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Units and Measurements question

2025 · 2 Apr · Shift 2 · Q54

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Units and Measurements
2025 · 2 Apr · Shift 2 · Q54

Units and Measurements question

2025 · 2 Apr · Shift 2 · Q54

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

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 :

A
$\mathrm{T}^2 / \mathrm{L}$
B
$\mathrm{L}^2 / \mathrm{T}^2$
C
$\mathrm{T}^2 / \mathrm{L}^2$
D
$\mathrm{L} / \mathrm{T}^2$

View written solutionFree

Correct answer: B

Use the known electromagnetic relation:

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

Squaring both sides,

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

Now, the dimension of speed $c$ is:

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

Therefore,

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

Compare with the given options:

A: $T^2/L$ ❌
B: $L^2/T^2$ ✅
C: $T^2/L^2$ ❌
D: $L/T^2$ ❌

Hence, the correct option is B.

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