Question

What is the dimensional formula of \( \frac{1}{\mu_0 \epsilon_0} \) (where \( \mu_0 \) is permeability and \( \epsilon_0 \) is permittivity of free space)?

• A. \( \text{LT}^{-1} \)
• B. \( \text{L}^2 \text{T}^{-1} \)
• C. \( \text{MLT}^{-1} \)
• D. \( \text{ML}^2 \text{T}^{-2} \)

Hint:

To calculate dimensional formulas involving constants, break down each constant's dimensional formula and perform the necessary arithmetic operations.

Answer 1

The Correct Option is D

The expression \( \frac{1}{\mu_0 \epsilon_0} \) involves the permeability of free space \( \mu_0 \) and the permittivity of free space \( \epsilon_0 \). - The dimensional formula for \( \mu_0 \) is \( \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 \), where \( A \) represents electric current. - The dimensional formula for \( \epsilon_0 \) is \( \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 \). Now, for \( \frac{1}{\mu_0 \epsilon_0} \), the dimensional formula becomes: \[ \left( \frac{1}{\mu_0 \epsilon_0} \right) = \left( \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 \right)^{-1} = \text{M} \text{L}^3 \text{T}^{-4} \text{A}^{-2} \] This simplifies to the dimensional formula of \( \frac{1}{\mu_0 \epsilon_0} \) as \( \text{ML}^2 \text{T}^{-2} \), hence the correct answer is option (4).