Question

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

Hint:

The dimension of \( \mu_0 \) and \( \epsilon_0 \) can be combined to give the dimension of inductance, which is \( \text{L}^2 \text{T}^{-2} \).

Answer 1

The Correct Option is C

Step 1: Write the dimensional formula for \( \mu_0 \) and \( \epsilon_0 \).
The vacuum permeability \( \mu_0 \) and the vacuum permittivity \( \epsilon_0 \) have the following dimensional formulas: \[ [\mu_0] = \text{M}^{-1} \text{L} \text{T}^{-2} \text{A}^2, \quad [\epsilon_0] = \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2. \]
Step 2: Calculate the dimension of \( \sqrt{\frac{\mu_0}{\epsilon_0}} \).
Now, substitute the dimensions of \( \mu_0 \) and \( \epsilon_0 \) into the expression \( \sqrt{\frac{\mu_0}{\epsilon_0}} \): \[ \left[\sqrt{\frac{\mu_0}{\epsilon_0}}\right] = \sqrt{\frac{\text{M}^{-1} \text{L} \text{T}^{-2} \text{A}^2}{\text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2}}. \]
Step 3: Simplify the dimensional formula.
Simplifying the above expression: \[ \left[\sqrt{\frac{\mu_0}{\epsilon_0}}\right] = \sqrt{\text{M}^0 \text{L}^4 \text{T}^{-6} \text{A}^0} = \text{L}^2 \text{T}^{-2}. \]
Step 4: Identify the physical quantity.
The dimension \( \text{L}^2 \text{T}^{-2} \) corresponds to the dimension of inductance.
Thus, the dimension of \( \sqrt{\frac{\mu_0}{\epsilon_0}} \) is equal to that of inductance.