Let $[ε_{0}]$ denote the dimensional formula for the permittivity of the vacuum and $[μ_{0}]$ that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current, then

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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

$\begin{array}{l} F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}} \\ [ε_{0}] = \frac{q_{1} q_{2}}{F r^{2}} = \frac{I T I T}{M L T^{- 2} L^{2}} = [M^{- 1} L^{- 3} T^{4} I^{2}] \\ C = \frac{1}{\sqrt{μ_{0} ε_{0}}} \Rightarrow μ_{0} = \frac{1}{C^{2} ε_{0}} = \frac{1}{L^{2} T^{- 2} M^{- 1} L^{- 3} T^{4} I^{2}} = [M L T^{- 2} I^{- 2}] \end{array}$