JEE Physics 2025 Question with Solution

Given: We need the dimension of $\left(\dfrac{1}{\mu_0 \epsilon_0}\right)$, where $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of free space.

Find: The dimensional formula of $\left(\dfrac{1}{\mu_0 \epsilon_0}\right)$.

The solution states the electromagnetic relation

$$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$

Hence,

$$\frac{1}{\mu_0 \epsilon_0} = c^2$$

The dimension of speed is

$$[c] = [L\,T^{-1}]$$

Therefore,

$$\left[\frac{1}{\mu_0 \epsilon_0}\right] = [c^2] = [L^2 T^{-2}]$$

The step-by-step dimensional multiplication in the solution also gives

$$[\mu_0] = [M^1 L^1 T^{-2} A^{-2}]$$ $$[\epsilon_0] = [M^{-1} L^{-3} T^{4} A^{2}]$$

So,

$$[\mu_0 \epsilon_0] = [L^{-2} T^2]$$

and taking the reciprocal,

$$\left[\frac{1}{\mu_0 \epsilon_0}\right] = [L^2 T^{-2}]$$

Therefore, the dimension is $[L^2 T^{-2}]$. The solution marks the correct option as B, but the listed listed options do not contain $L^2 T^{-2}$, so there is a discrepancy between the solution working and the available option texts.

Given: $c = \dfrac{1}{\sqrt{\mu_0 \epsilon_0}}$.

Find: The dimension of $\dfrac{1}{\mu_0 \epsilon_0}$.

Square the relation:

$$c^2 = \frac{1}{\mu_0 \epsilon_0}$$

Now use the dimensional formula of speed:

$$[c] = [LT^{-1}]$$

So,

$$[c^2] = [L^2T^{-2}]$$

Hence,

$$\left[\frac{1}{\mu_0 \epsilon_0}\right] = [L^2T^{-2}]$$

Thus the required dimension is the square of velocity, namely $[L^2 T^{-2}]$.