Let $ε_{0}$ denotes the dimensional formula of 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.

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$μ_{0} = [M L^{2} T^{- 1} I]$

$ε_{0} = [M^{- 1} L^{- 1} T^{- 1} I]$

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

$μ_{0} = [M L T^{- 2} I^{- 2}]$

answer is B, C.

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Detailed Solution

$F = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{2}} \Rightarrow [ε_{0}] = [\frac{q^{2}}{F r^{2}}] = [\frac{T^{2} I^{2}}{M L T^{- 2} L^{2}}] = [M^{- 1} L^{- 3} T^{4} I^{2}] Further, \frac{1}{{\sqrt{ε_{0} μ}}_{0}} = speed of light c ∴ [μ_{0}] = \frac{1}{[ε_{0} c^{2}]} = \frac{1}{[M^{- 1} L^{- 3} T^{4} I^{2} L^{2} T^{- 2}]} = [{MLT}^{- 2} I^{- 2}]$