Q 11-01-242JEE MainJEE Main 2025 (2 Apr, Shift 2)Easy
If $\mu_0$ and $\varepsilon_0$ are the permeability and permittivity of free space, respectively, then the dimension of $\left(\dfrac{1}{\mu_0\varepsilon_0}\right)$ is:
Answer: (B) $L^2/T^2$
The speed of light in vacuum is $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$, so
$$\frac{1}{\mu_0\varepsilon_0} = c^2 \Rightarrow \left[\frac{1}{\mu_0\varepsilon_0}\right] = L^2T^{-2} = \frac{L^2}{T^2}$$
Solution by Sreeraj P, M.Sc Physics