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    				| 1. | Q.25. Q 15) Derive an expression of kinetic energy of a body of mass m and moving with velocity vusing dimensional analysis. | 
| Answer» We know, dimension of energy is, [E] = [M1L2T-2] Dimension of mass is [m] = [M1L0T0] Dimension of velocity is [v] = [M0L1T--1] Suppose, [E] = k [m]x[v]y ‘k’ is a proportionality constant which is a dimensionless quantity. Therefore, [M1L2T-2] = [M1L0T0]x[M0L1T--1]y => [M1L2T-2] = [MxLyT-y] Thus, x = 1, y = 2 So, we have, E = k mv2 It is found that, k = ½ So, E = ½ mv2 Derived. | |