If force $(F)$, velocity $(V)$ and time $(T)$ are taken as fundamental units, then the dimensions of

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If force $(F)$, velocity $(V)$ and time $(T)$ are taken as fundamental units, then the dimensions of energy are

Professor PC Thomas Classes · Answer

(d) : Let energy $E \propto F^{a} V^{b} T^{c}$ or $E=k F^{a} V^{b} T^{c}$\\ where $k$ is a dimensionless constant and $a, b$ and $c$ are the exponents.\\ Writing dimensions on both sides, we get\\ $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}ight]=\left[\mathrm{MLT}^{-2}ight]^{a}\left[\mathrm{LT}^{-1}ight]^{b}[\mathrm{~T}]^{c}$\\ $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}ight]=\left[\mathrm{M}^{a} \mathrm{~L}^{a+b} \mathrm{~T}^{-2a-b+c}ight]$ \\ Applying the principle of homogeneity of dimensions, we get \\ $a=1 \quad \ldots(\mathrm{i}), \quad a+b=2 \quad \ldots(	ext{ii}), \quad-2a-b+c=-2 \quad \ldots (	ext{iii})$ \\ Solving eqns. (i), (ii) and (iii), we get $a=1, b=1, c=1$\\ Therefore, $[E]=\left[\mathrm{FVT}ight]$