Question:medium

If $\mu_0$ is the permeability of free space and $\varepsilon_0$ is the permittivity of free space, then the dimension for $(\mu_0 \varepsilon_0)$ is:

Show Hint

Instead of deriving the complicated individual dimensions of $\mu_0$ and $\varepsilon_0$ separately and multiplying them, look for an equation that links them together. The speed of light relation $c^2 = \frac{1}{\mu_0\varepsilon_0}$ makes this a $5$-second problem!
Updated On: May 16, 2026
  • $[L^{-1}T]$
  • $[MLT^{-1}]$
  • $[L^{-2}T^{2}]$
  • $[ML^{-1}T]$
Show Solution

The Correct Option is C

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