Let $[ε_{0}]$ denote the dimensional formula of the permittivity of the vacuum and $[μ_{0}]$ that of the permeability of the vacuum. If $M = m a s s,$ $L = l e n g t h,$ $T = t i m e$ $a n d$ $I = e l e c t r i c c u r r e n t$ .

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$[ε_{0}] = [M^{- 1} L^{- 3} T^{4} I^{2}]$

$[μ_{0}] = [M L T^{- 2} I^{- 2}]$

$[μ_{0}] = [M L^{2} T^{- 1} I]$

$[ε_{0}] = [M^{- 1} L^{- 3} T^{2} I]$

answer is B, C.

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Detailed Solution

Electrostatic Force, $F = \frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{r^{2}}$

$Permittivity of free space ϵ_{o} = \frac{q_{1} q_{2}}{4 π {Fr}^{2}}$

$Thus dimensional formula of [ϵ_{0}] = \frac{[AT] [AT]}{[{MLT}^{- 2}] [L^{2}]}$

$\Rightarrow [ϵ_{0}] = [M^{- 1} L^{- 3} T^{4} A^{2}]$

Magnetic Force, $F = \frac{μ_{0}}{4 π} \frac{m_{1} m_{2}}{r^{2}}$

${Permeability of free space, μ}_{o} = \frac{4 {πFr}^{2}}{m_{1} m_{2}}$

$Thus dimensional formula of [μ_{0}] = \frac{[{MLT}^{- 2}] [L^{2}]}{[AL] [AL]}$

$\Rightarrow [ϵ_{0}] = [M^{1} L^{1} T^{- 2} A^{- 2}]$

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