The dimension of √((μ₀)/(ε₀)) is equal to that of: ( μ₀ = Vacuum permeability and ε₀ = Vacuum permittivity) (a) Voltage (b) Inductance (c) Resistance (d) Capacitance

Answer: (c). μ₀ = ML T⁻²A⁻² and ε₀ = M⁻¹L⁻³T⁴A². μ₀/ε₀ = M²L⁴T⁻⁶A⁻⁴, so √(μ₀/ε₀) = ML²T⁻³A⁻². This is [R] = [V/I]; it is the impedance of free space ( ≈ 377 Ω). So the answer is (c) Resistance.

The dimension of √((μ₀)/(ε₀)) is equal to that of: ( μ₀ = Vacuum… | JEE Main 2025 (7 Apr, Shift 2)

Q11JEE Main 2025 · 7 APR · SHIFT 2Easy

Units & Dimensions

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

( $\mu_0$ = Vacuum permeability and $\epsilon_0$ = Vacuum permittivity)

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Answer: (c)

$\mu_0 = \mathrm{ML T^{-2}A^{-2}}$ and $\epsilon_0 = \mathrm{M^{-1}L^{-3}T^4A^2}$ .
$\mu_0/\epsilon_0 = \mathrm{M^2L^4T^{-6}A^{-4}}$ , so $\sqrt{\mu_0/\epsilon_0} = \mathrm{ML^2T^{-3}A^{-2}}$ .
This is $[R] = [V/I]$ ; it is the impedance of free space ( $\approx 377\ \Omega$ ).
So the answer is (c) Resistance.

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