Q 11-01-218JEE MainJEE Main 2019 (8 Apr, Shift 1)Easy
In SI units, the dimensions of $\sqrt{\dfrac{\varepsilon_0}{\mu_0}}$ is
Answer: (B) $A^2T^3M^{-1}L^{-2}$
$[\varepsilon_0] = M^{-1}L^{-3}T^4A^2$ (from Coulomb's law) and $[\mu_0] = MLT^{-2}A^{-2}$ (from $F = \mu_0 I^2 l/2\pi d$).
$$\frac{\varepsilon_0}{\mu_0} = M^{-2}L^{-4}T^6A^4 \Rightarrow \sqrt{\frac{\varepsilon_0}{\mu_0}} = A^2T^3M^{-1}L^{-2}$$
Solution by Sreeraj P, M.Sc Physics