The dimension of \sqrt\frac\mu_0\epsilon_0 is equal to that of: ( \mu_0 = Vacuum permeability and \epsilon_0 = Vacuum permittivity)

Units and Measurements PYQ — JEE Main 2025 · 7 Apr · Shift 2 · Q51 with Solution

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Correct answer: C

We need the dimension of

\sqrt{\frac{\mu_0}{\epsilon_0}}

where $\mu_0$ is vacuum permeability and $\epsilon_0$ is vacuum permittivity.

Use the electromagnetic relation:

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

So,

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

Also, the quantity

\sqrt{\frac{\mu_0}{\epsilon_0}}

is known as the intrinsic impedance of free space.

Let us verify its dimensions directly.

Permeability:

[\mu_0]=\frac{\text{N}}{\text{A}^2}

Since

[\text{N}] = MLT^{-2}, \qquad [A]=I

we get

[\mu_0]=MLT^{-2}I^{-2}

Permittivity: from Coulomb's law,

\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}=F

So,

[\epsilon_0]=\frac{q^2}{Fr^2}

Now,

[q]=IT, \qquad [F]=MLT^{-2}, \qquad [r]=L

Hence,

[\epsilon_0]=\frac{(IT)^2}{(MLT^{-2})(L^2)} = M^{-1}L^{-3}T^4I^2

Therefore,

\left[\frac{\mu_0}{\epsilon_0}\right] =\frac{MLT^{-2}I^{-2}}{M^{-1}L^{-3}T^4I^2} = M^2L^4T^{-6}I^{-4}

Taking square root,

\left[\sqrt{\frac{\mu_0}{\epsilon_0}}\right] =ML^2T^{-3}I^{-2}

Now compare with dimensions of the given options.

Voltage:

[V]=\frac{\text{work}}{\text{charge}}=\frac{ML^2T^{-2}}{IT}=ML^2T^{-3}I^{-1}

Not same.

Inductance: Using

\text{Energy}=\frac{1}{2}LI^2

[L]=\frac{ML^2T^{-2}}{I^2}=ML^2T^{-2}I^{-2}

Not same.

Resistance: Using Ohm's law $R=V/I$ ,

[R]=\frac{ML^2T^{-3}I^{-1}}{I}=ML^2T^{-3}I^{-2}

This matches.

Capacitance: Using $C=q/V$ ,

[C]=\frac{IT}{ML^2T^{-3}I^{-1}}=M^{-1}L^{-2}T^4I^2

Not same.

Hence,

\sqrt{\frac{\mu_0}{\epsilon_0}}

has the dimensions of resistance.

Therefore, the correct option is C.

More from Units and Measurements

View written solutionFree

Correct answer: C

We need the dimension of

\sqrt{\frac{\mu_0}{\epsilon_0}}

where $\mu_0$ is vacuum permeability and $\epsilon_0$ is vacuum permittivity.

Use the electromagnetic relation:

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

So,

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

Also, the quantity

\sqrt{\frac{\mu_0}{\epsilon_0}}

is known as the intrinsic impedance of free space.

Let us verify its dimensions directly.

Permeability:

[\mu_0]=\frac{\text{N}}{\text{A}^2}

Since

[\text{N}] = MLT^{-2}, \qquad [A]=I

we get

[\mu_0]=MLT^{-2}I^{-2}

Permittivity: from Coulomb's law,

\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}=F

So,

[\epsilon_0]=\frac{q^2}{Fr^2}

Now,

[q]=IT, \qquad [F]=MLT^{-2}, \qquad [r]=L

Hence,

[\epsilon_0]=\frac{(IT)^2}{(MLT^{-2})(L^2)} = M^{-1}L^{-3}T^4I^2

Therefore,

\left[\frac{\mu_0}{\epsilon_0}\right] =\frac{MLT^{-2}I^{-2}}{M^{-1}L^{-3}T^4I^2} = M^2L^4T^{-6}I^{-4}

Taking square root,

\left[\sqrt{\frac{\mu_0}{\epsilon_0}}\right] =ML^2T^{-3}I^{-2}

Now compare with dimensions of the given options.

Voltage:

[V]=\frac{\text{work}}{\text{charge}}=\frac{ML^2T^{-2}}{IT}=ML^2T^{-3}I^{-1}

Not same.

Inductance: Using

\text{Energy}=\frac{1}{2}LI^2

[L]=\frac{ML^2T^{-2}}{I^2}=ML^2T^{-2}I^{-2}

Not same.

Resistance: Using Ohm's law $R=V/I$ ,

[R]=\frac{ML^2T^{-3}I^{-1}}{I}=ML^2T^{-3}I^{-2}

This matches.

Capacitance: Using $C=q/V$ ,

[C]=\frac{IT}{ML^2T^{-3}I^{-1}}=M^{-1}L^{-2}T^4I^2

Not same.

Hence,

\sqrt{\frac{\mu_0}{\epsilon_0}}

has the dimensions of resistance.

Therefore, the correct option is C.

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