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Dimensional Analysis

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Published in: Physics
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Dimensional Analysis in units and measurement chapter

Paridhi G / Indore

4 years of teaching experience

Qualification: M.Sc. Physics, B.Ed. , CTET

Teaches: Mathematics, Science, Physics

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  1. Units And Measurement Part 6 Dimensional Analysis Easy Learning By Paridhi Gupta SUBSCRIBE A
  2. DIMENSIONS OF PHYSICAL QUANTITIES: The dimensions of a physical quantity are the powers to which the base quantities are raised to represent that quantity. The nature of a physical quantity is described by its dimension. Length Dimensions:- Mass Time Electric current Temperature Luminous intensity Amount of substance [M] [cd] [mol]
  3. Volume = m3 = [L] Force = mass x acceleration = [M] Velocity= [L]/[T] DIMENSIONAL FORMULAE AND DIMENSIONAL EQUATIONS Force = [M] = [M L T-2] Volume = [MO L3 TO] Density = [M L-3 TO] Velocity= [MO L T-1]
  4. Dimensional Analysis : The study of the relationship between physical quantities with the help of dimensions and units of measurement is termed as Dimensional Analysis. Checking the Dimensional Consistency of Equations :- The magnitudes of physical quantities may be added together or subtracted from one another only if they have the same dimensions. velocity cannot be added to force, or an electric current cannot be subtracted from the thermodynamic temperature. This simple principle called the principle of homogeneity of dimensions in an equation is extremely useful in checking the correctness of an equation.
  5. Homogeneity Principle: Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another. Example : Check the correctness of physical equation s = ut + h at2. In the equation, s is the displacement, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time in which change occurs. Solution: We know that L.H.S = s and R.H.S = ut + 1/2at2 The dimensional formula for the L.H.S can be written as S = [LIMOTO] — .(1) We know that R.H.S is ut + h at2 , simplifying we can write R.H.S as ..........(2) From (1) and (2), we have [L.H.S] = [R.H.S] Hence, by the principle of homogeneity, the given equation is dimensionally correct.
  6. Example: Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions. The dependence of time period T on the quantities l, g and m as a product may be written as T = k IXgYmz (where k is dimensionless constant and x, y and z are the exponents). By considering dimensions on both sides, we have [LO MOT [L T-2]Y [M = LX+Y T-2Y W On equating the dimensions on both sides, we have
  7. x + y = O; —2y = 1; and z so that, x = h, y = -1/2, z = O Then, T = k lh g-h T = 1<41/g k = 271, T = 2n 41/g Applications of Dimensional Analysis • To check the consistency of a dimensional equation To derive the relation between physical quantities in physical phenomena To change units from one system to another

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