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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 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 Physical Quantity = Numerical value x unit Q = Physical Quantity n= Numerical value u = Standard unit e.g. Mass of stool = 15 kg Mass = Phys 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. e¢ 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 x Breadth = Length x Length = (Length) Speed =Distance / Time =Length / Time The units for derived quantities are called Derived Units, Definition of Basic and Supplementary Unit of S.1. 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 x 10° 7Newton 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 10'? 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 @. 6 =1/r; Lis 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 Q. Q=As/r SOME IMPORTANT ABBREVIATIONS Symbol _| Prefix Multiplier | Symbol | Prefix Multiplier D Deci 10" da deca 10" c centi 10? h hecto 10° m milli 107 k kilo 108 u micro 10° M mega 10° n nano 10° G giga 10° P Pico 10°? T tera 10! femto 10'5 Pecta 10" a atto 10°68 E exa 10!8 = Some Important Units of Length: (i) 1 micron = 10° m= 10cm (ii) 1 angstrom = 1A= 10" ms 10% em (iii) 1 fermi = 1 fm = 107% m (iv) 1 Light year = 1 ly = 9.46 x 10'Sm (v) 1 Parsee = Ipe = 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° gram 1 centigram = 1/100 gram = 10° gram 1 decigram = 1/10 gram 1 quintal = 100 kg 1 metric ton = 1000 kilogram 14 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 = CL! <(L']=(27] 5 IML) Power (0,2,0) of fundamental units are called dimensions of area in mass, length and time respectively. eg, Density = mass/volume =(M}{L*] =[M!'L31° 1.5 DIMENSIONAL FORMULAE AND SI UNITS OF PHYSICAL QUANTITIES 19 | Density Mass/ volume [M'L°T°} Kg/m' 20. | Speed or velocity Distance/ time IM’L'T'] | m/s 21 | Acceleration Velocity/time [M°L'T?] m/s? 22 | Pressure Force/area [M'L'T?| N/m? 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. ew, 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 ={MJ/[L*] = [M'L3T°] (ii) Power = Work/Time =Force x Distance/Time =[M'L'T?] x [LT] =(M'L’T*} Force x Distance Area x Velocity Mass x Acceleration x Distance x time (iii) Co-efficient of viscosity = length x length x Displacement =[M] x [LT?] x (L] [TV{L?] x (L] =[M'L'T"] (iv) Angle = arc (length)/radius (length) = [LVL] =[M°L°T®] = 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 = [M'L’?T?] Dimension of power = work / time = [M'L?T*] Work and Power have not the same dimensions. (ii) Dimension of stress = force / area = [M'L'T?]{L’] = [M'L'T?] Dimension of pressure = force / area = [M'L'T?}/[L?] = [M'L'T?] Stress and pressure have the same dimension. (iii) Dimension of momentum = mass x velocity= [M'L'T"] Dimension of impulse = force x time =(M'L'T"] 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+%at? 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={L'] Velocity, u=[L'T'] Time, t=(T'] (ii) t = 2nv1/g Here, Dimensions of L.H.S, t = [T'] = [M°L°T'] Dimensions of the terms on R.H.S Dimensions of (length) = [L'] Dimensions of g (acc due to gravity) = [L'T?] 2m being constant have no dimensions. Hence, the dimensions of terms @nvi/g on R.H.S = (LY LIT?) )!2 = [TY] = [M°L°T}] Thus, the dimensions of the terms on both sides of the relation are the same i.e., [M°L°T']. Therefore, the relation is correct. Example 4. Check the correctness of the following equation on the basis of dimensional E analysis, V = v5 Here V is the velocity of sound, E is the elasticity and d is the density of the medium. Sol. Here, Dimensions of the term on L.H.S V =[M’L'T"] Dimensions of elasticity, E = [M'L'T?] & Dimensions of density, d = [M'L*T°| Therefore, Dimensions of the terms on R.H.S rE 7 (M'L!T?/ MIL! T2]!2 = [M°L!T!] 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 /rgCos@. Sol. The given formula is, h = 2Td /rgCos@. Dimensions of term on L.H.S Height (h) = [M°L'T®] Dimensions of terms on R.H.S T= surface tension = [M'L°T*] D= density = [M'L°T°] r =radius = [M°L'T°] g=ace.due to gravity = [M°L'T?] Cos@ = [M°L°T]= no dimensions So, Dimensions of 2Td/rgCos6 = [M'L°T?] x [M'L? T°] / (M°L!T°) x [M°L'T?] - [M2L°T°} 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 + mv’; (ii) vi-w? = 2as*. Sol. (i) E=mgh+%mvy- Here,dimensions of the term on L.H.S. Energy, E = [M'L’T?] Dimensions of the terms on R.H.S, s of the term, mgh = [M] »[LT2] x (L] = [M'L2T?] Dimensions of the term, ¥ mv?= [M] x [LT']?= [M'L?T?] 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, vi-u?= 2as” Dimensions of the terms on L.H.S v> =[M°] x [LT PB= [M°LAT} w= [M°] x [LT'P= [M°L?P?] Dimensions of the terms on R.H.S 2as® = [M°] x [LT?] x[L}? = [M°L?T] Substituting the dimensions in the relations, ve-u?= 2as” We get, [M°L3T?] - [M°L?T?] = [M°L3T?} 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+ bitte What are the dimensions of a, b and c? Sol. Dimensional formula for L.H.S V=[L'T"] In the R.H.S dimensional formula of At (T]= [L'T"] 11 Thus knowing [MiLiT)], [M2L2T2] a, b, ¢ and m, we can calculate n2 Example 9, Conyert a force of 1 Newton to dyne, Sol. To convert the force from MKS system to CGS system, we need the equation Q=nyuy=n2u2 Thus n, = th My Here ni =!, w=1N, uw=dyne [ML7,"| my = a [Mbt | nan { 10008m Je 2 ' gm cm s nm, = 1(1000)(100) n, =10° Thus IN=10° 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 Ih uy ny nisl uj=erg (CGS unit of work) u2= joule (SI unit of work) n, =100°)(10)? ny =10°7 Thus, 1 erg= 10’ 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. 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. It fails to derive relations of quantities involving exponential and trigonometric functions. 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 + % at’ ete. However, such relations can be derived indirectly. 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. AR kk EXERCISES Multiple Choice Questions 1, [ML"'T?] is the dimensional formula of (A) Force (B) Coefficient of friction (C) Modulus of elasticity (D) Energy. 2. 10°Fermi is equal to (A) | meter (B) 100 micron (C) | Angstrom (D) 1mm 3. rad / see 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 
