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Q. The dimensional formula of $\mu _{0}$ $\epsilon _{0}$ is

NTA AbhyasNTA Abhyas 2020Physical World, Units and Measurements

Solution:

$C = \frac{1}{\sqrt{\mu _{0} \epsilon _{0}}} \Rightarrow \mu _{0} \, \epsilon _{0} = \frac{1}{C^{2}}$
$C \rightarrow M^{0}L^{1}T^{- 1}$
$\Rightarrow $ $\frac{1}{C^{2}} \rightarrow M^{0}L^{- 2}T^{2}$