Which of the following combinations has the dimension of electrical resistance (ε0 is the permittivity of vacuum and μ0 is the permeability of vacuum)?

Steps to Solve

1. Identify the dimensions of permittivity and permeability

   The permittivity of vacuum ($\epsilon_0$) has the dimension of capacitance per unit length, which is given by: $$ [\epsilon_0] = \frac{[C^2]}{[N \cdot m^2]} = \frac{[Q^2]}{[M \cdot L^2 \cdot T^{-4}]} $$

   The permeability of vacuum ($\mu_0$) has the dimension of inductance per unit length: $$ [\mu_0] = \frac{[N \cdot s^2]}{[C^2]} = \frac{[M \cdot L]}{[T^2 \cdot Q^2]} $$

2. Express the dimensions of electrical resistance

   The electrical resistance ($R$) has the dimension: $$ [R] = \frac{[V]}{[I]} = \frac{[M \cdot L^2 \cdot T^{-3}]}{[Q]} $$

3. Evaluate each option

   Compute the dimensions for each option and check for dimensional equivalence with resistance.

   • Option (a): $$ \sqrt{\frac{\mu_0}{\epsilon_0}} = \sqrt{\frac{\frac{[M \cdot L]}{[T^2 \cdot Q^2]}}{\frac{[Q^2]}{[M \cdot L^2 \cdot T^{-4}]}}} = \sqrt{\frac{[M \cdot L^3 \cdot T^{-4}]}{[Q^4]}} $$

   • Option (b): $$ \frac{\mu_0}{\epsilon_0} = \frac{\frac{[M \cdot L]}{[T^2 \cdot Q^2]}}{\frac{[Q^2]}{[M \cdot L^2 \cdot T^{-4}]}} = \frac{[M^2 \cdot L^3 \cdot T^{-4}]}{[Q^4]} $$

   • Option (c): $$ \sqrt{\frac{\epsilon_0}{\mu_0}} = \sqrt{\frac{\frac{[Q^2]}{[M \cdot L^2 \cdot T^{-4}]}}{\frac{[M \cdot L]}{[T^2 \cdot Q^2]}}} = \sqrt{\frac{[Q^4]}{[M^2 \cdot L^3 \cdot T^{-2}]}} $$

   • Option (d): $$ \frac{\epsilon_0}{\mu_0} = \frac{\frac{[Q^2]}{[M \cdot L^2 \cdot T^{-4}]}}{\frac{[M \cdot L]}{[T^2 \cdot Q^2]}} = \frac{[Q^4 \cdot T^2]}{[M^2 \cdot L^3]} $$

4. Select the correct option

   Identify which of the derived expressions matches the dimension of electrical resistance.