Dimension | JEE Main 2025 Solution

JEE Advanced 2025 - Mathematics

02-04-2025 SHIFT-2

Question

If $\mu_{0}$ and $\epsilon_{0}$ are the permeability and permittivity of free space, respectively, then the dimension of $\left(\frac{1}{\mu_{0} \epsilon_{0}}\right)$ is:

Select the correct option:

$T^{2} / L^{2}$

$L / T^{2}$

$T^{2} / L$

$L^{2} / T^{2}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$C=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \Rightarrow \frac{1}{\mu_{0} \varepsilon_{0}}=C^{2}=L^{2} T^{-2}$

Question Tags

JEE Main

Physics

Easy

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