The dimension of 0 0 is equal to that of: (Where 0 is the vacuum permeability and 0 is the vacuum permittivity)

Given: We need the dimension of 0 0, where 0 is vacuum permeability and 0 is vacuum permittivity. Find: Which physical quantity has the same dimensions. From the solution: - The dimension of permeability 0 is [M L T^-2 A^-2]. - The dimension of permittivity 0 is [M^-1 L^-3 T^4 A^2]. Now evaluate the ratio: 0 0 = [M L T^-2 A^-2][M^-1 L^-3 T^4 A^2] = [M^2 L^4 T^-6 A^-4] Taking square root, [M^2 L^4 T^-6 A^-4] = [M L^2 T^-3 A^-2] This dimensional formula corresponds to inductance. Therefore, the correct option is C.

The dimension of 0 0 is equal to that of: (Where 0 is the vacuum… — Solution | JEE Physics

MCQEasyJEE 2025Dimensions & Dimensional Analysis

The dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is equal to that of:

(Where $\mu_0$ is the vacuum permeability and $\epsilon_0$ is the vacuum permittivity)

A
Voltage
B
Capacitance
C
Inductance
D
Resistance

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: We need the dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ , where $\mu_0$ is vacuum permeability and $\epsilon_0$ is vacuum permittivity.

Find: Which physical quantity has the same dimensions.

From the solution:

The dimension of permeability $\mu_0$ is $[M L T^{-2} A^{-2}]$ .
The dimension of permittivity $\epsilon_0$ is $[M^{-1} L^{-3} T^{4} A^{2}]$ .

Now evaluate the ratio:

\frac{\mu_0}{\epsilon_0} = \frac{[M L T^{-2} A^{-2}]}{[M^{-1} L^{-3} T^{4} A^{2}]} = [M^{2} L^{4} T^{-6} A^{-4}]

Step-by-Step Identification

Taking square root,

\sqrt{[M^{2} L^{4} T^{-6} A^{-4}]} = [M L^{2} T^{-3} A^{-2}]

This dimensional formula corresponds to inductance.

Therefore, the correct option is C.

Common mistakes

Using the incorrect dimensional formula for $\mu_0$ or $\epsilon_0$ . This changes the exponents and gives the wrong physical quantity. Write both formulas carefully before simplifying.
Forgetting to divide dimensions properly. When dividing, subtract the exponents of the denominator dimensions from those of the numerator instead of dividing the symbols directly.
Not taking the square root of the final dimensional expression. After obtaining the dimension of $\frac{\mu_0}{\epsilon_0}$ , halve all exponents to get the dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ .

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Step-by-step solution

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