Maharashtra State Board Class 12 Physics 2020-2021 PDF

The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 30.01.2020 and it has been decided to implement it from academic year 2020-21 PHYSICS...

The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 30.01.2020 and it has been decided to implement it from academic year 2020-21 PHYSICS Standard XII Download DIKSHA App on your smartphone. If you scan the Q.R.Code on this page of your textbook, you will be able to access full text and the audio-visual study material relevant to each lesson, provided as teaching and learning aids. 2020 Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. First Edition : © Maharashtra State Bureau of Textbook Production and 2020 Curriculum Research, Pune - 411 004. Reprint : 2022 The Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and Curriculum Research, ‘Balbharati’, Senapati Bapat Marg, Pune 411004. Committee: Illustration Dr. Chandrashekhar V. Murumkar, Chairman Shri. Shubham Chavan Dr. Dilip Sadashiv Joag, Convener Cover Shri. Vinayak Shripad Katdare, Co- Convener Shri. Vivekanand S. Patil Dr. Pushpa Khare, Member Typesetting Dr. Rajendra Shankar Mahamuni, Member DTP Section, Textbook Bureau, Pune Dr. Anjali Kshirsagar, Member Co-ordination : Dr. Rishi Baboo Sharma, Member Shri. Rajiv Arun Patole Shri. Ramesh Devidas Deshpande, Member Special Officer - Science Section Shri. Rajiv Arun Patole, Member Secretary Physics Paper Study group: 70 GSM Creamwove Dr. Umesh Anant Palnitkar Print Order Dr. Vandana Laxmanrao Jadhav Patil Dr. Neelam Sunil Shinde Printer Dr. Radhika Gautamkumar Deshmukh Shri. Dinesh Madhusudan Joshi Smt. Smitha Menon Production Shri. Govind Diliprao Kulkarni Shri. Sachchitanand Aphale Chief Production Officer Smt. Pratibha Pradeep Pandit Shri. Prashant Harne Shri. Prashant Panditrao Kolase Production Officer Dr. Archana Balasaheb Bodade Dr. Jayashri Kalyanrao Chavan Publisher Smt. Mugdha Milind Taksale Shri. Vivek Uttam Gosavi Dr. Prabhakar Nagnath Kshirsagar Controller Shri. Ramchandra Sambhaji Shinde Maharashtra State Textbook Bureau, Prabhadevi, Shri. Brajesh Pandey Mumbai - 400 025  The Constitution of India Preamble WE, THE PEOPLE OF INDIA, having solemnly resolved to constitute India into a SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC and to secure to all its citizens: JUSTICE, social, economic and political; LIBERTY of thought, expression, belief, faith and worship; EQUALITY of status and of opportunity; and to promote among them all FRATERNITY assuring the dignity of the individual and the unity and integrity of the Nation; IN OUR CONSTITUENT ASSEMBLY this twenty-sixth day of November, 1949, do HEREBY ADOPT, ENACT AND GIVE TO OURSELVES THIS CONSTITUTION. NATIONAL ANTHEM  Preface Dear Students, With great pleasure we place this detailed text book on basic physics in the hands of the young generation. This is not only a textbook of physics for XIIth standard, but contains material that will be useful for the reader for self study. This textbook aims to give the student a broad perspective to look into the physics aspect in various phenomena they experience. The National Curriculum Framework (NCF) was formulated in the year 2005, followed by the State Curriculum Framework (SCF) in 2010. Based on the given two frameworks, reconstruction of the curriculum and preparation of a revised syllabus has been undertaken which will be introduced from the academic year 2020-21. The textbook incorporating the revised syllabus has been prepared and designed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, (Balbharati), Pune. The objective of bringing out this book is to prepare students to observe and analyse various physical phenomena is the world around them and prepare a solid foundation for those who aspire for admission to professional courses through competitive examinations. Most of the chapters in this book assume background knowledge of the subject covered by the text book for XIth Standard, and care has been taken of mentioning this in the appropriate sections of the book. The book is not in the form of handy notes but embodies a good historical background and in depth discussion as well. A number of solved examples in every chapter and exercises at the end of each one of them are included with a view that students will acquire proficiency and also will get enlightened after solving the exercises. Physics is a highly conceptual subject. Problem solving will enable students understand the underlying concepts. For students who want more, boxes entitled ‘Do you know?’ have been included at a number of places. If you read the book carefully and solve the exercises in each chapter, you will be well prepared to face the challenges of this competitive world and pave the way for a successful career ahead. The efforts taken to prepare the textbook will prove to be worthwhile if you read the textbook and understand the subject. We hope it will be a wonderful learning experience for you and an illuminating text material for teachers too. (Vivek Gosavi) Pune Director Date : 21 February, 2020 Maharashtra State Bureau of Texbook Bhartiya Saur : 2 Phalguna, 1941 Production and Curriculum Research, Pune  - For Teachers - Dear Teachers, introduced in a graded manner to facilitate We are happy to introduce the revised knowledge building. textbook of Physics for XIIth standard. This 3 'Error in measurements' is an important book is a sincere attempt to follow the topic in physics. Please ask the students to maxims of teaching as well as develop a use this in estimating errors in their ‘constructivist’ approach to enhance the measurements. This must become an quality of learning. The demand for more integral part of laboratory practices. activity based, experiential and innovative 3 Major concepts of physics have a scientific learning opportunities is the need of the base. Encourage group work, learning hour. The present curriculum has been through each other’s help, etc. Facilitate restructured so as to bridge the credibility peer learning as much as possible by gap that exists between what is taught and reorganizing the class structure frequently. what students learn from direct experience 3 Do not use the boxes titled ‘Do you know?’ in the outside world. Guidelines provided or ‘Use your brain power’ for evaluation. below will help to enrich the teaching- However, teachers must ensure that students learning process and achieve the desired read this extra information and think about learning outcomes. the questions posed. 3 To begin with, get familiar with the 3 For evaluation, equal weightage should be textbook yourself, and encourage the assigned to all the topics. Use different students to read each chapter carefully. combinations of questions. Stereotype 3 The present book has been prepared for questions should be avoided. constructivist and activity-based teaching, 3 Use Q.R. Code given in the textbook. Keep including problem solving exercises. checking the Q.R. Code for updated 3 Use teaching aids as required for proper information. Certain important links, websites understanding of the subject. have been given for references. Also a list 3 Do not finish the chapter in short. However, of reference books is given. Teachers as well in the view of insufficient lectures, standard as the students can use these references for derivations may be left to the students for extra reading and in-depth understanding of self study. Problem sloving must be given the subject. due importance. Best wishes for a wonderful teaching 3 Follow the order of the chapters strictly as experience! listed in the contents because the units are References: 1. Fundamentals of Physics - Halliday, Resnick, Walker; John Wiley (Sixth ed.). 2. Sears and Zeemansky's University Physics - Young and Freedman, Pearson Education (12th ed.) 3. Physics for Scientists and Engineers - Lawrence S. Lerner; Jones and Bartlett Publishers, UK. Front Cover : Picture shows part of Indus 2, Synchrotron radiation source (electron accelerator) at RRCAT, Department of Atomic Energy, Govt. of India, Indore. Indus offers several research opportunities. The photoelectron spectroscopy beamline is also seen. Picture credit : Director, RRCAT, Indore. The permission to reproduce these pictures by Director, RRCAT, DAE, Govt. of India is gratefully acknowledged. Back Cover : Transmission Electron Microscope is based on De Broglie's hypothesis. TEM picture shows a carbon nanotube filled with water showing the miniscus formed due to surface tension. Other picture shows crystallites of LaB6 and the electron diffraction pattern (spot pattern) of the crystallite. Picture credit : Dr. Dilip Joag, Savitribai Phule Pune University. Pune DISCLAIMER Note : All attempts have been made to contact copy right/s (©) but we have not heard from them. We will be pleased to acknowledge the copy right holder (s) in our next edition if we learn from them.  Competency Statements : Standard XII Area/ Competency Statements Unit/ After studying the content in Textbook students would be able to.... Lesson Ratational Motion and Mechanical Distinguish between centrifugal and centripetal forces. Visualize the concepts of moment of inertia of an object. Relate moment of inertia of a body with its angular momentum. Properties of fluids Differentiate between translational and rotational motions of rolling objects. Relate the pressure of a fluid to the depth below its surface. Unit I Explain the measurement of atmospheric pressure by using a barometer. Use Pascal's law to explain the working of a hydraulic lift and hydraulic brakes. Relate the surface energy of a fluid with its surface tension. Distinguish between fluids which show capillary rise and fall. Identify processes in daily life where surface tension plays a major role. Explain the role of viscosity in everyday life. Differentiate between streamline flow and turbulent flow. Relate various gas laws to form ideal gas equation. Distinguish between ideal gas and a real gas. Visualise mean free path as a function of various parameters.. Kinetic theory and Thermodynamics Obtain degrees of freedom of a diatomic molecule. Apply law of equipartition of energy to monatomic and diatomic molecules. Unit II Compare emission of thermal radiation from a body with black body radiation. Apply Stefan’s law of radiation to hot bodies. Identify thermodynamic process in every day life. Relate mechanical work and thermodynamic work. Differentiate between different types of thermodynamic processes. Explain the working of heat engine, refrigerator and air conditioner. Identify periodic motion and simple harmonic motion. Oscillations and waves Obtain the laws of motion for simple pendulum. Visualize damped oscillations. Apply wave theory to understand the phenomena of reflection, refraction, interference and Unit III diffraction. Visualize polarized and unpolarized light. Apply concepts of diffraction to calculate the resolving power. Distinguish between the stationary waves in pipes with open and closed ends. Verify laws of vibrating string using a sonometer. Explain the physics involved in musical instruments. Electrostatics and electric current Use Gaus's law to obtain the electric field for a charge distribution. Relate potential energy to work done to establish a charge distribution. Determine the electrostatic potential for a given charge distribution. Distingusih between conductors and insulators. Unit IV Visualize polarization of dielectrics. Categorize dielectrics based on molecular properties. Know the effect of dielectric material used between the plates of a capacitor on its capacitance. Apply Kirchhoff’s laws to determine the current in different branches of a circuit. Find the value of an unknown resistance by using a meter bridge. Find the emf and internal resistance of a cell using potentiometer. Convert galvanometer into voltmeter and ammeter by using a suitable resistor.  Realize that Lorentz force law is the basis for defining unit of magnetic field. Visualize cyclotron motion of a charged particle in a magnetic field. Analyze and calculate magnetic force on a straight and arbitrarily shaped current carrying wires and a closed wire circuit. Apply the Biot-Savart law to calculate the magnetic field produced by various distributions of currents. Use Ampere’s law to get magnetic fields produced by a current distribution. Compare gravitational, magnetic and electrostatic potentials. Magnetism Distinguish between paramagnitic, diamagnetic and ferromagnetic materials. Unit V Relate the concept of flux to experiments of Faraday and Henry. Relate Lenz’s law to the conservation of energy. Visualize the concept of eddy currents. Determine the mutual inductance of a given pair of coils. Apply laws of induction to explain the working of a generator. Establish a relation between the power dissipated by an AC current in a resistor and the value of the rms current. Visualize the concept of phases to represent AC current. Explain the passage of AC current through circuits having resistors, capacitors and inductors. Explain the concept of resonance in LCR circuits. Establish validity of particle nature of light from experimental results. Determine the necessary wavelength range of radiation to obtain photocurrent from given metals. Visualize the dual nature of matter and dual nature of light. Apply the wave nature of electrons to illustrate how better resolution can be obtained with an electron microscope. Modern Physics Check the correctness of different atomic models by comparing results of various experiments. Identify the constituents of atomic nuclei. Unit VI Differentiate between electromagnetic and atomic forces. Obtain the age of a radioactive sample from its activity. Judge the importance of nuclear power. Explain use of p-n junction diode as a rectifier. Find applications of special purpose diodes for every day use. Explain working of solar cell, LED and photodiode. Relate the p-n junction diode and special purpose diodes. Realize transistor as an important building block of electronic circuits, analyze situations in which transistor can be used. Sr. No Title Page No 1 Rotational Dynamics 1-25 2 Mechanical Properties of Fluids 26-55 3 Kinetic Theory of Gases and Radiation 56-74 CONTENTS 4 Thermodynamics 75-108 5 Oscillations 109-130 6 Superposition of Waves 131-157 7 Wave Optics 158-185 8 Electrostatics 186-213 9 Current Electricity 214-229 10 Magnetic Fields due to Electric Current 230-250 11 Magnetic Materials 251-264 12 Electromagnetic induction 265-287 13 AC Circuits 288-305 14 Dual Nature of Radiation and Matter 306-323 15 Structure of Atoms and Nuclei 324-343 16 Semiconductor Devices 344-364  1. Rotational Dynamics Can you recall? of the right hand along the sense of rotation, with the thumb outstretched. The outstretched 1. What is circular motion? G thumb then gives the direction of Z. 2. What is the concept of centre of mass? 3. What are kinematical equations of motion? 4. Do you know real and pseudo forces, their origin and applications? 1.1 Introduction: Fig. 1.1: Directions of angular velocity. Circular motion is an essential part of our If T is period of circular motion or periodic daily life. Every day we come across several 2 time and n is the frequency,   2 n  revolving or rotating (rigid) objects. During T Uniform circular motion: During circular revolution, the object (every particle in the motion if the speed of the particle remains object) undergoes circular motion about some constant, it is called Uniform Circular Motion point outside the object or about some other object, while during rotation the motion is about (UCM). In this case, only the direction of its an axis of rotation passing through the object. velocity changes at every instant in such a way 1.2 Characteristics of Circular Motion: that the velocity is always tangential to the 1) It is an accelerated motion: As the path. The acceleration responsible for this is G 2G direction of velocity changes at every the centripetal or radial acceleration a r   r instant, it is an accelerated motion. For UCM, its 2magnitude is constant and it v 2) It is a periodic motion: During the motion, is a   2 r   v. It is always directed the particle repeats its path along the same r towards the centre of the circular motion trajectory. Thus, the motion is periodic in G (along r ), hence called centripetal. space. 1.2.1 Kinematics of Circular Motion: As seen in XIth Std, in order to describe a circular motion, we G use the quantities angular displacement T , angular velocity G G G d G d  and angular acceleration   dt dt which are analogous G to displacement G G G ds G dv s , velocity v dt and acceleration a Fig. 1.2: Directions of linear velocity and dt used in translational motion. acceleration. Also, the tangential velocity is given by Illustration: Circular motion of any particle G G G G v    r where Z is the angular G velocity. of a fan rotating uniformly. Here, the position vector r is the radius Non-uniform circular motion: When a fan is vector from the centre G of the circular motion. switched ON or OFF, the speeds of particles The magnitude of v is v = Z r. G of the fan go on increasing or decreasing Direction of Z is always along the axis of for some time, however their directions are rotation and is given by the right-hand thumb G always tangential to their circular trajectories. rule. To know the direction of Z , curl the fingers 1 During this time, it is a non-uniform circular G always change only the direction of Z and motion. As the velocity is still tangential, the G never its magnitude thereby continuously centripetal or radial acceleration a r is still changing the plane of rotation. (This is there. However, for non-uniform circular G G similar to an acceleration a perpendicular motion, the magnitude of a r is not constant. G to velocity v changing only its direction). The acceleration responsible for changing G the magnitude of velocity is directed along If the angular acceleration D is constant or opposite to the velocity, hence always and along the axis of rotation, all and tangential and is called as tangential will be directed along the axis. This makes it G acceleration a T. possible to use scalar notation and write the G kinematical equations of motion analogous to As magnitude of tangential velocity v is changing during a non-uniform circular those for translational motion as given in the G motion, the corresponding angular velocity Z Table 1.1 at the end of the Chapter. is also changing at every instant.G This is due to Example 1.1 : A fan is rotating at 90 rpm. G d the angular acceleration   It is then switched OFF. It stops after 21 dt Though the motion is non-uniform, the rotations. Calculate the time taken by it to particles are still in the same plane. Hence, stop assuming that the frictional torque is G the direction of D is still along the axis of constant. rotation. For increasing speed, it is along the Solution: G direction of Z while during decreasing speed, 90 rpm 1.5 rps 0 G s it is opposite to that of Z. The angle through which the blades of the fan move while stopping is T  = 2SN = 2S (21) = 42 S rad, Z = 0 (fan stops). Using equations analogous to kinematical equations of motion   0  2  02 Fig. 1.3: Direction of angular acceleration.   t 2 2 Do you know? 0  3 0  3  t  28 s G t 2 42 If the angular acceleration D is along any direction other than axial, it will have Remark: One can also use the unit a component perpendicular to the axis. ‘revolution’ for angle and get rid of S G Thus, it will change the direction of Z also, throughout (for such data). In this case, which will change the plane of rotation as 0  1.5 rps and   21 rev. G Z is always perpendicular to the plane of 1.2.2 Dynamics of Circular Motion rotation. G (Centripetal Force and Centrifugal Force): If D is i) Centripetal force (CPF): As seen above, constant in the acceleration responsible for circular magnitude, motion is the centripetal or radial acceleration but always G G a r   2 r. The force providing this perpendicular G acceleration is the centripetal or radial force, to Z , it will 2G CPF  m r 2  It must be understood that centrifugal Remember this force is a non-real force, but NOT an imaginary force. Remember, before the merry- (i) The word centripetal is NOT the name go-round reaches its uniform speed, you were or type of that force (like gravitational really experiencing an outward pull (because, force, nuclear force, etc). It is the centrifugal force is greater than the resultant adjective or property of that force force towards the centre). A force measuring saying that the direction of this force instrument can record it as well. is along the radius and towards centre On reaching the uniform speed, in the (centre seeking). frame of reference of merry-go-round, this (ii) While performing circular or rotational centrifugal force exactly balances the resultant motion, the resultant of all the real of all the real forces. The resultant force in forces acting upon the body is (or, must that frame of reference is thus zero. Thus, only be) towards the centre, hence we call this resultant force to be centripetal in such a frame of reference we can say that force. Under the action of this resultant the centrifugal force balances the centripetal force, the direction of the velocity is force. It must be remembered that in this case, always maintained tangential to the centrifugal force means the ‘net pseudo force’ circular track. and centripetal force means the ‘resultant of The vice versa need not be true, all the real forces’. i.e., the resultant force directed towards There are two ways of writing force the centre may not always result into equation for a circular motion: a circular motion. (In the Chapter 7 G you will know that during an S.H.M. Resultant force  m 2 r or  also the force is always directed to the m 2r real orces 0 centre of the motion). For a motion to be circular, correspondingly matching tangential velocity is also essential. Activity (iii) Obviously, this discussion is in an Attach a suitable mass to spring balance so inertial frame of reference in which we are observing that the body is that it stretches by about half is capacity. performing a circular motion. Now whirl the spring balance so that the (iv) In magnitude, centripetal force mass performs a horizontal motion. You will notice that the balance now reads more mass mv 2  mr 2   mv for the same mass. Can you explain this? r ii) Centrifugal force (CFF): 1.3 Applications of Uniform Circular Motion: Visualize yourself on a merry-go-round 1.3.1 Vehicle Along a Horizontal Circular rotating uniformly. If you close your eyes, you Track: will not know that you are performing a circular Figure 1.4 shows vertical section of a car motion but you will feel that you are at rest. In on a horizontal circular track of radius r. Plane order to explain that you are at rest, you need of figure is a vertical plane, perpendicular to to consider a force equal in magnitude to the the track but includes only centre C of the resultant real force, but directed opposite, i.e., 2G track. Forces acting on the car (considered away from the centre. This force, m r is to be a particle) are (i) weight mg, vertically the centrifugal (away from the centre) force. It is a pseudo force arising due to the centripetal downwards, (ii) normal reaction N, vertically acceleration of the frame of reference. upwards that balances the weight mg and (iii) 3 force of static friction fs between road and the but above it. Thus, the frictional force tyres. This is static friction because it prevents and the centrifugal force result into a the vehicle from outward slipping or skidding. torque which may topple the vehicle This is the resultant force which is centripetal. (even a two wheeler). (ii) For a two wheeler, it is a must for the rider to incline with respect to the vertical to prevent toppling. Use your brain power (I) Obtain the condition for not toppling for a four-wheeler. On what factors Fig. 1.4: Vehicle on a horizontal road. does it depend, and in what way? Think While working in the frame of reference about the normal reactions – where are attached to the vehicle, it balances the those and how much are those! What centrifugal force. mv 2 is the recommendation on loading the mg  N andf s  mr 2  r vehicle for not toppling easily? If a f s r 2 v 2 vehicle topples while turning, which   N g rg wheels leave the contact? Why? How For a given track, radius r is constant. For does it affect the tyres? What is the given vehicle, mg = N is constant. Thus, as the recommendation for this? speed v increases, the force of static friction fs (II) Determine the angle to be made with also increases. However, fs has an upper limit the vertical by a two wheeler rider while f s max  s. N , where P s is the coefficient of turning on a horizontal track. static friction between road and tyres of the Hint: For both (I) and (II) above, find the vehicle. This imposes an upper limit to the torque that balances the torque due speed v. to centrifugal force and torque due to At the maximum possible speed vs, we can static friction force. write (III)We have mentioned about static friction fs 2 between road and the tyres. Why is it v max max  s  v max  s rg static? What about the kinetic friction N rg between road and the tyres? (IV) What do you do if your vehicle is Do you know? trapped on a slippery or a sandy road? (i) In the discussion till now, we had What is the physics involved? assumed the vehicle to be a point. 1.3.2 Well (or Wall) of Death: (0L&>A): In reality, if it is a four wheeler, the This is a vertical cylindrical wall of radius resultant normal reaction is due to all r inside which a vehicle is driven in horizontal the four tyres. Normal reactions at all circles. This can be seen while performing the four tyres are never equal while stunts. undergoing circular motion. Also, the As shown in the Fig. 1.5, the forces acting centrifugal force acts through the centre on the vehicle (assumed to be a point) are (i) of mass, which is not at the ground level, Normal reaction N acting horizontally and 4  due to the weight. What about a four- wheeler? (iv) In this case, the angle made by the road surface with the horizontal is 90q, i.e., if the road is banked at 90q, it imposes a lower limit on the turning speed. In the Fig. 1.5: Well of death. previous sub-section we saw that for an towards the centre, (ii) Weight mg acting unbanked (banking angle 0) road there vertically downwards, and (iii) Force of static is an upper limit for the turning speed. friction fs acting vertically upwards between It means that for any other banking vertical wall and the tyres. It is static friction angle (0 < T < 90q), the turning speed because it has to prevent the downward will have the upper as well as the lower slipping. Its magnitude is equal to mg, as this limit. is the only upward force. Example 1.2: A motor cyclist (to be treated Normal reaction N is thus the resultant as a point mass) is to undertake horizontal centripetal force (or the only force that can circles inside the cylindrical wall of a well balance the centrifugal force). Thus, in of inner radius 4 m. Coefficient of static magnitude, mv 2 friction between the tyres and the wall is N  mr  2 andmg  f s 0.4. Calculate the minimum speed and r frequency necessary to perform this stunt. Force of static friction f s is always less than (Use g = 10 m/s2) or equal to P s N. Solution:  mv 2  fs  sN mg  s  rg 4  10  r  v min    10 m s 1 s 0.4 and v2 rg g s v2  v min 10 r nmin    0.4 rev s 1 s 2 r 2    4 rg v min  1.3.3 Vehicle on a Banked Road: s As seen earlier, while taking a turn on Remember this a horizontal road, the force of static friction mv 2 between the tyres of the vehicle and the road (i) N should always be equal to provides the necessary centripetal force (or 2 mv min mg r N min   balances the centrifugal force). However, the r s frictional force is having an upper limit. Also, (ii) In this case, fs = PsN is valid only for the its value is usually not constant as the road minimum speed as fs should always be surface is not uniform. Thus, in real life, we equal to mg. should not depend upon it, as far as possible. (iii) During the derivation, the vehicle is For this purpose, the surfaces of curved roads assumed to be a particle. In reality, it is not so. During revolutions in such are tilted with the horizontal with some angle a well, a two-wheeler rider is never T. This is called banking of a road or the road horizontal, else, the torque due to her/ is said to be banked. his weight will topple her/him. Think Figure 1.6 Shows the vertical section of of the torque that balances the torque a vehicle on a curved road of radius r banked 5  Use your brain power As a civil engineer, you are given contract to construct a curved road in a ghat. In order to obtain the banking angle T , you need to decide the speed limit. How will you decide the values of speed v and radius r ? Fig 1.6: Vehicle on a banked road. are (i) weight mg acting vertically downwards at an angle T with the horizontal. Considering and (ii) normal reaction N acting perpendicular the vehicle to be a point and ignoring friction to the road. As seen above, only at this speed, (not eliminating) and other non-conservative the resultant of these two forces (which is forces like air resistance, there are two forces Nsin T ) is the necessary centripetal force (or acting on the vehicle, (i) weight mg, vertically balances the centrifugal force). In practice, downwards and (ii) normal reaction N, vehicles never travel exactly with this speed. perpendicular to the surface of the road. As For speeds other than this, the component of the motion of the vehicle is along a horizontal force of static friction between road and the circle, the resultant force must be horizontal tyres helps us, up to a certain limit. and directed towards the centre of the track. It means, the vertical force mg must be balanced. Thus, we have to resolve the normal reaction N along the vertical and along the horizontal. Its vertical component Ncos T balances weight mg. Horizontal component Nsin T being the resultant force, must be the necessary centripetal force (or balance the centrifugal force). Thus, in magnitude, Fig 1.7: Banked road : lower speed limit. N cos   mg  and mv 2 v2 N sin   mr 2  tan   --- (1.1) r rg (a) Most safe speed: For a particular road, r and T are fixed. Thus, this expression gives us the expression for the most safe speed (not a minimum or a maximum speed) on this road as v s  rg tan  (b) Banking angle: While designing a road, this expression helps us in Fig 1.8: Banked road : upper speed limit. knowing the angle of banking as mv12 For speeds v1  rg tan  ,  N sin   v2  r   tan 1   --- (1.2) (or N sin T is greater than the centrifugal force rg   mv12 ). In this case, the direction of force of (c) Speed limits: Figure 1.7 and 1.8 show r vertical section of a vehicle on a rough static friction fs between road and the tyres curved road of radius r, banked at an angle is directed along the inclination of the road, T. If the vehicle is running exactly at the speed upwards (Fig. 1.7). Its horizontal component v s  rg tan  , the forces acting on the vehicle is parallel and opposite to Nsin T. These two 6 forces take care of the necessary centripetal Example 1.3: A racing track of radius of force (or balance the centrifugal force). curvature 9.9 m is banked at tan 1 0.5 mg  f s sin  N cos  and. Coefficient of static friction between mv12 the track and the tyres of a vehicle is 0.2.  N sin   f s cos  r Determine the speed limits with 10 % For minimum possible speed, fs is margin. (Take g = 10 m/s2) maximum and equal to PsN. Using this in the Solution: equations above and solving for minimum  tan   s  possible speed, we get v min  rg    1 s tan    tan   s  v1  v min  rg   --- (1.3)  0.5  0.2   1 s tan   min  9.9 10    1 0.2  0.5   For s  tan  ,vmin = 0. This is true for most of the rough roads, banked at smaller angles.  27  5.196 m / s mv 22 Allowed vmin should be 10% higher than (d) For speeds v 2  rg tan  ,  N sin  r this. (or N sinT is less than the centrifugal force 110 v min allowed  5.196  mv 22 100 ). In this case, the direction of force r m of static friction fs between road and the  5.716 s tyres is directed along the inclination of the road, downwards (Fig. 1.8). Its horizontal  tan   v max  rg  s  component is parallel to Nsin T. These two  1  s tan   forces take care of the necessary centripetal  0.5 0.2  force (or balance the centrifugal force).  9.9 10    1  0.2  0.5 mg  N cos   f s sin  and   mv 22  77  8.775 m / s  N sin  f s cos  r Allowed vmax should be 10% lower than For maximum possible speed, f s is this. maximum and equal to P s N. Using this in the 90 ? v max allowed  8.775   7.896 m / s equations above, and solving for maximum 100 possible speed, we get Use your brain power  tan   v2  v max  rg  s  --- (1.4)  1  s tan   max If friction is zero, can a vehicle move on If Ps = cot T , vmax = f. But ( Ps )max = 1. the road? Why are we not considering the friction in deriving the expression Thus, for T t 45q, vmax = f. However, for for the banking angle? heavily banked road, minimum limit may What about the kinetic friction between be important. Try to relate the concepts used the road and the tyres? while explaining the well of death. 1.3.4 Conical Pendulum: (e) For Ps = 0, both the equations 1.3 and 1.4 A tiny mass (assumed to be a point object give us v  rg tan  which is the safest speed and called a bob) connected to a long, flexible, on a banked road as we don’t take the help of massless, inextensible string, and suspended friction. to a rigid support is called a pendulum. If the 7 string is made to oscillate in a single vertical g sin  plane, we call it a simple pendulum (to be 2  r cos  studied in the Chapter 5). Radius r of the circular motion is r  L sin . We can also revolve the string in such a If T is the period of revolution of the bob, way that the string moves along the surface of 2 g a right circular cone of vertical axis and the   point object performs a (practically) uniform T L cos  horizontal circular motion. In such a case the L cos  Period T  2 --- (1.7) system is called a conical pendulum. g Frequency of revolution, 1 1 g --- (1.8) n  T 2 L cos  In the frame of reference attached to the bob, the centrifugal force should balance the resultant of all the real forces (which we call CPF) for the bob to be at rest. Fig. 1.9 (a): In an inertial frame ? T0 sin T = mr Z 2 --- (in magnitude). This is the same as the Eq. (1.5) Remember this (i) For a given set up, L and g are constant. Thus, both period and frequency Fig. 1.9 (b): In a non- inertial frame depend upon T. (ii) During revolutions, the string can Figure 1.9 shows the vertical section of NEVER become horizontal. This can a conical pendulum having bob (point mass) be explained in two different ways. of mass m and string of length L. In a given (a) If the string becomes horizontal, position B, the forces acting on the bob are (i) the force due to tension will also be its weight mg directed vertically downwards horizontal. Its vertical component will and (ii) the force T0 due to the tension in the then be zero. In this case, nothing will string, directed along the string, towards be there to balance mg. the support A. As the motion of the bob is a (b) For horizontal string, T= 90q. This will horizontal circular motion, the resultant force indicate the frequency to be infinite and the period to be zero, which are must be horizontal and directed towards the impossible. Also, in this case, the tension centre C of the circular motion. For this, all mg the vertical forces must cancel. Hence, we T0  in the string and the kinetic cos  shall resolve the force T0 due to the tension. 1 1 2 2 energy  mv  mr  of the bob 2 If T is the angle made by the string with the 2 2 vertical, at any position (semi-vertical angle will be infinite. of the cone), the vertical component T0 cos T balances the weight mg. The horizontal component T0 sinT then becomes the resultant Activity force which is centripetal. A stone is tied to a string and whirled T0 sin   centripetalforce  mr 2 --- (1.5) such that the stone performs horizontal Also, T0 cos T = mg --- (1.6) circular motion. It can be seen that the string Dividing eq (1.5) by Eq. (1.6), we get, is NEVER horizontal. 8  Example 1.4: A merry-go-round usually mv 2 Solution: N sin   mg and N cos   consists of a central vertical pillar. At the r top of it there are horizontal rods which can rg v 2 tan  tan   2  r  rotate about vertical axis. At the end of this v g horizontal rod there is a vertical rod fitted v 2max tan  like an elbow joint. At the lower end of rmax   0.3m g each vertical rod, there is a horse on which v  r  2 rn the rider can sit. As the merry-go-round is set into rotation, these vertical rods move If we go for the lower away from the axle by making some angle limit of the speed (while with the vertical. rotating), v  0 r  0 , but the frequency n increases. Hence a specific upper limit is not possible in the case of frequency. Thus, the practical limit on the frequency of rotation is its lower limit. It will be possible The figure above shows vertical section for r rmax of a merry-go-round in which the ‘initially v 1 nmin  max   1rev / s vertical’ rods are inclined with the vertical 2 rmax 0.3 at T = 370, during rotation. Calculate the frequency of revolution of the merry-go- round. (Use g = S2 m/s2 and sin 37q = 0.6) Activity Solution: Length of the horizontal rod, Using a funnel and a marble or a ball bearing H = 2.1 m try to work out the situation in the above Length of the ‘initially vertical’ rod, question. Try to realize that as the marble V = 1.5 m, T = 37q goes towards the brim, its linear speed ? Radius of the horizontal circular motion increases but its angular speed decreases. of the rider = H + V sin 37q = 3.0 m When nearing the base, it is the other way. If T is the tension along the inclined rod, T cos T = mg and T sin T = mr Z 2 = 4S2 mrn2 1.4 Vertical Circular Motion: 4 2 rn 2 Two types of vertical circular motions are tan    g commonly observed in practice: tan  1 (a) A controlled vertical circular motion such n  revs 1  as g   2 as a giant wheel or similar games. In this 4r 4 Example 1.5: Semi-vertical angle of the case the speed is either kept constant or conical section of a funnel is 370. There is a NOT totally controlled by gravity. small ball kept inside the funnel. On rotating (b) Vertical circular motion controlled only the funnel, the maximum speed that the ball by gravity. In this case, we initially can have in order to remain in the funnel is 2 supply the necessary energy (mostly) at m/s. Calculate inner radius of the brim of the the lowest point. Then onwards, the entire funnel. Is there any limit upon the frequency kinetics is governed by the gravitational of rotation? How much is it? Is it lower or force. During the motion, there is upper limit? Give a logical reasoning. (Use interconversion of kinetic energy and g = 10 m/s2 and sin 370 = 0.6) gravitational potential energy. 9 1.4.1 Point Mass Undergoing Vertical realized with minimum possible energy), Circular Motion Under Gravity: TA 0 v A min  rg --- (1.10) Case I: Mass tied to a string: Lowermost position (B): Force due to the The figure 1.10 shows a bob (treated as tension, TB is vertically upwards, i.e., towards a point mass) tied to a (practically) massless, the centre, and opposite to mg. In this case also inextensible and flexible string. It is whirled their resultant is the centripetal force. If vB is along a vertical circle so that the bob performs the speed at the lowermost point, we get, a vertical circular motion and the string rotates mv 2B TB  mg  --- (1.11) in a vertical plane. At any position of the bob, r there are only two forces acting on the bob: While coming down from the uppermost to A the lowermost point, the vertical displacement is 2r and the motion is governed only by gravity. Hence the corresponding decrease in the gravitational potential energy is converted into the kinetic energy. 1 1 mg 2r  mv 2B  mv 2A  2 2 v B  v A  4 rg 2 2 --- (1.12) B Using this in the eq (1.11), and using v A min from Eq. (1.10) we get, v B min  5rg --- (1.13) Subtracting eq (1.9) from eq (1.11) , we can Fig 1.10: Vertical circular motion. write, (a) its weight mg, vertically downwards, which m 2 is constant and (b) the force due to the tension TB  TA  2 mg  v B  v A2 --- (1.14) r along the string, directed along the string and Using eq (1.12) and rearranging, we get, towards the centre. Its magnitude changes TB  TA  6 mg --- (1.15) periodically with time and location. Positions when the string is horizontal (C As the motion is non uniform, the resultant and D): Force due to the tension is the only of these two forces is not directed towards force towards the centre as weight mg is the center except at the uppermost and the perpendicular to the tension. Thus, force due lowermost positions of the bob. At all the other to the tension is the centripetal force used to positions, part of the resultant is tangential and change the direction of the velocity and weight is used to change the speed. mg is used only to change the speed. Uppermost position (A): Both, weight mg and Using similar mathematics, it can be shown force due to tension TA are downwards, i.e., that TC  TA  TD  TA  3mg and towards the centre. In this case, their resultant is used only as the centripetal force. Thus, if v C min   v D min  3rg vA is the speed at the uppermost point, we get, Arbitrary positions: Force due to the tension mv 2A and weight are neither along the same line, mg TA  --- (1.9) r nor perpendicular. Tangential component of Radius r of the circular motion is the weight is used to change the speed. It decreases length of the string. For minimum possible the speed while going up and increases it while speed at this point (or if the motion is to be coming down. 10  1.4.2 Sphere of Death (0Cš1AK4): Remember this This is a popular show in a circus. During this, two-wheeler rider (or riders) undergo 1. Equation (1.15) is independent of v and r. rounds inside a hollow sphere. Starting with 2. TA can never be exactly equal to zero in small horizontal circles, they eventually the case of a string, else, the string will perform revolutions along vertical circles. The slack. ? TB > 6 mg. dynamics of this vertical circular motion is 3. None of the parameters (including the the same as that of the point mass tied to the linear and angular accelerations) are string, except that the force due to tension T is constant during such a motion. Obviously, kinematical equations given in the table1 replaced by the normal reaction force N. are not applicable. If you have seen this show, try to visualize 4. We can determine the position vector or that initially there are nearly horizontal circles. velocity at any instant using the energy The linear speed is more for larger circles but conservation. But as the function of the angular speed (frequency) is more for smaller radius vector is not integrable (definite circles (while starting or stopping). This is as integration is not possible), theoretically per the theory of conical pendulum. it is not possible to determine the period 1.4.3 Vehicle at the Top of a Convex Over- or frequency. However, experimentally Bridge: the period can be measured. 5. Equations (1.10) and (1.13) give only the respective minimum speeds at the uppermost and the lowermost points. Any higher speeds obeying the equation (1.14) are allowed. 6. In reality, we have to continuously supply some energy to overcome the air resistance. Fig. 1.11: Vehicle on a convex over-bridge. Case II: Mass tied to a rod: Consider a bob Figure shows a vehicle at the top of a (point mass) tied to a (practically massless and convex over bridge, during its motion (part rigid) rod and whirled along a vertical circle. of vertical circular motion). Forces acting on The basic difference between the rod and the the vehicle are (a) Weight mg and (b) Normal string is that the string needs some tension at reaction force N, both along the vertical line all the points, including the uppermost point. (topmost position). The resultant of these two Thus, a certain minimum speed, Eq. (1.10), is must provide the necessary centripetal force necessary at the uppermost point in the case (vertically downwards) if the vehicle is at the of a string. In the case of a rod, as the rod is uppermost position. Thus, if v is the speed at rigid, such a condition is not necessary. Thus the uppermost point, (practically) zero speed is possible at the mv 2 mg  N  uppermost point. r Using similar mathematics, it is left to the As the speed is increased, N goes on readers to show that decreasing. Normal reaction is an indication v lowermost min  4 rg  2 rg of contact. Thus, for just maintaining contact, vmin at the rod horizontal position = 2rg N = 0. This imposes an upper limit on the speed Tlowermost  Tuppermost  6 mg as v max rg 11  1 2 Do you know? ? 0.02 8  K.E.min  0.02 10 1.8  2 Roller coaster is a common event in the K.E.min  0.28 J amusement parks. During this ride, all 1 2 K.E.max  0.02 8  0.02 10 1.8 the parts of the vertical circular motion 2 described above can be experienced. The 1 2 major force that we experience during this is K.E.max  mvmax  1J 2 the normal reaction force. Those who have 2(K. E.) max experienced this, should try to recall the ? v max 10 m s-1 changes in the normal reaction experienced m by us during various parts of the track. at the lowermost position, for which T = 0. mv 2 T  mg cos   --- at any angle T , Use your brain power r where the speed is v. What is expected to happen if one travels Thus, if T = mg, we get, mv 2 fast over a speed breaker? Why? mg  mg cos    How does the normal force on a concave r suspension bridge change when a vehicle rg 1  cos   v 2 --- (A) is travelling on it with constant speed? Vertical displacement at the angular position T is r 1 cos . Thus, the energy Example 1.6: A tiny stone of mass 20 g is equation at this position can be written as tied to a practically massless, inextensible, 1 2 1 flexible string and whirled along vertical m 10  mv 2  mg  r 1  cos   2 2 circles. Speed of the stone is 8 m/s when By using Eq. A, we get the centripetal force is exactly equal to the 1 force due to the tension. 50  rg 1  cos   rg 1  cos  Calculate minimum and maximum kinetic 2 3 energies of the stone during the entire circle. 50  rg 1  cos  Let T = 0 be the angular position of the 2 string, when the stone is at the lowermost 23 cos     1480 25'  position. Determine the angular position of 27 the string when the force due to tension is numerically equal to weight of the stone. 1.5 Moment of Inertia as an Analogous Use g = 10 m/s2 and length of the string = Quantity for Mass: 1.8 m In XIth Std. we saw that angular Solution: When the string is horizontal, the displacement, angular velocity and force due to the tension is the centripetal force. Thus, vertical displacements of the angular acceleration respectively replace bob for minimum and maximum energy displacement, velocity and acceleration for positions are radius r each. various kinematical equations. Also, torque is If K.E.max and K.E.min are the respective an analogous quantity for force. Expressions of kinetic energies at the uppermost and the linear momentum, force (for a fixed mass) and lowermost points, kinetic energy include mass as a common term. 1 2 K.E.max  m 8  mgr  and In order to have their rotational analogues, we 2 need a replacement for mass. 1 2 m 8  K.E.min  mgr  If we open a door (with hinges), we give a 2 certain angular displacement to it. The efforts 12  If I  mi ri replaces mass m and angular needed for this depend not only upon the mass 2 of the door, but also upon the (perpendicular) speed Z replaces linear speed v, rotational distance from the axis of rotation, where we 1 2 apply the force. Thus, the quantity analogous K.E.  I  is analogous to translational 2 to mass includes not only the mass, but also 1 takes care of the distance wise distribution of K.E. mv 2. Thus, I is defined to be the 2 the mass around the axis of rotation. To know rotational inertia or moment of inertia (M.I.) the exact relation, let us derive an expression of the object about the given axis of rotation. for the rotational kinetic energy which is the It is clear that the moment of inertia of an sum of the translational kinetic energies of all object depends upon (i) individual masses and the individual particles. (ii) the distribution of these masses about the given axis of rotation. For a different axis, it will again depend upon the mass distribution around that axis and will be different if there is no symmetry. During this discussion, for simplicity, we assumed the object to be consisting of a finite number of particles. In practice, usually, it Fig. 1.12: A body of N particles. is not so. For a homogeneous rigid object of Figure 1.12 shows a rigid object rotating mathematically integrable mass distribution, with a constant angular speed Z  about an the moment of inertia is to be obtained by axis perpendicular to the plane of paper. integration as I  r 2 dm. If integrable mass For theoretical simplification let us consider distribution is not known, it is not possible to the object to be consisting of N particles obtain the moment of inertia theoretically, but of masses m1, m2, …..mN at respective it can be determined experimentally. perpendicular distances r1, r2, …..rN from the axis of rotation. As the object rotates, all these particles perform UCM with the same angular speed Z , but with different linear speeds v1  r1 ,v 2  r2 ,   v N  rN. Translational K.E. of the first particle is 1 1 Fig. 1.13: Moment of Inertia of a ring. K.E.1  m1v12  m1r12 2 2 2 1.5.1 Moment of Inertia of a Uniform Ring: Similar will be the case of all the other An object is called a uniform ring if particles. Rotational K.E. of the object, is its mass is (practically) situated uniformly the sum of individual translational kinetic on the circumference of a circle (Fig 1.13). energies. Thus, rotational K.E. Obviously, it is a two dimensional object of 1 1 1 negligible thickness. If it is rotating about its  m1r12 2 m2 r22 2  mN rN2 2 2 2 2 own axis (line perpendicular to its plane and RotationalK.E. passing through its centre), its entire mass M 1 1 is practically at a distance equal to its radius  m1r12 m2 r22  mN rN2  2  I  2 2 2 R form the axis. Hence, the expression for the N moment of inertia of a uniform ring of mass M Where I  m1r12 m2 r22  mN rN2   mi ri 2 i 1 and radius R is I = MR2. 13 1.5.2 Moment of Inertia of a Uniform Disc: of inertias of objects of several integrable Disc is a two dimensional circular object geometrical shapes can be derived. Some of of negligible thickness. It is said to be uniform those are given in the Table 3 at the end of the if its mass per unit area and its composition is topic. the same throughout. The ratio !  m  mass 1.6 Radius of Gyration: is called the surface density. A area As stated earlier, theoretical calculation Consider a uniform disc of mass M and of moment of inertia is possible only for radius R rotating about its own axis, which is mathematically integrable geometrical shapes. the line perpendicular to its plane and passing However, experimentally we can determine M the moment of inertia of any object. It depends through its centre ! .  R2 upon mass of that object and how that mass As it is a uniform circular object, it can is distributed from or around the given axis be considered to be consisting of a number of rotation. If we are interested in knowing of concentric rings of radii increasing from only the mass distribution around the axis of (practically) zero to R. One of such rings of rotation, we can express moment of inertia mass dm is shown by shaded portion in the of any object as I MK 2 , where M is mass Fig. 1.14. of that object. It means that the mass of that object is effectively at a distance K from the given axis of rotation. In this case, K is defined as the radius of gyration of the object about the given axis of rotation. In other words, if K is radius of gyration for an object, I MK 2 is the moment of inertia of that object. Larger the value of K, farther is the mass from the axis. Fig.1.14: Moment of Inertia of a disk. Consider a uniform ring and a uniform Width of this ring is dr, which is so small disc, both of the same mass M and same that the entire ring can be considered to be radius R. Let Ir and Id be their respective of average radius r. (In practical sense, dr is moment of inertias. less than the least count of the instrument that If Kr and Kd are their respective radii measures r, so that r is constant for that ring). of gyration, we can write, dm 2 Area of this ring is A = 2Sr.dr  !   Ir = MR2 = MK r ?Kr = R and 2 r.dr 1 2 R ? dm = 2SVr.dr. Id = MR2 = MK d ? Kd = ? Kd x2. The vertical forces acting on the below the surface as shown in Fig. 2.8. Let p0 cylinder are:   be the atmospheric pressure at the surface, i.e., 1. Force F1 acts downwards at the top surface at x1. Then, substituting x1 = 0, p1 = p0, x2 = -h, of the cylinder, and is due to the weight of and p2 = p in Eq. (2.6) we get, the water  column above the cylinder. p = p0+ hUg --- (2.7) 2. Force F2 acts upwards at the bottom The above equation gives the total surface of the cylinder, and is due to the pressure, or the absolute pressure p, at a depth water below the cylinder. h below the surface of the liquid. The total 3. The gravitational force on the water pressure p, at the depth h is the sum of: enclosed in the cylinder is mg, where m is 1. p0, the pressure due to the atmosphere, the mass of the water in the cylinder. As which acts on the surface of the the water is in static equilibrium, the forces liquid, and on the cylinder are balanced. The balance 2. hUg, the pressure due to the liquid at depth of these forces in magnitude is written as, h. F2= F1+ mg --- (2.3) p1and p2 are the pressures at the top and bottom surfaces of the cylinder respectively due to the fluid. Using Eq. (2.1) we can write F1 = p1A, and F2 = p2A --- (2.4) Also, the mass m of the water in the cylinder can be written as, m = density u volume = UV Fig. 2.8. Pressure at a depth h below the surface ?m = UA(x1-x2) --- (2.5) of a liquid. 29  In general, the difference between the base of the vessel B and the liquid from vessel absolute pressure and the atmospheric pressure C would rise into the vessel B. However, is called the gauge pressure. Using Eq. (2.7), it is never observed. Equation 2.2 tells that gauge pressure at depth h below the liquid the pressure at a point depends only on the surface can be written as, height of the liquid column above it. It does p - p0 = hUg --- (2.8) not depend on the shape of the vessel. In this Eq. (2.8) is also applicable to levels above case, height of the liquid column is the same the liquid surface. It gives the pressure at a for all the vessels. Therefore, the pressure of given height above a liquid surface, in terms liquid column in each vessel is the same and of the atmospheric pressure p0 (assuming that the system is in equilibrium. That means the the atmospheric density is uniform up to that liquid in vessel C does not rise in to vessel B. height). To find the atmospheric pressure at a distance d above the liquid surface as shown in Fig. 2.9, we substitute x1 = d, p1 = p, x2 = 0, p2 = p0 and U = Uair in Eq. (2.6) we get, p = po - dUair g --- (2.9) (a) (b) Fig. 2.10: Hydrostatic paradox. Consider Fig. 2.10 (b). The arrows indicate the forces exerted against the liquid by the walls of the vessel. These forces are perpendicular to walls of the vessel at each point. These forces can be resolved into vertical and horizontal components. The vertical components act in the upward direction. Weight of the liquid in Fig. 2.9: Change of atmospheric pressure section B is not balanced and contibutes the with height. pressure at the base. Thus, it is no longer a paradox! Can you tell? 2.3.5 Pascal’s Law: The figures show three containers filled Pascal’s law states that the pressure with the same oil. How will the pressures at applied at any point of an enclosed fluid at the bottom compare? rest is transmitted equally and undiminished to every point of the fluid and also on the walls of the container, provided the effect of gravity is neglected. (a) (b) (c) Experimental proof of Pascal’s law. 2.3.4 Hydrostatic Paradox: Consider a vessel with four arms A, B, C, Consider the inter connected vessels and D fitted with frictionless, water tight as shown in Fig. 2.10 (a). When a liquid is pistons and filled with incompressible fluid poured in any one of the vessels, it is noticed as shown in the figure given. Let the area of that the level of liquids in all the vessels is the cross sections of A, B, C, and D be a, 2a, 3a, same. This observation is somewhat puzzling. and a/2 respectively. If a force F is applied It was called 'hydrostatics paradox' before on the piston A, the pressure exerted on the the principle of hydrostatics were completely liquid is p = F/a. It is observed that the other understood. three pistons B, C, and D move outward. One can feel that the pressure of the base In order to keep these three pistons B, C, of the vessel C would be more than that at the 30  transmitted undiminished to the bigger piston and D in their original positions, forces 2F, S2. A force F2 = pA2 will be exerted upwards 3F, and F/2 respectively are required to be on it. applied on the pistons. Therefore, pressure A  on the pistons B, C, and D is: F2  F1  2  --- (2.10)  A1  Thus, F2 is much larger than F1. A heavy load can be placed on S2 and can be lifted up or moved down by applying a small force on S1. This is the principle of a hydraulic lift. Observe and discuss Blow air in to a flat balloon using a cycle 2F F pump. Discuss how Pascal’s principle is on B, pB   applicable here. 2a a 3F F ii) Hydraulic brakes: Hydraulic brakes are on C, pC   and used to slow down or stop vehicles in motion. 3a a F/2 F It is based on the same principle as that of a on D, p D   hydraulic lift. a/2 a Figure 2.12 shows schematic diagram i.e. pB = pC = pD = p, this indicates that the of a hydraulic brake system. By pressing the pressure applied on piston A is transmitted brake pedal, the piston of the master cylinder equally and undiminished to all parts of the is pushed in forward direction. As a result, fluid and the walls of the vessel. the piston in the slave cylinder which has a Applications of Pascal’s Law: much larger area of cross section as compared i) Hydraulic lift: Hydraulic lift is used to lift a to that of the master cylinder, also moves in heavy object using a small force. The working forward direction so as to maintain the volume of this machine is based on Pascal’s law. of the oil constant. The slave piston pushes the friction pads against the rotating disc, which is connected to the wheel. Thus, causing a moving vehicle to slow down or stop. Fig. 2.11 Hydraulic Lift. As shown in Fig. 2.11, a tank containing a fluid is fitted with two pistons S1 and S2. S1 has a smaller area of cross section, A1 while Fig. 2.12 Hydraulic brake system (schematic). S2 has a much larger area of cross section, The master cylinder has a smaller area of A2 (A2 >> A1). If we apply a force F1 on the cross section A1 compared to the area A2 of the smaller piston S1 in the downward direction it slave cylinder. By applying a small force F1 will generate pressure p = (F1/A1) which will be 31 to the master cylinder, we generate pressure i) Mercury Barometer: An instrument that p = (F1/A1). This pressure is transmitted measures atmospheric pressure is called a undiminished throughout the system. The force barometer. One of the first barometers was F2 on slave cylinder is then,