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Dimensions of physical quantities

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In a new system of units mass, acceleration and frequency are taken as fundamental units. If unit of mass is 100 g, unit of acceleration is 2 m/s² and unit of frequency is 4 sec^-1 in the new system of units. Then the value of 0.1 J in this system is

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Solution

$\begin{array}{l} Energy \propto m^{x} a^{y} f^{z} \\ [M L^{2} T^{- 2}] = [M]^{x} {[L T^{- 2}]}^{y} {[T^{- 1}]}^{z} \\ 1 = x \\ 2 = y \\ - 2 = - 2 y - z \Rightarrow z = - 2 \\ Energy \propto \frac{m a^{2}}{f^{2}} \\ 0.1 J = x J^{'} \\ x = (0.1) \frac{J}{J^{'}} = (0.1) {(\frac{M_{1}}{M_{2}})}^{1} {(\frac{a_{1}}{a_{2}})}^{2} {(\frac{f_{1}}{f_{2}})}^{- 2} \\ = (0.1) (\frac{1 kg}{100 g}) {(\frac{1 m / s^{2}}{2 m / s^{2}})}^{2} {(\frac{1 \sec^{- 1}}{4 \sec^{- 1}})}^{- 2} \\ = (0.1) (10) (\frac{1}{4}) (16) = 4 \end{array}$

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