Page 2 :
INDEX, , 1, , Physical World, , 1, , 2, , Units and Measurement, , 3, , 3, , Motion in A Straight Line, , 11, , 4, , Motion in A Plane, , 17, , 5, , Laws of Motion, , 24, , 6, , Work, Energy and Power, , 30, , 7, , Systems of Particles and Rotational Motion, , 36, , 8, , Gravitation, , 44, , 9, , Mechanical Properties of Solids, , 50, , 10, , Mechanical Properties of Fluids, , 54, , 11, , Thermal Properties of Matter, , 65, , 12, , Thermodynamics, , 71, , 13, , Kinetic Theory, , 79, , 14, , Oscillations, , 86, , 15, , Waves, , 92

Page 3 :
CHAPTER ONE, PHYSICAL WORLD, 1.1 INTRODUCTION, Science- Originates from the Latin word “scientia” meaning 'to know'., Physics – originates from the Greek word “Fusis” meaning nature. physics is the study of, the basic laws of nature and their manifestations., 1.2 BRANCHES OF PHYSICS, 1. Classical physics, 2. Modern physics, 1. Classical Physics, physics that was recognized and developed before the beginning of the 20 th century, Branch, Major focus, Classical mechanics, , The study of forces acting on bodies whether at rest or in, motion, , Thermodynamics, , The study of the relationship between heat and other forms, of energy, , Optics, , The study of light, , Electricity and magnetism, , The study of electricity and magnetism and their mutual, relationship, , 2.Modern Physics, Refers to the concepts in physics that have surfaced since the beginning of the 20 th century., Branch, Major focus, Quantum mechanics, , The study of the discrete nature of phenomena at the atomic, and subatomic levels, , Atomic physics, , The branch of physics which deals with the structure and, properties of the atom, , Nuclear physics, , The branch of physics which deals with the structure,, properties and reaction of the nuclei of atoms., , 1.3 FUNDAMENTAL FORCES IN NATURE, Name, , Relative, strength, , Range, , Operates among, , Gravitational force 10-39, , Infinite, , Weak nuclear force 10-13, , Very short, Sub-nuclear Some elementary particles,particularly, size, electron and neutrino, , All objects in the universe, , Electromagnetic, force, , 10-2, , Infinite, , Charged particle, , Strong nuclear, force, , 1, , Short, nuclear, size, , Nucleons, heavier elementary particles, , KAMIL KATIL VEETIL, , 1, , @, , SOHSS AREEKODE

Page 4 :
PREVIOUS QUESTIONS, 1.Which one of the following fundamental forces in nature binds Protons and neutrons, in a nucleus? ( IMP 2016), A) Gravitational force, B) Electromagnetic force, C) Strong nuclear force, D) Weak nuclear force, , (1), , 2. Which one of the following is present between all object in universe? (March 2016), A)Electromagnetic force, B)Magnetic Force, C)Gravitational force, D)Strong nuclear force, , (1), , 3.Choose the WRONG statement from the following statements. (IMP 2015), A)Electromagnetic force is the force between charged particles., B) Electrostatic force can be attractive or repulsive., C)Nuclear force binds protons and neutrons in a nucleus., D) Gravitational force is one of the strongest forces among fundamental forces in nature., , (1), , 4. Choose the correct answer from the bracket. Weakest force in nature is..................., (1), (Strong nuclear force, Electromagnetic force, Gravitational force, Weak nuclear force) (MAR 2015), 5. Pick the odd one out among the following forces, (MAR 2017), ( Gravitational force , Viscous force , Week nuclear force, Electromagnetic force ), , (1), , 6. The branch of physics that was developed to understand and improve the working of heat engine, is..........................., (MAR 2018), (1), (Optics , Thermodynamics ,electronics, Electrodynamics ), 7. The ratio of electrostatic force and the gravitational force between two protons kept fixed, distance is........................, (IMP 17), (1), -19, 19, 36, -36, ( 10 , 10 , 10 , 10 ), 8. The week nuclear force is stronger than gravitational force. State whether the statement is true or, false, (MAR 19), (1), , KAMIL KATIL VEETIL, , 2, , @, , SOHSS AREEKODE

Page 5 :
CHAPTER 2, , UNITS AND MEASUREMENTS, 2.1 PHYSICAL QUANTITY, A quantity, which can be measured directly or, indirectly, is called a physical quantity., Ex : length, mass, time, speed, force, volume,, etc., 2.2 UNITS, measurement of a physical quantity involves, comparison with a reference standard. This, reference standard is called unit., Ex: metre is the unit of length, kilogram is the, unit of mass, second is the unit of time., , Thus in the measurement of a physical, quantity, two things are involved – a number, and a unit. The unit is the standard quantity, with which the measured quantity is compared, and the number shows how many times the, measured quantity is greater than the unit., A good unit will have the following, characteristics., It should be (a) well defined (b) internationally, acceptable (c) invariable (d) reproducible., 2.3 FUNDAMENTAL QUANTITIES AND, FUNDAMENTAL UNITS, A set of physical quantities which are, independent of each other are known as, fundamental quantities. The units of, fundamental quantities are called fundamental, units., 2.4 DERIVED QUANTITIES AND, DERIVED UNITS, Physical quantities, which can be expressed in, KAMIL KATIL VEETIL, , 3, , terms of fundamental quantities, are called, derived quantities. The units of derived, quantities are called derived units., Eg: 1. Volume is measured in metre3 or m3 ., 2. Speed can be expressed in meter per, second (m/s), 2.5 SYSTEMS OF UNITS., A complete set of units for all physical, quantities with particular basic units is called a, system of units., The commonly used systems are:, (a) The FPS system: It is the British, Engineering system of units, which uses foot,, pound and second as the three basic units for, measuring length, mass and time respectively., (b) The c.g.s system: Which uses centimetre,, gram and second as the three basic units for, measuring length, mass and time respectively., (c) The MKS system:, (d) SI units(Metric System): In 1960,, International Committee for Weights Measures, adopted a system of units for all fundamental, physical quantities and is called International, system of units or SI units., In SI system, there are seven Fundamental, (basic) units and two Supplementary units., 2.6 FUNDAMENTAL QUANTITIES AND, THEIR UNITS IN SI SYSTEM, Fundamental quantity, , Unit, , Symbol, , Mass, , kilogram, , kg, , Length, , metre, , m, , Time, , second, , s, , Temperature, , kelvin, , K, , Electric current, , ampere, , A, , Luminous intensity, , candela, , cd, , Amount of substance mole, 2.7 SUPPLEMENTARY, AND THEIR UNITS, , mol, QUANTITIES, , Plane Angle, , radian, , rad, , Solid angle, , steradian, , sr, , @, , SOHSS AREEKODE

Page 6 :
NOTE: These two quantities have units but no, dimension., Advantages of SI units:, 1) It is comprehensive., 2) The system is coherent., 3) It is internationally accepted., 2.8 DESCRIPTION OF PLANE ANGLE, (dθ ), , The Radian (rad): One radian is the angle, subtended at the centre of a circle by an arc, length equal to the radius of the circle., 2.9 DESCRIPTION OF SOLID ANGLE, (dΩ ), , The Steradian (sr): One steradian is the solid, angle subtended at the centre of a sphere by a, surface of the sphere, which is equal in area, to, the square of radius of the sphere ., Relations between radian , degree and, minute, 0, =>, π radian=180, 10= π radian, , multip prefix symbo submu Prefix symbo, le, l, ltiple, l, 101, , deca, , 10-1, , deci, , d, , 102, , hecto h, , 10-2, , centi, , c, , 103, , kilo, , 10-3, , milli, , m, , 106, , mega M, , 10-6, , micro μ, , 109, , giga, , 10-9, , nano, , n, p, , da, k, G, , 10, , 12, , -12, , terra, , T, , 10, , pico, , 1015, , peta, , P, , 10-15, , femto f, , 1018, , exa, , E, , 10-18, , atto, , 1021, , zetta, , Z, , 10-21, , zepto z, , 1024, , yotta, , Y, , 10-24, , yocto y, , a, , 2.10, MEASUREMENT, OF, BASIC, QUANTITES, 1.Measurement of Length, Distances ranging from 10 −5m to 10 2m can be, measured by direct methods., Eg: Metre Scale - To measure the distance, from 10−3 m to 1m, Vernier Calipers - up to 10−4 m, Screw Gauge - up to 10−5 m, The atomic and astronomical distances cannot, be measured by any of the above mentioned, direct methods., Measurement of large distances, Parallax method, Parallax is the name given to the apparent, change in the position of an object with, respect to the background, when the object is, seen from two different positions., See Fig., , 180, , ∠LOR is called the parallax angle, or parallactic angle., , 10= 60' (minute of arc), 1'(minute of arc) = 60'' (seconds of arc), , Taking LR as an arc of length b and, radius LO = RO = x, , 2.10 MULTIPLES AND SUB MULTIPLES, OF UNITS., , we get, , To express magnitude of physical quantities,, which are very large or small, we use prefixes, to the unit., KAMIL KATIL VEETIL, , 4, , θ=, , b, x, , where b-basis ,, , x-unknown distance., Knowing ‘b’ and measuring θ, we can, @, , SOHSS AREEKODE

Page 7 :
calculate x., * HW: Example 2.2, 2.3, 2.4, 2.Measurement of Mass, Common objects - Common balance or spring, balance or electronic balance, Larger masses like that of planets, stars etc Gravitational methods, Small masses like atomic/subatomic particles, etc - Mass spectrograph., 3.Measurement of time, A clock is used to measure the time interval., An atomic standard of time, is based on the, periodic vibration produced in a Cesium atom., 6.11 THEORY OF ERRORS, The result obtained from any measurement, will contain some uncertainty. Such an, uncertainty is termed error., Accuracy And Precesion, Accuracy is a measure of how close the, measured value is to the true value of the, quantity., Precision refers to the closeness of two or, more measurements to each other., , Errors in Measurement, 1.Systematic error, 2. Random error ., 1.Systematic Errors:, The systematic errors are those errors that end, to be in one direction, either positive or, negative. Systematic errors can be classified as, follows., (a). Instrumental Error : These errors are due, to the defect of the instrument., Eg: 1. Zero error in screw gauge and vernier., 2.Faulty calibration of thermometer, metre, scale etc., (b). Error due to Imperfection : This is due to, KAMIL KATIL VEETIL, , 5, , imperfection of the experimental procedure., Eg: A thermometer placed under the armpit, will always give a temperature lower than the, actual value of the body temperature., (c). Personal Error : This is due to the mode of, observation of the person taking the reading ., Eg: Parallax error., (d). Errors due to external causes : The change, in the external conditions during an, experiment can cause error in measurement., Eg: changes in temperature,or pressure during, measurements may affect the result of the, measurement., (e). Least Count Error : LC is the smallest, value that can be measured by the measuring, instrument, and the error due to this, measurement is least count error. The, instrument’s resolution hence is, the cause of this error. Least count error can be, reduced by using a high precision instrument, for the, measurement., 2. Random Error :, The random errors are those errors, which, occur irregularly and hence are random with, respect to sign and size., Eg:, 1.unpredictable, fluctuations, in, temperature, voltage supply etc become source, of error 2. when the same person repeats the, same observation, it is very likely that he may, get different readings every time., ERROR ANALYSIS, 1. Absolute Error, The magnitude of difference between the true, value and the measured value of a quantity is, called absolute error. If a1 , a2 , a3 , ..........an, are the measured values of any quantity ‘a’ in, an experiment performed n times, then the, arithmetic mean of these values is called the, true value (am ) of the quantity. Ie,, am =, , a1 +a2 +a 3+............. a n, n, , The absolute error in measured values is given, by, |Δ a1|=|am−a1|, |Δ a2|=|am−a2|, @, , SOHSS AREEKODE

Page 8 :
Similarly, , two decimal place, (ii) Absolute error , |ΔT|=|T m−t|, , |Δ an|=|am −an|, , 2. Mean Absolute error ( |Δ am| ), The arithmetic mean of absolute errors, in all the measurements is called the mean, absolute error. Ie,, Δ a m=, , |Δ T 1|=|2.62−2.63|=0.01, |Δ T 2|=|2.62−2.56|=0.06, |Δ T 3|=|2.62−2.42|=0.20, |Δ T 4|=|2.62−2.71|=0.09, |Δ T 5|=|2.62−2.80|=0.18, , |Δ a1|+|Δ a2|+|Δ a3|+..............+|Δ an|, n, , (iii) Mean absolute error,, , 3. Relative Error, Relative error =, , |Δ T m|=, , Mean absolute error, Mean value, , = 0.11 s ( Rounded off to two decimal, places), , Eg: A driver’s speedometer shows that his car, is travelling at 60 km/h when it is actually, moving at 62 km/h. Then absolute error of, speedometer is 62 km/h-60 km/h= 2 km/h., Relative error of the measurement is, 2 km/ h, =0.322 km/h, 62 km/h, , (iv) Relative error,, Relative error =, , =, , =4%, (vi) Time period of simple pendulum,, T =(2.62±0.11)s, , NOTE: If we do a single measurement, the, value we get may be in the range am ± ∆am, Problem 1: In a series of successive, measurements in an experiment, the readings, of the period of oscillation of a simple, pendulum were found to be 2.63s, 2.56 s,, 2.42s, 2.71s and 2.80s. Calculate (i) the mean, value of the period of oscillation (ii) the, absolute error in each measurement (iii) the, mean absolute error (iv) the relative error, (v)the percentage error. Express the result in, proper form., Solution, , =>, , 0.11, =0.0419=0.04, 2.62, , P . E=R . E x 100=0.04 x 100, , |Δ am|, Percentage error =, x 100, am, , =>, , Mean absolute error, Mean value, , (v) Percentage error ,, , 4. Percentage Error, , (i), , 0.01+ 0.06+0.20+ 0.09+ 0.18, =0.108 s, 5, , t 1 +t 2+t 3 +t 4 +t 5, 5, 2.63+2.56+2.42+2.71+2.80, T m=, 5, T m=2.624 s = 2.62 s (Rounded off to, T m=, , KAMIL KATIL VEETIL, , 6, , COMBINATION OF ERRORS, The error in the final result depends on, (i) The errors in the individual measurements, (ii) On the nature of mathematical operations, performed to get the final result. So we should, know the rules to combine the errors., (a) Error of a sum or a difference, *Let ΔA and ΔB be the absolute errors in the, two quantities A and B respectively. Then,, Measured value of A = A ± ΔA, Measured value of B = B ± ΔB, Consider the sum, Z = A + B, The error ΔZ in Z is then given by, Z ±Δ Z=( A±Δ A)+(B±Δ B), = ( A+ B)±( Δ A + Δ B), = Z ±( Δ A +Δ B), , @, , SOHSS AREEKODE

Page 9 :
=>, , Δ Z=(Δ A +Δ B), , We will get the same result even if we take the, difference. Ie,, The maximum possible error in the sum of, two quantities or difference of two, quantities is equal to the sum of the, absolute errors in the individual quantities., Examples, 1)Two resistances R1 = (100 ± 3) Ω and R2 =, (150 ± 2) Ω are connected in series. What, is, their, equivalent, resistance?, Soln: (250 ± 5) Ω, 2) Do example 2.8 NCERT, (b) Error of a product or a quotient, Let ΔA and ΔB be the absolute errors in the, two quantities, A and B, respectively. Then,, Measured value of A = A ± ΔA, Measured value of B = B ± ΔB, Consider the product , Z = A B, , As, their, , product, , are both small quantities,, term, , ΔA ΔB, A B, , can, , We know, , Δ A Δl Δb, = +, A, l, b, , =>, , Δ A=A, , =>, , ( Δl l + Δbb ), 0.1 0.2, Δ A=19.4 (, +, 5.7 3.4 ), = 1.48 = 1.5, , Therefore , Area with error limit,, 2, , A=(19.4±1.5)cm, , (c) Error in the power of a quantity, , ΔZ, ΔB Δ A Δ A Δ B, =1±, ±, ±, Z, B, A, A B, ΔB, ΔA, ,, B, A, , Area with error limit = A ± ΔA = ?, Area A = l × b = 5.7 × 3.4, = 19.38 = 19.4 cm 2, , 2) Do Example 2.9 NCERT, 3) Example 2.10 NCERT [HW], 4) The voltage across a wire is (100 ± 5)V and, the current passing through it is (10±0.2) A., Find the resistance of the wire. [HW], , The error ΔZ in Z is given by, Z ± ΔZ = (A ± ΔA) (B ± ΔB), The error ΔZ in Z is given by, Z ± ΔZ =(A ± ΔA) (B ± ΔB), = (AB) ± (A ΔB) ± (B ΔA) ± (ΔA . ΔB), Dividing L.H.S by Z and R.H.S by AB,, we get,, 1±, , Examples, 1)The length and breadth of a rectangle are, (5.7 ± 0 . 1 ) cm and (3.4 ± 0 . 2 ) cm, respectively. Calculate the area of the, rectangle with error limits., Soln) Length l= (5.7 ± 0 . 1 ) cm ,, Breadth b =(3.4 ± 0 . 2 ) cm, , be, , neglected., So the maximum relative error in Z is ,, , The relative error in a physical quantity, raised to the power k is the k times the, relative error in the individual quantity., Examples, 1) Find the relative error in Z, if, Z = A 4 B 1/3 /CD 3/2 ., Solution, , ΔZ, Δ A 1 Δ B ΔC 3 Δ D, =4, +, +, +, Z, A 3 B, C 2 D, , ΔZ Δ A ΔB, =, +, Z, A, B, , 2) Example 2.12 NCERT [HW], , This is true for division also. Ie,, The maximum fractional error in the, product of two quantities or quotient of, two quantities is equal tothe sum of the, fractional errors in the individual, quantities., , SIGNIFICANT FIGURES, The digits that are known reliably plus the first, uncertain digit are known as significant figures, or significant digits., , KAMIL KATIL VEETIL, , 7, , @, , SOHSS AREEKODE

Page 10 :
RULES FOR COUNTING THE NUMBER, OF SIGNIFICANT FIGURES, 1) All non-zero digits are significant, Ex: 1342 has 4 significant figures, 2) All zeros between two non-zero digits are, significant, no matter where the decimal, point is., Ex: 1002 has 4 significant figures, 20.003 has 5 significant figures, 3) If the number is less than 1, the zero (s), on the right of the decimal point but to left, of the first non zero digit are not significant., Ex: 0.00345 has 3 significant figures, 4) The terminal or trailing zero(s) in a, number without a decimal point are not, significant., Ex: 123 m = 12300 cm = 123000 mm, has three significant figures, 5) The trailing zero(s) in a number with a, decimal point are significant., Ex: 3.500 has 4 significant figures, 0.06900 has 4 significant figures, 6) The number of significant figures does, not depend on the system of units used, Ex:123 m = 12300 cm = 123000 mm, has three significant figures, 7) The power of 10 is irrelevant to the, determination of significant figures., Ex: = 5.70 m = 5.70 × 102 cm =, 5.70 × 10 3 mm = 5.70 × 10 −3 km have 3, significant figures ., RULES FOR ROUNDING OFF, 1) If the digit to be dropped is smaller than, 5, then the preceding digit should be left, unchanged., i) 7.32 is rounded off to 7.3, , increased by 1, i) 17.26 is rounded off to 17.3, ii) 11.89 is rounded off to 11.9, 3) If the digit to be dropped is 5 followed by, digits other than zero, then the preceding digit, should be raised by 1, , i)7.352 is rounded off to 7.4, ii)18.159 is rounded off to 18.2, 4) If the digit to be dropped is 5 or 5, followed by zeros, then the preceding digit, is not changed if it is even, i) 3.45 is rounded off to 3.4, ii) 8.250 is rounded off to 8.2, 5)If the digit to be dropped is 5 or 5, followed by zeros, then the preceding digit, is raised by 1 if it is odd, i) 3.35 is rounded off to 3.4, ii) 8.350 is rounded off to 8.4, RULES FOR ARITHMETIC, OPERATIONS WITH SIGNIFICANT, FIGURES., 1) In addition and subtraction, the final result, should retain as many decimal places as there, are in the number with the smallest number of, decimal, , places., 1)3.1 + 1.780 + 2.046 = 6.926, Here the least number of significant, digits after the decimal is one. Hence the, result will be 6.9., 2)12.637 – 2.42 = 10.217 Hence the, , result will be 10.22, 2) In multiplication or division, the final, result should retain as many significant, figures as there are in the original number, with smallest number of significant figures., 1) 1.21 × 36.72 = 44.4312 = 44.4, 2) 36.72 ÷ 1.2 = 30.6 = 31, DIMENSION OF PHYSICAL, QUANTITIES, , ii) 8.94 is rounded off to 8.9, 2) If the digit to be dropped is greater than, 5, then the preceding digit should be, , Dimensions of a physical quantity are the, powers to which the fundamental units be, , KAMIL KATIL VEETIL, , 8, , @, , SOHSS AREEKODE

Page 11 :
raised in order to represent that quantity., Eg: Velocity=, , ,where m is the mass of the body, v its, velocity, g is the acceleration due to gravity, and h is the height. Check whether this, equation is dimensionally correct., , Displacement [ L], 0, −1, =, =[ M L T ], time, [T ], , Hence the dimensions of velocity is 0 in mass,, 1 in length and -1 in time., Problem: ( Do yourself), Find the dimensional formula of the following, physical quantities., (a) Acceleration (b) Force (c) Momentum, (d) Kinetic energy (e) Mass per unit length, , Solution, , DIMENSIONAL FORMULA AND, EQUATION, , Since both term has the same dimension , the, given equation is dimensionally correct., , Dimensional formula is an expression which, shows how and which of the fundamental units, are required to represent the unit of a physical, quantity., Eg: [M 0 LT −2 ] is the dimensional formula of, acceleration., , Eg: Do Examples 2.16 NCERT, , When the dimensional formula of a physical, quantity is expressed in the form of an, equation, such an equation is known as the, dimensional equation., Eg: Acceleration = [M 0 LT −2 ]., Principle of homogeneity of dimensions, The principle of homogeneity of dimensions, states that the dimensions of all the terms in, a physical expression should be the same., Eg: v 2 = u 2 +2as , the dimensions of v 2 , u 2, and 2as have the same dimension [L 2 T −2 ]., APPLICATIONS OF DIMENSIONAL, ANALYSIS, 1.To check the correctness of an equation, Checking the correctness of the equation using, principle of homogeneity., Here we use the principle of homogeneity to, check the correctness of the equation, Problem:, Let us consider an equation, , 1, m v 2=mgh, 2, , The dimension of, , = [M L2 T-2], The dimension of mgh = [M][LT-2][L], = [M L2 T-2], , 2.To establish the relation among various, physical quantities, Problem, Consider a simple pendulum, having a bob, attached to a string, that oscillates under the, action of the force of gravity. Suppose that the, period of oscillation of the simple pendulum, depends on its length (l), mass of the bob (m), and acceleration due to gravity (g). Derive the, expression for its time period using method of, dimensions., Solution, *The dependence of time period T on the, quantities l, g and m as a product may be, written as :, T=klxgymz, where k is dimensionless constant and x, y, and z are the exponents., By considering dimensions on both sides, we, have, [M o L o T 1 ]=[L 1 ]x [L 1 T –2 ]y [M 1 ]z, = L x+y T –2y M z, On equating the dimensions on both sides,, we have, x + y = 0; –2y = 1; and z = 0, So x=, , KAMIL KATIL VEETIL, , 9, , 1, 2, m v = [M][LT-1]2, 2, , @, , 1, ,, 2, , y=, , −1, ,, 2, , z=0, , SOHSS AREEKODE

Page 12 :
Then, T =k l1/ 2 g−1 /2, , => T =k, , Actually, k = 2π =>, , √, , (a) 3215, (e) 3100, , l, g, , T =2 π, , √, , (b) 11.01, (f) 007, , (c) 0.003 (d) 1.000, , 6. Report the result of the following with, correct significant figures, , l, g, , Limitations of Dimensional Analysis., 1) This method gives us no information about, dimensionless constants., 2) We cannot use this method if the physical, quantity depends on more than three other, physical quantities., 3) This method cannot be used if the left hand, side of the equation contains more than one, term., 4) Often it is difficult to guess the parameters, on which the physical quantity depends., , (a) 30.1 + 2.760 + 9.046, (c) 1.21 × 36.72, , 7. If the velocity of sound through a medium, depends on the bulk modulus , 'B' of the, medium and density, 'ρ' of the medium , obtain, the relation connecting the velocity of sound,, bulk modulus and the density of the medium., (Given that the bulk modulus has the, dimension of force per area), 8. “ A dimensionally correct equation may not, be a physically correct equation” . Comment, the statement., , ADDITIONAL QUESTIONS, 1 . a) A boy recalls the relativistic mass, wrongly, , as, , m=, , m0, , √(1−V 2 ), , ., , Using, , dimensional method put the missing ‘C’ at the, proper place., b) Name and state the principle used in solving, the above problem., [ imp. 2012], 2. If the percentage error in calculating the, radius of the sphere is 2% , what will be the, percentage error in calculating the volume?, [imp 2013], 3. Which of the following measurement is, more accurate? Why?, i. 500.00kg., ii. 0.0005kg., iii.6.00kg., , √, , 4. Check whether the equation T =2 π m is, g, , dimensionally correct. Where,, T - > Time period of a simple Pendulum., m - > mass of the bob., g -> acceleration due to gravity., , 5 . Find out the number of significant figures, in the followig, KAMIL KATIL VEETIL, , 10, , (b) 12.637 – 2.42, (d) 36.72 ÷ 1.2, , @, , SOHSS AREEKODE

Page 13 :
CHAPTER THREE, MOTION IN A STRAIGHT LINE, 3.1 INTRODUCTION, Point object : If the size of the object is much smaller than the distance it moves in a reasonable, duration of time, then the object is said to be point object., Reference point : In order to specify the position of an object, we need to use a reference point, and a set of axes. For convenience we use a rectangular coordinate system consisting three, mutually perpendicular axes (X,Y,Z) as set of axes and the point of intersection of these three, axes (origin ) as the reference point., 3.2 POSITION, PATH LENGTH AND DISPLACEMENT, , 3.2.1 Position, To specify position we fix a origin (O). Positions to the right of O are taken as positive and, to the left of O, as negative ( convention), Thus the position co ordinates of P ,Q , R are +360 m , +240 m , -120 m respectively., 3.2.2 Path length (Distance), It is the total distance covered., Suppose a car moves from O to P and then moves back from P to Q , then the total path length, (distance) covered = OP+PQ = 360 m+120 m= 480 m, 3.2.3 Displacement (Δx), It is the shortest distance between the final and initial positions, Suppose a car moves from O to P and then moves back from P to Q , then Δx = xQ-xO = + 240 m, If a car moves from O to P and then moves back from P to R , then Δx = xR-xO = -120 m, If a car moves from O to P and then moves back from P to O , then Δx = xO-xO = 0, NOTE:, * Distance is a scalar quantity ( only magnitude , no direction) while displacement is a vector, quantity ( it has both magnitude and direction), * Distance can only have positive values while Displacement can be positive, negative and even, zero., * The magnitude of displacement may or may not be equal to the path length traversed by an, object., Problem 1, A point P is the contact point of a wheel on ground which rolls on, ground without slipping. What is the displacement of the point P, when wheel completes half of rotation (if radius of wheels is 1m), Solution, , Displacement , S =, , √(π R)2 +( 2 R)2, , Since R=1 m,, , S=, , √ π2 + 4, , = R √ π2 +4, , m, , Motion: If a body changes its position with time , then the body is said to be in motion., , KAMIL KATIL VEETIL, , @, 11, , SOHSS AREEKODE

Page 14 :
3.3 SPEED AND VELOCITY, 3.3.1 Average speed, The average speed is defined as the ratio of the total path length covered by the particle to the total, time taken, , Average speed=, , total path length, Total time interval, , NOTE, * If a particle travels distances S 1 , S2 , S 3 with speeds v 1 , v 2 , v 3 respectively in same direction ., Then,, Total distance travelled=S1 +S 2 +S 3, , Therefore ,, , ,, , Total time taken=, , The average speed =, , S1 S2 S3, + +, v 1 v 2 v3, , S1 + S2 + S3, S1 S2 S3, + +, v1 v2 v3, , * If a particle travels with speeds v 1 , v 2 , v 3 during the time interval t1 , t2 , t 3 respectively., Total time taken=t 1+ t 2 +t 3, Then, Total distance travelled=v1 t 1 + v 2 t 2+ v 3 t3 ,, Therefore ,, , The average speed=, , v 1 t 1 +v 2 t 2+v 3 t 3, t 1+ t 2 +t 3, , 3.3.2 Average velocity, It is the ratio of displacement to the time taken, The average velocity=, , Δ x x 2−x 1, =, Δ t t 2−t 1, , 3.3.3 Instantaneous velocity (Velocity), The time rate of change of position with time at any instant is called Instantaneous velocity, Uniform motion (Uniform velocity), If a body covers equal displacement in equal interval of time , then the body is said to be in, uniform motion. During uniform motion magnitude and direction of the velocity remains constant., Non uniform velocity, If a body covers unequal displacements in equal interval of time or equal displacements in unequal, interval of time, then the body is said to be in non uniform motion. During non uniform motion, either magnitude or direction of the velocity changes., 3.4 Position – time graph, NOTE: The slope of position, time graph (Δx/Δt) gives the, velocity. In Fig, (a) Slope = (x2-x1)/(t2-t1) = 0, (b)Slope =(x2-x1)/(t2-t1) = +ve, (c) Slope =(x2-x1)/(t2-t1) = -ve, , KAMIL KATIL VEETIL, , @, 12, , SOHSS AREEKODE

Page 15 :
3.5 Velocity - time graph for uniform motion., NOTE :, Area under the velocity - time graph is equal, the displacement of the particle., Here the displacement covered by the particle, between 6s and 12s is 40 x 6 = 240m, , 3.6 Acceleration, The rate of change of velocity of an object is called acceleration., Or, , a=, , V −u, t, , NOTE:If an object is slowing down, then its acceleration is in the opposite direction of its, velocity. When an object is speeding up, the acceleration is in the same direction of the velocity., Ex : Raju throws a stone up and it comes back. Give sign for displacement, velocity and, acceleration during its (a) Upward motion (b) Downward motion, soln: I fix upward direction as positive and downward direction as negative ( Reverse can be, taken), (a) Upward motion: displacement= positive , velocity= positive, acceleration = negative ( during, upward motion body slows down , so acceleration is in the opposite direction of velocity), (b) Downward motion: displacement= negative , velocity= negative, acceleration = negative, ( during downward motion body speeds up , so acceleration is in the same direction of velocity), Instantaneous acceleration, , 3.6.1 Uniform acceleration, If the velocity of an object changes by equal amounts in equal intervals of time (however small the, intervals may be) , then the body is in uniform acceleration., Eg: A freely falling stone has an acceleration 9.8 m/s2 means velocity of the stone increases 9.8m/s, in each second., 3.6.2 Position-time graph for uniformly accelerated motion, Fig(a):Uniform acceleration, ( Note the slope of x-t graph, is increasing ,velocity, increases), , KAMIL KATIL VEETIL, , Fig(b):Uniform deceleration, ( Note the slope of x-t graph is, decreasing ,velocity decreases), , @, 13, , SOHSS AREEKODE

Page 16 :
3.6.3 Velocity-time graph for uniformly accelerated motion, Fig(a) Motion in positive, direction with positive, acceleration, , Fig(b) Motion in positive, direction with, negative acceleration, , Fig(c) Motion in negative, direction with negative, acceleration, , Fig(d) Motion of an, object with negative, acceleration that changes, its direction at time t1 ., , NOTE : 1) The slope of velocity time graph gives acceleration, 2)Area under the velocity time graph gives the displacement, 3.7 KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION, Let, u- Initial velocity, v- Final velocity after time t, s- displacement covered in time t, a- Uniform acceleration, First Equation Of Motion, Acceleration = Slope of velocity-time graph, v−u, t, v −u=at, v =u+a t, a=, , =>, =>, ........................ (1), This is the first equation of motion., , Second Equation Of Motion, Displacement = Area under the velocity-time graph, => S= Area of Δ ABC + Area of □ ACOt, , 1, 2, v, =u+a, t in (2), Substitute, (2) => S=ut +1/2(u+at−u) t, 1, => S=ut + a t 2 .....................(3), 2, , => S= (v−u) t+ ut ....................(2), , (OR), , 1, S= (u+v )t, 2, , This is the second equation of motion, , KAMIL KATIL VEETIL, , @, 14, , SOHSS AREEKODE

Page 17 :
Third equation of motion, From (1) v =u+a t, Squaring both sides , we get v 2=(u+ at )2=u2 +2u at +(a t )2, 1, 2, , = u2 +2 a(ut + a t 2 ), 2, , =>, , 2, , ................(4), This is the third equation of motion, v =u + 2 aS, , Summary, 1 2, S=ut + a t, 2, , v =u+a t, , 2, , 2, , v =u + 2aS, , 1, S= (u+v )t, 2, , Stopping distance of vehicles, When brakes are applied to a moving vehicle, the distance it travels before stopping is called, stopping distance. It is an important factor for road safety and depends on the initial velocity (u), and the braking capacity, or deceleration, –a that is caused by the braking., Expression for stopping distance, Let the distance travelled by the vehicle before it stops be d s ., Then, using equation of motion v 2=u 2+ 2 aS and noting that v = 0, S= d s , the equation, becomes,, 0=u2+ 2 a d s => −u 2=+2 a d s, ie, the stopping distance, , ds =, , −u2, 2a, , Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial, velocity increases the stopping distance by a factor of 4 (for the same deceleration)., 3.8 RELATIVE MOTION, Consider two objects A and B moving uniformly with velocities v A and v B in one, dimension(along x-axis) with respect to the ground . If x A (0) and x B (0) are positions of objects A, and B, respectively at time t = 0, their positions x A (t) and x B (t) at time t are given by:, Then, the displacement from object A to object, B is given by, ...........................(5), (5) => Final distance between objects=initial distance between objects+ Relative velocity x time interval ., = vBA is the velocity of object B relative to object A . ie, relative velocity of B wrto A., Similarly (vA – vB) = vAB is the velocity of object A relative to object B . ie, relative velocity of A, wrto B., Special cases, Case 1: when vA= vB from (4) xBA(t)= xB(0) – xA(0), i.e. the distance between the two objects remains the same at all times., The position time graph for such motion is as shown in fig., , KAMIL KATIL VEETIL, , @, 15, , SOHSS AREEKODE

Page 18 :
Case 2: when v A > v B ( v A and v B in the same direction), v B - v A is -ve., From (4), we get x B (t) - x A (t) < x B (0) - x A (0), ie, the value of x B (t) - x A (t) first decreases, becomes zero, and then increases in magnitude. The position time graph for such, motion is as shown in fig., Case 3: when v A < v B ( v A and v B in the same direction), v B - v A is +ve., From (4), we get x B (t) - x A (t) > x B (0) - x A (0), ie, the value of x B (t) - x A (t) increase as time passes. The position time, graph. for such motion is as shown in fig., Case 4 : When v A and v B are in opposite directions ( say v A in +ve X, direction and v B in the -ve X direction) .The position time graph, for such motion is as shown in fig., , Do Example 3.1, 3.4, 3.6, 3.8 ,3.9, , Previous Questions, 1.Velocity is defined as the rate of change of displacement., a) Distinguish between average velocity and instantaneous velocity., b) When does the average velocity becomes equal to the instantaneous velocity?, c) A car travels from A to B at 60 km/hr and returns to A at 90 km/hr. What is its average velocity, and Average speed?, .............................................................................................................................................................., 2. The figure shows the position – time, graph of a body moving along a straight line., a)Draw the velocity-time graph of the body., From the graph, b) find the displacement in 20 seconds., .............................................................................................................................................................., 3.Acceleration – time graph of a body starts from rest as shown below:, ( a in m/s2 and t in seconds), a) Draw the velocity – time graph using the above graph., b) Find the displacement in the given interval of time from 0 to 30 seconds., , KAMIL KATIL VEETIL, , @, 16, , SOHSS AREEKODE

Page 19 :
CHAPTER FOUR, , (Note that v1≠ 2 v2 but the magnitude of v1 is, equal to 2 x magnitude of v2), , MOTION IN A PLANE, 4.1 INTRODUCTION, In order to describe motion of an object in two, dimensions (a plane) or three dimensions, (space), we need to use vectors., Example for motion in a plane, Motion of a coin on a carom board,An insect, crawling over the floor of a room, projectile, motion, etc., 4.2 ELEMENTARY CONCEPTS OF, VECTOR ALGEBRA, Scalar quantity and vector quantity, Scalar quantity: A quantity with magnitude, only. It is specified completely by a single, number, along with the proper unit., Eg : mass, temperature , distance, The rules for combining scalars are the rules of, ordinary algebra., Vector quantity: A quantity that has both, magnitude and direction and obeys vector, algebra . It is specified by giving its magnitude, by a number and its direction., Eg: displacement, velocity, acceleration , force, Representation of vectors, 1. Geometrical (Graphical ) representation, 2. Analytical representation, Geometrical representation of vectors, A vector represented by an arrow. Length of the, arrow indicates its magnitude and arrow head, indicates its direction, , 4.2.1 TYPES OF VECTORS, 1.Equal vectors: Two vectors A and B are said, to be equal when they have, equal magnitude and same, direction and represent the, same physical quantity., 2. Collinear vectors: Those vectors which act, along the same line. The angle between them, can be 0° or 180°., 3.Parallel Vectors: If two vectors A and B act, in the same direction along the same line or on, parallel lines( the angle between them is 00), , 4.Anti–parallel vectors: Two vectors A and B, are said to be anti–parallel when they are in, opposite directions along the same line or on, parallel lines (the angle between them is 180 0), , 5.Unit vector: A vector divided by its, magnitude is a unit vector. The unit vector for, A is denoted by  ( A cap). It has a magnitude, equal to unity or one., , The given vector is toward east direction, 6.Orthogonal unit vectors:, Let i , j , k be three unit, vectors which specify the, directions along positive x–, axis, positive y–axis and, positive z–axis respectively., These three unit vectors are, directed perpendicular to each, other., These three vectors are orthogonal unit vectors., , F1 = 3 N towards east, F2 = 6 N towards east, ( Note that F2=2 F1), , v1= 10 m/s towards north, v2= 5 m/s towards north, east, , KAMIL KATIL VEETIL, 17, , @, , SOHSS AREEKODE

Page 20 :
4.2.2 MULTIPLICATION OF VECTORS, BY REAL NUMBERS, Multiplying a vector A with a positive number, λ gives a vector whose magnitude is changed, by the factor λ but the direction is the same as, that of A, Step1: Two vectors A and B with their tails, brought to a common origin. (Fig. a), Step 2: The sum R= A + B is obtained using, the parallelogram method . (Fig b), , Fig (a) : Vector A and the resultant vector after, multiplying A by a positive number 2., Fig (b) : Vector A and resultant vectors after, multiplying it by a negative number –1and –1.5., , NOTE, The parallelogram, method of vector, addition is equivalent, to the triangular, method. (Fig. c), 4.2.4 SUBTRACTION OF VECTORS, , 4.2.3 ADDITION OF VECTORS, (Geometrical method), (1) Triangular law of addition method (HT ), (2) Parallelogram law of addition method (TT), (1)Triangular Law of addition method, The, tail, of, the, , Fig a: Two vectors A and B , – B is also shown., Fig, b:, , second vector B is connected to the head of the, first vector A . Then the resultant is the vector, connecting the tail of the first vector A to the, head of the second vector B., , Subtracting vector B from vector A, (R 2= A-B) . For comparison, addition of, vectors A and B ( R 1= A+B) is also shown., NOTE, A+B=B+A ( Commutative), (A+B)+C=A+(B+C) (Associative), (2) Parallelogram law of addition method, , Magnitude and direction of the resultant of, two vectors A and B in terms of their, magnitudes and angle θ between them, , KAMIL KATIL VEETIL, 18, , @, , SOHSS AREEKODE

Page 21 :
In 2D , a vector A can be represented as, ⃗, A = A x ^i+ A y ^j, , Magnitude, From the figure For ΔOBN, we have, AN, B, BN, sin θ=, B, cos θ=, , =>, , AN =B cos θ, , =>, , BN =B sin θ, , From the diagram cos θ=, , OB 2=ON 2+ BN 2, , Ax, A, , ,, , sin θ=, , Ay, A, , --------> x component of A, --------> y component of A, , A x =A cosθ, A y = A sinθ, , where ‘A’ is the magnitude (length) of the, vector A and θ is the angle between A and the, x component of A . Also we will get, A= √ A 2x + A 2y, , This is law of cosines, tan θ=, , Direction, If R makes an angle α with A , then in, ΔOBN,, , Ay, Ax, , ,, , θ=tan, , −1, , Ay, Ax, , Thus the vector A is resolved into two, perpendicular components Ax and Ay along x, axis and y axis respectively., Problem1: Resolve the given force vector, along x and y direction and write the vector in, analytical form, , Do Example 4.3 NCERT, Analytical representation of vectors, (Resolution of vectors), In 3D a vector can be represented as, , Fy, , F=36N, , =300, Fx, , ⃗, A= A x ^i + A y ^j+ A z k^, , Soln ) Fx= F cosθ, =36 cos30 = 18√3, Fy= F sinθ, = 36 sin30 = 18, So the vector can be analytically represent as, F = 18√3 i + 18 j, , KAMIL KATIL VEETIL, 19, , @, , SOHSS AREEKODE

Page 22 :
4.3 VECTOR ADDITION – ANALYTICAL, METHOD, Let ⃗, A= A X ^i + A y ^j+ A z k^ and, , The position vector of r2 = x2 i +y2 j, Then the displacement vector , ∆r = r2 - r1, , = (x2 – x1) i + ( y2 –y1) j, , ⃗, B =B X ^i+ B y ^j+B z k^, R= ⃗, A+⃗, B =>, Then ⃗, , => ∆r = ∆x i + ∆ y j, 4.4 MOTION IN A PLANE, , ⃗, R=( A x + B x ) ^i +( A y + B y ) ^j+( A z +B z ) k^, , POSITION VECTOR, It is a vector which, denotes the position of, a particle at any instant, of time with respect to, some reference point., The position vector of, r is given by ⃗r =x ^i+ y ^j+ z k^, problem2:, Determine the, position, vectors for the, following, particles which, are located at, points P, Q, R,, S., Soln) rp= 3 i, rQ= 5 i + 4 j, rR= -2 i, rS= 3 i - 6 j, , VECTOR, Position vector ( r ), , EXPRESSION, r= x i + y j, , Displacement vector (dr) dr = dx i + dy j, Velocity vector (v), v=, , dr, dt, , Acceleration vector, dv, dt, , a=, , DISPLACEMENT VECTOR, , dr d, = ( x i+ y j), dt dt, dx dy, i+, j, =, dt dt, v = v x i+v y j, dv, dt, d, ( v i+ v y j), =, dt x, a= a x i+a y j, , a=, , 4.5 KINEMATIC EQUATIONS FOR, UNIFORM ACCELERATION FOR, MOTION IN A PLANE, Along x direction, V x =ux +a x t, , .....................(1), , 1, 2, S x =x=u x t+ ax t ..................(2), 2, , The position vector of r1 = x1 i +y1 j, , V 2x =u2x +2 a x S x ...................(3), , KAMIL KATIL VEETIL, 20, , @, , SOHSS AREEKODE

Page 23 :
Along y direction, , V x =ucos θ+0 x t, , 1, 2, S y = y =u y t+ a y t, 2, 2, y, , 2, y, , V =u + 2a y S y, , ..............(9), , V x =u cosθ, , ..................(4), , V y =u y + a y t, , Similarly V y =u y + a y t, V y =u sin θ−g t ............(10), , ................(5), ..................(6), , 4.6 PROJECTILE MOTION, When an object is thrown in the air with, some initial velocity (NOT just upwards), and, then allowed to move under the action of, gravity alone, the object is known as a, projectile., The path followed by the particle is called its, trajectory ., Motion of a projectile is an example for motion, in a plane with constant acceleration ., Fig. shows Motion of an object projected with, velocity u at angle θ ., , NOTE:, 1) x-component of velocity remains constant, throughout the motion , only the y- component, of velocity changes, 2) At maximum height vy =0, 3) The velocity at any instant is given by the, relation V = √V 2x +V 2y, 4) At maximum height V=vx=ucosθ since, vy=0 at maximum height, 4.6.1 Equation of path of a projectile, From (8) x=u x t=ucosθ t, x, .............(11), u cos θ, 1, 2, y=u y t + a y t, 2, , =>, , t=, , From (5), , For projectile u y =u sinθ and a=−g, Therefore, , 1 2, y=usin θ t− g t ..............(12), 2, , Substitute (11) in (12), =>, After the object has been projected, the, acceleration acting on it is acceleration due to, gravity alone , which is directed vertically, downward., Ie, a x =0 , a y =−g ..............(7), The components of initial velocity are, u x =u cosθ, u y =u sinθ ...........(8), , usinθ x, g x2, −, ucos θ 2cos 2 θ, , y=tanθ x−, , g, x2, 2, 2 cos θ, , This is of the form y = a x + b x 2 , in which a, and b are constants. This is the equation of a, parabola, i.e. the path of the projectile is a, parabola., 4.6.2 Time of maximum height (tm ), We know V y =u sin θ−g t, (From eqn 10), At maximum height V y =0, Therefore ,, (10) => 0=u sinθ−g t m, , The components of velocity at any time t can be, obtained using Eq (1) , (4) , (7) and (8), ie,, , =>, , y=, , V x =ux +a x t, , KAMIL KATIL VEETIL, 21, , @, , SOHSS AREEKODE

Page 24 :
=>, , g t m =u sinθ, , =>, , t m=, , ie, Rmax=, , u sinθ, g, , ................(13), , Do Example 4.7 ,4.8 NCERT, 4.7 UNIFORM CIRCULAR MOTION, When an object follows a circular path at a, constant speed, the motion of the object is, called uniform circular motion., , 4.6.3 Time of flight (TF ), Th e total time taken by the projectile, from the point of projection till it hits the, horizontal plane is called time of flight., TF = 2 tm, T F=, , 2 usinθ, g, , u2, g, , 4.7.1 Some basic terms, 1. Angular displacement (θ), The angle swept over by the radius vector in a, given interval of time is called angular, displacement (θ)., , ..............(14), , 4.6.3 Maximum height of a projectile (hm), The maximum vertical distance travelled, by the projectile during its journey is called, maximum height., We know V 2y =u 2y + 2a y S y, ( From (6)), When the projectile is at maximum height, , ,, , S y =hm, , V y =0, , Therefore the above equation becomes, , =>, =>, , 0=(usinθ )2−2 g h m, , Unit : radian, , 2 g h m=u2 sin2 θ, , 2) Angular velocity (ω), Angular velocity is the rate of change of, angular displacement. It is a vector quantity., , 2, , hm =, , 2, , u sin θ, 2g, , ................(15), , Angular velocity , ω =, , =>, , =>, , 4) Frequency (f), Number of revolutions per second, , 2 usinθ, g, 2, u 2sin θ cos θ, R=, g, , R=(u cosθ), , R=, , dθ, dt, , 3) Time period (T), Time taken for one complete revolution, , 4.6.4 Horizontal range of a projectile (R), It is the horizontal distance travelled by a, projectile during its time of flight., R=V x X T F, Therefore, =>, , Dimension :No dimension, , 1, T, , f=, , Also ω=, , dθ, 2π, =, dt, T, , Relation between linear velocity and angular, velocity, , u2, sin 2 θ, g, , Velocity=, , NOTE : Horizontal range is maximum when, 2θ=90o . ie, θ=45o for a particular 'u', , ie,, , KAMIL KATIL VEETIL, 22, , @, , ds, =, dt, , d, dθ, =, (r θ) =r, dt, dt, v =r ω, , SOHSS AREEKODE, , rω

Page 25 :
Angular acceleration (α), It is the rate of change of angular velocity, , α=, , dω, d v, 1 dv, =, =, ( ) =, dt, dt r, r dt, , ie,, , 2, Fc= m v = m ω2 r, , ie,, , r, , The centripetal force is always towards the, centre of the circle., 4.8 RELATIVE VELOCITY, , a, r, , a=r α, , Do Problem 4.8 NCERT, Add.Problem 1, A stone of mass 2 kg tied to the end of a, string 80 cm long is whirled in a horizontal, circle with a constant speed . If the stone makes, 14 revolutions in 25 seconds., (a) What is the angular velocity of the sone, (b) Centripetal acceleration, (c) Centripetal force, [Ans a) 3.52 rad/ s b) 9.9 m/s2 c) 19.8 N ], Add.Prob 2, Calculate the angular velocity of, (a) Second hand, (b) Minute hand and, (c) Hour hand of a clock, , 4.7.2 centripetal acceleration, Consider a particle executing uniform, circular motion with speed 'v' around a circle of, radius 'r'., Let the directions of position and velocity, vectors shift through the same angle θ in a, small interval of time ' Δt ' as shown in Figure., For uniform circular motion ,, , r = |r⃗1| = |r⃗2|, , and, , v= |v⃗1| = |v⃗2|, , Previous Questions, 1a) Find whether the given vectors 2 i +3 j + 4, k and 4 i + 6 j+ 8 k are parallel or not., b) What are orthogonal unit vectors?, , The magnitudes of the displacement Δr and, of Δv satisfy the following relation., Δv, v, , Δr, r, , | |=| |, =>, , Δv, , 2a) Obtain expression for Time of flight, for a projectile motion., b) What is the angle of projection for, maximum horizontal range?, c) The ceiling of a long hall is 25m, high. What is the maximum horizontal, distance that the ball thrown with a speed, of 40 m/s can go without hitting the ceiling, of the hall?, , = θ, , = v Δr, , r, 2, Δv, v Δr, a =, =, = v, Δt, r Δt, r, 2, v, a =, ie,, r, v =r ω, Also, ω2 r 2, a =, Therefore,, = ω2 r, r, , 3 . A stone is thrown upward from a moving, train., a) Name the path followed by the stone., b) ) A man throws a stone up into air at an, angle ' θ ' with the horizontal. Draw the, path of the projectile and mark directions, of velocity and acceleration at the highest, position, , 4.7.3 Centripetal force (Fc), The centripetal force is given by, F c=ma =, , mv 2, = m ω2 r, r, , KAMIL KATIL VEETIL, 23, , @, , SOHSS AREEKODE

Page 26 :
equal to three scalar equations., , CHAPTER FIVE, LAWS OF MOTION, 5.1 INTRODUCTION, Is an external force required to keep a, body in uniform motion?, Aristotle’s fallacy :An external force is required, to keep a body in motion., Galileo arrived at a new insight on motion, from his inclined plane experiments which, corrected Aristotle's idea., Inertia: Inertia is the inability of an object to, change its velocity by itself. This includes changes, to the object's speed, or direction of motion., , 5.2 NEWTON'S LAWS OF MOTION, 5.2.1 Newton’s First Law ( Law of inertia), Every body continues to be in the state of rest, or in the state of uniform motion (constant, velocity) unless there is external force acting, on it., , Problem : example 5.1 NCERT, Momentum (p) :Momentum of a body is the, product of its mass m and velocity, v, P=mv, 5.2.2 Newton’s second Law, The force acting on an object is equal to, the rate of change of its momentum, ═>, , Unit of Force ═> newton, Thus One Newton is defined as the force which, acts on 1 kg of mass to give an acceleration, 1 ms−2 in the direction of the force., NOTE, Newton’s laws are vector laws. The equation, F= ma is a vector equation and essentially it is, , By comparing both sides, the three scalar, equations are, Fx= max . The acceleration along the x direction, depends only on the component of force acting, along the x-direction. And, Fy= may .Th e acceleration along the y, direction depends only on the component, of force acting along the y-direction., , FZ= maZ, From the above equations, we can infer that the, force acting along y direction cannot alter the, acceleration along x direction. In the same way,, F z cannot affect a y and a x ., Problem : example 5.2 NCERT, 5.2.3 Newton’s third Law, Newton’s third law states that for every action, there is an equal and opposite reaction., Action and reaction pair of forces do not act on, the same body but on two different bodies. So, they don't cancel each other. Any one of the, forces can be called as an action force and, the other the reaction force. These actionreaction forces are not cause and effect forces., 5.3 IMPULSIVE FORCE, The force which acts for a very short, interval of time is called impulsive force., Eg: - (i) The force on a ball when hit with a bat., (ii) Force exerted on a bullet when fired from a, gun., 5.4 IMPULSE (I), Impulse of a force is the product of the, force and the time for which the force acts on, the body. Impulse is a measure of the total, effect of force., i.e, Impulse, I = force x time = Change in, momentum, i.e, I = F dt = dp where dt is the time for which, the force F acts., Problem : example 5.4 NCERT, Problem:, Show that Impulse= Change in momentum, using Newton's second law., Soln) According to second law, , KAMIL KATIL VEETIL, 24, , @, , SOHSS AREEKODE

Page 27 :
NOTE, The graphical representation of constant, force impulse and variable force impulse ( Area, under the graph give impulse), Soln) (a) (i)Downward gravitational force along, negative y direction (mg) ii) Tension (T) along, the two strings, (b) since the boy is in, equilibrium (rest),, 2T cosθ = mg, Problem:An object of mass 10 kg moving with, a speed of 15 m s −1 hits the wall and comes to, rest within a) 0.03 second, b) 10 second ., Calculate the impulse and average force acting, on the object in both the cases., [ Ans : a) Impulse= -150 N s Fav=-5000 N, b) Impulse= -150 N s Fav=-15 N ], 5.5 EQUILIBRIUM OF FORCES, , when the net external force on the, particle is zero , then the body is in equilibrium., ie, Even under action of forces the object will, be at rest . If the lines of forces are acting at a, common point , then the collection of forces is, said to be concurrent forces ( Fig Below), Prob: A baby is playing in a swing which is, hanging with the help of two identical chains is, at rest.(a) Identify the forces acting on the, baby. (b)find out the tension acting on the, chain, , => T= mg/ 2cosθ, NOTE, When θ = 0 °, the strings are, vertical and the tension on each string is, T = mg/2, 5.6 CONSERVATION OF MOMENTUM, If there are no external forces acting on, the system, then the total linear momentum of, the system ( p tot ) is always a constant vector., In other words, the total linear momentum of, the system is conserved in time. Here the word, ‘conserve’ means that p 1 and p 2 can vary, in, such a way that p 1 +p 2 is a constant vector., When two particles interact with each other,, they exert equal and opposite forces on each, other., Let F 21 → Force acting on 2 by 1, F 12 →Force acting on 1 by 2, Then by Newton's third law, ⃗, F21=−⃗, F 12 ......................(1), In terms of momentum of particles, the, force on each particle (Newton’s second law), can be written as, ⃗, F12=, , (1)=>, =>, =>, , KAMIL KATIL VEETIL, 25, , @, , d⃗, P1, dt, , and, , ⃗, F21=, , d⃗, P2, ..............(2), dt, , d⃗, P1 −d ⃗, P2, =, dt, dt, d⃗, P1 d ⃗, P2, +, =0, dt, dt, d ⃗ ⃗, ( P + P )=0, dt 1 2, , SOHSS AREEKODE

Page 28 :
ie,, p 1 + p2 = constant vector, Example 1 : Recoil of gun., , (always)., , System = Gun and bullet, M – mass of Gun, m – mass of bullet, U – inial velocity of gun, u– inial velocity of, bullet, Let momentum of the gun before, firing = p1=MU=0, momentum of the bullet before, firing = p2=mu=0, Total momentum of the system before, firing =p1 + p2 = 0, After firing bullet moves with a velocity, v, forward. According to the law of conservation, of linear momentum, total linear momentum, has to be zero after the firing also., Ie,, p1I + p2I = 0, => p1! + mv =0, => p1I = -mv, is the recoil momentum, , => MV= -mv, => V= -mv/M, , is the recoil velocity., The - ve sign shows that the gun is recoiling., Example 2 : Rocket Propulsion, In case of rocket, a fuel burnt in the, combustion chamber produces hot gas, which is, allowed to escape through a nozzle at the back, of the rocket. This produces a backward, momentum on the gas and the rocket acquires, an equal forward momentum. Thus the rocket, moves forward, Problem : A shell of mass 0.020 kg is fired by, a gun of mass 100 kg. If the muzzle speed of, the shell is 80 m/s, what is the recoil speed of, the gun?, , 1) Static friction, 2) Kinetic friction., 5.7.1 Static friction (fs), Static friction is the force which opposes, the initiation of motion of an object on the, surface., If some external force F is applied on an, object parallel to the surface on which the, object is at rest, the surface exerts exactly an, equal and opposite force on the object to resist, its motion and tries to keep the object at rest ., But if the external force is increased, after a, particular limit, the surface cannot provide, sufficient opposing frictional force to balance, the external force on the object, then the object, starts to slide. The maximum value of frictional, force before the body just slides over the, surface of another body is called limiting, friction or maximum static friction (fsmax)., Experimentally, it is found that the, magnitude of the maximum static friction, fsmax α Normal force, N, OR, where µs- coefficient of static friction, NOTE: Law of static friction, The static friction does not depend upon the, area of contact. And, 5.7.2 Kinetic Friction, When an object slides, the surface exerts, a frictional force called kinetic friction fk (also, called sliding friction or dynamic friction)., Experimentally it is found that, NOTE: Since μk< μs , starting of a motion is, more difficult than maintaining it., , [Ans = 0.016 m/s], 5.7 FRICTION, Frictional force is the force which, always opposes the relative motion between an, object and the surface where it is placed., Frictional force always acts on the object, parallel to the surface on which the object is, placed. There are two kinds of friction namely, , When, , KAMIL KATIL VEETIL, 26, , @, , relative, , motion, , has, , SOHSS AREEKODE, , begun,, , the

Page 29 :
acceleration of the body according to the, second law is ( F – fk )/m. For a body moving, with constant velocity, F = fk .If the applied, force on the body is removed, its acceleration is, – fk /m and it eventually comes to a stop., 5.7.3 Angle of Friction OR Angle of repose, Consider an inclined plane on which an object, is placed, as, shown in, Figure, Let the angle, which this, plane makes, with the, horizontal be θ. For small angles of θ , the, object may not slide down. As θ is increased,, for a particular value of θ , the object begins to, slide down. This value is called angle of repose, or angle of friction., Expression, From the above Fig. Since the body is in, equilibrium, ...............(1), ..............(2), ..................... (3), Equating the right hand side of equations, (2) and (3), we get, ie,, 5.7.4 Rolling Friction, In rolling motion when a wheel moves, on a surface, the point of contact with surface is, always at rest. Since, the point of contact is, at rest, there is no, relative, motion, between the wheel, and surface. Hence, the frictional force is, very less., However , Due to the elastic nature of, the surface at the point of contact there will be, some deformation on the object at the point on, , the wheel or surface. Due to this deformation,, there will be minimal friction between wheel, and surface. It is called ‘rolling friction’. In, fact, ‘rolling friction’ is much smaller than, kinetic friction., Do Examples 5.7,5.8,5.9 NCERT, 5.8 APPARENT WEIGHT (Motion of a body, in a lift), , Case (i) : When the lift is stationary or moving, up or down with a uniform velocity (a=0), Then Force, R - mg = 0, => R = mg (real weight), Case (ii) : When the lift moves up with an, acceleration, a, R - mg = ma =>R = mg +ma, => R = m(g +a), Case (iii) When the lift moves down with an, acceleration , a, R - mg = - ma =>R = mg - ma, => R = m(g – a), Case (iv) When the lift falls down freely., R=m (g-g)=0, 5.9 DYNAMICS OF CIRCULAR MOTION, 5.9.1 Centripetal Force, The centripetal acceleration of a particle, in the circular motion is given by ,, From Newton's second law the, centripetal force is given by, Note that , to execute, circular motion centripetal force is essential., The origin of the centripetal force can be, gravitational force, tension in the string,, frictional force, Coulomb force etc. Any of, these forces can act as a centripetal force., 1)In the case of whirling motion of a stone tied, , KAMIL KATIL VEETIL, 27, , @, , SOHSS AREEKODE

Page 30 :
to a string, the centripetal force on the particle, is provided by the tensional force on the string., 2) In motion of satellites around the Earth,the, centripetal force is given by Earth’s, gravitational force on the satellites., 3)When a car is moving on a circular track the, centripetal force is given by the frictional force, between the road and the tyres., 5.9.2 Motion of a car on a level road, , Fig. Forces acting on the vehicle on a leveled, circular road, Th ere are three forces acting on the vehicle, when it moves as shown in the Figure, 1.Gravitational force (mg) acting downwards, 2.Normal force (mg) acting upwards, 3. Frictional force (fs ) acting horizontally, inwards along the road, As there is no acceleration in the vertical, direction, N = mg ..............(1), , Consider a vehicle of mass m moving on, a banked curve. Then various forces acting on, the car are, 1) The gravitational force (weight) mg, downward, 2) Normal force N normal to the road, 3) Frictional force f s , acting parallel to the, road., Since there is no acceleration along the vertical, direction, the net force along this direction must, be zero. Hence,, N cos θ = mg + f sin θ ..........(1), The centripetal force is provided by the, horizontal components of N and f., ..............(2), , Then Eqn (1) and (2) become, , The centripetal force is provided by the, force of static friction fs between the tyre and, surface of the road. ie, Static friction opposes, the impending motion of the car moving away, from the circle., Ie,, , ................(3), , ..............(4), From (3), ..................., , (5), , Substituting value of N in Eq. (4), we get, This shows that for a given value of μ s and R,, there is a maximum speed of circular motion of, the car possible, namely, ..................(6), maximum possible speed of a car on a banked, road is greater than that on a flat road., , Note: vmax is independent of mass of the car., 5.9.3Motion of a car on a banked road, , KAMIL KATIL VEETIL, 28, , @, , SOHSS AREEKODE

Page 31 :
For μs = 0 , v0= (R g tanθ )1/2, At this speed, frictional force is not needed at, all to provide the necessary centripetal force., Driving at this speed on a banked road will, cause little wear and tear of the tyres., Do Example 5.10 NCERT, Problem 1 : Consider a circular levelled road of, radius 10 m having coefficient of static friction, 0.81. Three cars (A, B and C) are travelling with, speed 7 m/s , 8 m/s and 10 m/s respectively., Which car will skid when it moves in the, circular level road?(g =10 m/s2), Ans : car C, , KAMIL KATIL VEETIL, 29, , @, , SOHSS AREEKODE

Page 32 :
CHAPTER SIX, , soln), , WORK, ENERGY AND POWER, 6.1 INTRODUCTION, In physics work means mechanical work., Work is said to be done by a force, when the, force applied on a body displaces it., work is also defined as ‘the product of, displacement and the component of force in the, direction of displacement., Energy is the capacity to do work. So it, has the same unit and dimension of work., Power is the rate of change of work done., 6.2 SCALAR PRODUCT ( DOT PRODUCT), If there are two vectors A and B having, an angle θ between them, then their scalar, product is defined as A ⋅ B = AB cosθ . Here, A, and B are magnitudes of A and B ., Properties, 1.The product quantity A ⋅ B is always a scalar., 2.The scalar product is commutative,, ie, A ⋅ B= B . A, 3.The vectors obey distributive law . ie,, A . (B + C) = (A . B ) + (A . C), 3.The angle between the vectors, , NOTE: Geometrical interpretation of dot product, , (a) The scalar product of two vectors A and B is a, scalar : A. B = A B cos θ (b) B cos θ is the, projection of B onto A (c) A cos θ is the projection, of A onto B., ie. A.B= AB cos θ => magnitude of A x projection, of B onto A, B.A = BA cosθ => magnitude of B x projection of, A onto B, , 4. A . B= AB , if θ= 0o, A . B= 0 , if θ= 90o, A . B= -AB , if θ= 180o, 5. For unit vectors, , 6. In terms of components the scalar product of, A and B can be written as, , 7. The magnitude of vector A is given by, , A . A = Ax Ax + Ay Ay + Az Az = AA=A2, ie ,, problem1, , 6.2 WORK, The work done by the force is defined to, be the product of component of the force in the, direction of the displacement and the magnitude, of this displacement., ie, W= Fcosθ x d, NOTE, Work done is zero in the following cases., 1) When the force is zero (F = 0)., 2) When the displacement is zero (d = 0)., Eg : when force is applied on a rigid wall it, does not produce any displacement., 3)When the force and displacement are, , KAMIL KATIL VEETIL, 30, , @, , SOHSS AREEKODE

Page 33 :
perpendicular (θ = 90 o ) to each other., Eg:when a body moves on a, horizontal direction, the, gravitational force (mg) does no, work on the body, since it acts at, right angles to the displacement ., In circular motion the centripetal, force does not do work on the, object moving on a circle as it is, always perpendicular to the, displacement., Problem 2:A box is pulled with a force of, 25 N to produce a displacement of 15 m. If the, angle between the force and displacement, is 30 o , find the work done by the force., [Ans : 324.76 J], , NOTE : The work-kinetic energy theorem, implies the following., 1) If the work done by the force on the body is, positive then its kinetic energy increases., 2) If the work done by the force on the body is, negative then its kinetic energy decreases., 3) If there is no work done by the force on the, body then there is no change in its kinetic, energy, which means that the body has moved, at constant speed provided its mass remains, constant., Relation between Momentum and Kinetic, Energy, OR, Do Example 6.2 , 6.3 NCERT, , HW 6.4, , 6.5WORK DONE BY A VARIABLE FORCE, Problem 3 : Find the angle between force, F=( 3i+4j-5k ) unit and displacement d=, ( 5i+4j+3k ) unit, [ Ans : cos-10.32], , A plot of varying force, in one dimension shown, in figure., , 6.3 KINETIC ENERGY, Kinetic energy is the energy possessed, by a body by virtue of its motion. All moving, objects have kinetic energy., , If the displacement Δx is small , we can take, F(x) is approximately constant and work done, is then, , 6.4 WORK–KINETIC ENERGY, THEOREM (For Constant Force), For constant acceleration ( constant force) , we, know, .............. (1), , This is, illustrated in this, Fig ., , where u and v are the initial and final speeds, and s the distance traversed. Multiplying both, sides by m/2, we have, , Adding successive rectangular areas in Fig. we, get the total work done as, , .............(2), ie, Final Kinetic Energy- Initial Kinetic, Energy= Work done, , where the summation is from the initial position, x i to the final position x f ., , Kf – Ki =ΔK =W, ie, The change in kinetic energy of a particle is, equal to the work done on it by the force. This, is called work-kinetic energy theorem., , If the displacements are allowed to approach, zero, then the number of terms in the sum, increases without limit, but the sum approaches, a definite value equal to the area under the, , KAMIL KATIL VEETIL, 31, , @, , SOHSS AREEKODE

Page 34 :
curve., Then the work done is, , The gravitational potential energy (U) at some, height h is equal to the amount of work required, to take the object from ground to that height h, with constant velocity., The gravitational potential energy, (U) at some height h is equal to the, amount of work required to take the, object from the ground to that height, h at constant velocity., Fg = -mg j (downward), Fa = mg j ( upward), d= h j (upward ), , = Area under the curve, Do example 6.5 NCERT, Problem4:A variable force F = k x 2 acts on a, particle which is initially at rest. Calculate the, work done by the force during the, displacement of the particle from x= 0 m to x, =4 m. (Assume the constant k= 1 N m -2, , 6.5.1 Work Energy theorem for variable, force( see last page) *, 6.6 POTENTIAL ENERGY, The energy possessed by a body by, virtue of its position( in a field ) or, configuration ( in a state of strain) is called, potential energy., Potential energy of an object at a point P, is defined as the amount of work done by an, external force in moving the object at constant, velocity from the point O (initial location) to, the point P (final location). At initial point O, potential energy can be taken as zero., We have various types of potential energies., 1. Gravitational potential energy: The energy, possessed by the body due to gravitational force, gives rise to gravitational potential energy., 2. Elastic potential energy :The energy due to, spring force and other similar forces give rise to, elastic potential energy., 3. Electrostatic potential energy: The energy, due to electrostatic force on charges gives rise, to electrostatic potential energy.( Next year), 1.Gravitational Potential energy near the, surface of the Earth, , U= Fa .d, = mg j. h j = mgh cos0 = mgh, ie, U = mgh, 2. ELASTIC POTENTIAL ENERGY, ( POTENTIAL ENERGY OF A SPRING), When a spring is elongated, it develops a, restoring force. The potential energy possessed, by a spring due to a deforming force which, stretches or compresses the spring is termed as, elastic potential energy. The work done by the, applied force against the restoring force of the, spring is stored as the elastic potential energy in, the spring., , Fs =- Kx, Fa = Kx, , Force-displacement graph for a spring, The elastic potential, energy can be easily, calculated by drawing a, F - x graph. The shaded, area (triangle) is the work, done by the spring, force., , KAMIL KATIL VEETIL, 32, , @, , SOHSS AREEKODE

Page 35 :
6.7 CONSERVATIVE FORCE, A force is said to be a conservative force, if the work done by the force in moving the, body depends only on the initial and final, positions of the body and not on the nature of, the path followed between the initial and final, positions., There will be a potential energy, associate with a conservative force., Total mechanical energy remains, constant in a conservative force field., Eg :Elastic spring force, Electrostatic, force, Magnetic force, Gravitational force, etc., 6.8 NON CONSERVATIVE FORCE, A force is said to be non-conservative if, the work done by or against the force in moving, a body depends upon the path between the, initial and final positions., Eg: Frictional forces, The force due to, air resistance, viscous force , etc., 6.9 THE CONSERVATION OF, MECHANICAL ENERGY (IN CASE OF A, FREELY FALLING BODY), U=mgh, K=0,, E=U+K=mgh, , reaction, heat is released and the reaction is said to, be an exothermic reaction. If the reverse is happens,, heat is absorbed and the reaction is endothermic., 3.Electrical Energy: Energy is associated with an, electric current, 4.The Equivalence of Mass and Energy (Nuclear, energy):, Albert Einstein, E=mc2, The energy released from the nuclear reactions,, either fission or fusion, is called as nuclear energy., Nuclear fusion and fission are manifestations of the, equivalence of mass and energy., , 6.11 POWER, Power is a measure of how fast or slow a, work is done. Power is defined as the rate of, work done or energy delivered., , Unit: watt, 1 W = 1 J /s, 1 hp = 746 w, 1 kW = 1000 W, Note: kWh is the unit of energy not of power, 1 kWh= 3.6×10 6 J, Average power, Instantaneous power, , U=mg(h-y) ,, =1/2 m(2gy)= mgy, E=U+K= mgh, , 6.11.1 Relation between Power and velocity, , The work dW done by a force F for a, displacement dr is dW = F.dr, , U=0, K= 1/2 mv2, =1/2 m(2gh)=mgh, E= U+K = mgh, , Therefore, , [NOTE : In case of spring , the conservation of, mechanical energy will be discussed in the, chapter 14 , oscillations], 6.10 VARIOUS FORMS OF ENERGY, 1. Heat: The work done by friction is not ‘lost’,, but is transferred as heat energy., 2.Chemical Energy: Chemical energy arises from, the molecules participating in the chemical reaction, have different binding energies. If the total energy, of the reactants is more than the products of the, , Do Example 6.11 NCERT, 6.12 COLLISIONS, A collision is said to have taken place if two, moving objects strike each other or come close, to each other such that the motion of one of, them or both of them changes suddenly. Total, momentum will be conserved in all types of, collision., , KAMIL KATIL VEETIL, 33, , @, , SOHSS AREEKODE

Page 36 :
Types of collisions, (1) Elastic collision, (2) Inelastic collision, (1) Elastic collision, Elastic collision is one in which both, momentum and kinetic energy is conserved., Total kinetic energy before collision = Total, kinetic energy after collision, (2) Inelastic collision, A collision in which total kinetic energy is not, conserved. ( Note:Total energy is conserved in, all types of collision), Total kinetic energy before collision ≠, Total kinetic energy after collision, completely inelastic collision., A collision in which the two particles move, together after the collision is called a, completely inelastic collision., 6.12.1 Elastic collisions in one dimension, Two elastic bodies of masses m1 and m2, moving in a straight line (along positive x, direction) on a frictionless horizontal surface as, shown in Figure., , ie,, , KE i = KE f, , =>, , └ ---------(4), (4)/(3)=>, , =>, =>, This means that for any elastic head on, collision, the relative speed of the two bodies, after the collision has the same magnitude as, before collision but in opposite direction., To find the final velocities v1 and v2, From the above equation, .................... (5), , In order to have collision, u1 >, , u2, , For elastic collision, the total linear momentum, and kinetic energies of the two bodies before, and after collision must remain the same., , ..................... (6), Substitute (6) in (3) =>, , Total linear momentum before collision,, .................(1), Total linear momentum after collision,, ..................(2), From the law of conservation of linear, momentum, pi = pf, , .............(3), , └---------> (7), Substitute (7) in (6) =>, , since the collision is elastic,Total kinetic energy, also will be conserved, , ..............(8), , KAMIL KATIL VEETIL, 34, , @, , SOHSS AREEKODE

Page 38 :
CHAPTER SEVEN, , SYSTEM OF PARTICLES AND, ROTATIONAL MOTION, Rigid Body, A rigid body is a body with a perfectly, definite and unchanging shape. The distance, between any two points in the body always, remains same even under the action of, external forces., What kind of motion can a rigid body, have?, Pure translation: In pure translational motion, at any instant of time every particle of the, body has the same velocity. ( Fig (a),(b)), , Fig(a):Translational (sliding) motion of a block down, an inclined plane. (Any point like P 1 or P 2 of the, block moves with the same velocity at any instant of, time.), , Fig(d): Motion of a rigid body which is a combination, of translation and rotation., , Rotation about a fixed axis, In rotation of a rigid body about a fixed, axis,every particle of the body moves in a, circle, which lies in a plane perpendicular to, the axis and has its centre on the axis., , Fig : Rotation about a fixed axis : A ceiling fan and, a potter’s wheel ., , Rotation(Axis is not fixed,but a point on, the axis is fixed), , Fig(b): Motion of a rigid body which is pure, translation., , Translation plus Rotation, A system is said to be in rotational plus, translational motion, when the particle is, rotating with some angular velocity about a, movable axis. (Fig c ,d), Fig(c): Rolling, motion of a, cylinder. It is not, pure translational, motion. Points P 1 ,, P 2 , P 3 and P 4 have, different velocities, (shown by arrows), at any instant of time.In fact, the velocity of the point, of contact P 3 is zero at any instant, if the cylinder, rolls without slipping., , Fig:A spinning top (The point of contact of the top, with the ground, its tip O, is fixed.) and An oscillating, table fan with rotating blades. The pivot of the fan,, point O, is fixed. The blades of the fan are under, rotational motion, whereas, the axis of rotation of the, fan blades is oscillating., , Note, The motion of a rigid body which is not, pivoted or fixed in some way is either a pure, translation or a combination of translation and, rotation. The motion of a rigid body which is, pivoted or fixed in some way is rotation., , KAMIL KATIL VEETIL, , 36, , @, , SOHSS AREEKODE

Page 39 :
Centre Of Mass, The centre of mass of a system is the point, where all the mass of the system may be, assumed to be concentrated and where the, resultant of all the external forces acts., Centre of mass of a two particle system, Consider a system of two particles with, masses m1 and m2 and their position vectors, be r1 and r2 respectively from some arbitrary, origin O., , Z=, , m 1 z 1 +m2 z 2 +m3 z3 +........, m 1+ m2 +m3 +........, , (* Do Example 7.1 ,7.2,7.3 NCERT), Motion of centre of Mass, From (1) ,, =>, , m1 r 1+m2 r 2 +m3 r 3 +........, m1 +m2 +m3 +........, MR=m1 r 1 +m2 r 2+m3 r 3 +........, R=, , Differentiating wrto time ,, =>, , dr, dr, dr, dR, =m1 1 + m 2 2 +............. mn n, dt, dt, dt, dt, MV =m1 v 1 +m2 v 2+ ...........mn v n, , M, , =>, , Again differentiating wrto time,, =>, , dv, dv, dv, dV, =m1 1 +m2 2 +............. mn n, dt, dt, dt, dt, MA=m1 a1 +m2 a 2+........... m n an, , =>, , MA=F 1+ F 2 +........... F n, , =>, The centre of mass will be at a point C whose, position vector is given by, R=, , m 1 r 1 +m 2 r 2, m1 +m 2, , If the particles have the same mass, m1 = m2 =, m , then,, R=, , m r 1+ mr 2, m+ m, , =>, , R=, , r 1 +r 2, 2, , Thus, , if the particles are having same mass, the, centre of mass lies exactly midway between, them., Centre of mass of an N particle system, Consider a system of N particles of masses, m 1 , m 2 , m 3 ........ having position vectors, r 1 , r 2 , r 3 ,.........., The total mass of the system, M = m 1 + m 2 +, m 3 +........., m1 r 1+ m2 r 2 +m3 r 3 +........, ...........(1), m1 +m2 +m3 +........, mr, ∑ mi ri, =>, R=, R=∑ i i, mi, M, , Then R=, or, , In terms of co ordinates (the co ordinates of, the centre of mass),, X=, , m1 x 1 +m2 x 2 +m3 x 3 +........, m1 +m2+m3 +........, , Y=, , m1 y 1 +m2 y 2+ m3 y 3 +........, m 1+m2 +m3 +........, , M, , NOTE: The force F1 on the first particle is not, a single force,but the vector sum of all the, forces on the first particle ; likewise for the, second particle etc. Among these forces on, each particle there will be external forces, exerted by bodies outside the system and also, internal forces exerted by the particles on one, another. We know from Newton’s third law, that these internal forces occur in equal and, opposite pairs and in the sum of forces in the, above equation, their contribution is zero., Only the external forces contribute to the, equation. We can then rewrite the above, equation as, MA=F ext ................(2), where A is the acceleration of the centre of, mass and Fext represents the sum of all, external forces acting on the particles of the, system., Eqn (2) states that the centre of mass of, a system of particles moves as if all the mass, of the system is concentrated at the centre of, mass and all the external forces are applied, at that point., , KAMIL KATIL VEETIL, , 37, , @, , SOHSS AREEKODE

Page 40 :
Examples of centre of mass motion, (1) Explosion of a shell in flight., Consider a shell projected upwards. This shell, will follow a parabolic path. Now let the shell, explode during flight. After explosion, the, fragments travel in their own parabolic path., Since the force of explosion are all internal,, the centre of mass of the system will continue, to follow the same parabolic path of the shell, before, explosion., (2) Motion of earth – moon system, Moon moves round the earth in circular orbit,, and earth moves round the sun in an elliptical, orbit. Or we can say that the centre of mass of, the earth – moon system moves in an elliptical, orbit round the sun. Hence the force of, attraction between earth and moon is internal, to earth – moon system while sun’s attraction, on both earth and moon are external., Linear momentum of a system of particles, For a system of n particles, the total linear, momentum of the system is equal to the, vector sum of momentum of all individual, particles. Ie,, P= p1 + p2 +............ pn, , Now from Newton’s second law, F=, , dP, dt, , where F is the total external force., If F=0,, , dP, =0 . Therefore, P will be a, dt, , constant i.e, linear momentum will be, conserved., Examples:1) Decay of nucleus: Consider the decay of a, parent nuclei at rest into two fragments. They, move in opposite directions with different, velocities. Here the momentum is conserved, and the centre of mass of decay product, continuous to be at rest., [Fig.A heavy nucleus, (Ra) splits into a, lighter nucleus (Rn), and an alpha particle, (He). The CM of the, system is in uniform, motion.], , [Fig. The same splitting of the heavy, nucleus(Ra) with the centre of mass at, rest. The two product particles fly back, to back.], , 2) Motion of binary stars: In case of motion of, binary stars, if no external forces act, the, centre of mass moves like a free particle., Centre of mass of a rigid body, The centre of mass of a rigid body is a fixed, point with respect to the body as a whole., Depending on its shape and mass distribution,, the centre of mass of a rigid body may or may, not be a point within the body., Centre of mass of some regular bodies, A uniform rod – at the geometric center., A ring or a uniform disc – at the center., A uniform cylinder – at the center of its axis, of symmetry., A triangle – at the point of intersection of the, medians., ANGULAR VELOCITY AND ITS, RELATION WITH LINEAR VELOCITY, We know the average angular velocity= Δ θ, Δt, dθ, The instantaneous angular velocity ω=, dt, , The magnitude of linear velocity of a particle, moving in a circle relates with angular, velocity,, V =ωr, , In vector form V =ω x r, ANGULAR ACCELERATION AND ITS, RELATION WITH LINEAR, ACCELERATION, The angular acceleration , α =, , KAMIL KATIL VEETIL, , 38, , @, , dω, dt, , SOHSS AREEKODE

Page 41 :
Also, , dV, dω, =r, dt, dt, , •, , (if the particle moves in a constant radius), a=r α, ie,, , •, , *[ See appendix before proceeding further., Cross product of vectors discussed there ], , A pair of equal and opposite forces, with different lines of action is known, as a couple., A couple produces rotation without, translation., , principle of moments, In rotational equilibrium, total anti clockwise, moment = total clockwise moments. This is, called principle of moments., For a lever , the principle of moment, expressed as ,, , TORQUE ( τ ), The torque or moment of force about a point, is the turning effect of force about that point, and is measured as the product of force and, the perpendicular distance between the point, and the line of action of the force., , Load x Load arm=Effort x Effort arm, , ie ,, , Consider a particle P, whose position vector, with respect to origin O is r ., , F1 d 1=F 2 d 2, , note: The ratio, , F1 d2, =, is called the, F2 d1, , mechanical advantage., ANGULAR MOMENTUM OF A, PARTICLE ( L ), It is the rotational analogue of linear, momentum. It could also be referred to as, moment of linear momentum., , Let a force F act on the particle in a direction, making an angle θ with the direction of r ., , ⃗L=⃗r X ⃗, P, , Then the torque about the point O ,, τ= F x ON = F x r sinθ, 'r sinθ ' is called moment arm of the force, ie, τ= F (r sinθ), (or), , Consider a particle of mass 'm' and linear, momentum 'p' at a position r relative to the, origin O .The angular momentum is defined, to be, , ⃗τ =⃗r X ⃗, F, , note, • Torque is the rotational analogue of, force, • The direction of torque is, perpendicular to both r and F, • If Στ =0, then the body is in rotational, equilibrium, • If ΣF=0 along with Στ =0,then the, body is in mechanical equilibrium., , Angular momentum of a rotating body, The sum of the moments of linear momentum, of all the particles of the body about the axis, of rotation is called its angular momentum, about that axis., Consider a body rotating about an axis., This body is made up of a large number of, particles. Let one such particle of mass mi, be situated at a distance r i from the axis., Thenlinear velocity of particle , v i =r i ω, Linear momentum of particle=mi v i=mi r i ω, , Moment of linear momentum of the particle, , KAMIL KATIL VEETIL, , 39, , @, , SOHSS AREEKODE

Page 42 :
about the axis = mi r i ω r i=mi r 2i ω, Therefore Total moment of momentum of the, whole body about the axis = ∑ m i r 2i ω=I ω, ie, Angular momentum,, , L=Iω, , Where, I --> Moment of inertia of the body, 2, and, I =∑ m i r i, ( Moment of inertia will be discussed later in detail), , Relation between angular momentum and, torque, We know L= r x p, differentiate with respect to time, , *The kinetic energy of motion of this particle, is, , d⃗, L d, = ( r⃗ x ⃗, P), dt dt, d r⃗ ⃗, d⃗, P, x P + r⃗ x, =, dt, dt, F, = ⃗v x m ⃗v + r⃗ x ⃗, = 0+ τ, , 1, 1, 2, 2 2, K i= m i v i = m i r i ω, 2, 2, , *Therefore the total kinetic energy of the, body, 1, 2 2, K=∑ K i=∑ m i r i ω, 2, , d⃗, L, = ⃗τ, dt, , ie,, , Thus the time rate of change of angular, momentum of a particle is equal to the torque, acting on it., Torque and angular acceleration, We know, =>, , * Inertia in linear motion is the inability of a, body to change its state of rest or of uniform, motion in a straight line, without the help of, an external force., * A corresponding property of the body in, rotational motion is known as moment of, inertia., * We know that for a body rotating about a, fixed axis, each particle of the body moves in, a circle with linear velocity v given by the, eqn, v=r ω, * For a particle p at a distance ri from the axis,, v i=r i ω, the linear velocity is, , d⃗, L, dt, d⃗, L d, dω, ⃗τ =, = ( I ω)=I, =I α, dt dt, dt, ⃗τ =, , *Note ω is the same for all particles. Hence,, taking ω out of the sum,, 1 2, 1, 2, K= ω ∑ mi r i, 2, 2, , 5, We define a new parameter characterising the, rigid body, called the moment of inertia I ,, given by, I =∑ m i r i, , 2, , ⃗τ =I α, ⃗, ie,, where α is the angular acceleration., , Conservation of angular momentum, If τ =0 ,, , 1, 2, K= I ω, 2, , Therefore, , Thus “ The moment of inertia of a body about, a given axis is defined as the property of the, body by virtue of which it is unable to change, its position of rest or of uniform rotational, motion without the help of external torque”., Note, • I is independent of the magnitude of, the angular velocity, • It depends on two factors: 1) mass of, the body. 2) distribution of mass about, the axis of rotation, , dL, =0, dt, , ie,, L= a constant, Thus if the total external torque on a system, of particles is zero, then the total angular, momentum of the system is conserved., MOMENT OF INERTIA (I), * What is the analogue of mass(measure of, inertia ) in rotational motion?, , KAMIL KATIL VEETIL, , 40, , @, , SOHSS AREEKODE

Page 43 :
hands, moment of inertia increases and angular, velocity decreases., , THE MOMENT OF INERTIA OF, SOME RIGID BODIES, Body, , Axis, , I, , Thin circular, ring, radius R, , Perpendicular to, plane, at centre, , MR, , Thin circular, ring, radius R, , Diameter, , M R2, 2, , Circular disc,, radius R, , perpendicular to, disc , at centre, , MR, 2, , 2, , Circular disc,, radius R, , Diameter, , MR, 4, , 2, , Fig(b) : when the diver wants to do a triple, tuck under 2 seconds ,she can’t change the, angular momentum, but she can reduce the, moment of inertia by pulling the legs and, arms closer to the point of rotation. So angular, velocity increase., , 2, , Hollow cylinder, Axis of cylinder, radius R, , M R2, , Solid cylinder,, radius R, , Axis of cylinder, , M R2, 2, , Solid sphere,, radius R, , Diameter, , 2MR, 5, , Thin rod,, length L, , Perpendicular to, rod, at mid point, , Radius of gyration (K), Radius of gyration. The radius of gyration of a, body about an axis may be defined as the, distance from the axis of a mass point whose, mass is equal to the mass of the whole body, and whose moment of inertia is equal to the, moment of inertia of the body about the axis., ie, I =M K2, THEOREMS OF MOMENT OF INERTIA, 2, , 1) Perpendicular axes theorem, , M L2, 12, , Example for conservation of Angular, momentum, It states that “ the moment of inertia of a plane, lamina about an axis perpendicular to its plane, is equal to the sum of the moment of inertia, about two mutually perpendicular axes lying, in its plane and intersecting each other at the, point where the perpendicular axis passes, through the lamina”., , Fig (a), , If IX and IY are the moments of inertia of the, lamina about perpendicular axes OX and OY, in the plane of lamina, then moment of inertia, of lamina about Z axis (OZ) ,, , Fig(b), , I Z =I X + I Y, , 2) Parallel axes theorem, , When there is no external torque in action ,, angular momentum will remain conserve., Fig(a) : When the dancer stretches out her, KAMIL KATIL VEETIL, , 41, , @, , SOHSS AREEKODE

Page 44 :
“Moment of inertia of a body about any axis, is equal to the sum of the moment of inertia of, the body about a parallel axis through its, centre of mass and the product its mass and, square of the distance between the two axes”., If ICM is the moment of inertia of the body, about an axis passing through the centre of, mass, then the moment of inertia of the body, about a parallel axis distant a from the axis, through centre of mass,, I =I CM + M a, , 2, , where M is the mass of the body, , dW =τdθ and Instantaneous power, P=τω, , ., KINETIC ENERGY OF ROLLING, MOTION (WHEEL ROLLING OVER A, SMOOTH HORIZONTAL SURFACE), A wheel rolling over a horizontal smooth, surface without slipping has two types of, motion: - (i) Rotational motion about an axis, passing through its centre; and (ii) Linear, motion in the horizontal direction of the, surface. Therefore,, Total KE=Translational KE+ Rotational KE, , Do Example 7.10,11,12, WORK AND POWER IN ROTATIONAL, MOTION, Consider a force F acting at the rim of a, pivoted wheel of radius R. During the action,, the wheel rotates through a small angle dθ. If, this angle is small enough, the direction of, force remains constant. Let the corresponding, displacement is dx within a time interval dt., * Then work done, dw=F dx .=F Rdθ ., {Arc length = angle x radius}, , 1, 1, 2, 2, Total KE= mV cm + I ω, 2, 2, 1, 1, 2, 2 2, Total KE= mV cm + m k ω, 2, 2, , =>, =>, , 1, 1 2 2, 2, Total KE= mV cm + m k ω, 2, 2, , =>, , 1, k2, Total KE= mV 2cm ( 1+ 2 ), 2, R, , Do example 7.16, COMPARISON OF TRANSLATIONAL, AND ROTATIONAL MOTION, , But F R is the torque, τ ., * Therefore work done, dW =τdθ, , and, , * The work done to turn from θ1 to θ2 is given, by, , Translational, motion, , Rotational motion, about a fixed axis, , Displacement , x, , Angular displacement θ, , Velocity ,v, , Angular velocity , ω, dθ, ω=, dt, , V=, , θ2, , W =∫ τdθ=τ (θ 2−θ1 )=τ θ, θ1, , * The power, , ie,, , P=, , dW d, dθ, = (τ θ)=τ, dt dt, dt, , P=τ ω, , (, , dθ, =ω ), dt, , Therefore Work done by torque for a small, angular displacement dθ is given by, , Acceleration, a, dV, a=, dt, , Angular acceleration α, dω, α=, dt, , Mass , M, , Moment of inertia , I, , Force F = Ma, , Torque τ = I α, , Work dW = F ds, , Work dW = τ d θ, , Kinetic energy , K, 1, 2, K= M V, 2, , Kinetic energy , K, 1 2, K= I ω, 2, , KAMIL KATIL VEETIL, , 42, , dx, dt, , @, , SOHSS AREEKODE

Page 45 :
Power , P, P=F v, , Power , P, P=τω, , Linear momentum , p, P=mV, , Angular momentum , L, L=I ω, , KINEMATICS OF ROTATIONAL, MOTION ABOUT A FIXED AXIS, , F, We know ⃗τ =⃗r X ⃗, , The kinematical quantities in rotational, motion, angular displacement ( θ ), angular, velocity ( ω ) and angular acceleration ( α ), respectively, correspond, to, kinematic, quantities in linear motion, displacement (x ),, velocity (v) and acceleration (a)., LINEAR MOTION, , ROTATIONAL, MOTION, , V =u+ at, , ω=ω 0+α t, , 1 2, S=ut + a t, 2, , 1 2, θ=ω0 t + α t, 2, , 2, , 2, , Problem, Find the torque of a force ⃗, F =^i− ^j + k^ about, the origin. The force acts on a particle whose, position vector is ⃗r =7 ^i +3 ^j−5 k^, Solution, Given ⃗, F =^i− ^j+ k^ and ⃗r =7 ^i +3 ^j−5 k^, , | |, , ^i ^j k^, ⃗r x ⃗, F = 7 3 −5, 1 −1 1, , ⃗r x ⃗, F =^i (3 x 1−−5 x−1)− ^j(7 x 1−−5 x 1), + k^ (7 x −1−3 x 1), , => ⃗r x ⃗, F =^i (3−5)− ^j(7+5)+ k^ (−7−3), ie ⃗τ =⃗r X ⃗, F =−2 ^i−12 ^j−10 k^, , ω2=ω20 +2 α θ, , V =u +2 as, , APPENDIX, Vetor Product (Cross Product), A and ⃗, B are two vectors, θ be the, If ⃗, angle between them . Then the vector product, ⃗, A x⃗, B is defined as, ⃗, A x⃗, B= A Bsinθ n^, , Where n^ is the unit vector perpendicular to, A and ⃗, B, both ⃗, In analytical form, ⃗, A= A x ^i+ A y ^j+ A z k^, , , ⃗, B =Bx ^i+ B y ^j+ B z k^, , | |, ^i ^j k^, , Then ⃗, A x⃗, B= A x A y A z, Bx B y Bz, , Or, ⃗, A x⃗, B= ^i ( A y Bz −A z B y )− ^j( A x B z −A z B x ), + k^ ( A x B y − A y B x ), KAMIL KATIL VEETIL, , 43, , @, , SOHSS AREEKODE

Page 46 :
CHAPTER EIGHT, , planet. ie,, , GRAVITATION, 8.1 INTRODUCTION, Claudius Ptolemy ( AD 100 ) : Proposed, geocentric model. According to this model,, the Earth is at the centre of the universe and all, other celestial objects orbit the Earth., Nicholas Copernicus (1473-1543) : proposed, Heliocentric model , the Sun is at the centre of, the solar system and all planets including the, Earth orbit the Sun in circular orbits., Galileo Galilee (1564-1642) : Strongly, supported heliocentric model., Tycho Brahe (1546-1601) : Observed and, recorded The stellar and planetary positions, with his naked eye., Johannes Kepler (1571–1630) : Used Brahe's, observations and proposed Kepler's laws for, planetary motion., Isaac Newton (1642–1726) : Proposed, universal law of gravitation, 8.2 KEPLER’S LAWS OF PLANETARY, MOTION, 1. Law of orbits: All planets moves around the, Sun in an elliptical orbit with the Sun at one of, the foci., AP – Major axis, BC – Minor axis, S , S' – Foci, A-Farthest point,, aphelion, P – Closest point,, Perihelion, 2.Law of area: The radial vector (line joining, the Sun to a planet) sweeps equal areas in, equal intervals of time., The shaded area is the, area ∆ A swept out in a, small interval of, time ∆ t., , T2 α a3, , 8.3 UNIVERSAL LAW OF GRAVITATION, Every body in the universe attracts every other, body with a force which is directly proportional, to the product of their masses and inversely, proportional to the square of the distance, between them., Where the G is the, Universal gravitational, constant,, G=6.67 x 10 -11 Nm2/kg2, Definition of G, If m 1 = m 2 = 1 Kg and r = 1 m, Then F= G . ie,, Universal gravitational constant is numerically, equal to the force of attraction between two, unit masses kept at a distance of 1 m apart, 8.3.1 Determination of G, First done by English scientist HenryCavendish, in 1798., , He used a torsional balance for this .The, bar AB has two small lead spheres having mass, 'm' attached at its ends. Two equal heavy lead, spheres of mass ‘M’ each kept at a distance of, ‘d’ from each ‘m’ mass, on opposite sides. Due, to the gravitational attraction , the smaller ones, move slightly towards the larger ones and the, wire gets twisted through a small angle 'θ', The gravitational force on each pair of masses, (M, m) is, , 3. Law of period: The, square of the time period of revolution of a, planet is proportional to the cube of the semimajor axis of the ellipse traced out by the, , If L is the length of the bar AB , then the, torque arising out of F , τ = FL ., At equilibrium, this is equal to the restoring, torque and hence, , KAMIL KATIL VEETIL, , 44, , @, , SOHSS AREEKODE

Page 47 :
Cθ, where 'Cθ' is the restoring torque , C is the, torque for unit twist, Knowing M, m, L ,d , C , θ the value of G can, be determined., 8.4 ACCELERATION DUE TO GRAVITY, OF EARTH, Acceleration due to gravity on the surface, The gravitational force acting on a body on, the surface of earth is, M- Mass of Earth, ...........(1), m – Mass of body, R – Radius of earth, , (2) / (1) =>, , =>, gh, =, g, , =>, , R, , 2, 2, , h, 2, R (1+ ), R, , =>, , The weight experience by the body, F = mg .................(2), comparing (1) and (2), , If h << R, , mg, , Therefore, NOTE : The mass of the earth can be calculated, using the values of acceleration due to gravity,, G and radius of earth. That is why the statement, “Cavendish weighed the earth”., 8.5 VARIATION OF ACCELERATION, DUE TO GRAVITY, The value of acceleration due to gravity (g), varies as we go above or below the surface of, the earth. It also varies from place to place, on the surface of the earth., 1. Variation of 'g' with altitude (height), The acceleration due to gravity, at the surface of the earth, given by, ..............(1), , ie,, , ↑, , ...............(2), , ↓, , ↓, , 2. Variation of 'g' with depth, The acceleration due to, gravity at the surface of the, earth given by, ..............(1), If ρ is the density of the earth,, then, , The acceleration due to gravity, at a height 'h' above earth's, surface is given by, , ↑, , As h, 2 h/ R, (1- 2h/R), gh, Thus as we go above the surface of the earth, acceleration due to gravity decreases., , ρ=M/V, , where M- mass of earth , V – Volume of earth, Therefore M =, , KAMIL KATIL VEETIL, , 45, , @, , ρV, , SOHSS AREEKODE

Page 48 :
Ans) gh ≈ g, , 2) At what height above Earth’s surface is the, value of acceleration due to gravity half of its, value of the Earth’s surface? (RE=6400km), , ..................(2), , [Ans :2650 km] dont use h<<R approximation, , Substitute (2) in (1) we will get ' g ' at the, surface of the earth in terms of density of the, earth. ie ,, , Problem 3: a) At what height above the earth's, surface the value of g is same in a mine of, 100km deep?, b) What is the value of g at the centre of the, earth?, [ Ans : a) 50 km, b) 0 ], , ................(3), Now, consider a point mass m at a depth d, below the surface of the earth which is at a, distance (R – d) from centre of the earth., The outer spherical shell, whose thickness is d,, will not exert any force on the body ., Let M ! be the mass of the earth of portion of, radius (R-d) then, , 8.5 GRAVITATIONAL POTENTIAL, ENERGY, The gravitational force is a conservative, force and hence we can define a gravitational, potential energy associated with this conservative force field., Gravitational potential energy of a mass, is defined as the work done in bringing the, mass from infinity to a point in the, gravitational field of another body ., , ............(4), , Let the body of mass ‘m’ be at a distance x, from the mass ‘M’., Now the gravitational force on ‘m’ is, , (4) / (3) =>, , F=, , The work done to displace the body through a, distance ' dx ' is, , ie,, As, , d, , ↑, , d, R, , GMm, x2, , ↑, , 1−, , d, R, , ↓, , ↓, , gd, , Problem1: Calculate the value of g if a mango, of mass 1⁄2 kg falls from a tree from a height, of 15 meters, what is the acceleration due to, gravity when it begins to fall?, , dW = F.dx =, , GMm, dx, x2, , Therefore the total work done to bring the body, from infinity to the point P is given by,, , KAMIL KATIL VEETIL, , 46, , @, , SOHSS AREEKODE

Page 49 :
r, , W=, , r, , ∫ Fdx, , =, , ∞, , ∫ GMm, x2, , ie,, , dx, , ∞, , 8.5.1 Gravitational Potential ( V ), It is the work done in bringing a body of, unit mass from infinity to a point in the gravitational field., , r, , =GMm, , ∫ x12, , ΔU = mgh, , dx, , ∞, , V=, , 1 r, = −GMm [ ], x ∞, 1 1, = −GMm [ − ∞ ], r, , =>, , W=, , − GMm, r, , This work done is stored in the body as its, gravitational potential energy U., , 8.6 ESCAPE VELOCITY, It is the minimum velocity that a body must, attain to escape out of the gravitational field., Since the gravitational field is a conservative, field , the total energy will be conserved., [ Let M – Mass of earth , m – Mass of, projecting body , R – Radius of the earth , hheight above the surface of earth ], TE ( Initial) = TE ( Final) ..........(1), , − GMm, U=, r, , Therefore, , TE (initial) = KE ( initial) + PE (initial), , 8.5.1 potential energy near the surface of, earth, If a body of mass ‘m’ is taken from the surface, of earth to a height ‘h’ , then the gain in, potential energy, ΔU = U(R+h) – U(R), =>, , GMm, −GMm, +, R, (R+ h), , =, , −GMmR +GMm( R+ h), R( R+ h), , =, , GMmh, R ( R+h), , ................(1), , =, , − GMm, 1, mvi2 +, ......(2), R +h, 2, , =, , GMm, 1, mvi2 –, ...........(3), R+ h, 2, , TE (Final) = KE ( Final) + PE (Final), , −GMm, −GMm, ΔU =, −, R, ( R+ h), , =, , − GM, r, , =, =, , − GMm, 1, mvf2 +, R∞ + h, 2, 1, 2, mvf + 0 .................(4), 2, , Therefore,, (1) =>, , − GMm, 1, mvi2+, R +h, 2, , If vi = Ve, Therefore,, , ,, , GMm, 1, mve2 –, =0, R+ h, 2, , (5) =>, , GMmh, R2, , =>, , 1, mve2 =, 2, , GM, h, R2, , =>, , Ve2 =, , => ΔU = m, , KAMIL KATIL VEETIL, , 47, , @, , 1, mvf2 .......(5), 2, , vf = 0, , Since h is very small (R+h) ≈ R, Therefore, (1) => ΔU =, , =, , GMm, .............(6), R+ h, , 2 GM, R+ h, , SOHSS AREEKODE

Page 50 :
=>, , Ve =, , √, , 2 GM, .............(7), R+h, , ie,, , If we are projecting the body from the surface, of the earth , h=0, (7) => Ve =, =>, , √, , 2 GM, =>, R, , √, , 2 GMR, R2, , √, , V=, , GM, ....................(8), R+ h, , Special case : If the satellite is very close to the, earth (R+h) ≈ R, therefore (8) => V =, , Ve = √ 2 gR = 11.2 km/s, , NOTE : Moon has no atmosphere. Why ?, The escape speed for the moon is 2.3, km/s .The rms velocities of gas molecules on, the surface of moon are greater than its escape, speed. Therefore gas molecules escape from the, moon and hence moon has no atmosphere, 8.7 EARTH SATELLITES, Earth satellites are objects which revolve, around the earth., Moon - The one and only natural satellite of the, Earth., Artificial satellites – man made satellites- Use, for meteorology , telecommunication ,etc., Eg: Sputnik, Aryabatta, INSAT , Edusat ,etc., , 8.7.2 Time period of a satellite, It is the time taken by the satellite to, revolve once around the earth., Expression for time period, Distance travelled, =, Time taken, , Velocity =, , √, , =>, , The centripetal force is, provided, by, The, gravitational, force, between the earth and, the satellite., , From (10), , ie,, , Then (10)= >, , 2, , V =, , 2 π (R+h), T, , 2 π ( R+h), GM, =, ............ (9), T, R+h, , Squaring both sides, , =>, , √, , GMR, R2, , => V = √ gR, This velocity ( velocity in the minimum orbit)is, called first cosmic velocity, NOTE: Ve = √ 2V . That is escape velocity, is √ 2 times the first cosmic velocity., , 8.7.1 Orbital velocity of a satellite, Orbital velocity of a satellite is the velocity, with which it revolves around a planet in its, fixed orbit., Expression for orbital velocity, Consider a satellite of mass 'm' moving, with a velocity 'v' around the earth of mass 'M', and radius R. Let r = R+h is the orbital radius, of the satellite., , mV 2 GMm, =, R+ h (R+h)2, , √, , GM, =>, R, , 2, , 2, , 4 π ( R+h), GM, =, R+ h, T2, , (9)=>, =>, , T, , 2, , 2, = 4 π (R+ h)3 ............... (10), , GM, , NOTE, Equation (10) obeys Kepler's law of periods., T=, , 2π, 3/2, .........(11), ( R+ h), √GM, , Special case :If the satellite is very close to the, earth (R+h) ≈ R, 2, , 4π 3, R ............(12), GM, Dividing Numerator and denominator by R2, T2 =, , (12) =>, , GM, ( R+ h), , KAMIL KATIL VEETIL, , 48, , @, , T2 =, , 2, , 4π R, GM, 2, R, , SOHSS AREEKODE

Page 51 :
=>, , T, , 2, , =, , 4 π2 R, g, , => T = 2 π, , √, , 8.9.2. Polar Satellites, A satellite which revolves in polar orbit is, called a polar satellite. The polar orbit is in the, north-south direction, while earth spins below it, in the west-east direction. Thus a polar satellite, can scan the entire surface of earth. It is placed, at a distance of 500 to 800 km from the surface, of the Earth. The time period of a polar satellite, is nearly 100 minutes and the satellite, completes many revolutions in a day., Eg:IRS group of satellites - remote sensing,, meteorology,etc., , R, g, , 8.8 ENERGY OF AN ORBITING, SATELLITE, 1, 2, mV =, 2, , KE=, , 1, GM, m, 2 (R+h), , (From 8), , KE=, , GMm, ............(13), 2( R +h), , And we know, PE =, , −GMm, ............(14), ( R+ h), , ie,, , Therefore, TE = KE+PE =, , GMm, −GMm, +, 2( R +h), ( R+ h), , =>, , −GMm, ...............(15), 2 (R +h), , TE =, , 8.10 WEIGHTLESSNESS IN SATELLITES, In a satellite around the earth, every part of the, satellite has acceleration towards the centre of, the earth which is exactly the value of earth’s, acceleration due to gravity at that position., That is everything inside the satellite is in a, state of free fall. Thus everything inside the, satellite experience weightlessness., , 8.9 GEOSTATIONARY AND POLAR, SATELLITE, 8.9.1. Geostationary Satellite, If the orbit of the satellite is in the, equatorial plane of the earth and having the, same period as the period of rotation of the, earth about its own axis( T=24h) would appear, stationary viewed from a point on earth. Such, satellites are known as geostationary satellites, Orbital radius of a geostationary satellite, 2, , We know the relation T 2 = 4 π ( R+ h)3, GM, , 2, => (R+h)3 = GM T2, , Therefore, , 4π, GM T 2 1 /3, ), ( R+h) = (, 4 π2, , G= 6.67 X 10−11 Nm2/ kg2 ,M= 6 X 1024 kg, And T= 24 x 60 x 60 s = 86400 s, Therefore,, , h = 36000 km, , Problem : Pick the correct option and justify, your answer, Assertion: On satellites we feel weightlessness., Moon is also a satellite of earth. But we do not, feel weightlessness on moon., Reason : Mass of moon is considerable., A) Both assertion and reason are correct, B) Both assertion and reason are incorrect, C) Assertion is correct , reason is incorrect, Ans : A . On satellites, we feel weightless, because the gravitational force on us due to the, planet provides the centripetal force to keep us, revolving in the orbit. The same happens on the, moon also because moon is also a satellite, (natural satellite). . But, the difference is the, mass of the moon is very large compared to the, mass of the satellites. So, the weight that we, feel on the moon is the gravitational force with, which moon attracts us. For a man-made, satellite, the gravitational force between us and, the satellite is negligible hence we feel, weightless., , Eg: INSAT group for Telecommunications, , KAMIL KATIL VEETIL, , 49, , @, , SOHSS AREEKODE

Page 52 :
CHAPTER NINE, , MECHANICAL PROPERTIES OF, SOLIDS, Solids, * In solids, atoms or molecules are tightly, fixed., * Solids have a definite shape and size., Deforming force, * When suitable forces are applied on a body ,, it undergoes a change in length, volume or, shape( The body deforms), Restoring force, * When a body is subjected to a deforming, force, a restoring force is developed in the, body., Elasticity, * If a body regains its original shape and size, after the removal of deforming force, it is said, to be elastic and the property is called elasticity., * The deformation caused is known as Elastic, deformation., * Eg : Rubber, metals, steel ropes, etc., Plasticity, * If a body does not regain its original shape, and size after removal of the deforming, force, it is said to be a plastic body and the, property is called plasticity., * Eg : Glass , clay ,etc., Stress, * When a body is subjected to a deforming, force, a restoring force is developed in the, body. This restoring force is equal in magnitude, but opposite in direction to the applied force., * Stress is the restoring force per unit area., Stress, σ =, * Unit : N m−2, , F, A, , Dimension : M L−1 T −2, , Types of stress, 1.Longitudinal stress or linear stress: This, stress developed, when the applied force, produces a change in the length of the body., The change in length may be elongation( tensile, stress ) or compression (compressive stress), , 2.Shearing stress or tangential stress, This stress is developed when there is a, change in the shape of the body, , 3.Normal stress or hydraulic stress, This stress is developed in the body, when the, applied force produces a change in the volume, of the body., , Strain, * Strain measures how much an object is, stretched or deformed when a force is applied., Strain=, , Change i n dimension, Original dimension, , * Unit : Nil, Types of strain, , 1. Longitudinal strain, If the deforming force produces a, change in length, the strain produced in, the body is called longitudinal strain, Longitudinal strain =, Change i nlength, Original length, , Longitudinal strain =, , ΔL, L, , 2. Shearing Strain, If the deforming force, produces a, change in shape of the, body without, changing volume, the, strain produced, is called shearing strain., Shearing strain =, , KAMIL KATIL VEETIL, , Dimension: Nil, , 50, , @, , Δx, = tan θ ≈ θ, L, , SOHSS AREEKODE

Page 53 :
3) Volume strain, If the deforming force, produces a change in, volume, the strain, produced in the body is, called volume strain., Volume strain =, , Change i n volume, Original volume, , Volume strain =, , ΔV, V, , HOOKE'S LAW, For small deformations the stress is directly, proportional to strain., stress ∝ strain, stress = K × strain, where , K is the proportionality constant and is, known as modulus of elasticity., Note, * Modulus of elasticity depends on, nature of, the material of the body and temperature., * Unit : N m−2, Dimension : M L−1 T −2, , proportional. Still the body returns to its, original dimension when the load is removed, * If the load is increased further (beyond, elastic limit) the body cannot regain its, original dimension., * In the portion of the curve between C and D,, if the load is increased, strain increases rapidly, even for a small change in the stress. When the, load is removed at some point, say at C, between B and D, the body doesn’t regain its, original dimension. The material is said, to have a permanent set. The deformation is, said to be plastic deformation., * Beyond the point D, additional strain is, produced even by a reduced applied force. And, fracture occurs at E., NOTE, * If the ultimate strength and fracture points D, and E are close, the material is to be brittle., (Fig A), * If D and E are far apart, the material is said to, be ductile. (Fig B), , STRESS-STRAIN CURVE, * A graph drawn with strain along x-axis and, stress along y-axis., , Stress-strain curve for the elastic tissue of, Aorta, , * Point A – Proportional limit, * Point B – Elastic limit or yield point and, corresponding stress – Yield Stress , σy, Point D – The stress corresponding to D is, Ultimate tensile strength , σu, * Point E – Fracture point, * Region O A – Hook's law is valid, * Region AB, stress and strain are not, KAMIL KATIL VEETIL, , *Elastic region is very large, * The material does not obey Hooke’s law over, most of the region., * There is no well defined plastic region., * Substances like tissue of aorta, rubber etc., , 51, , @, , SOHSS AREEKODE

Page 54 :
which can be stretched to cause large strains, are called elastomers., , Y , G, B of some materails, Material, , MODULUS OF ELASTICITY, 1 . Young's Modulus ( Y ), * The ratio of longitudinal stress to the, longitudinal strain is defined as the Young’s, modulus, Y=, , Longitudinal stress, Longitudinal strain, , A, =, ΔL, L, , 20.0, , 8.0, , 15.8, , Aluminium, , 7.0, , 2.5, , 7.0, , Copper, , 12.0, , 4.0, , 12.0, , Iron, , 19.0, , 5.0, , 8.0, , Glass, , 7.0, , 3.0, , 3.6, , Do Example 9.1 ,2 ,3,4 ,5 NCERT, , FL, AΔL, , * If the material has circular cross- section, then, A= π r2, Then ,, , Y=, , FL, 2, πr Δ L, , 2. Rigidiy Modulus (G), * The ratio of shearing stress to the shearing, strain is called the Rigidity modulus or shear, modulus ., , G=, , Shearing stress, Shearing strain, F, F, A, θ = Aθ, , * Unit: N/m 2, , Compressibility ( K ), * The reciprocal of bulk modulus is called, compressibility., K=, , * Unit: N/m 2, Dimension: ML -1 T -2, (NO unit and dimension for strain), , G=, , −Δ V, PV, , * Unit: m 2/N, Dimension: M -1L T 2, * The bulk modulus for solids is much larger, than that for liquids,, *The bulk modulus of liquids are larger than, gases., * Thus solids are least compressible where as, gases are most compressible., DETERMINATION OF YOUNG’S, MODULUS OF THE MATERIAL OF A, WIRE, , Dimension: ML -1 T -2, , Note: For most material G ≈, , Y, 3, , 3. Bulk Modulus ( B ), * The ratio of the volume stress to the, corresponding volume strain is defined as bulk, modulus., Volume stress, Volume strain, F, FV, A, B=, =, A ΔV, ΔV, V, −P, −PV, B=, =, ΔV, ΔV, V, , B=, , Or, , * Unit: N/m 2, , B, 1010N/m2, , Steel, , F, , Y=, , Y, G, 1010N/m2 1010N/m2, , Dimension: ML -1 T -2, KAMIL KATIL VEETIL, , 52, , @, , SOHSS AREEKODE

Page 55 :
The Young’s modulus of the material of the, experimental wire is given by, , A bar of length l, breadth b, and depth d when, loaded at the centre by a load W=mg sags by an, amount given by, , mgL, π r2 Δ L, , Y=, , δ=, , Poisson's Ratio, The ratio of the lateral strain to the longitudinal, strain in a stretched wire is called Poisson’s, ratio., , mgl 3, 4 b d3 Y, , Δd, , d, , Poisson's ratio =, , ΔL, L, , SOME PRACTICAL APPLICATIONS, ELASTICITY, 1) To find the thickness required for a metal, rope, to be used in cranes to pull up heavy, objects., Eg:Make a crane which has a lifting capacity of, 10 tonnes or one metric tonne [10000Kg], (yield strength of steel , σy = 3× 10 8 N/m 2), Soln ) The load shouldn't deform the rope, permanently . Therefore the extension shouldn’t, exceed the elastic limit. Ie , the maximum stress, developed in the rope must be less than yield, strength., Stress=, , F, A, , =>, , Here the condition, sufficiently large., The minimum, , A=, F, , Therefore r2= π σ y, , A=, , F, Stress, , σ < σy , So A must be, F, σy, , => πr2 =, , =>, , r=, , √, , mg, π σy, , F, σy, , F, π σy, , -2, , factor of ten in the load) is provided. Thus a, thicker rope of radius about 3 cm is, recommended. A single wire of this radius, would practically be a rigid rod. So the ropes, are always made of a number of thin wires, braided together., 2 ) In designing a building, the structural, design of the columns, beams and, supports., KAMIL KATIL VEETIL, , To avoid this, large load bearing surface as, shown in figure (c) is used., This is called I-section of the beam or I- girder., 3)The maximum height of the mountain is, limited by the elastic properties of the, rock which hold the mountain., , = 1.03 x 10 m ≈ 1 cm, √, Generally a large margin of safety (of about a, ie , r =, , Note:* For a given material, increasing the, depth d rather than the breadth b is more, effective in reducing the bending., * Buckling: The bending of deep bars unless, the load is not in the proper position is called, buckling .(Shown in fig a ,b), , At the bottom of the mountain of height ‘h’, the, force per unit area due to the weight of the, mountain is hρg . This creates a shearing., Equating this with the elastic limit for a typical, rock is 3 × 10 8 N/m 2 ., h ρ g = 3 × 10 8, 8, h = 3 x 10 =, , ρg, , 8, , 3 x 10, 3, 3 x 10 x 9.8, , ≈ 10 km, which is more than the height of Mt. Everest!, , 53, , @, , SOHSS AREEKODE

Page 56 :
CHAPTER TEN, , the resultant horizontal forces should be zero, and the resultant vertical forces should balance, the weight of the element., , MECHANICAL PROPERTIES FLUIDS, FLUIDS, ➢ Anything which can flow is fluid, ➢ Gases and liquids are fluid, ➢ Fluids do not have definite shape, ➢ They offer less resistance to the shear, stress, Thrust, The total normal force acting on a surface., Unit : N, Density, Mass per unit volume. Density , ρ=, Unit : kg/m3, , m, v, , Relative density or specific gravity, The ratio of the density of a substance to the density, of water at 40C is the relative density of the, substance. It is a unitless quantity., , scalar quantity., , P=, , Unit : N/m2 or pascal, , F, A, , =>, , If 'ρ' is the density of the liquid,, and V be the volume of the liquid, element , then the mass of this, liquid element ,, m =ρV = ρAh ..........(2), Substitute (2) in (1), , =>, , When an object is submerged, in a fluid at rest, the fluid, exerts a force on its surface., This force is always normal to, the object’s surface., Do example 10.1 NCERT, , 10.1.1 variation of pressure with depth, Consider a fluid at, rest in a container., point 1 is at height h, above a point 2. The, pressures at points 1, and 2 are P 1 and P 2, respectively. Consider, a cylindrical element, of fluid having a base, area 'A' and height h., As the fluid is at rest, KAMIL KATIL VEETIL, , P2 – P1 = ρ g h ..............(3), , If the top of the liquid element is open to the, atmosphere, then P 1 = P a and let P 2 =P ., Then,, (3) => P – P a = ρ g h ................(4), , => P = P a + ρ g h ...............(5), , Dimension : ML-1T -2, , Note : atm , bar , torr are other units of pressure., 1 atm = 1.013×10 5 Pa ., , P2A – P1A= mg ..........(1), , (1) => ( P2 – P1 )A = ρAh g, , Dimension : ML-3, , 10.1 PRESSURE, Total normal force( Thrust) per unit area . It is a, , Ie, F2 – F1 = mg, , Thus the pressure increases with depth., The pressure P , at a depth below the surface of, a liquid open to the atmosphere is greater than, atmospheric pressure by an amount ρgh. This is, the absolute pressure ., The excess of pressure, P − P a , at depth ', h' is called gauge pressure at that point ., Hydrostatic paradox, From the above equation the liquid pressure is, the same at all points at the same horizontal, level (or same depth). Base area or the shape of, the container does not involve in the pressure., , On filling with water, the level in the three, vessels is the same( though they hold different, amounts of water) . This is because water at, @, , 54, , S O H S S AREEKODE

Page 57 :
the bottom has the same pressure below each, section of the vessel., , oil) for measuring small pressure differences, and a high density liquid (such as mercury) for, large pressure differences. One end of the tube, is open to the atmosphere and the other end is, connected to the system whose pressure we, want to measure, , Do Example 10.1, 10.2, 10.3, 10.4 NCERT, , 10.1.2Atmospheric Pressure, ➢, The pressure of the atmosphere at any, point is equal to the weight of air column of, unit cross-sectional area extending from that, point to the top of the atmosphere.., ➢, At sea level, the atmospheric pressure is, 1.013×10 5 Pa = 1 atm, ➢, It is maximum on the surface and it, decreases with altitude., Measurement of Pressure, 1. Barometer, 2. Manometer, 1. Barometer, ➢, Used to measure atmospheric pressure, ➢, Evangelista Torricelli (1608 –1647), , Pressure of the system , P = P at A, = P at B, P = Pa+ρ g h, ie,, P – Pa = ρ g h, , The space above the mercury, column in the tube contains, only mercury vapour whose, pressure P is so small that it, may be neglected. The, pressure inside the column at, a point B must be equal to the, pressure at point C which is at, the same level., , 10.1.3 Pascal's law & Applications, If the pressure in a liquid is changed at a, particular point, the change is transmitted, to the entire liquid without being diminished, in magnitude., Application – Hydraulic lift, , Atmospheric pressure , Pa = P at C, = P at B = ρ g h, ( B and C lies in same horizontal level), Where ρ – Density of mercury, g - Acceleration due to gravity, h – Height of the mercury column, ie, Atmospheric pressure, Pa = ρ g h, It is observed that at sea level the height of the, mercury column is 76 cm, ie, 1 atm = 76 cm of Hg, NOTE : In barometer mercury is used because, of it's High Density (So we can reduce the, height of the tube) and it does not wet glass. It, has a shining surface., 2. Manometer( Open tube manometer), An open tube manometer measures the pressure, difference .It consists of a U-tube containing a, suitable liquid i.e., a low density liquid (such as, KAMIL KATIL VEETIL, , A1 , A 2 –, The cross, sectional, areas of, the, cylinder (, A1 < A 2 ), Suppose a downward force F1 is applied on the, smaller piston, the pressure of the liquid under, this piston increases to P, , ( where P =, , F1, )., A1, , According to Pascal’s law this pressure is, transmitted throughout the liquid. Then the, upward force on the second piston,, , Thus the applied force has been increased by a, , @, , 55, , S O H S S AREEKODE

Page 58 :
factor of, , A2, is called mechanical, A1, , advantage, NOTE : hydraulic lift is a force multiplying, device not energy multiplying., Application – hydraulic brake, When we apply a small force on the pedal with, our foot the master piston moves inside the, master cylinder, and the pressure caused is, transmitted through the brake oil to act on a, piston of larger area. A large force acts on the, piston and is pushed down expanding the brake, shoes against brake lining. In this way, a small, force on the pedal produces a large retarding, force on the wheel, , 10.2.2 Equation of continuity, , Consider a pipe of varying cross sectional area, A1 , A2 such that A1 >A2 . A non-viscous and, incompressible liquid with density 'ρ' flows, steadily through the pipe, with velocities v 1, and v 2 in area A1 and A2 respectively as shown, in Figure., , Do Example 10.5 , 10.6 NCERT, , * The volume of the liquid going through the, cross sectional area A1 in a time interval 'Δt', = A1 v1 Δt, * The mass of the fluid, crossing the area A1 , M1= A1 v1 Δt ρ ......(1), , 10.2 FLUID DYNAMICS, The study of the fluids in motion is known, as fluid dynamics., , * Similarly The volume of the liquid going out, of the cross sectional area A2 in a time interval, 'Δt', = A2 v2 Δt, , 10.2.1 streamline flow ( Steady flow), * The flow of the fluid is said to be steady if at, any given point, the velocity of each passing, fluid particle remains constant in time. (velocity, at different points may not be same), * The path taken by a fluid particle under a, steady flow is a streamline., * A streamline is a curve whose tangent at, any point is the direction of the fluid velocity at, that point., * Two stream lines will never intersect., Because, If they intersect, there will be two, velocities at the point which is not possible for, a stream line flow., KAMIL KATIL VEETIL, , Fig : (a) A typical trajectory of a fluid particle., Fig : (b) A region of streamline flow., , * The mass of the fluid, crossing the area A2 , M2= A2 v2 Δt ρ ......(2), Since the liquid is incompressible we must have, A1 v1 Δt ρ = A2 v2 Δt ρ ............(3), OR, A1 v1 = A2 v2, .............(4), ie,, Av = a constant .................(5), This is the equation of continuity., (5) => V α, , 1, A, , . ie, at narrower portions of, , the pipe the velocity increases and vice versa, , @, , 56, , S O H S S AREEKODE

Page 59 :
Note: Equation of continuity is a statement of, conservation of mass, in the flow of, incompressible fluid., 10.2.2 Turbulent Flow, Steady flow is achieved at low flow speeds., Beyond a limiting value, called critical speed,, this flow loses steadiness and becomes, turbulent., Reynolds Number (Re), Reynolds number is a dimensionless number, which gives us the idea that whether the flow is, streamline or turbulent., ρVd, , Re=, η, ρ – density ,, V – velocity, d – diameter of pipe , η – Viscosity of the liquid, Note: If Re < 1000 – Stream line flow, Re > 2000 - Turbulent flow, 1000< Re < 2000 – Unsteady flow, 10.3 BERNOULLI’S PRINCIPLE, According to this principle, the sum of, pressure , kinetic energy per unit volume and, potential energy per unit volume of an, incompressible, non-viscous fluid in a, streamlined flow remains a constant., Mathematically,, 1, 2, P+ ρ V +ρ gh = a constant, 2, , and DE. In a very small interval of time Δ t, the, fluid at B moves to C at the same time fluid at, D moves to E., The volume between B and C = A1V1 Δt, The volume between D and E = A2V2 Δt, By the equation of continuity ,, A1V1 Δt = A2V2 Δt ........(1), The mass of this volume of liquid is ,, Δm= A1V1 Δt ρ = A2V2 Δt ρ ..........(2), Total Work done on part of the liquid just, considered, The Forces acting on this part of the liquid are, 1) P1A1 , by the liquid on the left, 2) P2A2 , by he liquid on the right, 3) Δm g , Gravitational force (weight) of the, liquid considered, 4) N , Normal force by the walls of the tube., In time Δt the point of application of P1A1 is, displaced by BC = V1 Δt . Thus the work done, by P1A1 in time Δt is, W1= ( P1A1) ( V1 Δt ) = P1, , ( Δρm ), , ..........(3), , Similarly the the work done by P2A2 in time, , It is just the application of work energy theorem, in the case of fluid flow., , Δt is, W2= (−P2A2) ( V2 Δt ) = −P2, , Proof, , ( Δρm ), , ..........(4), , The work done by the gravity, W3, = − ve change in potential energy, = − ( Δm g h2− Δm g h1), W3 = ( Δm g h1− Δm g h2 ) ...........(5), The contact force does no work because it is, perpendicular to the velocity., There fore,, Fig: - The flow of an ideal fluid of density ' ρ' in, a pipe of varying cross section., , The total work done on the liquid considered ,, in the time interval Δt is ,, , Consider the flow at the two regions BC, KAMIL KATIL VEETIL, , W4 = 0 ...........(6), , W = W1+W2+W3+W4, @, , 57, , S O H S S AREEKODE

Page 61 :
=, =>, , ie,, OR, , * By applying equation of continuity to the area, A and B , A V1 = a V2, , P−Pa +ρ gh, V 21=, , V1 =, V1 =, , 2( P−P a+ρ gh), ρ, , √, , OR, Then by, h1=h2), , 2(P−Pa +ρ gh), ρ, , √, , V 2=, , A, V ..................(1), a 1, , using Bernoulli’s equation (here, , 1, 1, 2, 2 A, P1+ ρ V 1 = P2+ ρ V 1, 2, 2, a, , 2, , ( ), , 2, ρ (P−Pa )+ 2gh, , This is the velocity of efflux, , ie,, , P1−P2 =, , 2, 1, 1, 2 A, 2, ρV1, − ρV 1, 2, a, 2, , Special cases, Case 1: ( If P>>Pa , 2gh can be ignored), This condition is used in rocket propulsion., , =>, , P1−P2 =, , 1, A 2, 2, ρV1, −1, 2, a, , =>, , ρm g h =, , 1, A 2, ρ V 21, −1, 2, a, , =>, , V 21 =, , V1 =, , √ 2ρ ( P−P ), a, , Now the speed of efflux is determined by, container pressure., Case 2: ( If P=Pa , ie, the tank is open to, atmosphere ), V1 =, , ( ), , [( ) ], [( ) ], , 2 ρm g h, ρ, , √2 g h, , ..........(2), , [( ) ], A 2, −1, a, , √ [( ), , 2ρm g h, , ie, speed of efflux from an open tank is given, by a formula identical to that of a freely falling, body. This is Torricelli's law., , =>, , Application 2 : Venturi-meter, * The Venturi-meter is a device to measure the, flow speed of incompressible fluid., * It consists of a tube with a broad diameter and, a small constriction at the middle as shown ., , The principle behind the venturi- meter is used, in the Carburetor of auto mobile, Bunsen, burner, Atomisers and Sprayers., , * A manometer in the form of a U-tube contains, a liquid of density ' ρm ' is also attached to it,, with one arm at the broad neck point of the tube, and the other at constriction., KAMIL KATIL VEETIL, , V1 =, , ρ, , A 2, −1, a, , ], , Fig: The spray gun. Piston forces air at high, speeds causing a lowering of pressure, at the neck of the container and liquid rises up, to a low pressure region., Application 3 :Blood Flow and Heart Attack, The artery may get constricted due to the, accumulation of plaque on its inner walls.The, @, , 59, , S O H S S AREEKODE

Page 62 :
speed of the flow of the blood in this region is, raised which lowers the pressure inside and the, artery may collapse due to the external, pressure. The heart exerts further pressure to, open this artery and forces the blood through., The internal pressure once again drops due to, same reasons leading to a repeat collapse. This, may result in heart attack., Application 4 : dynamic lift, Dynamic lift is the force that acts on a, body, such as airplane wing, a hydrofoil or a, spinningball, by virtue of its motion through a, fluid., (a)Ball moving with spin ( Magnus effect), , Viscous Force, when there is relative motion between the, layers of a liquid , a frictional force which, opposes the relative motion between the liquid, (fluid) layers. This force is the viscous force., Velocity distribution of layers, Suppose we consider a fluid like oil, enclosed between two glass plates as shown in, Fig (a). The bottom plate is fixed while the top, plate is moved with a constant velocity v, relative to the fixed plate (If oil is replaced by, honey, a greater force is required to move the, plate with the same velocity. Hence we say that, honey is more viscous than oil) The fluid in, contact with a surface has the same velocity as, that of the surfaces. Hence, the layer of the, liquid in contact with top surface moves with a, velocity v and the layer of the liquid in contact, with the fixed surface is stationary. Other layers, have velocity between zero and v . The velocity, distribution is shown in fig (a), , For a spinning ball as in diagram velocity of air, streamlines above it becomes larger than that, below it. So the pressure above it becomes, smaller than that below it. Thus the ball gets an, upward shift. This is called Magnus effect., (b)Aerofoil, The wings of an airplane (aerofoil) are so, designed that its upper surface is more curved, than the lower surface and the front edge is, broader than the real edge. As the aircraft, moves, the air moves faster above the aerofoil, than at the bottom as shown in Figure., , 10.4 VISCOSITY, Liquids, flow in the form of layers. This type of, flow is called laminar flow., , velocity distribution for viscous flow in a pipe, (Fig (b) ), , 10.4.1 Coefficient of viscosity( η ), η=, , Stress, Strain rate, , Refer fig (a) ,, KAMIL KATIL VEETIL, , @, , 60, , S O H S S AREEKODE

Page 63 :
η=, , F, A, =, Δx, /Δ t, l, , Or, , η=, , Fl, ..................(1), AV, , Or, , F=, , ηA V, l, , F, A, , the net upward force and hence the resultant, force on the sphere becomes zero. Thereafter, the body will move with constant velocity, called terminal velocity., , V, l, , Expression For terminal velocity, * Let the density of the material of the sphere, be ' ρ ' and the density of the fluid be' σ ' ., , ................(2), * When the body attains terminal velocity , Vt, , This is Newton's viscous formula, , W = U + F ..............(1), , Note: 1) Unit of viscosity : poiseiulle (Pl), Other units : N s m -2 Or Pa s, 2) The viscosity of liquids decreases with, temperature, while it increases in the case of, gases., 10.4.2 Stoke's law, * The viscous force on a body moving through, a fluid is proportional to the velocity of the, object and is opposite to the direction of, motion., * When a spherical body, having radius ' r ', moving through a fluid , the viscous drag force, is given by ,, F = 6 π r η V ....................(1), Note that this is proportional to V, 10.4.3 Terminal velocity( Vt), * Consider a sphere of radius ' r ' which falls, freely through a viscous fluid of coefficient of, viscosity η., * The forces acting on the body are ,, 1) Gravitational force (W), - downwards, 2) Buoyancy upthrust ( U ), - upwards, , =>, , 4 3, πr ρ g =, 3, , 4 3, π r σ g+6 π r ηV t .........., 3, , (2), =>, , 6 π r ηV t =, , =>, , 6 π r ηV t =, , =>, , Vt =, , 4 3, 4 3, π r ρ g− π r σ g, 3, 3, 4 3, π r (ρ−σ) g ...........(3), 3, , 2 r 2 (ρ−σ) g, ................(4), 9η, , NOTE : If σ is greater than ρ, then the term, (ρ - σ) becomes negative leading to a negative, terminal velocity. That is why air bubbles rise, up through water., 10.5 SURFACE TENSION, * For a molecule well inside the breaker the net, force on it is zero., * But for a molecule on the liquid meniscus, (free surface) there is a net downward force., * This asymmetric force distribution is, responsible for surface tension. ( Fig .a ), , 3) Viscous force (F), - upwards, * When it falls , initially W > U + F . So the, body will accelerate in the downward direction., The velocity of the body goes on increasing and, so viscous force also increases. Finally a stage, reaches at which the downward force balances, KAMIL KATIL VEETIL, , * The property due to which the free surface, of liquid tends to have minimum surface, area and behaves like a stretched membrane, is called surface tension., * imagine a line AB drawn on the surface of a, @, , 61, , S O H S S AREEKODE

Page 64 :
liquid (Fig .b) . The line divides the surface in, two parts . One to the left of the line and other, right to the right of the line. It is found that the, two parts of the surface pull each other with a, force proportional to the length of the line AB., These forces of pull are perpendicular to the, line separating the two parts and tangential to, the surface., * Let F be the common magnitude of the forces, exerted on each other by the two parts of the, surface across a line of length ' l ' . We define, the surface tension 'S ' of the liquid as, S=, , F, l, , ...............(1), , Unit : N/m, 10.5.1 Surface energy ( U ), * When a molecule is taken from the inside to, the surface layer, work is done against the, inward resultant force . The amount of work, done is stored as potential energy. This extra, energy that a surface layer has is called surface, energy ., 10.5.2 Relation between surface tension and, surface energy, , From (1) we know S =, (Or), , F, l, , F = Sl, , Since a soap film has two surfaces ,, F = 2 S l ..................(3), substitute this in (2), (2) => W =, , 2 S l x=S (2 lx) .........(4), , This work done is stored as potential energy of, the new surface., The increase in surface energy is, U = W = S (2 lx), Thus, , U, =S, 2 lx, , (Or), , U, =S, A, , That is the surface tension of a liquid is equal to, the increase in surface energy per unit area., NOTE: In this interpretation the unit of surface, tension may be written as J /m2, 10.5.3 Excess pressure inside a liquid drop, * Due to the force of surface tension, the drop, tends to contract. This create an excess, pressure inside than that of outside. So the drop, tends to expand. When the drop is in, equilibrium, these two forces will be equal and, opposite., * Let the excess pressure, Δ P = P i, , * Consider a U- shaped frame with a sliding bar, on its arm. Suppose it is dipped in soap solution, , taken out and placed in a horizontal position., * Suppose that we move the bar by a small, distance ' x ', * Since the area of the film increases a work, must be done for to move the bar., * The work done for that ,, W = F x ................(2), KAMIL KATIL VEETIL, , –Po, , * Suppose that the liquid drop expands under its, own pressure and its radius increases from ' R ', to ' R + ΔR ' and attains equilibrium., * The work done for this expansion,, , =>, , W = F ΔR, = (ΔP A) ΔR, W = (ΔP) (4 π R2) ΔR ...............(6), , * work done can be calculated using the idea of, surface tension also., @, , 62, , S O H S S AREEKODE

Page 65 :
We know surface tension , S =, Therefore ,, , U, =, A, , W, A, , 10.6 ANGLE OF CONTACT (θ ), * The angle between the tangent to the liquid, surface at the point of contact and solid surface, inside the liquid is termed as angle of contact., * The value of θ determines whether a liquid, will spread on the surface of a solid or it will, form droplets on it . ( It determines the, wettability), , W= SA, , ie, W = Surface tension X Increase in area, = >W = S [ 4 π (R +ΔR)2 - 4 π R2], = >W = S[4 π ( R 2+ 2R ΔR + (ΔR)2) - 4 π R2], = >W = S [ 8 π R ΔR + 4 π (ΔR)2], { Since ΔR is very small 4 π (ΔR)2 can be, neglected }, ie,, W = S ( 8 π R ΔR ) ..............( 7 ), Comparing (6) and (7), =>, (ΔP) (4 π R2) ΔR = S ( 8 π R ΔR ), =>, , ΔP =, , 2S, R, , .................(8), , 10.5.3 Excess pressure inside a bubble, * A bubble has two interfaces, thus by applying, the above argument, ΔP =, , 4S, R, , Note : For pure water and clean glass, angle, of contact, θ ≈ 0, 10.7 CAPILLARY RISE, * One consequence of the pressure difference, across a curved liquid-air interface is that water, rises up in a narrow tube against gravity, Expression for capillary rise, , ................. (9), , 10.5.3 Excess pressure inside a bubble in the, liquid, * A bubble in the liquid has one interface ,, ΔP =, , 2S, R, , .....................(10), , 10.5.4 Pressure on either side of a liquid, surface., 1. Plane liquid surface, The pressure inside and outside will be the, same. ( Fig .A), 2. Convex liquid surface, pressure inside is greater than that of the, outside.( Fig .B), 3.Concave liquid surface, pressure inside is less than that of the outside., ( Fig .C), , KAMIL KATIL VEETIL, , Let, , θ → Angle of contact, a → Radius of the capillary tube, r→ Radius of the meniscus, See the figure,, * Since the point 'A' and 'B' lies on the same, horizontal level ,, Pressure at A = Pressure at B, , @, , 63, , =>, , Pa =, , =>, , hρ g =, , =>, , h=, S O H S S AREEKODE, , P a−, , 2S, +h ρ g, r, , 2S, r, 2S, ............ (11), rρg

Page 66 :
From the figure,, cosθ =, , a, r, , therefore , (11) => h=, , ie,, , h=, , =>, , r=, , a, cos θ, , 2S, a, ρg, cos θ, 2 Scos θ, ..........(11), aρg, , 10.8 DETERGENTS AND SURFACE, TENSION, * Detergents (soap) reduce the surface tension, of water, * Soap molecule has two ends: One end of, detergent molecule form bond with water and, other end forms with oil dirt., * Dirt particles surrounded by detergent, molecule is removed on rinsing in water., , KAMIL KATIL VEETIL, , @, , 64, , S O H S S AREEKODE

Page 67 :
CHAPTER ELEVEN, , ie,, , THERMAL PROPERTIES OF MATTER, , Pα, , 1, V, , ...................(1), , Charles’ law : At constant pressure, the, volume of the gas is directly proportional to, absolute temperature., ie,, V α T ....................(2), , HEAT AND TEMPERATURE, Heat, * Heat is a form of energy transferred from one, body to another body by virtue of temperature, difference., * Unit : joule, , Ideal gas law : By combining these two, equations we have, PV α T, For N number of molecules, P V = N k T ..................(3), Where k =1.381×10 −23 JK −1 , the Boltzman, constant, * N can be represented in terms of number of, moles µ., ie,, N = µ NA, Where NA - Avogadro number, , A common misconception, There is a misconception that heat is a quantity, of energy. People often talk ‘this water has, more heat or less heat’. These words are, meaningless. Heat is energy in transit., ‘A hot cup of coffee has more heat’ is as, meaningless as ‘A lake has more rain’., When it rains, lake receives water from the, cloud. Once the rain stops, the lake will have, more water than before raining. Here ‘raining’, is a process which brings water from the cloud., Rain is water in transit. So the statement ‘lake, has more rain’ is wrong, instead the ‘lake has, more water’ will be appropriate., , Therefore (3) => P V = µ NA k T ...............(4), =>, , P V = µ R T ................(5), , Where R = 8.314 J /mol.K , the universal gas, constant, This is called the equation of state for an ideal, gas., , Temperature, * Temperature is the degree of hotness or, coolness of a body. Hotter the body higher, is its temperature., * Unit : kelvin, Measurement of temperature, * Temperature is measured with a thermometer., * Commonly used thermometer scales are:, (i) Celsius scale ( o C), (ii) Kelvin scale (K), (iii) Fahrenheit scale ( o F)., * The conversion of temperature from one scale, to other scale can be done by the following, formula., , Problem 1, A student comes to school by a bicycle whose, tire is filled with air at a pressure 240 kPa at, 27°C. She travels 8 km to reach the school and, the temperature of the bicycle tire increases to, 39°C. What is the change in pressure in the tire, when the student reaches school?(take the gas is, ideal), Solution, Here V1 = V2 = V, P 1 V = µ R T 1 ..................(1), , C, F−32, K −273, =, =, 5, 9, 5, , P 2 V = µ R T 2 ..................(2), (1), =>, (2), , BOYLE’S LAW, CHARLES’ LAW AND, IDEAL GAS LAW, Boyle's law : At constant temperature, the, pressure of the gas is inversely proportional to, the volume, , => P2=, , KAMIL KATIL VEETIL, , 65, , @, , P1, T, = 1, P2, T2, , P1T 2, T1, , 3, = 240 x 10 x 312, , 300, , SOHSS AREEKODE

Page 68 :
= 249 . 6 kPa, Therefore the change in pressure, =249.6 kPa – 240 kPa = 49.6 kPa, , 3. Volume Expansion, When temperature increases, the volume of a, substance (solid, liquid or gas) increases. This, called volume expansion., , ABSOLUTE ZERO ( 0 K), When temperature decreases, the pressure of a, low density gas decreases. At -273.15 0 C, ( 0 K) the pressure become zero. This, temperature is known as absolute zero., , ΔV, =αV Δ T, V, , Where α V - Coefficient of volume expansion., Problem 2, Eiffel tower is made up of iron and its height is, roughly 300 m. During winter season (January), in France the temperature is 2°C and in hot, summer its average temperature 25°C., Calculate the change in height of Eiffel tower, between summer and winter., Given αL of iron = 10 ×10 −6 per °C, Solution, ΔL, = αL Δ T, L, , or, , Δ L= α L L ΔT, , ie, ΔL = 10 × 10 −6 × 300 × 23 = 0.69 m=69 cm, THERMAL EXPANSION, Thermal expansion is the tendency of matter, to change in length, area, and volume due to a, change in temperature, 1 . Linear Expansion, When temperature increases, length of a solid, (rod like structure) increases. This is called, linear expansion., ΔL, = αL Δ T, L, , Problem 3 ( Do yourself – imp 2010), Railway lines are laid with gaps to allow for, expansion. If the gap between steel rails 66m, long is 3.63 cm at 100C, then at what, temperature will the lines just touch?, Coefficient of linear expansion for steel is, 11x10 -6 / 0 C, Do Example 11.1 ,11.2 NCERT, Relation connecting and α L and α A, We know, , ΔL, = αL Δ T, L, , Or, , α L=, , Similarly, , ΔA, =α A Δ T, A, , Or, , αA=, , Where α L - Coefficient of Linear expansion., 2 . Area Expansion, When temperature increases area of a solid, substance increases. This is called area, expansion., ΔA, =α A Δ T, A, , ΔL, ................. (3), LΔT, , ΔA, .................(4), A ΔT, , Δ A=(L+ Δ L)2−L2 =, 2, 2, 2, L + 2 L Δ L+Δ L −L, , Where α A - Coefficient of Area, expansion., KAMIL KATIL VEETIL, , 66, , @, , SOHSS AREEKODE

Page 69 :
( Δ L2 is very small), , => Δ A=2 L Δ L, Therefore (4) => α A =, , 2LΔ L 2Δ L, =, =2 α L, 2, L ΔT LΔT, , α A =2 α L, ie ,, Relation connecting and α L and α V, αV =3 α L, , Problem 4, Show that the coefficient of volume expansion, at constant pressure of an ideal gas is the, reciprocal of temperature, , ie, αV =, , 1, T, , Solution, The ideal Gas Equation is P V = μ R T ..........(1), At constant pressure, P ΔV = μ R ΔT ..........(2), Or, , ΔV =, , μR, ΔT, P, , =>, , ΔV =, , V, ΔT, T, , =>, , ΔV ΔT, =, V, T, , =>, , ΔV, 1, = αV =, V ΔT, T, , ANOMALOUS EXPANSION OF WATER, * Liquids expand on heating and contract on, cooling at moderate temperatures., * water exhibits an anomalous behaviour. It, contracts on heating between 0 C and 4 C then, it expands., * This means that the water has a maximum, density at 4 C ., , Important environmental effect ofanomalous, expansion of water., In cold countries during the winter season,, Lakes and ponds, freeze at the top first. As a, lake cools towards 40 C, water near the surface, become denser and sinks; the warmer, less, dense water near the bottom rises. However,, once the colder water on top reaches a, temperature below 40 C, it becomes less dense, and remains at the surface itself. So a water, body will not freeze from top to bottom. If, water did not have this property, lakes and, ponds would freeze from bottom to top., CHANGE OF STATE, * All matter exists normally in three states as, solids, liquids or gases. Matter can be changed, from one state to another either by heating or, cooling., * During change of state, the two different state, coexist in thermal equilibrium and temperature, remains constant until the completion of change, of state., , KAMIL KATIL VEETIL, , 67, , @, , SOHSS AREEKODE

Page 70 :
Melting point, * The temperature at which solid and liquid, coexist in thermal equilibrium with each other, is called melting point., * The melting point decreases with pressure., Boiling point., * The temperature at which liquid and vapour, state of substance coexist in thermal, equilibrium with each other is called boiling, point., * The boiling point increases with pressure ., Regelation, When pressure is applied, ice melts at low, temperature. If pressure is removed, water, refreezes. This refreezing is called regelation., , Molar Specific Heat of gas, There are two specific heats for gases., (i) Specific heat at constant volume CV, (ii) Specific heat at constant pressure CP, (i) Specific heat at constant volume ( CV ), * It is the amount of heat required to raise the, temperature of one mole of a gas at constant, volume by 1 0 C or 1K., (ii) Specific heat at constant pressure ( CP ), * It is the amount of heat required to raise the, temperature of one mole of a gas at constant, pressure by 1 0 C or 1K., Latent heat capacity ( L ), * There are two latent heat capacity, (i) Latent heat of fusion ( L f ), * The amount of heat energy required to change, a unit mass solid substance at its melting point, completely into liquid in the same temperature., , specific heat capacity ( s ), * It is the amount of heat required to raise the, temperature of 1kg substance by 1 0 C or 1K ., s =, , ΔQ, mΔ T, , Lf =, , * Unit : J/kg, (ii) Latent heat of vaporisation ( L v ), The amount of heat energy required to change a, unit mass liquid substance at its boiling point, completely into gas in the same temperature., , * Unit : J kg -1 K -1, , LV =, , Heat capacity (S), * It is the amount of heat required to raise the, temperature of a substance by 1 0 C or 1K., S=ms, Or, , S=, , Q, m, , Q, m, , * Unit : J/kg, Temperature versus heat for water, , ΔQ, ΔT, , * Unit : J/K, Molar specific heat capacity ( C ), * It is the amount of heat required to raise the, temperature of 1mole of a substance by 1 0C or, 1K., ΔQ, C=, μ ΔT, , μ – No of moles, , * Unit : J mol -1 K -1, , Problem 4 ( Do yourself), Find out the work done to convert 2 kg of ice at, -5 0 C to steam at 100 0 C . Specific heat capacity, of ice is 2100J/kg K, specific latent heat of, fusion of ice is 336x10 3 J/kg, Latent heat of, , KAMIL KATIL VEETIL, , 68, , @, , SOHSS AREEKODE

Page 71 :
vaporisation of water is 2250x10 3 J/kg and, specific heat capacity of water is 4200J/kg K., HEAT TRANSFER, * There are three modes of heat transfer:, 1)Conduction 2) Convection 3) Radiation., , 1) Conduction, * Conduction is the process of direct transfer, of heat through matter due to temperature, difference., * it is found experimentally that in the steady, state, the rate of flow of heat (or heat current) H, is proportional to the temperature difference, (T C – T D ) and the area of cross-section A and, is inversely proportional to the length L., ie ,, , A(T C −T D ), dQ, = H= K, dt, L, , Where K is a constant called Thermal, conductivity of material., * Unit : W m -1 K -1, Note :, *Silver is the best thermal conductor, * some cooking pots have copper coating on, the bottom - Being a good conductor of heat,, copper promotes the distribution of heat over, the bottom of a pot for uniform cooking., * Houses made of concrete roofs get very hot, during summer days because of considerable, thermal conductivity of concrete. People,, usually, prefer to give a layer of earth or, foam insulation on the ceiling so that heat, transfer is prohibited., * A brass tumbler feels much colder than a, wooden tray on a chilly day -Brass has greater, conductivity than wood ., Do Example 11.6 NCERT, 2) Convection, * Convection is a mode of heat transfer by, , actual motion of matter., * It is possible only in fluids., * When a fluid is heated from below, the hot, part expands and, therefore, becomes less, dense. Because of buoyancy, it rises and the, upper colder part replaces it. This again gets, heated, rises up and is replaced by the relatively, colder part of the fluid., * Natural convection is responsible for many, familiar phenomena such as sea breeze , land, breeze, etc., , 3) Radiation, * Radiation needs no medium for heat transfer., * In radiation, energy is transferred in the form, of electromagnetic radiation ., * The electromagnetic radiation emitted by a, body by virtue of its temperature, like radiation, by a red hot iron or light from a filament lamp, is called thermal radiation., LAWS OF HEAT TRANSFER, Blackbody, * A body which absorbs all the radiations, falling on it is called a black body. It is a, perfect radiator also., 1 ) Stefan Boltzmann law, * Stefan Boltzmann law states that, the total, amount of heat radiated per second per unit area, of a black body is directly proportional to the, fourth power of its absolute temperature., E = σT4, Where σ = 5.67 × 10 −8 W m −2 K −4 , the Stefan's, constant., 2) Wien’s displacement law, * Wien’s law states that, the wavelength of, maximum intensity of emission of a black body, radiation is inversely proportional to the, absolute temperature of the black body., , KAMIL KATIL VEETIL, , 69, , @, , SOHSS AREEKODE

Page 72 :
λm α, , 1, T, , or, , λ m=, , b, T, , Where, b is known as Wien’s constant. Its, value is 2.898× 10 -3 m K, * It implies that if temperature of the body, increases, maximal intensity wavelength ( λ m ), shift s towards lower wavelength (higher, frequency) of electromagnetic spectrum., , * initially the rate of cooling is higher and, decreases as the temperature of, the body falls., Do Example 11.8 NCERT, Problem 5 ( March 2010 ), You are in restaurant waiting for your friend, and ordered coffee. It has arrived. Do you add, sugar in your friend’s coffee and then wait for, him or do you add sugar after the arrives?, Explain with respect to the concept of cooling., Soln) For the coffee to be hotter when the, friend arrives, the better option is: first mix, sugar and then wait for the friend. When we, mix sugar with coffee the temperature of the, coffee decreases. So according to Newton’s law, of cooling the rate of loss of heat decreases., , NEWTON’S LAW OF COOLING, * Newton’s law of cooling states that the rate of, loss of heat of a body is directly proportional to, the difference in the temperature between that, body and its surroundings ., −dQ, α(T o −T s), dt, , * The negative sign indicates that the quantity, of heat lost by liquid goes on decreasing with, time., * Where, To = Temperature of the object, Ts = Temperature of the surrounding, Cooling of hot water with time, , KAMIL KATIL VEETIL, , 70, , @, , SOHSS AREEKODE

Page 73 :
CHAPTER TWELVE, , (2) Intensive state variables, Intensive variables do not depend on the size, THERMODYNAMICS, or mass of the system, Thermodynamics is the branch of physics Examples: Temperature, pressure, specific, which describes the laws governing the heat capacity, density etc., process of conversion of work into heat and, Equation of state, conversion of heat into work., * The equation which connects the state, variables in a specific manner is called, Thermodynamic system, equation of state., * Part of the universe under study, * It is a collection of large number of particles * For an ideal gas, the equation of state is, PV =μ R T, (atoms or molecules), * The system may exist in solid, liquid and, gaseous state. ( In this chapter we take a gas ZEROTH LAW OF THERMODYNAMICS, inside a cylinder attached with a piston as The zeroth law of thermodynamics states that, if two systems, A and B, are in thermal, system for the demonstrational convenience), equilibrium with a third system, C, then A, and B are in thermal equilibrium with each, Surrounding, * The remaining part of the universe except other., system., * The system and surrounding are separated, by a wall ( not a physical wall)., * Diathermic wall - A conducting wall that, allows heat flow through it, * Adiabatic wall- An insulating wall that does, INTERNAL ENERGY (U), not allow flow of heat, * The internal energy of a thermodynamic, system is the sum of kinetic and potential, Thermal equilibrium, energies of all the molecules of the system., * Two systems are said to be in thermal, * Since ideal gas molecules are assumed to, equilibrium with each other if they are at the, have no interaction with each other the, same temperature., internal energy consists of only kinetic energy, part , which depends on the temperature. So, Thermodynamic state variables, * The state of a thermodynamic system is internal energy of an ideal gas depends only, represented by a set of variables called on temperature., * Internal energy is a state variable. It depends, thermodynamic variables., Examples: Pressure, temperature, volume, only on the initial and final states of the, internal energy , heat capacity, specific heat thermodynamic system. For example, if the, temperature of water is raised from 30°C to, capacity , etc., * Note: Work and heat are not state variable. 40°C either by heating or by stirring, the final, internal energy depends only on the final, They are process variables, temperature 40°C and not the way it is arrived, at., * There are two types of thermodynamic, variables: Extensive and Intensive, FIRST LAW OF THERMODYNAMICS, (1) Extensive sate variables, * It states that the amount of heat given to a, Extensive variable depends on the size or, system is equal to the sum of the increase in, mass of the system., Examples: Volume, total mass, entropy, the internal energy of the system and the, external work done, internal energy, heat capacity etc., KAMIL KATIL VEETIL @, , 71, , SOHSS AREEKODE

Page 74 :
MOLAR SPECIFIC HEAT OF GAS., Unlike solids and liquids, gases have two, PΔV, molar specific heats: molar specific heat, * The first law of thermodynamics is a capacity at constant pressure ( C P ) and, molar specific heat capacity at constant, statement of law of conservation of energy., volume ( CV )., * We adopt the following sign convention, Δ Q=Δ U +Δ W, * At constant pressure ΔW =, Δ Q=Δ U + P Δ V, Now, , System gains heat, , Q is positive, , System loses heat, , Q is negative, , (1) Molar specific heat at constant, volume ( CV ), Molar specific heat of a gas at constant, volume is the amount of heat required to raise, the temperature of one mole of the gas, through 1K when its volume is kept constant., , Work done on the system W is negative, Work done by the system W is positive, Problem 1 :, Jogging every day is good for health. Assume, that when you jog a work of 500 kJ is done, and 230 kJ of heat is given off . What is the, change in internal energy of your body?, Solution, Work done by the system (body),, W = +500 kJ, Heat released from the system (body),, Q = –230 kJ, The change in internal energy of a body,, Δ U =Δ Q−Δ W =– 230 kJ – 500 kJ =– 730 kJ, , SPECIFIC HEAT CAPACITY (s), * Specific heat of a substance is the amount, of heat required to raise the temperature of, unit mass substance through one Kelvin, * Suppose an amount of heat ∆Q supplied to a, substance changes its temperature from T to, T + ∆T. We define heat capacity of the, substance, , S=, , (1) Molar specific heat at constant, pressure ( C P ), Molar specific heat of the gas at constant, pressure is defined as the amount of heat, required to raise the temperature of one mole, of the gas through 1K, when its pressure is, kept constant., Specific heat capacity of water, * The specific heat capacity of water varies, slightly with temperature., * Variation of specific heat capacity of water, with temperature is shown below., , ΔQ, ΔT, , *Unit: J / K, * Therefore the specific heat capacity of the, substance, , s=, , S 1 ΔQ, =, m m ΔT, , * One calorie is defined to be the amount of, heat required to raise the temperature of, 1g of water from 14.5 °C to 15.5 °C, , *Unit: J kg−1 K−1, , MOLAR SPECIFIC HEAT CAPACITY(C) * 1 calorie = 4.186 J, * it is the amount of heat required to raise the, temperature of one mole of a substance by 1K. * The specific heat capacity of water is, −1, −1, ( to raise from 14.5 °C to, 4186 J kg K, 1 ΔQ, s= μ, 15.5 °C), ΔT, −1, −1, *Unit: J mol K, KAMIL KATIL VEETIL @, , 72, , SOHSS AREEKODE

Page 75 :
Relation between C P and CV, relation), , (Mayer’s of 'dQ'. Hence C P is greater than CV, , ., , THERMODYNAMIC PROCESSES, (1) QUASI-STATIC PROCESS, , C P−C V =R, , Proof, * From first law of thermodynamics, Δ Q=Δ U +P Δ V, , We consider 1 mole of ideal gas, If ∆Q heat is absorbed at constant, volume, ∆V = 0, Then , CV =, , [ ] [ ] [ ] .......(1), ΔQ, ΔU, ΔU, =, =, ΔT V ΔT V ΔT, , The subscript 'v' can be dropped since U of an, ideal gas depends only on T., Similarly, C P=, , [ ] [ ] [ ], ΔQ, ΔU, =, ΔT P ΔT, , +P, , ΔV, ΔT, , .......(2), , It is a process in which a thermodynamic, system proceeds extremely slowly such that at, The subscript 'p' can be dropped from the, first term since U of an ideal gas depends only every instant of time, the temperature and, pressure are the same in all parts of the, on T., system., ΔQ, ΔU, ΔV, =, +P, So C P=, .......(3), Example, ΔT P ΔT, ΔT P, Consider a container of gas with volume V,, pressure P and temperature T. If we add sand, Now, for a mole of an ideal gas, particles one by one slowly on the top of the, PV =RT ................(4), piston, the piston will move inward very, Differentiating w.r.t. temperature (at constant, slowly. This can be taken as almost a quasipressure), static process. It is shown in the figure, P, , P, , [ ] [ ] [ ], , (4)=>, , P, , [ ], , ΔV, =R, ΔT P, , ...........(5), , (2) ISOTHERMAL PROCESS, A process in which the temperature remains, constant but the pressure and volume of the, thermodynamic system will change., , Sub (5) in (3), , [ ], , ΔU, + R ..........(6), ΔT, , (3)=>, , C P=, , (6) – (1) =>, , C P−C V =R, , The ideal gas equation is PV =μRT, Examples:, (i)When water is heated, at the boiling point,, even when heat flows to water, the, temperature will not increase unless the water, completely evaporates. Similarly, at the, freezing point, when the ice melts to water, the, temperature of ice will not increase even when, heat is supplied to ice., (ii) All biological processes occur at constant, body temperature (37°C)., Equation of state for isothermal process, , NOTE, C P is greater than CV, , . Why?, When one mole of the gas is heated at, constant volume, the heat is used only to, increase the internal energy of the gas. But, when the gas is heated at constant pressure,, the heat is used not only for increasing the, internal energy but also for doing external, work during expansion. So internal energy, change and hence the temperature change will, PV =Constant .............(1), be more at constant volume for same amount So for an isothermal change P1 V 1=P 2 V 2, KAMIL KATIL VEETIL @, , 73, , SOHSS AREEKODE

Page 76 :
PV Diagram for isothermal process, , =207.75 kPa, (ii), , W =μ RT ln (, , V2, ), V1, , W =0.5 x 8.31 x 300 ln (, , 6 x 10−3, ), 2 x 10−3, , = 1.369 kJ, (3) ADIABATIC PROCESS, * A process in which no heat flows into or out, Work done during isothermal process, Consider an ideal gas which is allowed to of the system (ΔQ=0)., expand isothermally at constant temperature, * The system is insulated with an adiabatic, T from initial state (P1,V1) to the final state wall or, * If the process occurs so quickly that there is, (P2 ,V2)., no time to exchange heat with surroundings, Then the work done, dW =PdV, even though there is no thermal insulation., V, Examples, (i)When the tyre bursts the air expands so, The total work done , W =∫ P dV .......(2), V, quickly that there is no time to exchange heat, μ RT, with the surroundings., But we know PV =μ R T (or) P=, V, (ii)propagation of sound through air., Substitute this in(1) ,, Equation of state for adiabatic process, V, V, 2, , 1, , μ RT, 1, (2) => W =∫ (, ) dV =μ RT ∫ ( )dV, V, V, V, V, 2, , 2, , 1, , 1, , PV γ =Constant ...........(5)where γ=, , => W =μ RT [lnV ]VV =μ RT (ln V 2−lnV 1 ), 2, 1, , =>, , V, W =μ RT ln ( 2 ), V1, , .............(3), , (or), , T V γ−1=Constant .............(6), , (or), , P, , 1−γ, , CP, CV, , γ, T =Constant ............(7), , Work done during an adiabatic process, for an isothermal process, V 2 P1, =, V 1 P2, , So, , P1 V 1=P 2 V 2, , V2, , =>, , W =∫ P dV ............(8), V1, , For an adiabatic process PV γ =Constant=K, W =μ RT ln(, , P1, ), P2, , K, .................(9), Vγ, V, K, Sub (9) in (8) => W =∫ ( γ ) dV, V, V, , (Or), , .............(4), , P=, , 2, , Problem 2, A 0.5 mole of gas at temperature 300K, expands isothermally from an initial volume, of 2 L to 6 L. (Given The value of gas, constant, R = 8.31 J mol -1K-1), (i) What is the fi nal pressure of the gas?, (ii) What is the work done by the gas?, Solution, (i)For an isothermal process, , 1, , V2, , 1, , Therefore(10)=>, , P1 V 1=P 2 V 2=μ R T, μ RT 0.5 x 8.31 x 300, Therefore P2=, =, V2, 6 x 10−3, , W=, , KAMIL KATIL VEETIL @, , 74, , [, , ], , V2, , −γ+1, => W =K ∫ ( 1γ )dV =K V, −γ +1 V, V V, K, −γ+1, −γ+1, => W =, [V 2 −V 1 ] ..............(10), 1−γ, But we know K=constant=P1 V γ1 =P 2 V 2γ, 1, , 1, γ −γ+1, γ −γ+1, [P V V, −P 1 V 1 V 1 ], 1−γ 2 2 2, , SOHSS AREEKODE

Page 77 :
=>, , W=, , 1, [P V −P1 V 1], 1−γ 2 2, , =>, , W=, , 1, [μ RT 2 −μ RT 1 ], 1−γ, , =>, , W=, , =>, , (4) ISOCHORIC PROCESS, * In an isochoric process, V is constant., * Thus work done on or by the system is zero., * The heat absorbed by the gas goes entirely, to change its internal energy and its, temperature, , μR, [T −T 1 ], 1−γ 2, μR, W=, [T −T 2 ], γ−1 1, , Problem3: Draw the indicator diagrams for, isothermal and adiabatic processes., , (5) CYCLIC PROCESS, * A process in which the system returns to its, initial state after undergoing a series of, changes., * Therefore Δ U =0 for a cyclic process, Note : that adiabatic curve is steeper than * Then by first law Δ Q=Δ W, isothermal curve. Th is is because γ > 1, PV diagram for a cyclic process ( Example), always., (3)ISOBARIC PROCESS, * A process in which pressure remains, constant., Equation of state for an isobaric process, P=Constant, , From the ideal gas equation, we have, PV =μ R T, , Or, , μR, V=, T, P, , or, , V αT, , ie, In an isobaric process the temperature is, directly proportional to volume., Work done during an an isobaric process, V2, , V2, , W =∫ P dV, V1, , => W =P [V ], , * The P - V diagram for cyclic process will be, closed loop and area of this loop gives work, done or heat absorbed by system., , V2, V!, , => W =P ∫ dV, V1, , =>, , W =P [V 2−V 1 ], , REVERSIBLE PROCESSE, A thermodynamic process can be considered, reversible only if it can be reversed such that, both the system and surroundings return to, their initial states, with no other change, anywhere else in the universe., Conditions for reversible process:, 1) The process should proceed at an extremely, , KAMIL KATIL VEETIL @, , 75, , SOHSS AREEKODE

Page 78 :
slow rate., For example, in the automobile engine,, 2) No dissipative forces such as friction, the cold reservoir is the surroundings at room, viscosity, electrical resistance should be temperature. The automobile ejects heat to, present, these surroundings through a silencer., HEAT ENGINE, Efficiency of heat engine (η), Q, Heat engine is a device which takes heat as, output W Q H −Q L, η=, =, =, =1− L, input and converts this heat in to work by, input Q H, QH, QH, undergoing a cyclic process., A heat engine has three parts:, Q, ie,, η=1− L, 1. Hot reservoir, QH, 2. Working substance, 3. Cold reservoir, SECOND LAW OF THERMODYNAMICS, Schematic diagram, Kelvin-Planck statement:, It is impossible to construct a heat engine that, operates in a cycle, whose sole effect is to, convert the heat completely into work. This, implies that no heat engine in the universe can, have 100% efficiency., Clausius statement, No process is possible whose sole result is the, transfer of heat from a colder object to hotter, object., Problem 3, According to first law of thermodynamics, in, an isothermal process the given heat is, completely converted into work (Q = W). Is it, a violation of the second law of, thermodynamics?, 1. Hot reservoir (or) Source:, Solution, * It supplies heatto the engine. It is always, No. For non-cyclic process like an isothermal, maintained at a high temperature T H, expansion, the heat can be completely, 2. Working substance:, converted into work. But Second law of, * It is a substance like gas or water, which, thermodynamics implies that ‘In a cyclic, converts the heat supplied into work., process only a portion of the heat absorbed is, * The working substance absorb Q H heat, converted into work’. All heat engines operate, from the source ( temperature T H ) and reject, in a cyclic process., Q L heat to the sink ( temperature T L) after, doing work W., CARNOT’S IDEAL HEAT ENGINE, A simple example of a heat engine is a, steam engine. The working substance in these, * A reversible heat engine operating in a cycle, is water which absorbs heat from the burning, between two temperatures in a particular way, of coal. The heat converts the water into, is called a Carnot Engine., steam. This steam is does work by rotating the, * The carnot engine has four parts, wheels of the train, 1) Source:, 3. Cold reservoir (or) Sink: The heat engine, It is the source of heat maintained at constant, ejects some amount of heat (Q L ) into cold, high temperature T H . Any amount of heat can, reservoir after it doing work. It is always, be extracted from it, without changing its, maintained at a low temperature T L ., KAMIL KATIL VEETIL @, , 76, , SOHSS AREEKODE

Page 79 :
temperature., 2) Sink:, It is a cold body maintained at a constant low, temperature T L . It can absorb any amount of, heat., 3) Insulating stand:, It is made of perfectly non-conducting, material. Heat is not conducted through this, stand., 4) Working substance:, It is an ideal gas enclosed in a cylinder with, perfectly non-conducting walls and perfectly, conducting bottom. A non-conducting and, frictionless piston is fitted in it, , Step 3: Isothermal compression of the gas, from (P3 ,V 3 , T L ) to ( P4 , V 4 , T L ) on placing, the cylinder on the sink., * The work done by the gas during this, process,, W C → D=μ R T L ln (, , V4, V, )=−μ R T L ln( 3 ) .....(3), V3, V4, , Step 4 : Adiabatic compression of the gas, from (P4 , V 4 , T L ) to (P1 ,V 1 , T H ), * The work done by the gas during this, process,, W D→ A=, , μR, −μ R, [T H −T L ]=, [T −T H ] ....(4), γ−1, γ−1 L, , The total work done (W), Carnot’s cycle:, W =W A →B +W B →C +W C→ D +W D → A, The working substance is subjected to four, successive reversible processes forming, V, V, Carnot’s cycle., => W =μ R T H ln ( 2 )−μ R T L ln ( 3 ) .....(5), V1, , V4, , Since B and C lie on same adiabatic, =>, , γ−1, T H V γ−1, 2 =T L V 3, T L V 2γ−1, =, ..............(6), T H V 3γ−1, , Since A and D lie on same adiabatic, γ−1, , γ−1, , T H V 1 =T L V 4, T L V 1γ−1, =, ..............(7), T H V 4γ−1, , Comparing (6) and (7), V 2 V1, =>, =, V3 V4, , V 2 V3, ..........(8), =, V 1 V4, , Step1 : Isothermal expansion of the gas from, (P1 ,V 1 , T H ) to ( P2 ,V 2 , T H ) by absorbing a Substitute (8) in (5), V, V, heat Q H from the source at TH K., (5) => W =μ R T H ln( 2 )−μ R T L ln( 2 ), V1, V1, * The work done by the gas during this, V, process,, => W =μ R ln ( 2 )[T H −T L ] ...........(6), W A → B=Q H =μ R T H ln(, , V1, , V2, ) .............(1), V1, , Step 2 :Adiabatic expansion of the gas from Efficiency of Carnot engine (η), placing, (P2 ,V 2 , T H ) to (P3 ,V 3 , T L ) by, work done, W, =>, η=, =, cylinder on insulating stand., heat absorbed Q H, * The work done by the gas during this, V, process,, μ R ln( 2 )[T H −T L ], [ T H −T L ], V1, μR, η=, =>, η=, W, =, [T −T ] ...............(2), B→ C, , γ−1, , L, , H, , μ R T H ln(, , KAMIL KATIL VEETIL @, , 77, , V2, ), V1, , SOHSS AREEKODE, , TH

Page 80 :
the cold reservoir at temperature T 2, some, external work W is done on it and heat Q H is, rejected to the hot reservoir at temperature TH., NOTE, 1) η is always less than 1 because T L is less Coefficient of performance (COP) (α), Q, QL, than T H . This implies the efficiency cannot be, α= L =, W Q H −QL, 100%. It can be 1 or 100% only when, T L = 0K (absolute zero of temperature) which, HEAT PUMPS, is impossible to attain practically., 2) The efficiency of the Carnot’s engine is The device used to pump heat into a portion of, independent of the working substance. It space (to warm-up room) is called heat pump., depends only on the temperatures of the Coefficient of performance (COP) (α), Q, source and the sink., α= H, 3) The entire process is reversible in the, W, Carnot engine cycle. So Carnot engine is itself, a reversible engine and has maximum, efficiency. But all practical heat engines like Problem 4, diesel engine, petrolengine and steam engine There are two Carnot engines A and B, have cycles which are not perfectly reversible. operating in two different temperature regions., So their efficiency is always less than the For Engine A the temperatures of the two, Carnot efficiency. This can be stated in the reservoirs are 150°C and 100°C. For engine B, form of the Carnot theorem., the temperatures of the reservoirs are 350°C, and 300°C. Which engine has lesser, Carnot theorem., efficiency? ( Do yourself), ‘Between two constant temperature reservoirs, Problem 5, only Carnot engine can have maximum One mole of an ideal gas initially kept in a, efficiency. All real heat engines will have cylinder at pressure 1 MPa and temperature, efficiency less than the Carnot engine’, 27°C is made to expand until its volume is, doubled., REFRIGERATOR, (a) How much work is done if the expansion is, A refrigerator is a heat engine working in the, (i) adiabatic (ii) isobaric (iii) isothermal?, reverse order, (b) Name the processes in which the heat, transfer is maximum and minimum., Solution, (a) Do yourself, (b)In an adiabatic process no heat enters into, the system or leaves from the system. In an, isobaric process the work done is more so heat, supplied should be more compared to an, isothermal process., =>, , η=1−, , TL, TH, , The working substance extracts heat Q L from, KAMIL KATIL VEETIL @, , 78, , SOHSS AREEKODE

Page 81 :
CHAPTER THIRTEEN, , KINETIC THEORY, 13.1 INTRODUCTION, , * Developed by – Maxwell , Boltzman and, others in 19th century., * Kinetic theory relates pressure and, temperature to molecular motion of sample of, a gas ., * It is a bridge between Newtonian mechanics, and thermodynamics., 13.2 BEHAVIOUR OF GASES, , * Gases at low pressure and high temperature, approximately satisfy the relation PV = NK BT, = µRT. Such gases are ideal gases., (P- Pressure, V- Volume, N-Total no of, molecules in the sample, KB- Boltzman, constant , T- Temperature in kelvin , µ – no of, moles, R- Universal gas constant., KB= 1.38 x 10-23 J/K R = 8.31 J mol-1K-1)., , Fig 1; A real gas approaches ideal gas at high, temperature and low pressure., 13.3 POSTULATES OF KINETIC THEORY, OF GASES, , 1. All the molecules of a gas are identical,, elastic spheres., 2. The molecules of different gases are, different., 3. The number of molecules in a gas is very, large and the average separation between them, is larger than size of the gas molecules., 4. The molecules of a gas are in a state of, continuous random motion., 5. The molecules collide with one another and, also with the walls of the container., 6. These collisions are perfectly elastic so that, there is no loss of kinetic energy during, , collisions., 7. Between two successive collisions, a, molecule moves with uniform velocity., 8. The molecules do not exert any force of, attraction or repulsion on each other except, during collision. The molecules do not possess, any potential energy and the energy is wholly, kinetic., 9. The collisions are instantaneous. The time, spent by a molecule in each collision is very, small compared to the time elapsed between, two consecutive collisions., 10. These molecules obey Newton’s laws of, motion even though they move randomly., 13.4 PRESSURE EXERTED BY A GAS, ( KINETIC INTERPRETATION), , Consider a monatomic gas of N molecules, each having a mass m inside a cubical, container of side L ., , The molecules of the gas are in random, motion. They collide with each other and also, with the walls of the container. The molecules, of the gas exert pressure on the walls of the, container due to collision on it. During each, collision, the molecules, impart certain momentum to the wall. Due to, transfer of momentum, the walls experience a, continuous force. The force experienced per, unit area of the walls of the container, determines the pressure exerted by the gas., * A molecule of mass m moving with a, velocity v having components (vx , vy ,vz ) hits, the right side wall. Since we have assumed that, the collision is elastic, the particle rebounds, with same speed and its x-component is, reversed. This is shown in the Figure. The, components of velocity of the molecule after, collision are (-vx , vy , vz )., , KAMIL KATIL VEETIL @, , 79, , SOHSS AREEKODE

Page 82 :
* The change in momentum of the molecule=, –mvx – (mvx ) = – 2mvx, , * Therefore The mean square speed is written, as, v 2=v 2x + v 2y +v 2z=3 v 2x, , * According to law of conservation of linear * Hence pressure P is, momentum, the change in momentum of, the wall = 2mvx, 1, 1N, 2, 2, P= n m v or P=, m v .............(1), 3, 3V, * In a small time interval Δt, a molecule with, N, as n=, x-component of velocity vx will hit the wall, V, if it is within the distance vx Δt from the, NOTE, wall. (see fig below), 1.Number density( n) : It implies that if the, , (, , * The maximum, number of molecules, that will hit the right, side wall in a time, interval ∆t = nA vx Δt, where n= N/V, But, on the average,, half of these are, moving towards the, wall and the other half away from the wall. (ie,, nAvxΔt /2 molecules only hit the wall), * The total momentum transferred to the, wall by these molecules in time Δt is, n, Δ P= A v x Δ t (2 m v x )=A v 2x mn Δ t, 2, , * The force exerted by the molecules on the, wall (Newton's second law) is, F=, , number density increases then pressure will, increase. For example when we pump air, inside the cycle tyre or car tyre essentially the, number density increases and as a result the, pressure increases., 2.Mass of the molecule : Since the pressure, arises due to momentum transfer to the wall,, larger mass will have larger momentum for a, fixed speed. As a result the pressure will, increase., 3.Mean square speed : For a fixed mass if we, increase the speed, the average speed will also, increase. As a result the pressure will increase., 4. Pressure of a Mixture Of Non- reactive, Gases:, For a mixture of non-reactive ideal gases, the, total pressure gets contribution from each gas, in the mixture. Ie, P=P1 + P2 + P3 ..............., 13.5 KINETIC INTERPRETATION OF, TEMPERATURE, 1, 1N, 2, 2, P= n m v (or) P=, m v ..............(2), 3, 3V, , ΔP, 2, =n m A v x, Δt, , * Therefore pressure , P is, P=, , ), , F, 2, =n m v x, A, , ie,, , 1, 2, PV = N m v ...............(3), 3, , * Since all the molecules are moving Comparing (3), with ideal gas equation, completely in random manner, they do not, PV =N k B T ,, have same speed. So we can replace the term, 2, v 2x by v x, 1, 1, 2, (3)=> N k B T = N m v 2 =>, k B T = mv, Therefore pressure P is P=n m v 2x, 3, 3, * Since the gas is assumed to move in random, direction, it has no preferred direction, of motion. It implies that the molecule has, same average speed in all the three direction., So v 2x =v2y =v 2z, , Multiply the above equation by 3/2 on both, sides, , This is the average kinetic energy of a single, molecule (ε)., , KAMIL KATIL VEETIL @, , 80, , 3, 1, 2, k BT = m v, 2, 2, , SOHSS AREEKODE

Page 83 :
square speed is defined as the square root of, the mean of the square of speeds of all, * By multiplying the total number of gas molecules. It is denoted by vrms.., 3k T, molecules with average kinetic energy of each, mean square speed , v 2= B, m, molecule, the internal energy of the gas is, Therefore, ,, obtained., That is the Internal energy of ideal gas, 3 k BT, 3, KE=ϵ= k B T, 2, , ie,, , root mean square speed , v=v rms =, , 1, 2, U=N ( m v ), 2, (or), , (or), , 3, U= N k B T, 2, , ie,, , v rms =, , √, , √, , 3 kB T, 3 RT, =, m, M, , √, , m, , Where m – mass of the molecule , M – molar, mass of the molecule., , 3, U= μ RT, 2, , Impact of v rms in nature : Moon has no, atmosphere: The escape speed of gases on the, surface of Moon is much less than the root, mean square speeds of gases due to low, gravity. Due to this all the gases escape from, the surface of the Moon., 2. Mean (or) average speed (v):It is defined, as the mean (or) average of all the speeds of, molecules, If v 1 , v 2 , v 3 .......... v n are the individual speed, 13.6 RELATION BETWEEN PRESSURE, of the molecules. Then, AND INTERNAL ENERGY, v + v +v +.......... v n, 8 kB T, 8 RT, The internal energy of the gas is given by, v= 1 2 3, =, =, NOTE, The average kinetic energy of each molecule, depends only on temperature of the gas not on, mass of the molecule. In other words, if the, temperature of an ideal gas is measured using, thermometer, the average kinetic energy of, each molecule can be calculated without, seeing the molecule through naked eye., , ( Or), (Or), , 3, U= N k B T, 2, 3, U= PV, 2, , P=, , √, , N, , πm, , √, , πM, , 3.Most probable speed (vmp ):It is defined as, the speed acquired by most of the molecules of, the gas., , 2U 2, = u, 3V 3, , v mp =, , √, , √, , 2 k BT, 2 RT, =, m, M, , ie, The pressure of the gas is equal to two thirds, 4. Mean Free Path (λ) : It is the average, of internal energy per unit volume, (U/V=u), , distance travelled by the molecule between, 13.7 RELATION BETWEEN PRESSURE AND two successive, collisions., AVERAGE KINETIC ENERGY, 1, 1N, 1 2, 2, 2, P= n m v =, mv = ρ v, 3, 3V, 3, , Expression for Mean Free Path, Consider a system of molecules each with, Multiply and divide R.H.S of equation by 2 , diameter d. Let n be the number of molecules, 2 ρ 2, per unit volume., P= ( v ), We get, 3 2, Assume that only one molecule is in, motion and all others are at rest as shown in, ie, pressure is equal to 2/3 of mean kinetic the Figure., energy per unit volume., 13.8 SOME IMPORTANT TERMS, 1.Root mean square speed (vrms ): Root mean, We know, , KAMIL KATIL VEETIL @, , 81, , SOHSS AREEKODE

Page 84 :
√, √, , 3 RT, v rms ( Ar ), M Ar, M Cl, 70.9, =, =, =, =1.33, M, 39.9, v rms (Cl), 3 RT, Ar, M Cl, , * If a molecule moves with average speed v in, a time t, the distance travelled is vt., * In this time t, consider the molecule to move, in an imaginary cylinder of volume πd2vt., * It collides with any molecule whose centre is, within this cylinder., * Therefore, the number of collisions is equal, to the number of molecules in the volume of, the imaginary cylinder. It is equal to πd2vtn ., , √ √, , 13.9 DEGREES OF FREEDOM (f), The total number of co-ordinates or, independent quantities required to completely, specify the position and configuration of a, system is called the, degrees of freedom of that system.( OR it is, the number of squared term in the expression, of total energy of a molecule), , 13.9.1 Degrees of freedom of a monatomic, * The total path length divided by the number molecule, of collisions in time t is the mean free path., A monatomic molecule has only three, translational degrees of freedom. The average, Distance travelled, kinetic energy of a monatomic molecule is, Mean free path, λ=, Number of collisions, vt, 1, λ=, =, 2, 2, n π d vt n π d, , given by, , A more exact treatment gives, λ=, , 1, 2 1, 2 1, 2, KE= m v x + m v y + m v z, 2, 2, 2, , So f =3, (total no of squared terms ), Eg: Helium, Neon, Argon, , 1, √2 n π d 2, , 13.9.2 Degrees of freedom of a diatomic, molecule, 1. At Normal temperature, Problem1 (NCERT example 13.5), A diatomic molecule at normal, A flask contains argon and chlorine in the temperature has 3 translational, ratio of 2:1 by mass. The temperature of the degrees of freedom + 2, Rotational degrees of freedom., mixture is 27 °C. Obtain the ratio of, moment of inertia about its, (i) Average translational kinetic energy per, own axis of rotation is, molecule of the two gases and, (ii) Root mean square speed v rms of the negligible (about y axis in the figure )., Therefore, it has only two rotational degrees of, molecules of the two gases., freedom (one rotation is about Z axis and, (Atomic mass of argon = 39.9 u; Molecular, another rotation is about X axis). Avg KE is, mass of chlorine = 70.9 u.), given, by, Solution, 1, 2 1, 2 1, 2 1, 2 1, 2, (i) Since argon and chlorine both have the, KE= m v x + m v y + m v z + I x ω + I y ω, 2, 2, 2, 2, 2, same temperature in the flask, the ratio of, average translational kinetic energy (per, So f = 5, (Total no of squared terms), molecule) of the two gases is 1:1., , √, , 2.At High Temperature, At a very high temperature such as 5000, K, the diatomic molecules possess additional, two degrees of freedom due to vibrational, , √, , 3 kB T, 3 RT, =, (ii) We know v rms =, m, M, , The ratio of the rms speed of molecules,, , KAMIL KATIL VEETIL @, , 82, , SOHSS AREEKODE

Page 85 :
motion[one due to kinetic energy of vibration, and the other is due to potential energy]., So f = 7, Eg: Hydrogen, Nitrogen, Oxygen, , 13.11 APPLICATION OF LAW OF, EQUIPARTITION ENERGY IN SPECIFIC, HEAT OF A GAS, 1) Monatomic molecule, , 13.9.3 Degrees of freedom of a Triatomic The average kinetic energy of a monatomic, 3, molecules, molecule ,, KE= k B T, 2, 1. Linear triatomic molecule, Therefore the total energy of a one mole gas,, At normal temperature f = 5, 3, 3, At high temperature f=7, =>, U= k B T N A, U= R T, 2, 2, Eg: Carbon dioxide., 2. Non Linear triatomic molecule, The molar specific heat at constant volume, CV, f=6, Eg: Water, Sulphur dioxide., CV =, , [ ], , dU d 3, 3, =, RT = R, dT dT 2, 2, , 13.10 LAW OF EQUIPARTITION OF, ENERGY, 3, 5, The average kinetic energy of system of, C P=C V + R= R+ R= R, 2, 2, molecules in thermal equilibrium at, temperature T is uniformly distributed to all, degrees of freedom so that each degree of The ratio of specific heats,, freedom will get, , 1, k T, 2 B, , of energy. Thus, , 5, R, CP 2, 5, γ= =, = =1.67, CV 3, 3, R, 2, , *The Average kinetic energy of a monatomic, molecule (with f=3), 1, 2, , = 3 x k BT, , =, , 3, k T, 2 B, , 2)Diatomic molecule, *The average kinetic energy of a diatomic The average kinetic energy of a diatomic, 5, molecule at low temperature (with f = 5), molecule at low temperature, KE= k B T, 2, 1, 5, = 5 x k BT =, k BT, Therefore the total energy of one mole of gas ,, 2, , 2, , 5, U= k B T N A, 2, , *The Average kinetic energy of a diatomic, molecule at high temperature (with f =7), =, , 7, k T, 2 B, , CV =, , 5, k T, 2 B, , 7, k T, 2 B, , The ratio of specific heats,, , *The Average kinetic energy of a non linear, triatomic molecule at high temperature, (with f = 6), , =, , [ ], , dU d 5, 5, =, RT = R, dT dT 2, 2, , 5, 7, C P=C V + R= R+ R= R, 2, 2, , *The Average kinetic energy of linear triatomic, molecule at high temperature (with f = 7), =, , 5, U= R T, 2, , The molar specific heat at constant volume, CV, , *The Average kinetic energy of linear triatomic, molecule at low temperature (with f = 5), =, , =>, , 7, CP 2 R 7, γ= =, = =1.40, CV 5, 5, R, 2, , 6, k T, 2 B, , Similarly Energy of a diatomic molecule at, KAMIL KATIL VEETIL @, , 83, , SOHSS AREEKODE

Page 86 :
5, 2, , high temperature , U= R T, The molar specific heat at constant volume, CV, CV =, , [ ], , dU d 7, 7, =, RT = R, dT dT 2, 2, , 13.13 MAXWELL-BOLTZMANN SPEED, DISTRIBUTION FUNCTION, To find how many gas molecules have the, range of speed from to v + dv. This is given by, Maxwell’s speed distribution function., , 7, 9, C P=C V + R= R+ R= R, 2, 2, , The ratio of specific heats,, , The above expression is graphically shown as, follows, , 9, CP 2 R 9, γ= =, = =1.28, CV 7, 7, R, 2, , 3)Triatomic molecules, Do yourself for linear and for non linear, molecules, 13.12 SPECIFIC HEAT CAPACITY OF, SOLIDS, *Consider a solid of N atoms, each vibrating, about its mean position., *An oscillation in one dimension has, It is clear that, for a given temperature the, 1, number of molecules having lower speed, average energy of 2× k B T =k B T, 2, increases, parabolically, but, decreases, *In three dimensions, the average energy is, exponentially after reaching most probable, 1, speed., 6 × k B T =3 k B T, 2, The rms speed, average speed and most, *For a mole of solid, N = NA, probable speed are indicated in the Fig. It can, Therefore total energy , U=3 N A k B T =3 R T, be seen that the rms speed is greatest among, the three. The area under the graph will give, *Now at constant pressure, the total number of gas molecules in the, ΔQ=ΔU + PΔV = ΔU, system, (since for a solid ΔV is negligible), PROBLEMS, Hence, , C=, , ΔQ ΔU, =, =3 R, ΔT ΔT, , 13.13 SPECIFIC HEAT CAPACITY OF, WATER, We treat water like a solid. For each atom, average energy is 3 k B T . Water molecule, has three atoms, two hydrogen and one, oxygen., So U= 3 x 3NAkBT=9RT, and C=9 R, , 1)A room contains oxygen and hydrogen, molecules in the ratio 3:1. The temperature of, the room is 27°C. The molar mass of 02 is 32 g, mol -1 and for H 2 is 2 g mol -1 . The value of, gas constant R is 8.32 J mol -1 K -1. Calculate, (a) rms speed of oxygen and hydrogen, molecule, (b) Average translational kinetic energy per, oxygen molecule and per hydrogen molecule, (c) Ratio of average internal energies of, , KAMIL KATIL VEETIL @, , 84, , SOHSS AREEKODE

Page 87 :
oxygen molecules and hydrogen molecules, Soln) (a), Oxygen, , No of molecules of H2=1000/2 x 6.023 x 1023, So No of H2 molecules, > No of H2, molecules, , Hydrogen, , (b)From ideal gas equation we can say, PV = nRT, P = nRT/V, so here pressure of the gas directly, proportional to its number of moles because, temperature and volume is same here in both, containers., so pressure of hydrogen gas will be more, iii) RMS speed of the gas is given as, , (b)The average translational kinetic energy per, molecule is 3/2KBT. It depends only on, absolute temperature of the gas and is So vrms of Hydrogen > vrms of Nitrogen, independent of the nature of molecules. Since, both the gas molecules are at the same, temperature, they have the same average, kinetic energy per, molecule., , (c) The internal energy of total oxygen, molecules =, , 3, N k T, 2 O B, , Average kinetic energy of total hydrogen, molecules =, , 3, N k T, 2 H B, , It is given that the number of oxygen, molecules is 3 times more than number of, hydrogen molecules in the room. So the ratio, of internal energy of oxygen molecules with, internal energy of hydrogen molecules is 3:1, 2) Two vessels of the same size are at the same, temperature. One of them holds 1 kg of N 2 gas., (molar weight 28g) and the other contains 1kg of, hydrogen (molar weight 2g), (a) Which of the vessels contains more molecules?, (b) Which of the vessels is under greater pressure, and why?, (c) In which vessel is the average molecular speed, greater? How many times is it greater?, , Soln) (a) No of moles of N2=1000/28, No of moles of H2=1000/2, No of molecules of N2=1000/28 x 6.023 x 1023, KAMIL KATIL VEETIL @, , 85, , SOHSS AREEKODE

Page 88 :
CHAPTER FOURTEEN, , OSCILLATIONS, PERIODIC MOTION, * Any motion, which repeats itself in regular, interval of time, is called a periodic motion., Examples, * Hands in pendulum clock, * Swing of a cradle, * The revolution of the Earth around the Sun, * Waxing and waning of Moon, etc., * Vibration of tuning fork, Types of periodic motion, 1) Rotatory motion, Particle completes the rotation in regular, interval of time, Example : The revolution of the Earth around, the Sun, 2) Oscillatory motion, The particle moves to and fro with less, frequency., Example : Swing of a cradle, 3) Vibratory motion, the particle moves to and fro with large, frequency., Example : Vibration of tuning fork, , Displacement Variable, The physical quantity which changes with, time in a Periodic motion is called, displacement Variable or displacement., Examples, * For an oscillating simple pendulum - the, angle from the vertical is the displacement., * Propagation of sound wave - pressure, variation is the displacement., * Alternating current – Voltage or current is, the displacement., SIMPLE HARMONIC MOTION (SHM), * Simple harmonic motion is a special type of, oscillatory motion in which there is a restoring, force acting on the particle which is directly, proportional to its displacement from a fixed, point and is always directed towards, that fixed point., Differential Equation for a simple harmonic, motion, Consider a particle vibrating to and fro about, the origin of an x-axis between the limits –A, and +A as shown in figure ., , Note, All oscillatory motion are periodic whereas all, periodic motions are need not be oscillatory., Period ( T ), * The smallest interval of time after which the, motion is repeated is called its period., * Unit : second, Frequency ( υ ), * Number of repetitions per second ., υ=, , 1, T, , Restoring force , F α – x ..............(1), Or, , F = - kx ...............(2), , Where k – The spring constant, From Newton's second law, , * Unit : hertz (Hz), Angular frequency ( ω ), ω=, , 2π, T, , or, , F=m, , ω=2 π υ, , d2 x, ..............(3), dt 2, , Comparing (2) and (3), , Do example 14.1, , m, , KAMIL KATIL VEETIL, 86, , @, , d2 x, = -kx, dt 2, , S O H S S AREEKODE

Page 90 :
Variation of displacement, velocity and, acceleration at different instant of time, , => K=, , T, 4, π, 2, , T, 2, , 3T, 4, , T, , π, , 3π, 2, , 2π, , 0, , A, , 0, , -A, , 0, , v, Aωcosωt, , Aω, , 0, , -Aω, , 0, , Aω, , a, 2, -Aω sinωt, , 0, , -Aω2, , 0, , Aω2, , 0, , t, , 0, , ωt, , 0, , x, Asinωt, , 1, 2, 2, k ( A −x ), 2, , .....................(17), , Potential energy ( U ), U=, , ∫ F ext dx, , =>, , U=, , ∫ kx dx, , =>, , U=, , 1 2, kx, 2, , Or U =, , =k, , ∫ x dx, , 2, =k x, , 2, , .....................(18), , 1, 2, 2, k A sin ωt ..................(19), 2, , Total energy ( E ), E=K+U, =>, , E=, , 1, 1 2, 2, 2, kx, k ( A −x ) +, 2, 2, , =>, , E=, , 1, 2, kA, 2, , ...............(20), , Variation of KE (K) ,PE (U) and TE (E), with the displacement, , 4) Energy in SHM, A particle executing simple harmonic motion, has kinetic and potential energies., Kinetic energy ( K ), K=, , 1, 2, mv =, 2, , => K =, , 1, 2, m(A ω cos ω t), 2, , Variation of KE (K) , PE (U) and TE (E) at, different instant of time, , 1, 2 2, 2, m A ω cos ω t ..............(16), 2, , 1, 2, 2, k A cos ω t, 2, 1, 2, 2, => K =, k A (1−sin ω t ), 2, , => K =, , (, , k, =ω 2 ), m, , Note : The kinetic energy and potential energy, both repeat after a period T/2. The total energy, remains constant at all t or x., , 1, => K = k ( A 2−A 2 sin2 ωt ), 2, , KAMIL KATIL VEETIL, 88, , @, , S O H S S AREEKODE

Page 91 :
5) Time period of oscillation, We know ω2=, Or, =>, , ω=, , √, , k, m, , k, m, , A on y axis at time t ,, Projection of ⃗, 2π, =, T, , =>, , T =2 π, , √, , √, , y(t)=A sinθ = A sinωt ..........(24), , k, m, , From (23) and (24) we can conclude that, a simple harmonic motion can be defined as, the projection of uniform circular motion on a, diameter of the circle., , m, ...........(21), k, , 6) Frequency of oscillation, 1, T, , EXAMPLES OF SIMPLE HARMONIC, MOTION, , √, , 1) Simple pendulum, , ν=, , ie, ν= 1, , k, ............(22), 2π m, , 7) Force ( F ), From equation (2), , F = - kx, , Slope =, , x 1, =, F k, , * From the diagram T = mg cosθ, * ie, mg cosθ cancel with the tension T in, the string., * Mg sinθ acts as the restoring force., * Therefore the restoring torque,, , SIMPLE HARMONIC MOTION AND, UNIFORM CIRCULAR MOTION, , τ = − mg sinθ x L ................(25), Negative sign shows that force acts to, reduce θ., We know τ = I α, ...............(26), Comparing (25) and (26), Consider a particle executing uniform circular, motion as shown in the diagram., From the diagram ,, , I α =− mg L sinθ .................(27), If θ is small, sinθ ≈ θ, , (θ in radian), , Therefore (25) => I α =− mg L θ, , A , The position vector of the, Let OA = ⃗, particle at time t, , =>, , A on x axis at time t ,, Projection of ⃗, x(t)=A cosθ = A cosωt ..........(23), similarly, , α=, , −mgL, θ ................(28), I, , But the moment of inertia of the bob is,, I =mL2, , KAMIL KATIL VEETIL, 89, , @, , S O H S S AREEKODE

Page 92 :
Therefore, , Differential Equation for a damped simple, harmonic motion, , −mgL, θ, m L2, , α=, , Or, * Damping force (Fd) is proportional to, velocity and opposite to the velocity, ie ,, Fd = − b v ...............(32), , −g, θ ..............(29), L, , α=, , * That is, the angular acceleration of the, pendulum is proportional to the angular, displacement θ but opposite in sign. Thus the, motion of a simple pendulum swinging, through small angles is SHM., * Period of simple pendulum, , * The restoring force F = − kx ...........(33), * Total force , F = −bv − kx, , −g, θ, L, 2π, g, =>, =, T, L, , Comparing a=−ω2 x with α =, , √, , we get ω2=, , g, L, , ie,, , T = 2π, , * Frequency, , => ω= g, , √, √, , L, g, , L, , Or, , √, , F = −b, , dx, − kx ................(34), dt, , According to Newton's second law,, 2, F = m d x2, , ................(30), , dt, , ...............(35), , Comparing (35) and (36), , 1 g, ................(31), υ=, 2π L, , m, , Note, * Time period is independent of Mass of the, bob (m) , Amplitude of oscillations(θ), depends only on length of the pendulum(L), and acceleration due to gravity(g), * Seconds pendulum, A pendulum whose period is 2s., * Length of the second pendulum, , √, , L, We know T = 2 π, g, 2, Therefore L= T g2, 4π, , For seconds pendulum L=, , dx, d2 x, = −b − kx, 2, dt, dt, , =>, , d2 x, dx, m 2 +b + kx=0, dt, dt, , =>, , d 2 x b dx k, +, + x=0, dt 2 m dt m, , =>, , d 2 x b dx, 2, +, +ω x=0 ...............(37), 2, dt m dt, , ...............(36), , This is the differential equation for a damped, SHM., , 22 x 9.8, = 1m, 4 π2, , Solution of the equation, , 2) Horizontal Oscillations of a block of, mass attached to a spring., Refer the above discussions, , x = A e −b t / 2m sin(ω! t+ ϕ), Where A e −b t / 2m is the amplitude of the, oscillation., , DAMPED SIMPLE HARMONIC, MOTION, * If the amplitude of oscillation continuously, decreases (due to dissipating forces like, frictional force ) is said to be damped simple, harmonic motion., , Energy , E =, , Also, , KAMIL KATIL VEETIL, 90, , @, , ω! =, , 1, k A 2 e− b t /m, 2, , √, , k, b2, −, m 4 m2, , S O H S S AREEKODE

Page 93 :
* Note, Amplitude of oscillation is not constant but, decreases exponentially with time . Similarly, energy also decreases exponentially with time., It is graphically shown below., , * Tacoma Bridge in Washington was destroyed, by resonance produced by wind., , ADDITIONAL DISCUSSION, COMBINATIONS OF SPRINGS, 1 . Series combination, , FORCED OSCILLATIONS, * Any oscillator driven by an external periodic, agency to overcome the damping is known, as forced oscillator or driven oscillator and the, oscillation is known as forced oscillation., , The effective spring constant , k s is given by, , Differential Equation for forced oscillations, , 2 . Parallel combination, , m, , k s=, , k1 k 2, k 1 +k 2, , d2 x, dx, +b + kx=F 0 sin ω d t .........(38), 2, dt, dt, , Where F0 - Amplitude of the driving force, ωd - Frequency of the driving force, Solution of the equation, The effective spring constant , k s is given by, , x= A ! sin( ωd t+ ϕ) ..............(39), , k s=k 1 + k 2, , Where the amplitude,, A !=, , F0, , √m (ω −ω ) +ω b, 2, , 2, , 2 2, d, , 2, d, , 2, , ..........(40), , RESONANCE, * From (40) it is clear that the amplitude of the, oscillation will be maximum when ω=ωd, The phenomenon of increase in amplitude of, oscillations when the driving frequency is, close to the natural frequency of the oscillator, is called resonance., Note, * Soliders are not allowed to march on a, bridge. This is to avoid resonant vibration of, the bridge., , KAMIL KATIL VEETIL, 91, , @, , S O H S S AREEKODE

Page 94 :
CHAPTER FIFTEEN, , WAVES, WAVE MOTION, * The propagation of disturbance which carries, energy and momentum from one point in space, to another point in space without the transfer, of the medium is known as a wave motion., Types of waves, 1. Mechanical waves, 2. Non mechanical waves, , PROGRESSIVE WAVES ( TRAVELLING, WAVES), A wave which travels from one point of the, medium to another is called a progressive, wave or travelling wave., [* See the appendix before proceeding further], , 1. Mechanical waves, * Waves which require a medium for propaga tion are known as mechanical waves., *Ex : Sound waves ,, Ripples formed on the surface of water, etc., , DISPLACEMENT RELATION IN A, PROGRESSIVE WAVE, * A transverse wave travelling in the, positive x direction can be represented as:, y (x,t) = A sin (kx – ωt + ϕ ) .........(1), , 2. Non mechanical waves, * Waves which do not require any medium for, propagation are known as non-mechanical, waves., * Ex : All electromagnetic waves (Radio, waves, micro waves , light ,etc.), , * Similarly A transverse wave travelling in the, negative x direction can be represented as:, y (x,t) = A sin (kx + ωt + ϕ ) .........(2), , TYPES OF MECHANICAL WAVES, a) Transverse waves, b) Longitudinal waves, , Note: The displacement , y (x,t) is a function, of x and t ., , a) Transverse wave, * In transverse wave motion, the constituents, of the medium vibrate about their mean, positions in a direction perpendicular to the, direction of propagation., * Ex : Harmonic wave travelling along a, stretched string., Where,, , b)Longitudinal wave, * In longitudinal wave motion, the constituent, of the medium vibrate about their mean, positions in a direction parallel to the direction, of propagation, * Ex : Sound waves travelling in air, , 2π, T, , and, , 2π, k= λ, , Graphical representation of the wave, Let us graphically represent the two forms, of the wave variation, (a) Space (or Spatial )variation graph, (b) Time (or Temporal) variation graph, , KAMIL KATIL VEETIL, 92, , ω=, , @, , S O H S S AREEKODE

Page 95 :
(a) Space (or Spatial )variation graph, , the displacement of the wave at this instant, is, y(xʹ, tʹ) = A sin[k (x + ∆x)- ω (t + ∆t)] .......(7), Since the shape of the wave remains the, same (i.e., the y- displacement of the point is a, constant), the phase of the wave remains, constant., , Graph of sinusoidal function y = A sin(kx) ., By keeping the time fixed, the change in, displacement with respect to x is plotted., , Ie,, , y(x',t') = y(x,t),, , k (x + ∆x)− ω (t + ∆t)= k x− ω t = constant, , From the graph ,, y = A sin(kx) = A sin(k(x + λ)), , => d( k x− ω t )= d(constant), , = A sin(kx + k λ) ...........(3), , The sine function is a periodic function with, period 2π. Hence,, y = A sin(kx + 2π) = A sin(kx) .........(4), , => d( k x− ω t )= 0, =0, => k dx = ω dt, , 2π, k = λ rad/m, , Or, , ...........(5), , v=, , =>, =>, , From (3) and (4) k λ = 2π, Or, , which implies, , k dx – ω dt, , dx, =v= ω .....(8), dt, k, , 2 π /T λ, = =υ λ, 2 π/λ T, , ..........(9), , Speed of a transverse wave on a stretched, string, , (b) Time variation graph, , √, , T, v= μ, , .............(10), , Where, T – Tension in the string, μ=, , Graph of sinusoidal function y =A sin(ωt)., By keeping the position fixed, the change in, displacement with respect to time is, plotted., Time period is given by, , mass of the string, length of the string, , This relation can be derived using dimensional, analysis( Do yourself), Do Example 15.3 NCERT, , 2π, T= ω s, , Speed of a Longitudinal Wave, (Speed of Sound), * The general formula for velocity of longitudinal waves in a medium is given by, , THE SPEED OF A TRAVELLING WAVE, , √, , B, v= ρ, , ...............(11), , * The speed of longitudinal waves in a solid, bar is given by, , The displacement of the wave at an instant t is, y(x,t) = A sin(k x− ω t) ..............(6), At the next instant of time tʹ = t + ∆t the, position of the point P is xʹ = x + ∆x. Hence,, , √, , Y, v= ρ, , KAMIL KATIL VEETIL, 93, , @, , ...............(12), , S O H S S AREEKODE

Page 96 :
Note : Liquids and Solids have higher mass, densities ( ρ ) than gases. But the, corresponding increase in the bulk modulus, (B)of solids and liquids is much higher. This is, the reason why the sound waves travel faster in, solids and liquids, Medium, Speed (m/s), Air ( 0 0C), , 331, , Air ( 20 0C), , 343, , Water ( 20 0C), , 1482, , Sea water, , 1522, , Steel, , 5941, , Aluminium, , 6420, , −VdP, dV, , =>, , P=, , P γ V γ−1 dV =−V γ dP, , =>, , −V γ dP −V γ dP, P γ= γ−1, = γ, =B, V dV V dV /V, , ie,, , γ p=B, , , the bulk modulus, , * Therefore velocity of sound is given by, v=, , √, , γP, ρ ................(14), , * For air γ = 7/5. Now using (14) to estimate, the speed of sound in air at STP, we get a value, 331.3 m/s , which agrees with the measured, speed., , Speed of sound in a gas, 1. Newton’s formula, *Newton assumed that when sound propagates, in air, the formation of compression and, rarefaction takes place in a very slow manner, so that the process is isothermal in nature., * For an isothermal process, P V = constant, => d( P V )=0, => P dV + V dP =0, =>, P dV = −V dP, => P =, , =>, , −dP, =B, dV / V, , THE PRINCIPLE OF SUPERPOSITION, OF WAVES, When a no. of waves meet at a particular point, in a medium, net displacement at the point is, the algebraic sum of the displacements due, to each wave., y(x,t) = y 1 (x,t) + y 2 (x,t) + .........., Superposition of waves going in same, direction, Consider two harmonic travelling waves on a, stretched string both having the same, frequency and same amplitude but with, different initial phase., , ie, P = B , the bulk modulus, * Therefore velocity of sound is given by, , √, , P, v = ρ ................(13), , This is Newton's formula ., * When we calculate the velocity of sound in, air at STP using the above formula, the, obtained value is 280m/s, which is about 15%, smaller as compared to the experimental value, of 331 m/s., * Laplace corrected this problem, Laplace formula ( Laplace correction), *Laplace assumed that the process is adiabatic., * For an adiabatic process, PV γ =constant => d ( P V γ )=0, , Let the waves are travelling along the positive, x-direction ., So, y 1 ( x , t)=A sin(kx −ωt ), y 1 ( x , t)=A sin(kx − ωt+ ϕ), , Suppose they move simultaneously in a, particular direction, then interference occurs., According to superposition principle, y= A sin( kx − ωt )+ A sin (kx −ωt +ϕ), , => Pd V γ +V γ dP=0, , => y= A [sin (kx −ωt )+sin( kx − ωt+ϕ)] .....(15), , => P γ V γ−1 dV + V γ dP=0, , We know, , KAMIL KATIL VEETIL, 94, , @, , S O H S S AREEKODE

Page 97 :
sin A +sin B=2 sin(, , A+ B, A−B, )cos (, ), 2, 2, , closed end, is reflected with a phase reversal., * ie, there will be a phase difference of π, between the incident and reflected waves., * The incident wave ,, , So equation (15) becomes, , ϕ, −ϕ, y= A [2 sin(kx − ωt + )cos (, )], 2, 2, , =>, , y i ( x ,t)=A sin(kx − ωt), , So the reflected wave,, , ϕ, ϕ, y (x ,t )=2 Acos ( )[sin (kx −ωt + )], 2, 2, , y r ( x , t)= A sin(kx +ωt +π), , *This is also a travelling wave in the positive, x-direction , with the same frequency and, wavelength. However, its initial, ϕ, phase angle is, ., , Or, , y r ( x , t)=−A sin( kx+ ωt), , 2, , * The amplitude of the resultant wave is, ϕ, A (ϕ)=2 Acos ( ) ..............(16), 2, , * Special cases, Case 1 : if ϕ=0 , the waves are in phase., , And, , (2) Reflection of wave at an open boundary, * A travelling wave ,at an open boundary is, reflected without any phase change., * So the reflected wave is ,, , ϕ, ϕ, y (x ,t )=2 Acos ( )[sin (kx −ωt + )], 2, 2, A (ϕ)=2 A, , y r ( x , t)= A sin(kx +ωt ), , i.e, the resultant wave has amplitude 2a, the, largest possible value for A . This is the, constructive interference., , Case 2 : if ϕ=π , the waves are completely, out of phase., y ( x ,t )=0, , and, , NOTE, * For longitudinal wave there is no phase, change on reflection at a rigid boundary(closed, end of a pipe) and a phase change of π at an, open boundary ( Open end of a pipe), , A (ϕ)=0, , ie, the resultant wave has zero displacement, everywhere at all times. This is the destructive, interference., , STANDING WAVES AND NORMAL, MODES, * When two waves of same amplitude and, frequency travelling in opposite direction super, impose , the resulting wave pattern does not, move to either sides. This pattern is called, standing waves., * To see mathematically , consider a wave, travelling along the positive x- direction and a, reflected wave of the same amplitude and, , REFLECTION OF WAVES, (1) Reflection of wave at a rigid boundary, * A travelling wave, at a rigid boundary or a, , KAMIL KATIL VEETIL, 95, , @, , S O H S S AREEKODE

Page 99 :
This frequency is called fundamental, frequency or first harmonic, The fundamental frequency can also be, expressed in terms of 'T 'and 'μ' of the string, as, υ1=, , √, , 1 T, ................(20), 2L μ, , Second mode of vibration, L=λ2, , =>, , λ 2=L, , Therefore frequency of vibration,, V V, υ2= λ = =2 υ1, L, 2, , .................(21), , This frequency is called second harmonic or, first overtone., The second harmonic can also be expressed in, terms of 'T 'and 'μ' of the string as, υ2=, , √, , 1 T, L μ, , ..............(22), , Third mode of vibration, L=, , 3 λ3, 2, , =>, , λ 3=, , 2L, 3, , First mode of vibration, , Therefore frequency of vibration,, V 3V, υ3= λ =, =3 υ1, 2L, 3, , L=, , .................(23), , υ1=, , The third harmonic can also be expressed in, terms of 'T 'and 'μ' of the string as, , √, , =>, , λ 1=4 L, , Therefore frequency of vibration,, , This frequency is called third harmonic or, second overtone., , 3 T, υ3=, 2L μ, , λ1, 4, , V, V, ...............(25), =, λ1 4 L, , This frequency is called fundamental, frequency or first harmonic., , ..............(24), , The first harmonic can also be expressed in, terms of 'P 'and 'ρ' of the gas as, , STANDING WAVES IN A CLOSED PIPE, * A closed pipe means one end closed and, other end opened. A PVC pipe partially filled, with water is an example, , υ1=, , KAMIL KATIL VEETIL, 97, , @, , √, , 1 γP, ..............(26), 4L ρ, , S O H S S AREEKODE

Page 100 :
Second mode of vibration, 3, L= λ 2, 4, , =>, , λ 2=, , First mode of vibration, , 4L, 3, , L=, , Therefore frequency of vibration,, υ2=, , λ1, 2, , =>, , λ 1=2 L, , Therefore frequency of vibration,, , V 3V, =, =3 υ1 ...............(27), λ2 4 L, , V V, ..............(31), υ1= λ =, 2L, 1, , This frequency is called third harmonic or, first overtone., , This frequency is called fundamental, frequency or first harmonic., , The third harmonic can also be expressed in, terms of 'P 'and 'ρ' of the gas as, , The fundamental harmonic can also be, expressed in terms of 'P 'and 'ρ' of the gas as, , υ2=, , √, , 3 γP, ..............(28), 4L ρ, , Third Mode of vibration, 5, L= λ 3, 4, , υ1=, , Second mode of vibration, L=λ2, =>, , 4L, λ 3=, 5, , =>, , V V, υ2= λ = =2 υ1 ..............(33), L, 2, , V 5V, υ3= =, =5 υ1 ...............(29), λ3 4 L, , This frequency is called second harmonic or, first overtone., , This frequency is called fifth harmonic or, second overtone., , The second harmonic can also be expressed in, terms of 'P 'and 'ρ' of the gas as, , The third harmonic can also be expressed in, terms of 'P 'and 'ρ' of the gas as, , √, , λ 2=L, , Therefore frequency of vibration,, , Therefore frequency of vibration,, , 5, υ3=, 4L, , √, , 1 γP, ..............(32), 2L ρ, , γP, ρ ..............(30), , υ2=, , 1, L, , √, , γP, ρ ..............(34), , Third mode of vibration, , * For a closed pipe only odd harmonics, are present., , 3, L= λ3, 2, , STANDING WAVES IN AN OPEN PIPE, , =>, , λ 3=, , 2L, 3, , Therefore frequency of vibration,, υ3=, , V 3V, =, =3 υ1 ..............(35), λ3 2 L, , This frequency is called third harmonic or, second overtone., , KAMIL KATIL VEETIL, 98, , @, , S O H S S AREEKODE

Page 102 :
Expression for apparent frequency, , *Now the listener also moving in the same, direction in which the sound is moving ., , *Consider a general case in which the source, of sound and the listener moving in a direction, of sound from the source to the listener. We, derive the general equation in 3 steps., V −Velocity of sound, Let, , In this case the apparent frequency received by, the listener is given by, υΙ=, , V S −Velocity of the source, V L −Velocity of the listener, , Relative velocity of sound wrto listener, apparent wavelength, , ie,, , * First let's assume that the source and the, listener at rest , and the source is emitting a, sound of frequency, υ . The sound moves, with a velocity , V through the medium., , υΙ=, , V −V L, λΙ, , =>, , υΙ=υ, , (Or), , (, , V −V L, V −V S, , υΙ=, , V −V L, V −V S, υ, , ), , Special Cases, 1. Source moves towards stationary listener, V L =0 , V S →+ ve ,, , υΙ=υ, , V, ( V −V, ), S, , 2.Source moves away from stationary, listener., V L =0 , V S →−ve ,, , (Or ) λ=, , V S =0 , V L →−ve ,, , V, λ= υ, , velocity of sound, Noof waves receiving per second, , V S =0 , V L →+ ve ,, , Ι, , Ι, , υ =υ, , (, , V +V L, V, , ), , υΙ=υ, , ( V −V, V ), L, , 5. Both approaching each other, V S →+ ve , V L →−ve, , υΙ=υ, , (, , V +V L, V −V S, , ), , 6.Both moves away from each other, V S →−ve , V L →+ ve, , Let λ Ι be the new wave length and is given, λ=, , ), , 4. Listener moves away from stationary, source, , *Now suppose the source moves with velocity,, V s towards the listener and because of this, motion of the source the wavelength get, reduced ( ie , listener receives more number of, waves per second), , by, , (, , V, V +V S, , 3. Listener moves towards stationary source, , In this case the listener will receive the real, frequency of the source., Therefore, , Ι, , υ =υ, , υΙ=υ, , (, , V −V L, V+VS, , V −V s, υ, , KAMIL KATIL VEETIL, 100, , @, , S O H S S AREEKODE, , )

Page 103 :
APPENDIX, Further, we can take a = vt and v = π , and, , Problem 1, Sketch y = x −a for different values of a., , 4, , sketching for different times t = 0s, t = 1s,, t = 2s etc., we once again observe that, y = sin(x−vt) moves towards the positive xdirection. Hence, y = sin(x−vt) is a travelling, wave moving towards the positive x direction., If y = sin(x+vt) then the travelling wave moves, towards the negative x-direction., , Solution, , * Thus, any arbitrary function of type y =, f(x−vt) characterising the wave must move, towards positive x -direction and similarly,, any arbitrary function of type y = f(x+vt), characterizing the wave must move towards, negative x-direction., , This implies, when increasing the value, of a, the line shifts towards positive, x- direction., Note : if y = x – vt and v= 1 m/s , then, y=x–t, This implies, when increasing the value, of ' t ' (t=1s , 2s, 3s.......) the line shifts, towards positive x- direction with a velocity, 1m/s., Problem 2, How does the wave y = sin(x − a) for a = 0,, , Problem 3, Check the dimension of the wave, y = sin(x−vt). If it is dimensionally wrong,, write the above equation in the correct form., , 3π, a= π , a= π , a=, , a=π look like?, 4, 4, 2, , Sketch the waves, Solution, , Solution, we know that (x−vt) must be a dimensionless, quantity but x−vt has dimension., The correct equation is y = sin (k x−ωt), where, k and ω have the dimensions of inverse, of length and inverse of time respectively., , From the above picture we observe that the, function y = sin (x−a) shifts towards positive, x- direction for those values of 'a' ., , KAMIL KATIL VEETIL, 101, , @, , S O H S S AREEKODE