JEE Main 2025 Apr 7 Shift 2, Physics Q4: Electric Potential

$L=\frac{{\mu }_{0}NA}{\ell }$

$C=\frac{{\epsilon }_{0}A}{d}$

$\therefore \frac{L}{C}\propto \frac{{\mu }_0}{{\epsilon }_0}$

$\Rightarrow\ \sqrt{\dfrac{\mu_0}{\varepsilon_0}} \propto \sqrt{\dfrac{L}{C}}$

Also, $\dfrac{L}{C} = \dfrac{\tau R}{\tau/R} = R^2$

$\therefore\ \sqrt{\dfrac{\mu_0}{\varepsilon_0}} \propto R$

Hence, $\boxed{\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}}$ has the dimensions of Resistance.