Physics · Units, Dimensions & Error Analysis

JEE Main 2025 — 2 April, Evening Shift — Question 52

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

Option A:
$T^{2} / L^{2}$
Option B:
$L^{2} / T^{2}$
Correct
Option C:
$T^{2} / L$
Option D:
$L / T^{2}$

Answer: B

Step-by-step solution

$C=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$

$\mu_{0} \varepsilon_{0} \equiv \frac{1}{C^{2}} \equiv\left[L^{-2} T^{2}\right]$

$\therefore\left[\frac{1}{\mu_{0} \varepsilon_{0}}\right]=\frac{L^{2}}{T^{2}}$

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Units and Dimensions Analysis

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Answer: B

Step-by-step solution

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