The dimensions of $(\mu_{0}\varepsilon_{0})^{-1/2}$ are: [PMT/NEET-2011]
a. $[L^{1/2}T^{-1/2}]$
b. $[L^{-1}T]$
c. $[LT^{-1}]$
d. $[L^{1/2}T^{1/2}]$

Answer: c. $ [LT^{-1}] $

Explanation:
To find the dimensions of $ (\mu_{0}\varepsilon_{0})^{-1/2} $, we need to determine the dimensions of both permeability of free space $ \mu_{0} $ and permittivity of free space $ \varepsilon_{0} $.

Dimensions of Permittivity ($ \varepsilon_{0} $) #

From Coulomb’s law: $ F = \frac{1}{4\pi\varepsilon_{0}} \frac{q_1 q_2}{r^2} $

Solving for $ \varepsilon_{0} $: $ \varepsilon_{0} = \frac{1}{4\pi} \frac{q_1 q_2}{Fr^2} $

• Product of charges: $ [I^2T^2] $
• Force: $ [MLT^{-2}] $
• Distance squared: $ [L^2] $

Therefore: $ [\varepsilon_{0}] = \frac{[I^2T^2]}{[MLT^{-2}][L^2]} = [M^{-1}L^{-3}T^4I^2] $

Dimensions of Permeability ($ \mu_{0} $) #

From Ampere’s law for magnetic field around a current-carrying wire: $ B = \frac{\mu_0 I}{2\pi r} $

Solving for $ \mu_{0} $: $ \mu_{0} = \frac{B \cdot 2\pi r}{I} $

• Magnetic field $ B $: $ [MT^{-2}I^{-1}] $
• Distance $ r $: $ [L] $
• Current $ I $: $ [I] $

Therefore: $ [\mu_{0}] = \frac{[MT^{-2}I^{-1}][L]}{[I]} = [MLT^{-2}I^{-2}] $

Finding the Product #

$
[\mu_{0}\varepsilon_{0}] = [MLT^{-2}I^{-2}][M^{-1}L^{-3}T^4I^2] = [L^{-2}T^2] $

Taking the Inverse Square Root #

$
[(\mu_{0}\varepsilon_{0})^{-1/2}] = ([L^{-2}T^2])^{-1/2} = [L^{1}T^{-1}] = [LT^{-1}] $

This represents the dimensions of velocity. Physically, $ (\mu_{0}\varepsilon_{0})^{-1/2} $ equals the speed of light in vacuum, which is why it has velocity dimensions.