, students and physics lovers have made valuable suggestions which I have incorporated in this work. It is not possible for me to acknowledge all of them individually. I take this opportunity to express my gratitude to them. However, to Dr S B Mathur, who took great pains in going through the entire manuscript and made valuable comments, I am specially indebted. I am also beholden to my colleagues Dr A Yadav, Dr Deb Mukherjee, Mr M M R Akhtar, Dr Arjun Prasad, Dr S K Sinha and others who gave me valuable advice and were good enough to find time for fruitful discussions. To Dr T K Dutta of B E College, Sibpur I am grateful for having taken time to go through portions of the book and making valuable comments. I thank my student Mr Shailendra Kumar who helped me in checking the answers. I am grateful to Dr B C Rai, Mr Sunil Khijwania & Mr Tejaswi Khijwania for helping me in the preparation of rough sketches for the book. Finally, I thank the members of my family for their support and encouragement. H C Verma  TO THE STUDENTS Here is a brief discussion on the organisation of the book which will help you in using the book most effectively. The book contains 47 chapters divided in two volumes. Though I strongly believe in the underlying unity of physics, a broad division may be made in the book as follows: Chapters 1–14: Mechanics 15–17: Waves including wave optics 18–22: Optics 23–28: Heat and thermodynamics 29–40: Electric and magnetic phenomena 41–47: Modern physics Each chapter contains a description of the physical principles related to that chapter. It is well supported by mathematical derivations of equations, descriptions of laboratory experiments, historical background, etc. There are "in-text" solved examples. These examples explain the equation just derived or the concept just discussed. These will help you in fixing the ideas firmly in your mind. Your teachers may use these in-text examples in the classroom to encourage students to participate in discussions. After the theory section, there is a section on Worked Out Examples. These numerical examples correspond to various thinking levels and often use several concepts introduced in that chapter or even in previous chapters. You should read the statement of a problem and try to solve it yourself. In case of difficulty, look at the solution given in the book. Even if you solve the problem successfully, you should look into the solution to compare it with your method of solution. You might have thought of a better method, but knowing more than one method is always beneficial. Then comes the part which tests your understanding as well as develops it further. Questions for Short Answer generally touch very minute points of your understanding. It is not necessary that you answer these questions in a single sitting. They have great potential to initiate very fruitful dicussions. So, freely discuss these questions with your friends and see if they agree with your answer. Answers to these questions are not given for the simple reason that the answers could have cut down the span of such discussions and that would have sharply reduced the utility of these questions. There are two sections on multiple-choice questions, namely OBJECTIVE I and OBJECTIVE II. There are four options following each of these questions. Only one option is correct for OBJECTIVE I questions. Any number of options, zero to four, may be correct for OBJECTIVE II questions. Answers to all these questions are provided. Finally, a set of numerical problems are given for your practice. Answers to these problems are also provided. The problems are generally arranged according to the sequence of the concepts developed in the chapter but they are not grouped under section-headings. I don’t want to bias your ideas beforehand by telling you that this problem belongs to that section and hence use that particular equation. You should yourself look into the problem and decide which equations or which methods should be used to solve it. Many of the problems use several concepts developed in different sections of the chapter. Many of them even use the concepts from the previous chapters. Hence, you have to plan out the strategy after understanding the problem. Remember, no problem is difficult. Once you understand the theory, each problem will become easy. So, don’t jump to exercise problems before you have gone through the theory, the worked-out problems and the objectives. Once you feel confident in theory, do the exercise problems. The exercise problems are so arranged that they gradually require more thinking. I hope you will enjoy Concepts of Physics. H C Verma  Table of Contents Chapter 1 Objective II 50 Exercises 51 Introduction to Physics 1 1.1 What Is Physics ? 1 Chapter 4 1.2 Physics and Mathematics 1 The Forces 56 1.3 Units 2 4.1 Introduction 56 1.4 Definitions of Base Units 3 4.2 Gravitational Force 56 1.5 Dimension 4 4.3 Electromagnetic (EM) Force 57 1.6 Uses of Dimension 4 4.4 Nuclear Forces 59 1.7 Order of Magnitude 6 4.5 Weak Forces 59 1.8 The Structure of World 6 4.6 Scope of Classical Physics 59 Worked Out Examples 7 Worked Out Examples 60 Questions for Short Answer 8 Questions for Short Answer 61 Objective I 9 Objective I 62 Objective II 9 Objective II 62 Exercises 9 Exercises 63 Chapter 2 Chapter 5 Physics and Mathematics 12 Newton’s Laws of Motion 64 2.1 Vectors and Scalars 12 5.1 First Law of Motion 64 2.2 Equality of Vectors 13 5.2 Second Law of Motion 65 2.3 Addition of Vectors 13 5.3 Working with Newton’s First and Second Law 66 2.4 Multiplication of a Vector by a Number 14 5.4 Newton’s Third Law of Motion 68 2.5 Subtraction of Vectors 14 5.5 Pseudo Forces 69 2.6 Resolution of Vectors 14 5.6 The Horse and the Cart 71 2.7 Dot Product or Scalar Proudct of Two Vectors 15 5.7 Inertia 71 2.8 Cross Product or Vector Product of Two Vectors 16 Worked Out Examples 72 dy 2.9 Differential Calculus : as Rate Measurer 17 Questions for Short Answer 76 dx Objective I 77 2.10 Maxima and Minima 18 Objective II 78 2.11 Integral Calculus 19 Exercises 79 2.12 Significant Digits 21 2.13 Significant Digits in Calculations 22 Chapter 6 2.14 Errors in Measurement 23 Worked Out Examples 24 Friction 85 Questions for Short Answer 27 6.1 Friction as the Component of Contact Force 85 Objective I 28 6.2 Kinetic Friction 86 Objective II 28 6.3 Static Friction 87 Exercises 29 6.4 Laws of Friction 88 6.5 Understanding Friction at Atomic Level 88 Chapter 3 6.6 A Laboratory Method to Measure Rest and Motion : Kinematics 31 Friction Coefficient 89 3.1 Rest and Motion 31 Worked Out Examples 91 3.2 Distance and Displacement 31 Questions for Short Answer 95 3.3 Average Speed and Instantaneous Speed 32 Objective I 96 3.4 Average Velocity and Instantaneous Velocity 33 Objective II 97 3.5 Average Acceleration and Instantaneous Acceleration 34 Exercises 97 3.6 Motion in a Straight Line 34 Chapter 7 3.7 Motion in a Plane 37 3.8 Projectile Motion 38 Circular Motion 101 3.9 Change of Frame 39 7.1 Angular Variables 101 Worked Out Examples 41 7.2 Unit Vectors along the Radius and the Tangent 102 Questions for Short Answer 48 7.3 Acceleration in Circular Motion 102 Objective I 49 7.4 Dynamics of Circular Motion 103  (xii) 7.5 Circular Turnings and Banking of Roads 104 10.6 Bodies in Equilibrium 172 7.6 Centrifugal Force 105 10.7 Bending of a Cyclist on a Horizontal Turn 172 7.7 Effect of Earth’s Rotation on Apparent Weight 106 10.8 Angular Momentum 173 Worked Out Examples 107 10.9 L = Iω 173 Questions for Short Answer 111 10.10 Conservation of Angular Momentum 173 Objective I 112 10.11 Angular Impulse 174 Objective II 113 10.12 Kinetic Energy of a Rigid Body Exercises 114 Rotating About a Given Axis 174 10.13 Power Delivered and Work Done by a Torque 175 Chapter 8 10.14 Calculation of Moment of Inertia 175 Work and Energy 118 10.15 Two Important Theorems on Moment of Inertia 178 8.1 Kinetic Energy 118 10.16 Combined Rotation and Translation 180 8.2 Work and Work-energy Theorem 118 10.17 Rolling 180 8.3 Calculation of Work Done 119 10.18 Kinetic Energy of a Body in Combined Rotation and Translation 182 8.4 Work-energy Theorem for a System of Particles 120 10.19 Angular Momentum of a Body 8.5 Potential Energy 121 in Combined Rotation and Translation 182 8.6 Conservative and Nonconservative Forces 121 10.20 Why Does a Rolling Sphere Slow Down ? 183 8.7 Definition of Potential Energy and Worked Out Examples 183 Conservation of Mechanical Energy 122 Questions for Short Answer 192 8.8 Change in the Potential Energy in a Rigid-body-motion 123 Objective I 193 8.9 Gravitational Potential Energy 124 Objective II 194 8.10 Potential Energy of a Compressed or Exercises 195 Extended Spring 124 Chapter 11 8.11 Different Forms of Energy : Mass Energy Equivalence 126 Gravitation 203 Worked Out Examples 126 11.1 Historical Introduction 203 Questions for Short Answer 130 11.2 Measurement of Gravitational Constant G 204 Objective I 131 11.3 Gravitational Potential Energy 206 Objective II 131 11.4 Gravitational Potential 207 Exercises 132 11.5 Calculation of Gravitational Potential 207 11.6 Gravitational Field 210 Chapter 9 11.7 Relation between Gravitational Field and Potential 210 Centre of Mass, Linear Momentum, Collision 139 11.8 Calculation of Gravitational Field 211 9.1 Centre of Mass 139 11.9 Variation in the Value of g 214 9.2 Centre of Mass of Continuous Bodies 141 11.10 Planets and Satellites 216 9.3 Motion of the Centre of Mass 142 11.11 Kepler’s Laws 217 9.4 Linear Momentum and its Conservation Principle 144 11.12 Weightlessness in a Satellite 217 9.5 Rocket Propulsion 144 11.13 Escape Velocity 217 9.6 Collision 145 11.14 Gravitational Binding Energy 218 9.7 Elastic Collision in One Dimension 147 11.15 Black Holes 218 9.8 Perfectly Inelastic Collision in One Dimension 148 11.16 Inertial and Gravitational Mass 218 9.9 Coefficient of Restitution 148 11.17 Possible Changes in the Law of Gravitation 219 9.10 Elastic Collision in Two Dimensions 148 Worked Out Examples 219 9.11 Impulse and Impulsive Force 149 Questions for Short Answer 223 Worked Out Examples 149 Objective I 224 Questions for Short Answer 156 Objective II 225 Objective I 157 Exercises 225 Objective II 158 Chapter 12 Exercises 159 Simple Harmonic Motion 229 Chapter 10 12.1 Simple Harmonic Motion 229 Rotational Mechanics 166 12.2 Qualitative Nature of Simple Harmonic Motion 229 10.1 Rotation of a Rigid Body 12.3 Equation of Motion of a Simple Harmonic Motion 230 about a Given Fixed Line 166 12.4 Terms Associated with Simple Harmonic Motion 231 10.2 Kinematics 167 12.5 Simple Harmonic Motion as a 10.3 Rotational Dynamics 168 Projection of Circular Motion 233 10.4 Torque of a Force about the Axis of Rotation 169 12.6 Energy Conservation in Simple Harmonic Motion 233 10.5 Γ = Iα 170 12.7 Angular Simple Harmonic Motion 234  (xiii) 12.8 Simple Pendulum 235 Questions for Short Answer 297 12.9 Physical Pendulum 237 Objective I 298 12.10 Torsional Pendulum 237 Objective II 300 12.11 Composition of Two Simple Harmonic Motions 238 Exercises 300 12.12 Damped Harmonic Motion 242 12.13 Forced Oscillation and Resonance 242 Chapter 15 Worked Out Examples 243 Wave Motion and Waves on a String 303 Questions for Short Answer 249 15.1 Wave Motion 303 Objective I 250 15.2 Wave Pulse on a String 303 Objective II 251 15.3 Sine Wave Travelling on a String 305 Exercises 252 15.4 Velocity of a Wave on a String 307 Chapter 13 15.5 Power Transmitted along the String by a Sine Wave 308 Fluid Mechanics 258 15.6 Interference and the Principle of Superposition 308 13.1 Fluids 258 15.7 Interference of Waves Going in Same Direction 309 13.2 Pressure in a Fluid 258 15.8 Reflection and Transmission of Waves 310 13.3 Pascal’s Law 259 15.9 Standing Waves 311 13.4 Atmospheric Pressure and Barometer 260 15.10 Standing Waves on a String Fixed 13.5 Archimedes’ Principle 261 at Both Ends (Qualitative Discussion) 312 13.6 Pressure Difference and Buoyant 15.11 Analytic Treatment of Vibration Force in Accelerating Fluids 262 of a String Fixed at Both Ends 314 13.7 Flow of Fluids 263 15.12 Vibration of a String Fixed at One End 315 13.8 Steady and Turbulent Flow 263 15.13 Laws of Transverse Vibrations of a 13.9 Irrotational Flow of an String : Sonometer 315 Incompressible and Nonviscous Fluid 264 15.14 Transverse and Longitudinal Waves 317 13.10 Equation of Continuity 264 15.15 Polarization of Waves 317 13.11 Bernoulli’s Equation 264 Worked Out Examples 318 13.12 Applications of Bernoulli’s Equation 266 Questions for Short Answer 321 Worked Out Examples 267 Objective I 322 Questions for Short Answer 270 Objective II 323 Objective I 271 Exercises 323 Objective II 272 Exercises 273 Chapter 16 Chapter 14 Sound Waves 329 Some Mechanical Properties of Matter 277 16.1 The Nature and Propagation of Sound Waves 329 14.1 Molecular Structure of a Material 277 16.2 Displacement Wave and Pressure Wave 330 14.2 Elasticity 279 16.3 Speed of a Sound Wave in a Material Medium 331 14.3 Stress 279 16.4 Speed of Sound in a Gas : Newton’s 14.4 Strain 280 Formula and Laplace’s Correction 332 14.5 Hooke’s Law and the Modulii of Elasticity 280 16.5 Effect of Pressure, Temperature and Humidity on the Speed of Sound in Air 333 14.6 Relation between Longitudinal Stress and Strain 281 16.6 Intensity of Sound Waves 333 14.7 Elastic Potential Energy of a Strained Body 282 16.7 Appearance of Sound to Human Ear 334 14.8 Determination of Young’s Modulus in Laboratory 283 14.9 Surface Tension 284 16.8 Interference of Sound Waves 335 14.10 Surface Energy 286 16.9 Standing Longitudinal Waves and Vibrations of Air Columns 336 14.11 Excess Pressure Inside a Drop 286 16.10 Determination of Speed of Sound in Air 339 14.12 Excess Pressure in a Soap Bubble 288 16.11 Beats 340 14.13 Contact Angle 288 16.12 Diffraction 342 14.14 Rise of Liquid in a Capillary Tube 289 16.13 Doppler Effect 342 14.15 Viscosity 290 14.16 Flow through a Narrow Tube : Poiseuille’s 16.14 Sonic Booms 344 Equation 291 16.15 Musical Scale 345 14.17 Stokes’ Law 291 16.16 Acoustics of Buildings 345 14.18 Terminal Velocity 292 Worked Out Examples 346 14.19 Measuring Coefficient of Viscosity Questions for Short Answer 351 by Stokes’ Method 292 Objective I 351 14.20 Critical Velocity and Reynolds Number 293 Objective II 352 Worked Out Examples 293 Exercises 352  (xiv) Chapter 17 19.6 Resolving Power of a Microscope and a Telescope 425 19.7 Defects of Vision 425 Light Waves 360 Worked Out Examples 427 17.1 Waves or Particles 360 Questions for Short Answer 430 17.2 The Nature of Light Waves 360 Objective I 431 17.3 Huygens’ Principle 362 Objective II 431 17.4 Young’s Double Hole Experiment 365 Exercises 432 17.5 Young’s Double Slit Experiment 365 17.6 Optical Path 366 Chapter 20 17.7 Interference from Thin Films 367 Dispersion and Spectra 434 17.8 Fresnel’s Biprism 369 20.1 Dispersion 434 17.9 Coherent and Incoherent Sources 369 20.2 Dispersive Power 434 17.10 Diffraction of Light 370 20.3 Dispersion without Average Deviation 17.11 Fraunhofer Diffraction by a Single Slit 371 and Average Deviation without Dispersion 435 17.12 Fraunhofer Diffraction by a Circular Aperture 372 20.4 Spectrum 436 17.13 Fresnel Diffraction at a Straight Edge 373 20.5 Kinds of Spectra 437 17.14 Limit of Resolution 373 20.6 Ultraviolet and Infrared Spectrum 438 17.15 Scattering of Light 374 20.7 Spectrometer 438 17.16 Polarization of Light 374 20.8 Rainbow 440 Worked Out Examples 376 Worked Out Examples 440 Questions for Short Answer 379 Questions for Short Answer 441 Objective I 379 Objective I 441 Objective II 380 Objective II 442 Exercises 380 Exercises 442 Chapter 18 Geometrical Optics 385 Chapter 21 18.1 Reflection at Smooth Surfaces 385 Speed of Light 444 18.2 Spherical Mirrors 385 21.1 Historical Introduction 444 18.3 Relation Between u, v and R for Spherical Mirrors 387 21.2 Fizeau Method 444 18.4 Extended Objects and Magnification 388 21.3 Foucault Method 445 18.5 Refraction at Plane Surfaces 388 21.4 Michelson Method 447 18.6 Critical Angle 389 Questions for Short Answer 447 18.7 Optical Fibre 389 Objective I 448 18.8 Prism 390 Objective II 448 18.9 Refraction at Spherical Surfaces 391 Exercises 448 18.10 Extended Objects : Lateral Magnification 392 18.11 Refraction through Thin Lenses 393 Chapter 22 18.12 Lens Maker’s Formula and Lens Formula 394 Photometry 449 18.13 Extended Objects : Lateral Magnification 395 22.1 Total Radiant Flux 449 18.14 Power of a Lens 396 18.15 Thin Lenses in Contact 396 22.2 Luminosity of Radiant Flux 449 18.16 Two Thin Lenses Separated By a Distance 397 22.3 Luminous Flux : Relative Luminosity 449 18.17 Defects of Images 398 22.4 Luminous Efficiency 450 Worked Out Examples 400 22.5 Luminous Intensity or Illuminating Power 450 Questions for Short Answer 410 22.6 Illuminance 450 Objective I 410 22.7 Inverse Square Law 451 Objective II 412 22.8 Lambert’s Cosine Law 451 Exercises 412 22.9 Photometers 451 Worked Out Examples 452 Chapter 19 Questions for Short Answer 453 Optical Instruments 419 Objective I 454 19.1 The Eye 419 Objective II 454 19.2 The Apparent Size 420 Exercises 455 19.3 Simple Microscope 420 APPENDIX A 457 19.4 Compound Microscope 421 APPENDIX B 458 19.5 Telescopes 422 INDEX 459  CHAPTER 1 INTRODUCTION TO PHYSICS 1.1 WHAT IS PHYSICS ? Physics goes the same way. The nature around us is like a big chess game played by Nature. The events The nature around us is colourful and diverse. It in the nature are like the moves of the great game. contains phenomena of large varieties. The winds, the We are allowed to watch the events of nature and sands, the waters, the planets, the rainbow, heating of guess at the basic rules according to which the events objects on rubbing, the function of a human body, the take place. We may come across new events which do energy coming from the sun and the nucleus …… there not follow the rules guessed earlier and we may have are a large number of objects and events taking place to declare the old rules inapplicable or wrong and around us. discover new rules. Physics is the study of nature and its laws. We Since physics is the study of nature, it is real. No expect that all these different events in nature take one has been given the authority to frame the rules of place according to some basic laws and revealing these physics. We only discover the rules that are operating laws of nature from the observed events is physics. For in nature. Aryabhat, Newton, Einstein or Feynman are example, the orbiting of the moon around the earth, great physicists because from the observations falling of an apple from a tree and tides in a sea on a available at that time, they could guess and frame the full moon night can all be explained if we know the laws of physics which explained these observations in Newton’s law of gravitation and Newton’s laws of a convincing way. But there can be a new phenomenon motion. Physics is concerned with the basic rules any day and if the rules discovered by the great which are applicable to all domains of life. scientists are not able to explain this phenomenon, no Understanding of physics, therefore, leads to one will hesitate to change these rules. applications in many fields including bio and medical sciences. 1.2 PHYSICS AND MATHEMATICS The great physicist Dr R. P. Feynman has given a The description of nature becomes easy if we have wonderful description of what is “understanding the the freedom to use mathematics. To say that the nature”. Suppose we do not know the rules of chess gravitational force between two masses is proportional but are allowed to watch the moves of the players. If to the product of the masses and is inversely we watch the game for a long time, we may make out proportional to the square of the distance apart, is some of the rules. With the knowledge of these rules more difficult than to write we may try to understand why a player played a particular move. However, this may be a very difficult m1m2 F∝ 2 ⋅ … (1.1) task. Even if we know all the rules of chess, it is not r so simple to understand all the complications of a game Further, the techniques of mathematics such as in a given situation and predict the correct move. algebra, trigonometry and calculus can be used to Knowing the basic rules is, however, the minimum make predictions from the basic equations. Thus, if we requirement if any progress is to be made. know the basic rule (1.1) about the force between two One may guess at a wrong rule by partially particles, we can use the technique of integral calculus watching the game. The experienced player may make to find what will be the force exerted by a uniform rod use of a rule for the first time and the observer of the on a particle placed on its perpendicular bisector. game may get surprised. Because of the new move Thus, mathematics is the language of physics. some of the rules guessed at may prove to be wrong Without knowledge of mathematics it would be much and the observer will frame new rules. more difficult to discover, understand and explain the 2 Concepts of Physics laws of nature. The importance of mathematics in and any changes in standard units are communicated today’s world cannot be disputed. However, through the publications of the Conference. mathematics itself is not physics. We use a language to express our ideas. But the idea that we want to Fundamental and Derived Quantities express has the main attention. If we are poor at There are a large number of physical quantities grammar and vocabulary, it would be difficult for us which are measured and every quantity needs a to communicate our feelings but while doing so our definition of unit. However, not all the quantities are basic interest is in the feeling that we want to express. independent of each other. As a simple example, if a It is nice to board a deluxe coach to go from Delhi to unit of length is defined, a unit of area is automatically Agra, but the sweet memories of the deluxe coach and obtained. If we make a square with its length equal the video film shown on way are next to the prime to its breadth equal to the unit length, its area can be goal of reaching Agra. “To understand nature” is called the unit area. All areas can then be compared physics, and mathematics is the deluxe coach to take to this standard unit of area. Similarly, if a unit of us there comfortably. This relationship of physics and length and a unit of time interval are defined, a unit mathematics must be clearly understood and kept in of speed is automatically obtained. If a particle covers mind while doing a physics course. a unit length in unit time interval, we say that it has a unit speed. We can define a set of fundamental 1.3 UNITS quantities as follows : (a) the fundamental quantities should be indepen- Physics describes the laws of nature. This dent of each other, and description is quantitative and involves measurement (b) all other quantities may be expressed in terms and comparison of physical quantities. To measure a of the fundamental quantities. physical quantity we need some standard unit of that It turns out that the number of fundamental quantities quantity. An elephant is heavier than a goat but is only seven. All the rest may be derived from these exactly how many times ? This question can be easily quantities by multiplication and division. Many answered if we have chosen a standard mass calling different choices can be made for the fundamental it a unit mass. If the elephant is 200 times the unit quantities. For example, one can take speed and time mass and the goat is 20 times we know that the as fundamental quantities. Length is then a derived elephant is 10 times heavier than the goat. If I have quantity. If something travels at unit speed, the the knowledge of the unit length and some one says distance it covers in unit time interval will be called that Gandhi Maidan is 5 times the unit length from a unit distance. One may also take length and time here, I will have the idea whether I should walk down interval as the fundamental quantities and then speed to Gandhi Maidan or I should ride a rickshaw or I will be a derived quantity. Several systems are in use should go by a bus. Thus, the physical quantities are over the world and in each system the fundamental quantitatively expressed in terms of a unit of that quantities are selected in a particular way. The units quantity. The measurement of the quantity is defined for the fundamental quantities are called mentioned in two parts, the first part gives how many fundamental units and those obtained for the derived times of the standard unit and the second part gives quantities are called the derived units. the name of the unit. Thus, suppose I have to study for 2 hours. The numeric part 2 says that it is 2 times Fundamental quantities are also called base of the unit of time and the second part hour says that quantities. the unit chosen here is an hour. SI Units Who Decides the Units ? In 1971 CGPM held its meeting and decided a system of units which is known as the International How is a standard unit chosen for a physical System of Units. It is abbreviated as SI from the quantity ? The first thing is that it should have French name Le Systéme International d′Unités. This international acceptance. Otherwise, everyone will system is widely used throughout the world. choose his or her own unit for the quantity and it will Table (1.1) gives the fundamental quantities and be difficult to communicate freely among the persons their units in SI. distributed over the world. A body named Conférence Générale des Poids et Mesures or CGPM also known as General Conference on Weight and Measures in English has been given the authority to decide the units by international agreement. It holds its meetings  Introduction to Physics 3 Table 1.1 : Fundamental or Base Quantities (a) Invariability : The standard unit must be Quantity Name of the Unit Symbol invariable. Thus, defining distance between the tip of the middle finger and the elbow as a unit of length is Length metre m not invariable. Mass kilogram kg (b) Availability : The standard unit should be Time second s easily made available for comparing with other Electric Current ampere A quantities. Thermodynamic Temperature kelvin K CGPM decided in its 2018 meeting that all the SI Amount of Substance mole mol base quantities will be defined in terms of certain Luminous Intensity candela cd universal constants and these constants will be assigned fixed numerical values by definition. In this case both the criteria of invariability and availability Besides the seven fundamental units two are automatically satisfied. The new definitions supplementary units are defined. They are for plane th became operative since 20 May 2019. We give below angle and solid angle. The unit for plane angle is the definitions of the these quantities. The fixed values radian with the symbol rad and the unit for the solid given to the universal constants will appear in the angle is steradian with the symbol sr. definitions only. The definitions carry certain physical SI Prefixes quantities and concepts that are beyond the scope of this book but you need not worry about it. The magnitudes of physical quantities vary over a wide range. We talk of separation between two Second – 15 protons inside a nucleus which is about 10 m and 1 second is the time that makes the unperturbed the distance of a quasar from the earth which is about  31 ground state hyperfine transition frequency Cs to be 10 m. The mass of an electron is 9.1  10 26 kg and 9192631770 when expressed in the unit Hz which is that of our galaxy is about 2.2  10 kg. The 41 1 equal to s. CGPM recommended standard prefixes for certain powers of 10. Table (1.2) shows these prefixes. Metre Table 1.2 : SI prefixes 1 metre is the length that makes the speed of light Power of 10 Prefix Symbol in vacuum to be 299792458 when expressed in the unit 1 ms , where the second is defined in terms of the 18 exa E caesium frequency Cs. 15 peta P 12 tera T Kilogram 9 giga G 6 mega M 1 kilogram is the mass that makes the Planck’s 34 3 kilo k constant h to be 6.62607015  10 when expressed in 2 1 2 hecto h the unit Js which is equal to kgm s , where the metre 1 deka da and the second are defined in terms of c and Cs. – 1 deci d Ampere – 2 centi c – 3 milli m 1 ampere is the current which makes the 19 – 6 micro  elementary charge e to be 1.602176634  10 when – 9 nano n expressed in the unit C which is equal to As, where – 12 pico p the second is defined in terms of Cs. – 15 femto f – 18 atto a Kelvin 1 kelvin is the temperature that makes the 23 Boltzmann constant to be 1.380649  10 when 1 1.4 DEFINITIONS OF BASE UNITS expressed in the unit JK which is equal to 2 2 1 Any standard unit should have the following two kgm s K , where kilogram, metre and second are properties : defined in terms of h, c and Cs. 4 Concepts of Physics Mole brackets to remind that the equation is among the dimensions and not among the magnitudes. Thus 1 mole of a substance is defined to contain exactly 2 23 equation (1.2) may be written as [force]  MLT. 6.02214076  10 elementary entities. This number is the fixed numerical value of the Avogadro constant Such an expression for a physical quantity in terms 1 of the base quantities is called the dimensional NA when expressed in the unit mol and is called formula. Thus, the dimensional formula of force is Avogadro number. 2 MLT. The two versions given below are equivalent Candela and are used interchangeably. (a) The dimensional formula of force is MLT  2. The candela is the SI unit of luminous intensity. 1 candela is the luminous intensity that makes the (b) The dimensions of force are 1 in mass, 1 in luminous efficacy of monochromatic radiation of length and –2 in time. 12 frequency 540  10 Hz, Kcd to be 683 when expressed Example 1.1 1 1 2 3 in the unit lmW which is equal to cdsrkg m s , Calculate the dimensional formula of energy from the where kilogram, metre and second are defined in terms 1 equation E  2 of h, c and Cs. 2 mv. 1 Solution : Dimensionally, E  mass  velocity , since 2 is 1.5 DIMENSION 2 a number and has no dimension. All the physical quantities of interest can be 2 derived from the base quantities. When a quantity is L 2  [E]  M     ML T. 2 expressed in terms of the base quantities, it is written T as a product of different powers of the base quantities. The exponent of a base quantity that enters into the 1.6 USES OF DIMENSION expression, is called the dimension of the quantity in A. Homogeneity of Dimensions in an Equation that base. To make it clear, consider the physical quantity force. As we shall learn later, force is equal An equation contains several terms which are to mass times acceleration. Acceleration is change in separated from each other by the symbols of equality, velocity divided by time interval. Velocity is length plus or minus. The dimensions of all the terms in an divided by time interval. Thus, equation must be identical. This is another way of saying that one can add or subtract similar physical force  mass  acceleration quantities. Thus, a velocity cannot be added to a force velocity or an electric current cannot be subtracted from the  mass  time thermodynamic temperature. This simple principle is length  time called the principle of homogeneity of dimensions in an  mass  equation and is an extremely useful method to check time 2 whether an equation may be correct or not. If the  mass  length  time.  (1.2) dimensions of all the terms are not same, the equation must be wrong. Let us check the equation Thus, the dimensions of force are 1 in mass, 1 in 1 2 length and –2 in time. The dimensions in all other x  ut  at 2 base quantities are zero. Note that in this type of for the dimensional homogeneity. Here x is the distance calculation the magnitudes are not considered. It is travelled by a particle in time t which starts at a speed equality of the type of quantity that enters. Thus, u and has an acceleration a along the direction of change in velocity, initial velocity, average velocity, motion. final velocity all are equivalent in this discussion, each [x]  L one is length/time. length For convenience the base quantities are [ut]  velocity  time   time  L time represented by one letter symbols. Generally, mass is 1 2 2 2 denoted by M, length by L, time by T and electric  at   [at ]  acceleration  time 2  current by I. The thermodynamic temperature, the amount of substance and the luminous intensity are velocity 2 length/time 2   time   time  L denoted by the symbols of their units K, mol and cd time time respectively. The physical quantity that is expressed Thus, the equation is correct as far as the dimensions in terms of the base quantities is enclosed in square are concerned.  Introduction to Physics 5 Limitation of the Method Thus, knowing the conversion factors for the base 1 2 quantities, one can work out the conversion factor for Note that the dimension of at is same as that 2 any derived quantity if the dimensional formula of the 2 of at. Pure numbers are dimensionless. Dimension derived quantity is known. does not depend on the magnitude. Due to this reason 2 C. Deducing Relation among the Physical Quantities the equation x = ut + at is also dimensionally correct. Thus, a dimensionally correct equation need not be Sometimes dimensions can be used to deduce a actually correct but a dimensionally wrong equation relation between the physical quantities. If one knows must be wrong. the quantities on which a particular physical quantity depends and if one guesses that this dependence is of Example 1.2 product type, method of dimension may be helpful in the derivation of the relation. Taking an example, Test dimensionally if the formula t = 2 π √F/x  m may be suppose we have to derive the expression for the time period of a simple pendulum. The simple pendulum correct, where t is time period, m is mass, F is force and has a bob, attached to a string, which oscillates under x is distance. the action of the force of gravity. Thus, the time period −2 Solution : The dimension of force is MLT. Thus, the may depend on the length of the string, the mass of dimension of the right-hand side is the bob and the acceleration due to gravity. We assume that the dependence of time period on these quantities  √ MLT ⁄ L √ M −2=  T 1 −2 =T⋅ is of product type, that is, a b c The left-hand side is time period and hence the t = kl m g … (1.3) dimension is T. The dimensions of both sides are equal where k is a dimensionless constant and a, b and c and hence the formula may be correct. are exponents which we want to evaluate. Taking the dimensions of both sides, a b −2 c a+c b − 2c B. Conversion of Units T = L M (LT ) = L M T. When we choose to work with a different set of Since the dimensions on both sides must be identical, units for the base quantities, the units of all the we have derived quantities must be changed. Dimensions can a+c=0 be useful in finding the conversion factor for the unit b=0 of a derived physical quantity from one system to and − 2c = 1 other. Consider an example. When SI units are used, giving a = 1 , b = 0 and c = − 1 ⋅ the unit of pressure is 1 pascal. Suppose we choose 2 2 1 cm as the unit of length, 1 g as the unit of mass and Putting these values in equation (1.3) 1 s as the unit of time (this system is still in wide use and is called CGS system). The unit of pressure will be different in this system. Let us call it for the time- t=k √gl ⋅ … (1.4) being 1 CGS pressure. Now, how many CGS pressure Thus, by dimensional analysis we can deduce that is equal to 1 pascal ? the time period of a simple pendulum is independent Let us first write the dimensional formula of of its mass, is proportional to the square root of the pressure. length of the pendulum and is inversely proportional F to the square root of the acceleration due to gravity at We have P= ⋅ the place of observation. A −2 [F] MLT −1 −2 Limitations of the Dimensional Method Thus, [P] = = 2 = ML T [A] L Although dimensional analysis is very useful in −1 −2 deducing certain relations, it cannot lead us too far. so, 1 pascal = (1 kg) (1 m) (1 s) First of all we have to know the quantities on which −1 −2 and 1 CGS pressure = (1 g) (1 cm) (1 s) a particular physical quantity depends. Even then the −1 −2 method works only if the dependence is of the product 1 pascal  1 kg   1 m   1 s  type. For example, the distance travelled by a Thus, =     1 CGS pressure  1 g   1 cm   1 s  uniformly accelerated particle depends on the initial  3  2 − 1 velocity u, the acceleration a and the time t. But the = 10 10    = 10 method of dimensions cannot lead us to the correct or, 1 pascal = 10 CGS pressure. expression for x because the expression is not of 6 Concepts of Physics product type. It is equal to the sum of two terms as of magnitude calculation. In this all numbers are 1 2 x = ut + at. b approximated to 10 form and the calculation is made. 2 Secondly, the numerical constants having no Let us estimate the number of persons that may dimensions cannot be deduced by the method of sit in a circular field of radius 800 m. The area of the dimensions. In the example of time period of a simple field is pendulum, an unknown constant k remains in equation A = πr = 3.14 × (800 m) ≈ 10 m. 2 2 6 2 (1.4). One has to know from somewhere else that this constant is 2π. The average area one person occupies in sitting −1 −1 ≈ 50 cm × 50 cm = 0.25 m = 2.5 × 10 m ≈ 10 m. 2 2 2 Thirdly, the method works only if there are as many equations available as there are unknowns. In The number of persons who can sit in the field is mechanical quantities, only three base quantities 6 10 m 2 7 length, mass and time enter. So, dimensions of these N≈ −1 2 = 10. three may be equated in the guessed relation giving 10 m 7 at most three equations in the exponents. If a Thus of the order of 10 persons may sit in the particular quantity (in mechanics) depends on more field. than three quantities we shall have more unknowns and less equations. The exponents cannot be 1.8 THE STRUCTURE OF WORLD determined uniquely in such a case. Similar constraints are present for electrical or other Man has always been interested to find how the nonmechanical quantities. world is structured. Long long ago scientists suggested that the world is made up of certain indivisible small 1.7 ORDER OF MAGNITUDE particles. The number of particles in the world is large but the varieties of particles are not many. Old Indian In physics, we come across quantities which vary philosopher Kanadi derives his name from this over a wide range. We talk of the size of a mountain proposition (In Sanskrit or Hindi Kana means a small and the size of the tip of a pin. We talk of the mass particle). After extensive experimental work people of our galaxy and the mass of a hydrogen atom. We arrived at the conclusion that the world is made up of talk of the age of the universe and the time taken by just three types of ultimate particles, the proton, the an electron to complete a circle around the proton in neutron and the electron. All objects which we have a hydrogen atom. It becomes quite difficult to get a around us, are aggregation of atoms and molecules. feel of largeness or smallness of such quantities. To The molecules are composed of atoms and the atoms express such widely varying numbers, one uses the have at their heart a nucleus containing protons and powers of ten method. neutrons. Electrons move around this nucleus in In this method, each number is expressed as special arrangements. It is the number of protons, b a × 10 where 1 ≤ a < 10 and b is a positive or negative neutrons and electrons in an atom that decides all the integer. Thus the diameter of the sun is expressed as properties and behaviour of a material. Large number 1.39 × 10 m and the diameter of a hydrogen atom as of atoms combine to form an object of moderate or large 9 − 10 size. However, the laws that we generally deduce for 1.06 × 10 m. To get an approximate idea of the these macroscopic objects are not always applicable to number, one may round the number a to 1 if it is less atoms, molecules, nuclei or the elementary particles. than or equal to 5 and to 10 if it is greater than 5. These laws known as classical physics deal with large The number can then be expressed approximately as b size objects only. When we say a particle in classical 10. We then get the order of magnitude of that physics we mean an object which is small as compared number. Thus, the diameter of the sun is of the order to other moderate or large size objects and for which 9 of 10 m and that of a hydrogen atom is of the order the classical physics is valid. It may still contain − 10 of 10 m. More precisely, the exponent of 10 in such millions and millions of atoms in it. Thus, a particle a representation is called the order of magnitude of of dust dealt in classical physics may contain about 18 that quantity. Thus, the diameter of the sun is 19 10 atoms. orders of magnitude larger than the diameter of a Twentieth century experiments have revealed hydrogen atom. This is because the order of magnitude 9 − 10 another aspect of the construction of world. There are of 10 is 9 and of 10 is − 10. The difference is perhaps no ultimate indivisible particles. Hundreds of 9 − (− 10) = 19. elementary particles have been discovered and there To quickly get an approximate value of a quantity are free transformations from one such particle to the in a given physical situation, one can make an order other. Nature is seen to be a well-connected entity.  Introduction to Physics 7 Worked Out Examples 1. Find the dimensional formulae of the following (c) Q  CV 1 3 1 2 quantities : or, IT  [C]ML I or, [C]  M 2 2 4 T L I T. (a) the universal constant of gravitation G, (d) V  RI (b) the surface tension S, 2 1 3 V ML I T 2 3 or, R  or, [R]   ML I 2 (c) the thermal conductivity k and T. I I (d) the coefficient of viscosity . Some equations involving these quantities are Gm1 m2 , grh 3. The SI and CGS units of energy are joule and erg F S , 2 2 respectively. How many ergs are equal to one joule ? r A 2  1 t v2  v1 Solution : Dimensionally, Energy  mass  velocity 2 Qk and F    A d x2  x1  length 2 2  mass    ML T. 2  where the symbols have their usual meanings.  time  2 Thus, 1 joule  1 kg 1 m 1 s 2 m1 m2 Solution : (a) F  G 2 r and 1 erg  1 g 1 cm 1 s 2 2 2 Fr 2 or, G 1 joule  1 kg   1 m  2 1 s m1m2      2 2 2 1 erg  1 g   1 cm  1 s [F]L MLT. L 1 3 2 2 or, [G]  2  M L T.  1000 g   100 cm    1000  10000  10. 7 M M 2    1 g   1 cm  grh 1 joule  10 erg. 7 (b) S So, 2 M L 2 [S]  [] [g]L  3  2  L  MT 2 2 or,. 4. Young’s modulus of steel is 19  10 10 2 L T N/m. Express it in dyne  cm. Here dyne is the CGS unit of force. 2 A 2  1 t (c) Qk d Solution : The unit of Young’s modulus is N/m. 2 Qd Force or, k  A2  1 t This suggests that it has dimensions of 2  distance 2 Here, Q is the heat energy having dimension [F] MLT 1 2 2 Thus, [Y]  2   ML T. ML T , 2  1 is temperature, A is area, d is 2 2 L L thickness and t is time. Thus, 2 N/m is in SI units. 2 2 ML T L 3 1 1 [k]  2  MLT K. So, 1 N/m  1 kg1 m 2 1 s 2 L KT 1 2 1 dyne/cm  1 g1 cm 1 s 2 v2  v1 and (d) FA 1 2 x2  x1 1 Nm 2  1 kg   1 m   1 s  2 so, 2       2 L/T L 1 dyne  cm  1 g   1 cm  1 s   []L  [] 2 or, MLT L T 1 or, 1 []  ML T. 1  1000   1  10 100 1 N/m  10 dyne/cm 2 2 or, 2. Find the dimensional formulae of 19  10 N/m  19  10 10 2 11 2 (a) the charge Q, or, dyne/c m.