Homogeneity of equations This is a method of checking if an equation is correct by looking at the units. An equation is homogeneous if, when the base units of all the quantities are written, they are the same on both sides of the equation. Dimension of physical quantity Dimension of a physical quantity is the relationship between the physical quantity and the basic physical quantity such as mass, M, length L, time T, electric current A, temperature π½ and amount of substance N. Square brackets are used to indicate the dimensions of M, L and T are used to denote mass, length and time when we are dealing with dimensions. Thus, the dimensions of force are; [πΉ] = [ππΏπ −2 ] Important point note: Two physical quantities can be added or subtracted if and only if their dimensions are the same. e.g. πΉ = ππ‘ + ππ‘ 2 From the equation (πΉ − ππ‘) it means (at) has the dimensions of (F) as well (ππ‘ 2 ) has the dimensions of (F). Examples 1. [π΄πππ] = [πππππ‘β × ππππππ‘β] = πΏ × πΏ = πΏ2 ππππ‘ πππ ππππ = π2 2. [πππππππ‘π¦] = [ πππ πππππππππ‘ π‘πππ πΏ ] = π = πΏπ −1 ππππ‘ πππ π£ππππππ‘π¦ = ππ −1 3. [π΄ππππππππ‘πππ] = [ πβππππ ππ π£ππππππ‘π¦ π‘πππ ]= πΏπ −1 π = πΏπ −2 ππππ‘ πππ ππππππππ‘πππ = π/π 2 4. [πΉππππ] = [πππ π × πππππππππ‘πππ] = ππΏπ −2 ππππ‘ πππ πππππ = π = ππππ −2 5. [ππππ] = [πππππ × πππ πππππππππ‘] = ππΏ2 π −2 ππππ‘ πππ π€πππ = π½ = πππ2 π −2 6. [πΈππππ‘πππ πβππππ] = [ππ’πππππ‘ × π‘πππ)= AT Unit for electric charge = C = A s 1 7. [πΉππππ’ππππ¦] = 8. ππ₯π‘πππ πππ Strain = ππππππππ ππππβπ‘ ππππππ = π −1 = πΏ πΏ which is dimensionless More examples Example - 01: 1 Check the correctness of physical equation, π = π’π‘ + 2 ππ‘ 2 , Where π’ is the initial velocity, vis the final velocity, π is the acceleration, π is the displacement and π‘ is the time in which the change occurs. Solution: 1 Given equation is, π = π’π‘ + 2 ππ‘ 2 πΏ. π». π. = π£, βππππ [πΏ. π». π. ] = [π ] = [πΏ1 π0 π 0 ] . . . . . . . . . . . . . . . . . . . (1) π . π». π = π’π‘ + 1 2 ππ‘ , βππππ [π . π». π] = [π’][π‘] + [π] [π‘]2 2 = [πΏ1 π0 π −1 ][πΏ0 π0 π 1 ] + [πΏ1 π0 π −2 ][πΏ0 π0 π 1 ]2 = [πΏ1 π0 π 0 ] + [πΏ1 π0 π −2 ][πΏ0 π0 π 2 ] = [πΏ1 π0 π 0 ] + [πΏ1 π0 π 0 ] = [πΏ1 π0 π 0 ] . . . . . . … … … … … … … … … … … … … … . . . . . . . . . . . . . (2) (1) and (2) we have πΏ. π». π. = [π . π». π. ] Hence by the principle of homogeneity, the given equation is dimensionally correct. Example - 02: Check the correctness of physical equation,πΉ = mπ£ 2 r , Where πΉ is the centripetal force acting on a body of mass m performing uniform circular motion along a circle of radius π with linear speed π£. Solution: Given the equation mπ£ 2 πΉ= r πΏ. π». π. = πΉ, βππππ [πΏ. π». π. ] = [πΉ] = [πΏ1 π1 π −2 ] . . . . . . . . . . . . . . . . … . . . . (1) mπ£ 2 π . π». π = , βππππ [π . π». π] = [π][π£]2 /[π] r = [πΏ0 π1 π 0 ][πΏ1 π0 π −1 ]2 [πΏ1 π0 π 0 ] = [πΏ0 π1 π 0 ][πΏ2 π0 π −2 ] [πΏ−1 π0 π 0 ] 2 = [πΏ1 π1 π −2 ] … … … … … … … . … … … … … … … … … . . . . . . . . . . . . . (2) Hence by the principle of homogeneity, the given equation is dimensionally correct. Example - 03: Check the correctness of physical equation. π£ 2 = π’2 + 2ππ 2 Solution: Given equation is π£ 2 = π’2 + 2ππ 2 πΏ. π». π. = π£2 , βππππ [πΏ. π». π. ] = [π£2 ] = [πΏ1 π0 π −1 ]2 = [πΏ2 π0 π −2 ] . . . . . . . . . . . . . . . (1) π . π». π = π’2 + 2ππ 2 = [πΏ1 π0 π −1 ] + [πΏ1 π0 π −2 ][πΏ1 π0 π 0 ]2 = [πΏ2 π0 π −2 ] + [πΏ1 π0 π −2 ][πΏ2 π0 π 0 ] = [πΏ2 π0 π −2 ] + [πΏ3 π0 π −2 ]. … … … … … … … … … … … … … … . . . . . . . . . . (2) πΉπππ (1) πππ (2) π€π βππ£π [πΏ. π». π. ] ≠ [π . π». π. ] Example - 04: Check the homogeneity of equation, when the periodic time, “T” of vibration of magnet of moment of inertia 'I', magnetic moment 'M' vibrating in magnetic induction 'B' is given by πΌ π = 2π√ ππ΅ π 0πΌ Where πΌ = mr 2 , π = and π΅ = 2ππ Given the equation; πΌ π = 2π√ ππ΅ πΏ. π». π = [π] = [πΏ0 π0 π1 ] . . . . . . . . . . . . . . . (1) [πΌ] π . π». π = √ [π][π΅] [π1 πΏ2 π 0 ] = √ 2 0 0 1 0 1 −2 −1 [πΏ π π πΌ ][πΏ π π πΌ ] [π1 πΏ2 π 0 ] = √ 2 1 −2 0 [πΏ π π πΌ ] π . π». π = √[πΏ0 π0 π 2 ] = [πΏ0 π0 π 1 ] … … … … … … … . . (2) πΉπππ (1)πππ (2)π€π βππ£π [πΏ. π». π. ] = [π . π». π. ] Hence by the principle of homogeneity, the given equation is dimensionally correct. 3 Example - 05: a) The period T of a tuning fork depends on the density π, Young modulus E and length π of the tuning fork. Which of the following equations can be used to relate T with the quantities mentioned? Nb: E has the units of pascals per second. π΄π √ππ 3 i). π= ii). π = π΄π√πΈ iii). π = π √π πΈ π π΄πΈ π Where A is a dimensionless constant and g is the acceleration due to gravity. b) The table below is obtained using various steel tuning forks which are geometrically identical in an experiment. Frequency /cycle π −1 (Hz) 256 288 320 384 480 Length I/cm 12.0 10.6 9.6 8.0 6.4 Use the above data to confirm the equations you have chosen. Hence determine the value of the constant A, if, for steel, E = 2·0 x 1011 Nπ−2 p = 8 500 kg π−3. [Unit for E = kg π−1 π −2] Solution a) Unit for period T = s For equation (i) π΄π √ππ 3 π= πΈ π΄π πππ−3 1 3 √ππ = ππππ‘ πππ π = (ππ −2 π3 )− ⁄2 −1 −2 πΈ πππ π = π Equation (i) is homogeneous and can be used. For equation (ii) π π = π΄π √ πΈ 1⁄ 2 πππ−3 ππππ‘ πππ π = π΄π √ = π ( ) πΈ πππ−1 π −2 π =π Equation (ii) is homogeneous and can be used. 4 For equation (iii) π= π΄πΈ π √ π π ππππ‘ πππ π = π΄πΈ π πππ−1 π −2 π √ = ( ) = π2 π −1 π π πππ−3 ππ −2 Which is different from the unit of ‘T’, so it cannot be used. b) Frequency f /cycle π−π 256 288 320 384 480 Period T/(x ππ−π s) 3·91 3.47 3·13 2·60 2·08 Length I/cm 12.0 10.6 9.6 8.0 6.4 π π π ⁄π /ππ ⁄π 41.6 34.5 29.7 22.6 16.2 (Hz) From the graphs above, the graph of π against π is a straight line passing through the origin. π Therefore, the equation π = π΄π√πΈ is correct. The graph or T against is a straight line but does not pass through the origin. Therefore, the equation π = π΄π πΈ √ππ 3 is wrong. 5 π From the equa1ion π = π΄π√πΈ and from the table. when l = 12·0 cm. T = 3 · 91 × 10− 3 π , π΄= π πΈ √ π π π΄= π πΈ √ π π (3 · 91 × 10− 3 ) 2.0 × 1011 √( ) 0.12 8 500 = 158 Example - 06: Check the homogeneity of equation, when the rate of flow of a liquid having coefficient of viscosity 'π' through a capillary tube of length 'π' and radius 'π' under pressure head ′π' is given by ππ πππ4 = ππ‘ 8ππ Nb: the unit of ′ π′ is the pascal per second. Solution Given the equation ππ πππ4 = ππ‘ 8ππ [π0 πΏ3 π 0 ] [ππ] [ππ] = = = [π0 πΏ3 π −1 ] . . . . . . . . . . . . . . . (1) [ππ‘] [ππ‘] [π0 πΏ0 π 1 ] [π1 πΏ−1 π −2 ] [π0 πΏ1 π 0 ]4 [π][π]4 π . π». π = = [π][π] [π0 πΏ1 π 0 ] [π1 πΏ−1 π −1 ] πΏ. π». π. = = [π1 πΏ−1 π −2 ][π0 πΏ4 π 0 ] [π1 πΏ0 π −1 ] [π1 πΏ3 π 2 ] = 1 0 −1 = [π0 πΏ3 π −1 ] … … … … … … … … … … . … … … … (2) [π πΏ π ] πΉπππ (1)πππ (2)π€π βππ£π [πΏ. π». π. ] = [π . π». π. ] Hence by the principle of homogeneity, the given equation is dimensionally correct. Example - 07: Check the homogeneity of equation, when the terminal velocity 'v' of a small sphere of radius 'a' and density ' π ' falling through a liquid of density ' π ' and coefficient of viscosity ' π ' is given by 2ππ2 (π − π) π£= 9π Given the equation 6 2ππ2 (π − π) 9π 0 1 −1 ] [π£] [π πΏ. π». π. = = πΏπ … … … … … … … … … … … . . . . . . . . . . . . . . . (1) π£= π . π». π = 2ππ2 (π − π) [π][π]2 ([π] − [π]) = 9π [π] [π0 πΏ1 π −2 ] [π0 πΏ1 π 0 ]2 ([π1 πΏ−3 π 0 ] − [π1 πΏ−3 π 0 ]) π . π». π = [π1 πΏ−1 π −1 ] π . π». π = [π0 πΏ1 π −2 ][π0 πΏ2 π 0 ]([π1 πΏ−3 π 0 ]) [π1 πΏ−1 π −1 ] [π1 πΏ0 π −2 ] π . π». π = 1 −1 −1 = [π0 πΏ1 π −2 ] … … … … … … … … … … … … … … . . (2) [π πΏ π ] πΉπππ (1)πππ (2)π€π βππ£π [πΏ. π». π. ] = [π . π». π. ] Hence by the principle of homogeneity, the given equation is dimensionally correct. Example - 08: The distance covered by a body in time tis given by the relation π = π + ππ‘ + ππ‘ 2 . What are the dimension of a,b,c? Also write the quantities they represent. Solution: By the principle of homogeneity, the dimensions on either side of the physical equation must be the same. Now two physical quantities can be added or subtracted if and only if their dimensions are the same. Dimensions of S = Dimensions of a [πΏ1 π0 π 0 ] = [π] Dimensions of 'a' are that of displacement. Hence 'a' represent the displacement. Dimensions of S = Dimensions of ππ‘ [πΏ1 π0 π 0 ] = [π][π‘] [πΏ1 π0 π 0 ] [π] = [π] [π] = [πΏ1 π0 π −1 ] Dimensions of 'a' are that of velocity. Hence 'b' represents the velocity. Dimensions of S = Dimensions of ππ‘ 2 . [πΏ1 π0 π 0 ] = [π] [πΏ1 π0 π 0 ] [π] = [π]2 [π] = [πΏ1 π0 π −2 ] 7 Dimensions of 'c' are that of acceleration. Hence 'c' represents the acceleration. Ans: The dimensional formula of 'a', 'b' and 'c' are [πΏ1 π0 π 0 ], [πΏ1 π0 π −1 ], and [πΏ1 π0 π −2 ]. The quantities represented by a, b and c are displacement, velocity and acceleration. Example - 09: πΎπ⁄ 1⁄2 π) , where P is the pressure, d is the density of a gas. What is the dimensional formula for πΎ? According to Laplace's formula, the velocity (V) of sound in a gas is given by π£ = ( Solution: πΎπ⁄ π 2 π£ π ⁄π ∴πΎ= ∴ π£2 = ∴πΎ= [π£ 2 ][π] [πΏ1 π0 π −1 ][πΏ−3 π1 π 0 ] = [πΏ−1 π1 π −2 ] [π] [πΏ−1 π1 π −2 ] ∴ πΎ = −1 1 −2 = [πΏ0 π0 π 0 ] [πΏ π π ] Example - 10: A force F is given by πΉ = ππ‘ + ππ‘ 2 , where π‘ is the time. Find the dimensional formula of 'π' and 'b'. Solution: By the principle of homogeneity, the dimensions on either side of the physical equation must be the same. Now two physical quantities can be added or subtracted if and only if their dimensions are the same. Dimensions of F = Dimensions of ππ‘ [πΏ1 π1 π −2 ] = [π][π‘] [π] = [πΏ1 π1 π −2 ] [π] [π] = [πΏ1 π1 π −3 ] Dimensions of F = Dimensions of ππ‘ 2 [πΏ1 π1 π −2 ] = [π][π]2 [πΏ1 π1 π −2 ] [π] = [π]2 [π] = [πΏ1 π1 π −4 ] The dimensional formula of a and b are [πΏ1 π1 π −3 ] and [πΏ1 π1 π −4 ] Example - 11: In an equation {π + π⁄π 2 }(π − π) = π π, where P is the pressure, V is the volume, T is the temperature and a, b, R are constants. What is the dimensional formula of π ⁄π? 8 Solution: Two physical quantities can be added or subtracted if and only if their dimensions are the same. Thus, the dimensions of P and π⁄π 2 are same [π] [π] = 2 [π] ∴ [π] = [π][π] 2 ∴ [π] = [πΏ−1 π1 π −2 ][πΏ3 ] 2 ∴ [π] = [πΏ−1 π1 π −2 ][πΏ6 ] ∴ [π] = [πΏ5 π1 π −2 ] Dimensions of b = Dimensions of V [π] = [πΏ3 π0 π 0 ] [a] [πΏ5 π1 π −2 ] Now the dimensions of is 3 0 0 = [πΏ2 π1 π −2 ] [πΏ π π ] [b] The dimensions of π⁄π is = [πΏ2 π1 π −2 ]4 Practice questions 1. What are the S. I. units of k so that the equation? π£ππππππ‘π¦ = π × ππππ ππ‘π¦ 2. A sphere of radius a moving with a velocity v under streamline conditions in a viscous fluid experiences a retarding force πΉ given by πΉ = πππ£ where π is a constant What are the S.I. units of π in terms of the base units? 3. The heat capacity C of a solid can be expressed as a function of temperature T to fit the expression: πΆ = πΌπ + π½π 3 a. Find the possible units of ′πΌ′ and ′π½′. b. Write down the base S.I. units of specific heat capacity. 4. Bernoulli’s equation. which applies to fluid flow. states that: 1 π + β π π + ππ£ 2 = π 2 where p is pressure, h height, p density, g acceleration due to gravity, v velocity and k a constant. Show that the equation is dimensionally consistent and state an S.I. unit for k. Deriving Physical equations Determining a physical quantity usually depends on a number of other physical quantities. For example, the period T of a simple pendulum depends on its length π and the acceleration due to gravity g. Using units or dimensions. an C4ua1ion can be derived to relate the period T with / and g. Suppose that ππΌ π π₯ πy 9 where.π₯ and π¦ are non-dimensional constants. Then π = π π π₯ πy where k is a dimensionless constant of proportionality. Since a physical equation must be dimensionally consistent, units on the left side of the equation must be the same as the units on the right side of the equation. Units of T = units of [π π π₯ πy ] π = π π₯ (ππ −2 )π¦ = π π₯+π¦ π −2 π¦ Equating indices of s: 1 = −2π¦. ∴ π¦ = − Equating indices of m: 0=π₯+π¦ 1 π₯ = −π¦ = 2 1 2 ∴π=ππ 1⁄ 2 π− 1⁄ 2 π ∴π=π√ π Example The speed of sound π£ in a medium depends on its wavelength π. the Young modulus E and the density π of the medium. Use the method of base units to derive a formula for the speed of sound in a medium. Unit of E: πππ−1 π −2. Solution: Suppose: π£ = π ππ₯ πΈ y π z where π, π₯, π¦ and π§ are non-dimensional constants. Units of v = units of [π ππ₯ πΈ y πz ] ππ −1 = π π₯ (πππ−1 π −2 )π¦ (πππ−3 )z = π π₯−π¦−3π§ π −2y ππy+z Equating indices of s: −1 = −2π¦. ∴ π¦ = Equating indices of kg: 0=π¦+π§ 1 2 10 π₯ = −π¦ = 1 1 ∴ π§ = −π¦ = − 2 2 Equating indices of m: 1 = π₯ − π¦ − 3π§ 1 3 =π₯− − 2 2 ∴π₯=0 ∴π=ππΈ 1⁄ 2 π− 1⁄ 2 πΈ ∴π£=π√ π Example Poiseuille assumed that the rate or now or a liquid through a horizontal tube under streamline flow depends a) a. radius of the tube. b) π viscosity of the liquid. and c) π π , the pressure gradient along the tube, where π = pressure difference across the length or the tube π = length of tube. Using Poiseuille's assumption. derive an expression for the rate or now of a liquid through a horizontal tube in terms of π, π, π and π. Solution π π’ππππ π π‘βπ πππ‘π ππ ππππ€, ππ π π¦ = ππ π₯ ( ) π π§ ππ‘ π π€βπππ π, π₯ π¦ πππ π§ πππ ππππππ ππππππ π ππππ π‘πππ‘π . ππππ‘π ππ ππ π π¦ = ππππ‘π ππ [ππ π₯ ( ) π π§ ] ππ‘ π π¦ 3 −1 π π ππππ −2 1 = π ×( × ) (πππ−1 π −1 ) π§ π2 π π₯ = ππy+z π π₯−2π¦−π§ π −2y−z 11