Q. Dimensions of $\frac{1}{\mu_{0}\epsilon_{0}}$, where symbols have their usual meaning, are
A) $\lbrack lT^{- 1}\rbrack$
B) $\lbrack L^{- 1}T\rbrack$
C) $\lbrack L^{- 2}T^{2}\rbrack$
D) $\lbrack L^{2}T^{- 2}\rbrack$
Explanation
$C = \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}} \Rightarrow \frac{1}{\mu_{0}\varepsilon_{0}} = c^{2} = \lbrack L^{2}T^{- 2}\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.