# dimensions

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Published on January 31, 2009

Author: sheikmohamed

Source: authorstream.com

Slide 1: Dimensions and Dimensional Formulae Dimensions and Dimensional Formulae : The dimensions of a physical quantity may be defined as the power raised to fundamental units to obtain the derived unit of that physical quantity. The expression representing the dimensions of a physical quantity is called the dimensional formula. If a physical quantity X has dimensions a in mass, b in length and c in time, then its dimensional formula may be expressed as [X] = [MaLbTc]. Dimensions and Dimensional Formulae Uses of Dimensional Equations : 1. Conversion of units of one system to another : This is based on the fact that the specification of a physical quantity requires a proper unit [u] and a numerical value representing the number (n) of units contained in that physical quantity and the product of numerical value contained in and the unit chosen of physical quantity always remains constant, whatever the system of units may be i.e. , n[u] = constant If a physical quantity X has dimensional formula [MaLbTc], and if (derived) units of that physical quantity in two systems are [M1aL1bT1c] and [M2aL2bT2c] respectively, then n1[u1] = n2[u2] i.e. , n1[M1aL1bT1c] = n2[M2aL2bT2c] therefore, n2 = n1[M1/M2]a[L1/L2]b[T1/T2]c Thus knowing the values of fundamental units in two systems and numerical value in one system, the numerical value in other system may be evaluated. Uses of Dimensional Equations Slide 4: ? Find the value of 60 joule/minute on a system which has 100g, 100cm and 1min as fundamental units. Solution: 60 joule/minute = joule/second = 1 joule/second = 1 watt SI n1 = 1, M1 = 1 kg, L1 = 1 m, T1 = 1 s New system n2 = ?, M2 =100 g, L2 =100 cm, T2 =1 min   Dimensional formula of power is [ML2T-3] So, a = 1, b = 2 and c = - 3 n2 = Slide 5: n2 = = on simplification, n2 = 2.16 × 106 Slide 6: 2. To check the correctness of a physical relation: This is based on the principle of homogeneity of dimensions. According to which the dimensions of all the terms on the two sides of an equation must be the same. ? Applying the principle of homogeneity of dimensions, check the dimensional accuracy of the relation V = Here V is the volume of liquid flowing out of a capillary tube per unit time, is the coefficient of viscosity of the liquid, p is the pressure difference across the two ends of the capillary tube, r is the radius of the capillary tube and l is the length of the capillary tube. Slide 7: Solution. Dimensional formula of V is [M0L3T-1] .…(i) Dimensional formula of p is [M1L-1T-2]. Dimensional formula of r4is [M0L4T0]. Dimensional formula of is [ML-1T-1]. Dimensional formula of is i.e., [M0L3T-1] ….(ii) It is clear from equations (i) and (ii) that dimensions of L.H.S. are equal to the dimensions of R.H.S. So, the given relation is dimensionally correct. Slide 8: 3. To derive the new relation : If a physical quantity X depends on other physical quantities P. Q and R (say), then we may write X PaQbRc …(1) where a, b and c are powers raised to physical quantities P, Q and R respectively. Then writing dimensional formula for X, P, Q and R and equating the dimensions on either sides, the values of a, b and c may be determined. The substitution of these values in (1) gives the new dimensional relation. ?Assuming that the escape velocity for a planet depends upon gravitational constant G, radius R of planet and also its density , derive formula for escape velocity from dimensional considerations.  Solution. if ve represents the escape velocity, then ve = KGaRbpc Slide 9: Writing dimensions, we get [M0L1T-1] = [M-1L3T-2]a[L]b[ML-3]c or [M0L1T-1]=[M-a+cL3a+b-3cT-2a] Equating corresponding powers, we get –a+c=0 ; 3a+b-3c=1 - 2a - 1 or a = ; - + c = 0 or c = Again, 3 × + b – 3 × = 1 or b = 1 From Eqn. (1) v2 = KG½ R1 ½ or ve=KR Limitations of Dimensional Analysis : It supplies no information about dimensionless constants. They have to be determined either by experiment or by mathematical investigation. This method is applicable only in the case of power functions. It fails in the case of exponential and trigonometric relations. This method fails to derive directly a relation which contains two or more than two quantities of like nature. It fails to derive the exact form of a physical relation, if a physical quantity depends upon more than three other physical quantities. This is because by equating powers of M,L and T, we can obtain only three equations. Three equations cannot determine more than three ‘unknowns’ Limitations of Dimensional Analysis Slide 11: If we cannot identify all the factors on which a physical quantity depends, then the method of dimensional analysis cannot be used to derive expression for a physical quantity. It can only check whether a physical relation is dimensionally correct or not. It cannot tell whether the relation is absolutely correct or not. Let us consider the equation: S=ut+ at2. Though the equation is dimensionally correct but it is actually is actually wrong equation because the correct equation is S=ut+ at2.

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