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Chapter 1 UNITS AND DIMENSIONS Learning objective:After going through this chapter, students will be able to; - understand physical quantities, fundamental and derived; - describe different systems of units; - define dimension...

Chapter 1 UNITS AND DIMENSIONS Learning objective:After going through this chapter, students will be able to; - understand physical quantities, fundamental and derived; - describe different systems of units; - define dimensions and formulate dimensional formulae; - write dimensionalequations and apply these to verify various formulations. 1.1 DEFINITION OF PHYSICS AND PHYSICAL QUANTITIES Physics: Physics is the branch of science, which deals with the study of nature and properties of matter and energy. The subject matter of physics includes heat, light, sound, electricity, magnetism and the structure of atoms. For designing a law of physics, a scientific method is followed which includes the verifications with experiments. The physics, attempts are made to measure the quantities with the best accuracy.Thus, Physics can also be defined as science of measurement. Applied Physics is the application of the Physics to help human beings and solving their problem, it is usually considered as a bridge or a connection between Physics & Engineering. Physical Quantities: All quantities in terms of which laws of physics can be expressed and which can be measured are called Physical Quantities. For example; Distance, Speed, Mass, Force etc. 1.2 UNITS: FUNDAMENTAL AND DERIVED UNITS Measurement: In our daily life, we need to express and compare the magnitude of different quantities; this can be done only by measuring them. Measurement is the comparison of an unknown physical quantity with a known fixed physical quantity. Unit: The known fixed physical quantity is called unit. OR The quantity used as standard for measurement is called unit. For example, when we say that length of the class room is 8 metre. We compare the length of class room with standard quantity of length called metre. Length of class room = 8 metre Q = nu 1  Physical Quantity = Numerical value × unit Q = Physical Quantity n = Numerical value u = Standard unit e.g. Mass of stool = 15 kg Mass = Physical quantity 15 = Numerical value Kg = Standard unit Means mass of stool is 15 times of known quantity i.e. Kg. Characteristics of Standard Unit: A unit selected for measuring a physical quantity should have the following properties (i) It should be well defined i.e. its concept should be clear. (ii) It should not change with change in physical conditions like temperature, pressure, stress etc.. (iii) It should be suitable in size; neither too large nor too small. (iv) It should not change with place or time. (v) It should be reproducible. (vi) It should be internationally accepted. Classification of Units: Units can be classified into two categories. Fundamental Derived Fundamental Quantity:The quantity which is independent of other physical quantities. In mechanics, mass, length and time are called fundamental quantities. Units of these fundamental physical quantities are called Fundamental units. e.g. Fundamental Physical Quantity Fundamental unit Mass Kg, Gram, Pound Length Metre, Centimetre, Foot Time Second Derived Quantity: The quantity which is derived from the fundamental quantities e.g. area is a derived quantity. Area = Length  Breadth = Length  Length = (Length)2 Speed =Distance / Time =Length / Time The units for derived quantities are called Derived Units. 2 1.3 SYSTEMS OF UNITS: CGS, FPS, MKS, SI For measurement of physical quantities, the following systems are commonly used:- (i) C.G.S system: In this system, the unit of length is centimetre, the unit of mass is gram and the unit of time is second. (ii) F.P.S system: In this system, the unit of length is foot, the unit of mass is pound and the unit of time is second. (iii) M.K.S: In this system, the unit of length is metre, unit of mass is kg and the unit of time is second. (iv) S.I System: This system is an improved and extended version of M.K.S system of units. It is called international system of unit. With the development of science & technology, the three fundamental quantities like mass, length & time were not sufficient as many other quantities like electric current, heat etc. were introduced. Therefore, more fundamental units in addition to the units of mass, length and time are required. Thus, MKS system was modified with addition of four other fundamental quantities and two supplementary quantities. Table of Fundamental Units Sr. No. Name of Physical Quantity Unit Symbol 1 Length Metre m 2 Mass Kilogram Kg 3 Time Second s 4 Temperature Kelvin K 5 Electric Current Ampere A 6 Luminous Intensity Candela Cd 7 Quantity of Matter Mole mol Table of Supplementary unit Sr. No Name of Physical Quantity Unit Symbol 1 Plane angle Radian rad 2 Solid angle Steradian sr Advantage of S.I. system: (i) It is coherent system of unit i.e. the derived units of a physical quantities are easily obtained by multiplication or division of fundamental units. (ii) It is a rational system of units i.e. it uses only one unit for one physical quantity. e.g. It uses Joule (J) as unit for all types of energies (heat, light, mechanical). (iii) It is metric system of units i.e. it’s multiples & submultiples can be expressed in power of 10. 3 Definition of Basic and Supplementary Unit of S.I. 1. Metre (m): The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second. 2. Kilogram (Kg) : The kilogram is the mass of the platinum-iridium prototype which was approved by the ConférenceGénérale des Poids et Mesures, held in Paris in 1889, and kept by the Bureau International des Poids et Mesures. 3. Second (s): The second is the duration of 9192631770 periods of the radiation corresponding to the transition between two hyperfine levels of the ground state of Cesium- 133 atom. 4. Ampere (A) : The ampere is the intensity of a constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2  10- 7 Newton per metre of length. 5. Kelvin (K): Kelvin is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water. 6. Candela (Cd): The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 x 1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian. 7. Mole (mol): The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of Carbon-12. Supplementary units: 1. Radian (rad): It is supplementary unit of plane angle. It is the plane angle subtended at the centre of a circle by an arc of the circle equal to the radius of the circle. It is denoted by 𝜃. 𝜃 = l / r; 𝑙 is length of the arcand 𝑟 is radius of the circle 2. Steradian (Sr): It is supplementary unit of solid angle. It is the angle subtended at the centre of a sphere by a surface area of the sphere having magnitude equal to the square of the radius of the sphere. It is denoted by Ω. Ω = ∆s / r2 SOME IMPORTANT ABBREVIATIONS Symbol Prefix Multiplier Symbol Prefix Multiplier D Deci 10-1 da deca 101 c centi 10-2 h hecto 102 m milli 10-3 k kilo 103 µ micro 10-6 M mega 106 n nano 10-9 G giga 109 4  P Pico 10-12 T tera 1012 f femto 10-15 P Pecta 1015 a atto 10-18 E exa 1018 Some Important Units of Length: (i) 1 micron = 10–6 m = 10–4 cm (ii) 1 angstrom = 1Å = 10–10 m = 10–8 cm (iii) 1 fermi = 1 fm = 10–15 m (iv) 1 Light year = 1 ly = 9.46 x 1015m (v) 1 Parsec = 1pc = 3.26 light year Some conversion factor of mass: 1 Kilogram = 2.2046 pound 1 Pound = 453.6 gram 1 kilogram = 1000 gram 1 milligram = 1/1000 gram = 10-3 gram 1 centigram = 1/100 gram = 10-2 gram 1 decigram = 1/10 gram 1 quintal = 100 kg 1 metric ton = 1000 kilogram 1.4 DEFINITION OF DIMENSIONS Dimensions: The powers, to which the fundamental units of mass, length and time written as M, L and T are raised, which include their nature and not their magnitude. For example Area = Length x Breadth = [ L1] × [L1] = [L2] = [M0L2T0] Power (0,2,0) of fundamental units are called dimensions of area in mass, length and time respectively. e.g. Density = mass/volume = [M]/[L3] = [ M1L-3T0] 1.5 DIMENSIONAL FORMULAE AND SI UNITS OF PHYSICAL QUANTITIES 5 Dimensional Formula:An expression along with power of mass, length & time which indicates how physical quantity depends upon fundamental physical quantity. e.g. Speed = Distance/Time = [L1]/[T1] =[M0L1T-1] It tells us that speed depends upon L & T. It does not depends upon M. Dimensional Equation: An equation obtained by equating the physical quantity with its dimensional formula is called dimensional equation. The dimensional equation of area, density & velocity are given as under- Area = [M0L2T0] Density = [M1L-3T0] Velocity = [M0L1T-1] Dimensional formula SI& CGS unit of Physical Quantities Sr. Physical Quantity Formula Dimensions Name of S.I unit No. 1 Force Mass × acceleration [M1L1T-2] Newton (N) 2 Work Force × distance [M1L2T-2] Joule (J) 3 Power Work / time [M1L2T-3] Watt (W) 4 Energy ( all form ) Stored work [M1L2T-2] Joule (J) 5 Pressure, Stress Force/area [M1L-1T-2] Nm-2 6 Momentum Mass × velocity [M1L1T-1] Kgms-1 7 Moment of force Force × distance [M1L2T-2] Nm 8 Impulse Force × time [M1L1T-1] Ns 9 Strain Change in dimension [M0L0T0] No unit / Original dimension 10 Modulus of Stress / Strain [M1L-1T-2] Nm-2 elasticity 11 Surface energy Energy / Area [M1L0T-2] Joule/m2 12 Surface Tension Force / Length [M1L0T-2] N/m 13 Co-efficient of Force × Distance/ [M1L-1T-1] N/m2 viscosity Area × Velocity 14 Moment of inertia Mass × (radius of [M1L2T0] Kg-m2 gyration)2 15 Angular Velocity Angle / time [M0L0T-1] Rad.per sec 16 Frequency 1/Time period [M0L0T-1] Hertz 17 Area Length × Breadth [M0L2T0] Metre2 18 Volume Length × breadth × [M0L3T0] Metre3 height 6  19 Density Mass/ volume [M1L-3T0] Kg/m3 20 Speed or velocity Distance/ time [M0L1T-1] m/s 21 Acceleration Velocity/time [M0L1T-2] m/s2 22 Pressure Force/area [M1L-1T-2] N/m2 Classification of Physical Quantity: Physical quantity has been classified into following four categories on the basis of dimensional analysis. 1. Dimensional Constant: These are the physical quantities which possess dimensions and have constant (fixed) value. e.g. Planck’s constant, gas constant, universal gravitational constant etc. 2. Dimensional Variable: These are the physical quantities which possess dimensions and do not have fixed value. e.g. velocity, acceleration, force etc. 3.DimensionlessConstant: These are the physical quantities which do not possess dimensions but have constant (fixed) value. e.g. e,𝜋 , 𝑛𝑢𝑚𝑏𝑒𝑟𝑠𝑙𝑖𝑘𝑒 1,2,3,4,5 etc. 4. Dimensionless Variable: These are the physical quantities which do not possess dimensions and have variable value. e.g. angle, strain, specific gravity etc. Example.1 Derive the dimensional formula of following Quantity & write down their dimensions. (i) Density (ii) Power (iii) Co-efficient of viscosity (iv) Angle Sol. (i) Density = mass/volume =[M]/[L3] = [M1L-3T0] (ii) Power = Work/Time =Force x Distance/Time =[M1L1T-2] x [L]/[T] =[M1L2T-3] Force x Distance (iii) Co-efficient of viscosity = Area x Velocity Mass x Acceleration x Distance x time length x length x Displacement =[M] x [LT-2] x [L] [T]/[L2] x [L] =[M1L-1T-1] (iv) Angle = arc (length)/radius (length) = [L]/[L] =[M0L0T0] = no dimension 7  Example.2 Explain which of the following pair of physical quantities have the same dimension: (i) Work &Power (ii) Stress & Pressure (iii) Momentum &Impulse Sol. (i) Dimension of work = force x distance = [M1L2T-2] Dimension of power = work / time = [M1L2T-3] Work and Power have not the same dimensions. (ii) Dimension of stress = force / area = [M1L1T-2]/[L2] = [M1L-1T-2] Dimension of pressure = force / area = [M1L1T-2]/[L2] = [M1L-1T-2] Stress and pressure have the same dimension. (iii) Dimension of momentum = mass x velocity= [M1L1T-1] Dimension of impulse = force x time =[M1L1T-1] Momentum and impulse have the same dimension. 1.6 PRINCIPLE OF HOMOGENEITY OF DIMENSIONS It states that the dimensions of all the terms on both sides of an equation must be the same. According to the principle of homogeneity, the comparison, addition & subtraction of all physical quantities is possible only if they are of the same nature i.e., they have the same dimensions. If the power of M, L and T on two sides of the given equation are same, then the physical equation is correct otherwise not. Therefore, this principle is very helpful to check the correctness of a physical equation. Example: A physical relation must be dimensionally homogeneous, i.e., all the terms on both sides of the equation must have the same dimensions. In the equation, S = ut + ½ at2 The length (S) has been equated to velocity (u) & time (t), which at first seems to be meaningless, But if this equation is dimensionally homogeneous, i.e., the dimensions of all the terms on both sides are the same, then it has physical meaning. Now, dimensions of various quantities in the equation are: Distance, S = [L1] Velocity, u = [L1T-1] Time, t = [T1] 8  Acceleration, a = [L1T-2] ½ is a constant and has no dimensions. Thus, the dimensions of the term on L.H.S. is S=[L1] and Dimensions of terms on R.H.S. ut + ½ at2 = [L1T-1] [T1] + [L1T-2] [T2] = [L1] + [L1] Here, the dimensions of all the terms on both sides of the equation are the same. Therefore, the equation is dimensionally homogeneous. 1.7 DIMENSIONAL EQUATIONS, APPLICATIONS OF DIMENSIONAL EQUATIONS; Dimensional Analysis: A careful examination of the dimensions of various quantities involved in a physical relation is called dimensional analysis. The analysis of the dimensions of a physical quantity is of great help to us in a number of ways as discussed under the uses of dimensional equations. Uses of dimensional equation: The principle of homogeneity & dimensional analysis has put to the following uses: (i) Checking the correctness of physical equation. (ii) To convert a physical quantity from one system of units into another. (iii) To derive relation among various physical quantities. 1. To check the correctness of Physical relations: According to principle of Homogeneity of dimensions a physical relation or equation is correct, if the dimensions of all the terms on both sides of the equation are the same.If the dimensions of even one term differs from those of others, the equation is not correct. Example 3. Check the correctness of the following formulae by dimensional analysis. (i) 𝐹 = 𝑚v 2 /r (ii)𝑡 = 2𝜋√𝑙/𝑔 Where all the letters have their usual meanings. Sol. 𝑭 = 𝒎𝐯 𝟐 /𝐫 Dimensions of the term on L.H.S Force, F = [M1L1T-2] Dimensions of the term on R.H.S 𝒎𝐯 𝟐 /𝐫 = [M1][L1T-1]2 / [L] =[M1L2T-2]/ [L] =[M1L1T-2] The dimensions of the term on the L.H.S are equal to the dimensions of the term on R.H.S. Therefore, the relation is correct. 9 (ii) 𝒕 = 𝟐𝝅√𝒍/𝒈 Here, Dimensions of L.H.S, t = [T1] = [M0L0T1] Dimensions of the terms on R.H.S Dimensions of (length) = [L1] Dimensions of g (acc due to gravity) = [L1T-2] 2𝜋 being constant have no dimensions. Hence, the dimensions of terms 2𝜋√𝑙/𝑔 on R.H.S = (L1/ L1T-2] )1/2 = [T1] = [M0L0T1] Thus, the dimensions of the terms on both sides of the relation are the same i.e., [M0L0T1].Therefore, the relation is correct. Example 4. Check the correctness of the following equation on the basis of dimensional 𝐸 analysis, 𝑉 = √. Here V is the velocity of sound, E is the elasticity and d is the density 𝑑 of the medium. Sol. Here, Dimensions of the term on L.H.S V =[M0L1T-1] Dimensions of elasticity, E = [M1L-1T-2] & Dimensions of density, d = [M1L-3T0] Therefore, Dimensions of the terms on R.H.S 𝑬 √ = [M1L-1T-2/ M1L-1T-2]1/2 = [M0L1T-1] 𝒅 Thus, dimensions on both sides are the same, therefore the equation is correct. Example 5. Using Principle of Homogeneity of dimensions, check the correctness of equation, h = 2Td /rgCos𝜃. Sol. The given formula is, h = 2Td /rgCos𝜃. Dimensions of term on L.H.S Height (h) = [M0L1T0] Dimensions of terms on R.H.S T= surface tension = [M1L0T-2] D= density = [M1L-3T0] r =radius = [M0L1T0] g=acc.due to gravity = [M0L1T-2] Cos𝜃 = [M0L0T0]= no dimensions So, Dimensions of 2Td/rgCos𝜃 = [M1L0T-2] x [M1L-3T0] / [M0L1T0] x [M0L1T-2] = [M2L-5T0] 10  Dimensions of terms on L.H.S are not equal to dimensions on R.H.S. Hence, formula is not correct. Example 6. Check the accuracy of the following relations: (i) E = mgh + ½ mv2; (ii) v3-u2 = 2as2. Sol. (i) E = mgh + ½ mv2 Here,dimensions of the term on L.H.S. Energy, E = [M1L2T-2] Dimensions of the terms on R.H.S, Dimensions of the term, mgh = [M] ×[LT-2] × [L] = [M1L2T-2] Dimensions of the term, ½ mv2= [M] × [LT-1]2= [M1L2T-2] Thus, dimensions of all the terms on both sides of the relation are the same, therefore, the relation is correct. (ii) The given relation is, v3-u2= 2as2 Dimensions of the terms on L.H.S v3 = [M0] × [LT-1]3= [M0L3T-3] u2 = [M0] × [LT-1]2= [M0L2T-2] Dimensions of the terms on R.H.S 2as2 = [M0] × [LT-2] ×[L]2 = [M0L3T-2] Substituting the dimensions in the relations, v3-u2 = 2as2 We get, [M0L3T-3] - [M0L2T-2] = [M0L3T-2] The dimensions of all the terms on both sides are not same; therefore, the relation is not correct. Example 7. The velocity of a particle is given in terms of time t by the equation v = At + b/t+c What are the dimensions of a, b and c? Sol. Dimensional formula for L.H.S V = [L1T-1] In the R.H.S dimensional formula of At [T]= [L1T-1] 11  A =[LT-1] / [T-1] = [L1T-2] t +c = time, c has dimensions of time and hence is added in t. Dimensions of t + c is [T] Now, b/t+c=v b = v(t + c) = [LT-1] [T] = [L] There dimensions of a= [L1T-2], Dimensions of b = [L] and that of c = [T] Example 8. In the gas equation (P + a/v2) (v – b) = RT, where T is the absolute temperature, P is pressure and v is volume of gas. What are dimensions of a and b? Sol. Like quantities are added or subtracted from each other i.e., (P + a/v2) has dimensions of pressure = [ML-1T-2] Hence, a/v2 will be dimensions of pressure = [ML-1T-2] a = [ML-1T-2] [volume]2= [ML-1T-2] [L3]2 a = [ML-1T-2] [L6]= [ML5T-2] Dimensions of a = [ML5T-2] (v – b) have dimensions of volume i.e., b will have dimensions of volume i.e., [L3] or [M0L3T0] 2. To convert a physical quantity from one system of units into another. Physical quantity can be expressed as Q = nu Let n1u1 represent the numerical value and unit of a physical quantity in one system and n2u2 in the other system. If for a physical quantity Q;M1L1T1be the fundamental unit in one system and M2L2T2 be fundamental unit of the other system and dimensions in mass, length and time in each system can be respectively a,b,c. u1 = [ M1aL1bT1c] and u2 = [ M2aL2bT2c] as we know n1u1 = n2u2 n2 =n1u1/u2  M1a L1bT1c  n2 = n1  a b c   M 2 L2T2     M  a  L  b  T  c  n2 = n1  1   1   1    M 2   L2   T2     While applying the above relations the system of unit as first system in which numerical value of physical quantity is given and the other as second system 12  Thus knowing [M1L1T1], [M2L2T2] a, b, c and n1, we can calculate n2. Example 9. Convert a force of 1 Newton to dyne. Sol. To convert the force from MKS system to CGS system, we need the equation Q=n1u1=n2u2 n1u1 Thus n2 = u2 Here n1 =1, u1=1N, u2=dyne  M1 L1T1−2  n2 = n1    M 2 L2T2  −2   −2  M  L  T  n2 = n1  1  1  1   M 2  L2  T2  −2  kg   m  s  n2 = n1      gm   cm  s  −2  1000 gm   100cm  s  n2 = n1      gm   cm  s  n2 = 1(1000)(100) n2 = 105 Thus 1N= 105 dynes. Example 10.Convert work of 1 erg into Joule. Sol: Here we need to convert work from CGS system to MKS system Thus in the equation n1u1 n2 = u2 n1=1 u1=erg (CGS unit of work) u2= joule (SI unit of work) n1u1 n2 = u2 M1L12T1−2 n2 = n1 M 2 L22T2−2 2 −2  M  L   T  n2 = n1  1  1   1   M 2  L2   T2  −2  gm   cm   s  2 n2 = n1       kg   m   s  −2  gm   cm   s  2 n2 = n1       1000 gm   100cm   s  13  n2 = 1(10−3 )(10−2 ) 2 n2 = 10−7 Thus, 1 erg= 10−7 Joule. Limitations of Dimensional Equation: The method of dimensionshas the following limitations: 1. It does not help us to find the value of dimensionless constants involved in various physical relations. The values, of such constants have to be determined by some experiments or mathematical investigations. 2. This method fails to derive formula of a physical quantity which depends upon more than three factors. Because only three equations are obtained by comparing the powers of M, L and T. 3. It fails to derive relations of quantities involving exponential and trigonometric functions. 4. The method cannot be directly applied to derive relations which contain more than one terms on one side or both sides of the equation, such as v= u + at or s = ut + ½ at2 etc. However, such relations can be derived indirectly. 5. A dimensionally correct relation may not be true physical relation because the dimensional equality is not sufficient for the correctness of a given physical relation. ****** EXERCISES Multiple Choice Questions 1. [ML-1T-2] is the dimensional formula of (A) Force (B) Coefficient of friction (C) Modulus of elasticity (D) Energy. 2. 105Fermi is equal to (A) 1 meter (B) 100 micron (C) 1 Angstrom (D) 1 mm 3. rad / sec is the unit of (A) Angular displacement (B) Angular velocity (C) Angular acceleration (D) Angular momentum 4. What is the unit for measuring the amplitude of a sound? 14  (A) Decibel (B) Coulomb (C) Hum (D) Cycles 5. The displacement of particle moving along x-axis with respect to time is x=at+bt2-ct3. The dimension of c is (A) LT-2 (B) T-3 (C) LT-3 (D) T-3 Short Answer Questions 1. Define Physics. 2. What do you mean by physical quantity? 3. Differentiate between fundamental and derived unit. 4. Write full form of the following system of unit (i) CGS (ii) FPS (iii) MKS 5. Write definition of Dimensions. 6. What is the suitable unit for measuring distance between sun and earth? 7. Write the dimensional formula of the following physical quantity - (i) Momentum (ii) Power (iii) Surface Tension (iv) Strain 8. What is the principle of Homogeneity of Dimensions? 9. Write the S.I & C.G.S units of the following physical quantities- (a) Force (b) Work 10. What are the uses of dimensions? Long Answer Questions 1. Check the correctness of the relation 𝜆 = h /mv; where𝜆is wavelength, h- Planck’s constant, m is mass of the particle and v - velocity of the particle. 2. Explain different types of system of units. 3. Check the correctness of the following relation by using method of dimensions (i) v = u + at (ii) F = mv / r2 (iii) v2 – u2 = 2as 4. What are the limitations of Dimensional analysis? 5. Convert an acceleration of 100 m/s2 into km/hr2. Answers to multiple choice questions: 1 (C) 2 (C) 3 (B) 4 (A) 5 (C) 15  Chapter 2 FORCE AND MOTION Learning objective: After going through this chapter, students will be able to; - understandscalar and vector quantities, addition of vectors, scalar and vector products etc. - State and apply Newton’s laws of motion. - describe linear momentum, circular motion, application of centripetal force. 2.1 SCALAR AND VECTOR QUANTITIES Scalar Quantities: Scalar quantities are those quantities which are having only magnitude but no direction. Examples: Mass, length, density, volume, energy, temperature, electric charge, current, electric potential etc. Vector Quantities: Vector quantities are those quantities which are having both magnitude as well as direction. Examples: Displacement, velocity, acceleration, force, electric intensity, magnetic intensity etc. Representation of Vector: A vector is represented by a straight line with an arrow head. Here, the length of the line represents the magnitude and arrow head gives the direction of vector. Figure:2.1 Typesof Vectors Negative Vectors: The negative of a vector is defined as another vector having same magnitude but opposite in direction. i.e., any vector 𝐴⃗ and its negative vector [–𝐴⃗] are as shown. Figure:2.2 16 Equal Vector: Two or more vectors are said to be equal, if they have same magnitudeand direction. If 𝐴⃗ and 𝐵⃗⃗ are two equal vectors then Figure:2.3 Unit Vector: A vector divided by its magnitude is called a unit vector. It has a magnitudeone unit and direction same as the direction of given vector. It is denoted by𝐴̂. 𝐴⃗ 𝐴̂ = 𝐴 Collinear Vectors: Two or more vectors having equal or unequal magnitudes, but having same direction are called collinear vectors. Figure:2.4 Zero Vector: A vector having zero magnitude and arbitrary direction (be not fixed) iscalled zero vector. It is denoted by O. 2.2 ADDITION OF VECTORS, TRIANGLE &PARALLELOGRAM LAW Addition of Vectors (i) Triangle law of vector addition. If two vectors can be represented in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant is represented in magnitude and direction, by third side of the triangle taken in the opposite order (Fig. 2.5). Magnitude of the resultant is given by Figure:2.5 𝑅 = √𝐴2 + 𝐵2 + 2𝐴𝐵 𝑐𝑜𝑠 𝜃 And direction of the resultant is given by 𝐵 𝑠𝑖𝑛 𝜃 𝑡𝑎𝑛 𝛽 = 𝐴 + 𝐵 𝑐𝑜𝑠 𝜃 17 (ii) Parallelogram (||gm) law of vectors: It states that if two vectors, acting simultaneously at a point, can have represented both in magnitude and direction by the two adjacent sides of a parallelogram, the resultant is represented by the diagonal of the parallelogram passing through that point (Fig. 2.6). Magnitude of the resultant is given by 𝑅 = √𝑃2 + 𝑄2 + 2𝑃𝑄 𝑐𝑜𝑠 𝜃 And direction of the resultant is given by 𝑄 𝑠𝑖𝑛 𝜃 𝑡𝑎𝑛  = 𝑃 + 𝑄 𝑐𝑜𝑠 𝜃 Figure:2.6 2.3 SCALAR AND VECTOR PRODUCT Multiplicationof Vectors (i) Scalar (or dot) Product: It is defined as the product of magnitude of two vectors and the cosine of the smaller angle between them. The resultant is scalar. The dot product of vectors 𝐴⃗ and 𝐵⃗⃗ is defined as Figure:2.7 (ii) Vector (or Cross) Product: It is defined as a vector having a magnitude equal to the product of the magnitudes of the two vectors and the sine of the angle between them and is in the direction perpendicular to the plane containing the two vectors. Thus, the vector product of two vectors A and B is equal to 𝐴⃗ × 𝐵 ⃗⃗ = 𝐴𝐵 𝑠𝑖𝑛𝜃 𝑛̂ 2.4 DEFINITION OF DISTANCE, DISPLACEMENT, SPEED, VELOCITY, ACCELERATION Distance: How much ground an object has covered during it motion. Distance is a scalar quantity. SI unit is meter. Displacement: The shortest distance between the two points is called displacement. It is a vector quantity. SI unit is meter. Dimension formula: [L] 18  Speed: The rate of change of distance is called speed. Speed is a scalar quantity. Unit: ms-1. Linear Velocity: The time rate of change of displacement. 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑣= 𝑡𝑖𝑚𝑒 Units of Velocity: ms-1 Dimension formula = [M0L1T-1] Acceleration: The change in velocity per unit time. i.e. the time rate of change of velocity. 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝐴= 𝑡𝑖𝑚𝑒 If the velocity increases with time, the acceleration ‘a’ is positive. If the velocity decreases with time, the acceleration ‘a’ is negative. Negative acceleration is also known as retardation. Units of Acceleration: C.G.S. unit is cm/s2 (cms-2) and the SI unit is m/s2 (ms-2). Dimension formula = [M0L1T-2] 2.5 FORCE AND ITS UNITS, CONCEPT OF RESOLUTION OF FORCE Force: Force is an agent that produces acceleration in the body on which it acts. Or it is a push or pull which change or tends to change the position of the body at rest or in uniform motion. Force is a vector quantity as it has both direction and magnitude.For example, (i) To move a football, we have to exert a push i.e., kick on the football (ii) To stop football or a body moving with same velocity, we have to apply push in a direction opposite to the direction of the body. SI unit is Newton. Dimension formula: [MLT-2] Resolutionof a Force The phenomenon of breaking a given force into two or more forces in different directions is known as 'resolution of force'. The forces obtained on splitting the given force are called components of the given force. If these are at right angles to each other, then these components are called rectangular components. Let a force F be represented by a line OP. Let OB (or Fx) is component of F along x-axis and OC (or Fy) is component along y-axis (Fig. 2.8). 19  Figure:2.8 Let force F makes an angle θ with x-axis. In Δ OPB 𝑃𝐵 sin𝜃 = 𝑂𝑃 PB = OP sin𝜃 Fy = F sin𝜃 𝑂𝐵 cos𝜃 = 𝑂𝑃 OB = OP cos𝜃 Fx = F cos𝜃 Vector𝐹⃗ = 𝐹⃗𝑥 + 𝐹⃗𝑦 Resultant: 𝐹 = √𝐹𝑥2 + 𝐹𝑦2 2.6 NEWTON'S LAWS OF MOTION Sir Isaac Newton gave three fundamental laws. These laws are called Newton's laws of motion. Newton’s First Law:It states that everybody continues in its state of rest or of uniform motion in a straight line until some external force is applied on it. For example, the book lying on a table will not move at its own. It does not change its position from the state of rest until no external force is applied on it. Newton’s Second law: The rate of change of momentum of a body is directly proportional to the applied force and the change takes place in the direction of force applied. Or Acceleration produced in a body is directly proportional to force applied. 20  Let a body of mass m moving with a velocity u. Let a force F be applied so that its velocity changes from u to v in t second. Initial momentum = mu Final momentum after time t second = mv Total change in momentum = mv-mu. Thus, the rate of change of momentum will be 𝑚𝑣 − 𝑚𝑢 𝑡 From Newton's second law 𝑚𝑣−𝑚𝑢 𝑚(𝑣−𝑢) 𝐹 or𝐹  𝑡 𝑡 𝑣 − 𝑢 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 but = = Acceleration(a) 𝑡 𝑇𝑖𝑚𝑒 Hence, we have F ma or F = k ma Where k is constant of proportionality, for convenience let k = 1. Then F = ma Units of force: Onedyne is that much force which produces an acceleration of 1cm/s 2 in a mass of 1 gm. 1 dyne = 1gm x 1 cm/s2 = 1gm.cm s-2 One Newton is that much force which produces an acceleration of 1 m/s 2 in a mass of 1kg. using F = ma 1N = 1kg x 1m/s 2 or = 1kgm/s2 IN =1000gm×100 cm/s 2 = 10 5 dyne Newton’s Third law: To every action there is an equal and opposite reaction or action and reaction are equal and opposite. When a body exerts a force on another body, the other body also exerts an equal force on the first, in opposite direction. From Newton's third law these forces always occur in pairs. FAB (force on A by B) = -FBA (force on B by A) 21 2.7 LINEAR MOMENTUM, CONSERVATION OF MOMENTUM, IMPULSE Linear Momentum (p): The quantity of motion contained in the body is linear momentum. It is given by product of mass and the velocity of the body. It is a vector and its direction is the same as the direction of the velocity. Let m is mass and vis the velocity of a body at some instant, then momentum is given by p = mv Example, a fast-moving cricket ball has more momentum in it than a slow moving one. But a slow-moving heavy roller has more momentum than a fast cricket ball. Units of momentum: The SI unit is kg m/s i.e. kg.ms-1. Dimension formula = [M1L1T-1]. Conservation of Momentum If external force acting on a system of bodies is zero then the total linear momentum of a system always remains constant. i.e. If F=0 𝑑𝑝 Thus,𝐹 = =0 𝑑𝑡 Hence, p (momentum) is constant. Recoil of the Gun: When a bullet is fired with a gun the bullet moves in forward direction and gun is recoiled/pushed backwards. Let m = mass of bullet u = velocity of bullet M = mass of gun v = velocity of gun The gun and bullet form an isolated system So the total momentum of gun and bullet before firing = 0 Total momentum of gun and bullet after firing=m.u+M.v Using law of conservation of momentum 0 = m.u+M.v M.v = -m.u −𝑚𝑢 𝑣 = 𝑀 This is the expression for recoil velocity of gun. Here negative sign shows that motion of the gun is in opposite direction to that of the 22  bullet. Also, velocity of gun is inversely proportional to its mass. Lesser the mass, larger will be the recoil velocity of the gun. Impulse Impulse is defined as the total change in momentum produced by the impulsive force. OR Impulse may be defined as the product of force and time and is equal to the total change in momentum of the body. F.t=p2– p1= total change in momentum Example. A kick given to a football or blow made with hammer. 2.8.CIRCULAR MOTION The motion of a body in a circle of fixed radius is called circular motion. For example, the motion of a stone tied to a string when whirled in the air is a circular motion. Angular Displacement: The angle described by a body moving in a circle is called angular displacement. Consider a body moves in a circle, starting from A toB so that ∠BOA is called angular displacement SI unit of angular displacement is radian (rad.) Figure:2.9 Angular Velocity: Angular velocity of a body moving in a circleis the rate of change of angular displacement with time. It is denoted by ω (omega) If θ is the angular displacement in time t then 𝜃 𝜔= 𝑡 SI unit of angular velocity is rad/s. Time Period: Time taken by a body moving in a circle to complete one cycle iscalled time period. It is denoted by T Frequency (n): The number of cycles completed by a body is called frequency. It is reciprocal of time period 1 𝑛= 𝑇 Angular Acceleration: The time rate of change of angular velocity of a body. It is denoted by α. Let angular velocity of a body moving in a circle change from ω 1 23 to ω2 in time t, then 𝜔1 − 𝜔2 𝛼= 𝑡 SI unit of ‘’ is rad/s2 Relationship between linear and angular velocity Consider a body moving in a circle of radius r Let it start from A and reaches to B after time t, so that ∠BOA = θ (Fig. 2.9). Now 𝑎𝑟𝑐 𝑎𝑛𝑔𝑙𝑒 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝐴𝐵 𝑆 𝜃= = 𝑂𝐴 𝑟 𝑆 = 𝑟𝜃 Divide both side by time (t) 𝑆 𝜃 =𝑟 𝑡 𝑡 𝑆 Here = 𝑣is linear velocity 𝑡 𝜃 And = 𝜔 is angular velocity 𝑡 Hence 𝑣 = 𝑟𝜔 2.9 CENTRIPETAL AND CENTRIFUGAL FORCES Centripetal Force The force acting along the radius towards the centre of circle to keep a body moving with uniform speed in a circular path is called centripetal force. It is denoted by FC. 𝑚𝑣 2 𝐹𝑐 = 𝑟 For example, a stone tied at one end of a string whose other end is held in hand, when round in the air, the centripetal force is supplied by the tension in the string. Centrifugal Force: A body moving in circle with uniform speed experience a force in a direction away from the centre of the circle. This force is called centrifugal force. For example, cream is separated from milk by using centrifugal force. When milk is rotated in cream separator, cream particles in the milk being lighter, and experience less centrifugal force. 2.10 APPLICATION OF CENTRIPETAL FORCE IN BANKING OF ROADS Banking of Roads: While travelling on a road, you must have noticed that,the outer edge of circular road is slightly raised above as compared to the inner edge of road. This is called banking of roads (Fig. 2.10). 24 Angle of Banking: The angle through which the outer edge of circular road is raised above the inner edge of circular roads is called angle of banking. Application of centripetal force in banking of roads Let m = mass of vehicle r=radius of circular road v=uniform speed (velocity) of vehicle θ = angle of banking At the body two forces act. (i) Weight (mg) of vehicle vertically downwards. Figure:2.10 (ii) Normal reaction (R). R makes an angle θ and divides the forces into two components (i) Rsinθ towards the centre (ii) Rcosθ vertically upwards and balance by weight of (mg) vehicle 𝑚𝑣 2 Rsinθprovides the necessary centripetal force ( ) 𝑟 𝑚𝑣 2 R Sinθ= - - - - - (1) 𝑟 and R Cosθ = mg - - - - -(2) Divide equation 1 by 2 𝑚𝑣 2 𝑅𝑆𝑖𝑛𝜃 = 𝑟 𝑅𝐶𝑜𝑠𝜃 𝑚𝑔 𝑣2 𝑡𝑎𝑛𝜃 = 𝑟𝑔 𝑣2 𝜃 = 𝑡𝑎𝑛−1 ( ) 𝑟𝑔 ****** 25  EXERCISES Multiple Choice Questions 1. What is the maximum possible number of components of a vector can have (A) 2 (B) 3 (C) 4 (D) Any number 2. Which of the following operations with two vectors can result in a scalar (A) Addition (B) Subtraction (C) Multiplication (D) None of these 3. The acceleration of the particle performing uniform circular motion is (A) 2/r (B) zero (C) vr (D) v2/r 4. Centripetal force always acts at 90 degrees to the velocity, and away from the centre of the circle. (A) True (B) False (C) can’t predict (D) none of these 5. Railway tracks are banked at the curves so that the necessary centripetal force may be obtained from the horizontal component of the reaction on the train (A) True (B) False (C) can’t predict (D) none of these 6. Which of the following is called a fictitious force? (A) Gravitational force (B) Frictional force (C) Centrifugal force (D) Centripetal force 7. At which place of the earth, the centripetal force is maximum (A) At the earth surface 26  (B) At the equator (C) At the north pole (D) At the south pole 8. The angle through which the outer edge is raised above the inner edge is called (A) angle of inclination (B) angle of repose (C) angle of banking (D) angle of declination 9. A model aeroplane fastened to a post by a fine thread is flying in a horizontal circle. Suddenly the thread breaks. What direction will the aeroplane fly? (A) In a circular path, as before (B) Directly to the centre of the circle (C) In a straight line at a tangent (D) Directly to the centre of the circle. 10. A force which acts for a small time and also varies with time is called: (A) Electrostatic force (B) Electromagnetic force (C) Impulsive force (D) Centripetal force Short Answer Type Questions 1. State and explain laws of vector addition. 2. What do you understand by resolution of a vector? 3. How is impulse related to linear momentum? 4. What do you mean by circular motion? Give examples? 5. What do you mean by banking of roads? 3. What are scalar and vector quantities? Give examples? 4. Define resolution and composition of forces. 5. What is impulse? 6. Why does a gun recoil when a bullet is fired? 7. Differentiate between centripetal and centrifugal forces? 8. An artificial satellite takes 90 minutes to complete its revolution around the earth. Calculate the angular speed of satellite. [Ans. 2700 rad/sec] 9. At what maximum speed a racing car can transverse an unbanked curve of 30 m radius? The co-efficient of friction between types and road is 0.6. [Ans. 47.8] 10. Justify the statement that Newton’s second law is the real law of motion. 11. Define Force. Give its units. 12. Define Triangle law of vector addition. 13. State parallelogram law of vector addition. 27 Long Answer Type Questions 1. Explain Newton’s Law of Motion. 2. Explain Banking of Roads. 3. What is conservation of momentum? 4. Derive relationship between linear and angular velocity. 5. Derive a relation between linear acceleration and angular acceleration. Answers to multiple choice questions: 1. (D) 2. (C) 3. (D) 4. (B) 5. (A) 6. (C) 7. (B) 8. (C) 9. (C) 10. (C) 28  Chapter 3 WORK, POWER AND ENERGY Learning objective: After going through this chapter, students will be able to; - understand work, energy and power, their units and dimensions. - describe different types of energies and energy conservation. - solve relevant numerical problems 3.1 WORK (DEFINITION, SYMBOL, FORMULA AND SI UNITS) Work Work is said to be done when the force applied on a body displaces it through certain distance in direction of applied force. Work= Force × Displacement   In vector form, it is written as; F × S = FS Cos It is measured as the product of the magnitude of force and the distance covered by the body in the direction of the force. It is a scalar quantity. Unit: SI unit of work is joule(J). In CGS system, unit of work is erg. 1 J = 107 ergs Dimension of work = [M1L2T–2] Example 1. What work is done in dragging a block 10 m horizontally when a 50 N force is appliedby a rope making an angle of 30° with the ground? Sol. Here, F = 50 N, S = 10 m, = 30 W = FS Cos θ W = 50 × 10 × Cos 30° W = 50x10x 3 2 = 612.4 J Example 2. A man weighing 50 kg supports a body of 25 kg on head. What is the work done whenhe moves a distance of 20 m. Solution. Total mass = 50 + 25 = 75 kg θ = 90° 29  Distance = 20 m W = FS × 0 (Cos 90o = 0) W = 0 Thus, work done is zero. Example 3. A man weighing 50 kg carries a load of 10 kg on his head. Find the work done whenhe goes (i) 15 m vertically up (ii) 15 m on a levelled path on the ground. Solution. Mass of the man,m1= 50 kg Mass carried by a man,m2 = 10 kg Total mass M = m1 + m2 = 50 + 10 = 60 kg. When the man goes vertically up, Height through which he rises, h = 15 m W =Mgh = 60 × 9.8 × 15 = 8820 J When the man goes on a levelled path on the ground. W= FS Cos θ As θ =90 , therefore, Cos 90o= 0 o Hence W= F×S×0 =0 3.2 ENERGY (DEFINITION AND ITS SI UNITS), EXAMPLES OF TRANSFORMATION OF ENERGY Energy Energy of a body is defined as the capacity of the body to do the work. Like work, energy is also a scalar quantity. Unit:SI system - Joule, CGS system - erg Dimensional Formula: [ML2 T–2]. Transformation of Energy The energy change from one form to another is called transformation of energy. Forexample. In a heat engine, heat energy changes into mechanical energy In an electric bulb, the electric energy changes into light energy. In an electric heater, the electric energy changes into heat energy. In a fan, the electric energy changes into mechanical energy which rotates the fan. In the sun, mass changes into radiant energy. In an electric motor, the electric energy is converted into mechanical energy. In burning of coal, oil etc., chemical energy changes into heat and light energy. In a dam, potential energy of water changes into kinetic energy, then K.E rotates the turbine which produces the electric energy. 30  In an electric bell, electric energy changes into sound energy. In a generator, mechanical energy is converted into the electric energy. 3.3 KINETIC ENERGY (FORMULA, EXAMPLES AND ITS DERIVATION) Kinetic Energy(K.E.): Energy possessed by the body by virtue of its motions is called kinetic energy. For example; (i) running water (ii) Wind energy; work on the K.E. of air (iii) Moving bullet. Expression for Kinetic Energy Consider F is the force acting on the body at rest (i.e., u = 0), then it moves in the direction of force to distance (s). Figure:3.1 Let v be the final velocity. Using relation v 2 − u 2 = 2aS v2 − u 2 =a 2S v2 − 0 =a 2S v2 =a --------------(1) 2S Now, work done, W= F.S or W= ma.S (using F =ma) ------------- (2) By equation (1) and (2) v2 W = m..S 2S or W = 12 mv2 This work done is stored in the body as kinetic energy.So kinetic energy possessed by the body is (K.E.) = 12 mv 2 3.4 POTENTIAL ENERGY (FORMULA, EXAMPLES AND ITS DERIVATION) Potential Energy (P.E.): Energy possessed by the body by virtue of its position iscalled 31 potential energy. Example (i) Water stored in a dam (ii) Mango hanging on the branch of a tree Expression for Potential Energy (P.E) It is defined as the energy possessed by the body by virtue of its positionabove the surface of earth. W=FxS Workdone = Force × height = mg × h = mgh h This work done is stored in the form of gravitational potential energy. Hence Potential energy =mgh. Figure:3.2 Law of Conservation of Energy Energy can neither be created nor be destroyed but can be converted from one form to another. 3.5 CONSERVATION OF MECHANICAL ENERGY OF A FREE FALLING BODY Let us consider K.E., P.E. and total energy of a body of mass m falling freely under gravity from a height h from the surface of ground. According to Fig. 3.3 At position A: Initial velocity (u)= 0 K.E = 12 mv 2 P. E. = mgh Total Energy=K.E + P.E =0 + mgh =mgh ------------- (1) At position B Figure:3.3 Potential energy=mg(h – x) Velocity at point B= u From equation of motion K.E. = 12 mu 2 As V 2 − U 2 = 2aS Hence u 2 − 02 = 2 gx or u 2 = 2 gx Putting this value we get, KE= 12 m(2 gx) 32  or K.E. = mgx Total Energy =K.E + P.E =mgx + mg(h – x) =mgh --------(2) At position C Potential energy = 0(as h = 0) Velocity at Point B= v From equation of motion K.E. = 1 2 mv 2 As V 2 − U 2 = 2aS Hence v 2 − 02 = 2 gh or v 2 = 2 gh Putting this value we get KE= 12 m(2 gh) or K.E. = mgh Total Energy =K.E + P.E =mgh + 0 =mgh ---------(3) From equations (1), (2) and (3), it is clear that total mechanical energy of freely falling body at all the positions is same and hence remain conserved. Example 3. A spring extended by 20 mm possesses a P.E. of 10 J. What will be P.E., if theextension of spring becomes 30 mm? Solution: h = 20 mm = 20 × 10–3m g = 9.8 ms–2, m = ? P.E =mgh = 10J i.e., m × 9.8 × 20 × 10–3 = 10 J 10 m= 9.8x20x10−3 m =51.02 Kg When extension is 30 mm i.e., 30 × 10–3 m, then P.E =mgh = 51.02 × 9.8×3 × 10–3= 15.0 J 3.6 POWER (DEFINITION, FORMULA AND UNITS) Power is defined as the rate at which work is done by a force. The work done per unit time is also called power. 33  If a body do work W in time t, then power is W P= t Units of Power: SI unit of power is watt (W) Power is said to be 1 W, if 1 J work is done in 1 s. 1J 1W = 1s Bigger units of power are: Kilowatt (KW) =103W Megawatt (MW) =106 W Horse power (hp) = 746 W Dimension of power = [M1 L2T-3] Example 4.A man weighing 65 kg lifts a mass of 45 kg to the top of a building 10 meters high in 12second. Find; (i) Total work done by him. (ii) The power developed by him. Solution. Mass of the man, m1= 65 kg Mass lifted m2= 45 kg Height through which raised h = 10 m Time taken t = 12 seconds. (i) Total work done by the man, W = mgh = 110 × 9.81 × 10 = 10791.0J W 10791J (ii) Power developed P= = = 899.25W t 12 s ****** 34  EXERCISES Multiple Choice Questions 1. Which of the following is not correct for the condition for work not to be done: (A) Force and displacement are perpendicular to each other (B) Force and displacement are at 180 degrees with each other (C) Displacement is zero, though force is non-zero (D) Force is zero 2. There are two bodies X and Y with equal kinetic energy but different masses m and 4m respectively. The ratio of their linear momentum is- (A) 1:2 (B) 4:1 (C) 1:√2 (D) 1:4 3. Which of the following statements is false: (A) Kinetic energy is positive (B) Potential energy is positive (C) Kinetic energy is negative (D) Potential energy is negative 4. How should the force applied on a body be varied with velocity to keep the power of force constant? (A) Force should be inversely proportional to the square root of the velocity of the body (B) Force should be inversely proportional to the velocity of the body (C) Force should be directly proportional to the velocity of the body (D) Force should not be varied. It should remain constant with the velocity 5. When does the potential energy of a spring increase? (A) Only when spring is stretched (B) Only when spring is compressed (C) When spring is neither stretched nor compressed (D) When spring is compressed or stretched 6. Which of the following force is non-conservative? (A) Restoring force of spring (B) Force between two stationary masses (C) Force between two stationary charges (D) Human push or pull 7. You are in a lift moving from the 3rd floor to the 12th floor, through a height H. If the elevator moves at a constant speed without stopping, what is the work performed on you by the elevator? Take your body mass as M. (A) MgH (B) Mg (C) -MgH (D) -Mg 35  8. Which of the following is not a kind of potential energy? (A) Gravitational potential energy (B) Magnetic potential energy (C) Electrostatic potential energy (D) Nuclear potential energy Short Answer Type Questions 1. Define the terms energy, potential energy and kinetic energy. 2. Define potential energy, Derive expression for gravitational potential energy. 3. Define work and write its unit. 4. Define the term power and write its unit. 5. State and prove principle of conservation of energy. 6. Define power. Give it S.I unit. 7. What is transformation of energy? 8. A person walking on a horizontal road with a load on his head does not work. Explain. 9. State kinetic energy. Write expression for kinetic energy of a body of mass m moving at a speed u. 10. Define potential energy of body. Give expression for it. 11. Give some examples of transformation of energy. 12. Define power. Give its units and dimensions. Long Answer Type Questions 1. Explain the law of conservation of energy for free falling body, show that mechanical energy remains same. 2. What is meant by positive work, negative work and zero work? Illustrate your answer with two examples of each type. 3. What are conservative and non-conservative forces, explain with examples. Mention some of their properties. 4. What is meant by power and energy? Give their units. 5. Explain meaning of kinetic energy with examples. Obtain an expression for kinetic energy of body moving uniformly? Answer to multiple choice questions: 1. (C) 2. (A) 3. (C) 4. (B) 5. (D) 6. (D) 7. (C) 8. (D) 36  Chapter 4 ROTATIONAL MOTION Learning objective: After going through this chapter, students will be able to; - define rotational motion and parameters like; torque, angular momentum and momentum conservation. - describe Moment of inertia and radius of gyration. - solve relevant numerical problems. 4.1 ROTATIONAL MOTION WITH EXAMPLES The rotation of a body about fixed axis is called Rotational motion. For example, (i) motion of a wheel about its axis (ii) rotation of earth about its axis. 4.2 DEFINITION OF TORQUE AND ANGULAR MOMENTUM Torque () It is measured by the product of magnitude of force and perpendicular distance of the line of action of force from the axis of rotation. It is denoted by τ,  = Fxr whereF is force and r is perpendicular distance. Unit: Newtons (N) Dimension Formula: [M1L2T-2] Figure:4.1 Angular Momentum (L) Angular momentum of a rotating body about its axis of rotation isthe algebraic sum of the linear momentum of its particles about the axis. It is denoted by L. L = Momentum × perpendicular distance L= p × r or L= mvr Unit: Kg m2/sec Dimensional Formula=[ML2T–1] 37 4.3 CONSERVATION OF ANGULAR MOMENTUM Law of Conservation of Angular Momentum When no external torque acts on a system of particles, then the total angular momentum of the system remains always a constant. Let I be moment of inertia and ω the angular velocity,then angular momentum is given as L=Iω Also the torque is given by dL = dt If no external torque is present on the body i.e., τ=0 dL Hence  = =0 dt which means L is constant (as derivative of constant quantity is zero). Hence, if no external torque acts on system, the total angular momentum remains conserved. Examples: (i) An ice skater who brings in her arms while spinning spins faster. Her moment of inertia is dropping (reducing the moment of arm) so her angular velocity increases to keep the angular momentum constant (ii) Springboard diver stretches his body in between his journey. 5.4 MOMENT OF INERTIA AND ITS PHYSICAL SIGNIFICANCE, RADIUS OF GYRATION Moment of Inertia Moment of Inertia of a rotating body about an axis is defined as the sum of the product of the mass of various particles constituting the body and square of respective perpendicular distance of different particles of the body from the axis of rotation. Expression for the Moment of Inertia: Let us consider a rigid body of mass M havingn number of particles revolving about any axis. Let m1, m2, m3..., mn be the masses of particles at distance r1, r2, r3... rn from the axis of rotation respectively (Fig. 4.2). Moment of Inertia of whole body I = m1r12 + m2r22 +... mnrn2 Figure:4.2 38  n or I =  mi ri 2 i =1 Physical Significance of Moment of Inertia It is totally analogous to the concept of inertial mass. Moment of inertia plays the same role in rotational motion as that of mass in translational motion. In rotational motion, a body, which is free to rotate about a given axis, opposes any change in state of rotation. Moment of Inertia of a body depends on the distribution of mass in a body with respect to the axis of rotation Radius of Gyration It may be defined as the distance of apoint from the axis of rotation at which whole mass of the body is supposed to be concentrated, so that moment of inertia about the axis remains the same. It is denoted by K If the mass of the body is M, the moment of inertia (I) of the body in terms of radius of gyration is given as, I = MK 2 ---------- (1) Expression for Radius of Gyration Let m1, m2, m3..., mn be the masses of particles at distance r1, r2, r3... rn from the axis of rotation respectively (Fig. 4.3). Then Moment of Inertia of whole body I = m1r12 + m2r22 +.......+mnrn2 If mass of all particles is taken same, then I = m(r12 + r22 +..........+rn2) Multiply and divide the equation by n (number of particle) mxn(r12 + r22 +............ + rn2 ) I= n M (r12 + r22 +............ + rn2 ) or I = ---------- (2) n (M=m×n, is total mass of body) Figure:4.3 Comparing equation (1) and (2) , we get M (r12 + r22 +............ + rn2 ) MK 2 = n (r12 + r22 +............ + rn2 ) Or K 2 = n (r12 + r22 +............ + rn2 ) K= n 39  Thus, radius of gyration may also be defined as the root mean square (r.m.s.) distance of particles from the axis of rotation. Unit: SI unit of radius of gyration is meter. Example 1. What torque will produce an acceleration of 2 rad/s2in a body if moment ofinertia is 500 kg m2? Solution. Here, I= 500 kg m2 α= 2 rad/s2 Now, torque τ = I× α = 500 kg m2 × 2 rad/s2 = 1000 kg m2 s–2 =1000 Nm or J Example 2. An engine is rotating at the rate of 1500 rev. per minute. Find its angular velocity. Solution. Here, Revolution per minute of engine, N= 1500 Angular velocity  = 2 N 22 1500 Or  = 2  7 60  = 157.1 rad/s Example 3. How large a torque is needed to accelerate a wheel, for which I = 2 kg m2, from rest to30 r.p.s in 20 seconds? Solution. Here, Moment of inertia,I= 2 kg m2 R.P.S after 20 sec, n = 30 Initial velocity, ω1 = 0 Final velocity, ω2 = 2 x π x 30 = 188.4 rad/s.  − 1 188.4 − 0 Angular acceleration= 2 = =9.43 rad/s2. t 20 Now, torque, τ = I× α = 2 kg m2× 9.43 rad/s2= 18.86 Nm or J Example 4. Ifa point on the rim of wheel 4 m in diameter has a linear velocity of 16 m/ s,find theangular velocity of wheel in rad/sec. Diameter 4 Solution. Radius of wheel (R) = = =2m 2 2 From the relation v = r 40  v 16  = = = 8 rad/s. r 2 Angular velocity of wheel is 8 rad/s. ***** EXERCISES Multiple Choice Questions 1. The radius of gyration of a ring of radius R about an axis through its centre and perpendicular to its plane is (A) R / √2 (B) R (C) R/2 (D) 5 R / √2 2. Two rings have their moment of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses will be: (A) 1:2 (B) 2:1 (C) 1:4 (D) 1:1 3. The moment of inertia of a body is independent of (A) Choice of axis of rotation (B) Its mass (C) Its shape and size (D) Its angular velocity 4. A ring has greater moment of inertia than a circular disc of same mass and radius, about an axis passing through its centre of mass perpendicular to its plane, because (A) All mass is at maximum distance from axis (B) Because the centre of the ring does not lie on it (C) Because the ring needs greater inertia to bend it (D) Because the moment produced in the ring is more 5. A person standing on a rotating platform with his hands lowered outstretches his arms. The angular momentum of the person (A) Become zero (B) Decreases (C) Remains constant (D) Increases 6. Relation between torque and angular momentum is similar to the relation between (A) Force and linear momentum 41  (B) Energy and displacement (C) Acceleration and velocity (D) Mass and moment of inertia 7. An earth satellite is moving around the earth in a circular orbit. In such case, what is conserved? (A) Force (B) Velocity (C) Angular momentum (D) Linear momentum 8. When no external Torque acts on a system, what is conserved (A) Energy (B) Force (C) Angular momentum (D) Linear momentum Short Answer Type Question 1. Define torque. 2. What is moment of inertia? 3. What is Radius of gyration? 4. What is rotational inertia or moment of inertia? Give its SI unit. 5. What is radius of gyration and mention its SI units? 6. What do you understand by kinetic energy of rotation with expression? 7. Derive an expression for torque in terms of moment of inertia. 8. Derive the relation between torque and angular momentum. Long Answer Type Question 1. Derive an expression for angular momentum in terms of moment of inertia. 2. State and prove law of conservation of angular momentum. 3. What is radius of gyration and derive its expression. 4. What is moment of inertia? Derive its expression and what is its physical significance? Answers to multiple choice questions: 1. (B) 2. (A) 3. (D) 4. (A) 5. (C) 6. (A) 7. (C) 8. (C) 42  Chapter 5 PROPERTIES OF MATTER Learning objective: After going through this chapter, students will be able to; - understandelasticity, deforming force, restoring force etc. - define stress, strain, Hook’s law, modulus of elasticity, pressure etc.. - describe surface tension, viscosity and effect of temperature on these. - understand fluid motion and nature of flow. 5.1 DEFINITION OF ELASTICITY, DEFORMING FORCE, RESTORING FORCE, EXAMPLE OF ELASTIC AND PLASTIC BODY Elasticity:It is the property of solid materials to return to their original shape and size after the forces deforming them have been removed. Deforming Forces: The forces which bring the change in configuration of the body are called deforming forces. Restoring Force: It is a force exerted on a body or a system that tends to move it towards an equilibrium state. Elastic Body: It is the body that returns to its original shape after a deformation.Examples are Golf ball, Soccer ball, Rubber band etc. Plastic Body: It is the body that do not return to its original shape after a deformation.Examples are Polyethylene (PE), Polypropylene (PP), Polystyrene (PS) and Polyvinyl Chloride (PVC). 5.2 DEFINITION OF STRESS AND STRAIN WITH THEIR TYPES Stress:It is defined as the restoring force per unit area of a material.Stress is of two types: 1. Normal Stress: If deforming force acts normal to the surface of the body then the stress is normal stress. 2. Tangential Stress: If deforming force acts tangentially to the surface of the body then the stress is tangential stress. Strain: It is defined as the ratio of change in configuration to the original configuration, when a deforming force is applied to a body. The strain is of three types: (i) Longitudinal strain: If the deforming force produces a change in length only, the strain produced is called longitudinal strain or tensile strain. It is defined as the ratio of change in length to the original length. Change in length(∆l) Longitudinal strain = original length(l) 43 (ii) Volumetric strain: It is defined as the ratio of the change in volume to the original volume. Change in volume(∆v) Volumetric strain = original volume(v) (iii) Shearing strain: It is defined as the ratio of lateral displacement of a surface under the tangential force to the perpendicular distance between surfaces Lateral Displacement Shearing strain = Distance between surfaces ∆𝐿 = = tan Ф 𝐿 Figure: 5.1 The shearing strain is also defined as the angle in radian through which a plane perpendicular to the fixed surface of a rectangular block gets turned under the effect to tangential force. Units of strain: Strain is a ratio of two similar physical quantities, it has no units and dimensions. 5.3 HOOK’S LAW, MODULUS OF ELASTICITY Hook’s law: Within elastic limits, the stress and strain are proportional to each other. Thus, Stress ∝ Strain Stress = E × Strain Where E is the proportionality constant and is known as modulus of elasticity. Modulus of Elasticity: The ratio of stress and strain is called modulus of elasticity. Young’s Modulus(Y): The ratio of normal stress to the longitudinal strain is defined as Young’s modulus and is denoted by the symbol Y. 𝐹/𝐴 𝐹×𝑙 Y= = 𝛥𝑙/𝑙 𝐴×𝛥𝑙 Since strain is a dimensionless quantity, the unit of Young’s modulus is the same as that of stress i.e., Nm–2 or Pascal (Pa) 44 Bulk Modulus (B): The ratio of normal (hydraulic) stress to the volumetric strain is called bulk modulus. It is denoted by symbol B. 𝐹/𝐴 𝐹×𝑉 B= = 𝛥𝑉/𝑉 𝐴×𝛥𝑉 SI unit of bulk modulus is the same as that of pressure i.e., Nm –2or Pa Shear Modulus or Modulus of rigidity (𝜂): The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material and is represented by 𝜂. It is also called the modulus of rigidity. 𝜂 = Tangential stress /Shear strain 𝐹/𝐴 𝐹×𝐿 𝜂= = 𝛥𝐿/𝐿 𝐴×𝛥𝐿 SI unit of shear modulus is Nm–2or Pa. 5.4 PRESSURE AND PASCALS LAW Pressure: It is defined as the force per unit area over the surface of a body. F P= A SI unit is Nm–2or Pa Pascal Law: A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container. 5.5 SURFACE TENSION AND EFFECT OF TEMPERATURE ON SURFACE TENSION Surface Tension: The property of a liquid due to which its free surface behaves like stretched membrane and acquires minimum surface area. It is given by force per unit length. 𝐹 𝑇= 𝑙 Surface tension allows insects (e.g. water striders), usually denser than water, to float and stride on a water surface. SI unit is N/m. Applications of surface tension It plays an important role in many applications in our daily life. Washing clothes Figure: 5.2 Cleaning Cosmetics Lubricants in machines 45  Spreading of ink, colours Wetting of a surface Action of surfactants Paints, insecticides Creating fuel-spray in automobile engines Passing of liquid in porous media Effect of Temperature on Surface Tension In general, surface tension decreases when temperature increases and vice versa. This is because cohesive forces decrease with an increase of molecular thermal activity. The influence of the surrounding environment is due to the adhesive action liquid molecules have at the interface. 5.6 VISCOSITY AND EFFECT OF TEMPERATURE ON VISCOSITY Viscosity: The property of liquid due to which it oppose the relative motion between the layers of fluid. It is also known as liquid friction. SI unit of viscosity is Pascal-second (Pa.s) and CGS unit is Poise. Effect of Temperature on Viscosity In liquids the source for Viscosity is considered to be atomic bonding. As we understand that, with the increase of temperature the bonds break and make the molecule free to move. So, we can conclude that the viscosity decreases as the temperature increases and vice versa. In gases, due to the lack of cohesion, the source of viscosity is the collision of molecules. Here, As the temperature increases the viscosity increases and vice versa. This is because the gas molecules utilize the given thermal energy in increasing its kinetic energy that makes them random and therefore resulting in more the number of collisions. 5.7 FLUID MOTION, STREAM LINE AND TURBULENT FLOW Fluid Motion: A liquid in motion is called fluid. There are two types of fluid motions; streamline and turbulent. Streamline Flow: Flow of a fluid in which its velocity at any point of given cross section is same. It is also called laminar flow. Turbulent flow: It is type of fluid (gas or liquid) flow in which the speed of the fluid at given cross section is continuously undergoing changes in both magnitude and direction. Figure: 5.3 46  EXERCISES Multiple Choice Questions 1) Elastic bodyis the body that returns to its original shape after a ………. (A) Restoration (B) Deformation (C) Elongation (D) Acceleration 2) Stress is defined as the ………………. per unit area of a material. (A) Force (B) Velocity (C) Distance (D) Displacement 3) ……………………is the ratio of change in dimensions to the original dimensions. (A) Strain (B) Surface Tension (C) energy (D) elasticity 4) For small deformations the stress and strain are proportional to each other. This is called ………………… (A) Hook’s Law (B) Pascal’s Law (C) Snell’s law (D) Newton’s law 5) Pressure is defined as the force per unit ………………. over the surface of a body. (A) Area (B) Volume (C) Line (D) Energy 6) A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid to the walls of its container. It is called…………………. (A) Hooks Law (B) Pascal’s Law (C) Snell’s law (D) Newton’s law 7) Elasticity is the property of solid materials to return to their original shape and size after the removal of deforming forces. (A) True (B) False 8) Restoring force is the force exerted on a body or a system that tends to move it towards an equilibrium state. (A) True 47  (B) False Short Answer Type Questions 1. Define Elasticity. 2. What is Viscosity? 3. What is Turbulent Flow? 4. Define Surface Tension. 5. What is Young’s Modulus of Elasticity? 6. State and explain Hooks Law. 7. State and explain Pascal’s Law. 8. What is the effect of temperature on surface tension? 9. What is the effect of temperature on viscosity? 10. Give any five applications of Surface Tension. 11. What is difference between elastic and plastic bodies? Long Answer Type Questions 1. Explain different kind of modulus of elasticity. 2. What is Surface Tension? Give formula, Units and Applications of Surface Tension. 3. Explain streamline flow, laminar flow and turbulent flow. 4. Explain different types of stress. 5. Explain Young’s modulus and its units. Answers to multiple choice questions: 1. (B) 2. (A) 3. (A) 4. (A) 5. (A) 6. (B) 7. True 8. True 48  Chapter 6 HEAT AND TEMPERATURE Learning Objectives:After going through this chapter, the students will be able to: - define heat and temperature, understand the difference between heat and temperature; - describe principles of measuring temperature and different temperature scales, - enlist properties of heat radiations and various modes of transfer of heat. 6.1 HEAT AND TEMPERATURE All objects are made of atoms or molecules. These molecules are always in some form of motion (linear, vibrational or rotational) and possess kinetic energy by virtue of their motion. The hotter an object is, faster will be the motion of the molecules inside it and hence more will be its kinetic energy. Heat of an object is the total energy of all the individual molecules of which the given object is made. It is a form of thermal energy. When the object is heated, its thermal energy increases, means its molecules begin to move more violently. Temperature,on theother hand, is a measure of the average heat or thermal energy of the molecules in a substance. Heat is the form of energy which produces the sensation of warmth or coldness. Conventionally, heat energy supplied to a body is taken as positive and the heat energy given out by a body is taken to be negative. The CGS unit of heat is the calorie (cal) - defined as the amount of heat required to raise the temperature of 1g of water through 1oC. The S.I. unit of heat energy is the Joule (J) – defined as the amount of work done when a force of one Newton acts through one meter parallel to itself. The relationship between two units is: 1 cal = 4.18 J. Heat on the basis of kinetic theory:According to the kinetic theory, heat of a body is total kinetic energy of all its molecules.If a body have ‘n’ number of molecule having mass m and velocities v1, v2, v3, --------, vn respectively, then Total heat energy in the body (H) = Sum of kinetic energy of all molecules 1 1 1 1  H = K  mv12 + mv22 + mv32 +........ + mvn2  ; where K is thermal constant. 2 2 2 2  When the body is heated, the kinetic energy of each molecule inside it increases due to increase in their velocity. This results in the increase of total kinetic energy of the body and in turn represents total heat of the body. Temperature 49  Temperature is the degree of hotness of the body. It is the average kinetic energy of all the molecules of which the given body is made and is given by the expression; 1 1 1 1  K  mv12 + mv22 + mv32 +........ + mvn2  T=   2 2 2 2 n Units of temperature are; Fahrenheit ( F), Celsius (oC) and Kelvin (K). Kelvin is the S.I. unitof o temperature. 6.2 DIFFERENCE BETWEEN HEAT AND TEMPERATURE: Heat Temperature Heat is energy that is transferred from Temperature is a measure of hotness one body to another as the result of a or coldness difference in temperature It is total kinetic energy of all the It is average kinetic energy of all the molecules molecules It depends on quantity of matter It does not depend on quantity of matter It is form of energy (Thermal) It is measure of energy S.I. unit is Joule S.I. unit is Kelvin 6.3 PRINCIPLES OF MEASUREMENT OF TEMPERATURE: Measurement of temperature depends on the principle that properties (physical/electrical/chemical) of material changes with change in temperature. A device that utilizes a property of matter to measure temperature is known as thermometer. Temperature is a principle parameter that needs to be monitored and controlled in most engineering applications such as heating, cooling, drying and storage. Temperature can be measured via a diverse array of sensors. All of them infer temperature by sensing some change in a physical characteristic; be it a thermal expansion, thermoelectricity, electrical resistance or thermal radiation. There are four basic types of thermometers, each working on a different principle: 1. Mechanical (liquid-in-glass, bimetallic strips, bulb & capillary, pressure type etc.) 2. Thermo-electric (Thermocouples) 3. Thermo-resistive (RTDs and thermistors) 4. Radiative (Infrared and optical pyrometers). Each produces a different scale of temperature which can be related to one another. Commonly used thermometers are mercury thermometer, platinum resistance thermometer, thermo-electric and pyrometers. Liquid thermometers can measure temperature upto 300oC. Resistance thermometers can go upto 1200oC while thermo-electrics are used for measuring temperature as high as 3000oC. For still higher temperatures pyrometers (very hot furnaces) are used. 50 6.4 DIFFERENT SCALES OF TEMPERATURE AND THEIR RELATIONSHIP In general, there are three scales of temperature measurement. The scales are usually defined by two fixed points; temperature at which water freezes and the boiling point of wateras defined at sea level and standard atmospheric pressure. a) Fahrenheit Scale: It was given by physicist Daniel Gabriel Fahrenheit in 1724. It uses the degree Fahrenheit (symbol: °F) as the unit. On this scale freezing point of water is taken as the lower fixed point (32°F) and boiling point of water is taken as upper fixed point (212°F). The interval between two points is divided into 180 equal parts. Each division is 1o F. This scale is used for clinical and meteorological purpose. b). Celsius Scale: This scale was given by Anders Celsius in 1742. The scale was called centigrade scale. However in 1948 it was given the name Celsius to honour Anders Celsius.On this scale freezing point of water is taken as the lower fixed point (marked 0°C) and boiling point of water is taken as upper fixed point (marked 100°C). The interval between two points is divided into 100 equal parts. Each division is 1oC.This scale is used for common scientific, clinical, meteorological and technological work. c). Kelvin Scale: In1954, the Celsius scale was redefined in terms of the absolute zero and the triple point of a specially purified water. This definition also precisely relates the Celsius scale to the Kelvin, which defines the SI base unit of temperature with symbol K. On this scale freezing point of water is taken as the lower fixed point (273K) and boiling point of water is taken as upper fixed point (373K). The interval between two points is divided into 100 equal parts. Each division is 1K. 1oC = 1 K This is the natural scale of temperature also called the absolute temperature scale. Absolute zero is the basis of the Kelvin scale. The scale is based on ideal gas thermometer. Figure 6.1 shows three temperature scales with maximum and minimum temperature points. Absolute Zero:Absolutezeroisthe temperature at which all molecular motions comes to stand still i.e. net kinetic energy becomes zero. It is taken as zero Kelvin (-273o C). At absolute zero temperature, the pressure (or volume) of the gas goes to zero. This may implies that if the temperature is reduced below -273.15°C, the Fig. 6.1 Three scales of Temperature 51 volume becomes negative which is obviously not possible. Hence -273.15°C is the lowest temperature that can be achieved and thereforecalled the absolute zero of temperature. The interval on the scale is the same as on the Celsius scale (1 K = 1 oC) and two scales can be related as. K = oC + 273.15 Thus on absolute scale of temperature, water freezes at 273.15K and boils at 373.15K. Triple Point of water:The triple point is that point on a pressure versus temperature graph which corresponds to the equilibrium among three phases of a substance i.e. gas, liquid and solid. Triple point of pure water is at 273.15K. It is unique and occurs at single temperature and single pressure. Relation among the Scales of Temperature Temperature of a body can be converted from one scale to the other. Let, L = lower reference point (freezing point) H = upper reference point (boiling point) T = temperature read on the given scale. T −L Now = Relative temperature w.r.t. both reference point. H −L This relative temperature should not change if we are measuring the temperature of a body by using different thermometers. Let us take a body whose temperature is determined by three different thermometers giving readings in oC, oF and K respectively. Let T1 = C = Temperature in o C, L1 = 0°C H 1 =100°C T2 = F = Temperature inoF, L2 = 32°F H 2 = 212°F T3 = K = Temperature Kelvin, L3 =273 K H 3 = 373K We can write, 52 6.5 MODES OF TRANSFER OF HEAT When two bodies having different temperatures a