Which of the following combinations has the dimension of electrical resistance ( $\epsilon_{0}$ is the permittivity of vacuum and $\mu_{0}$ is the permeability of vacuum)?

(D) $\sqrt{\frac{\mu_{0}}{\epsilon_{0}}}$

(A) $\sqrt{\frac{\epsilon_{0}}{\mu_{0}}}$

(B) $\frac{\mu_{0}}{\epsilon_{0}}$

(C) $\frac{\epsilon_{0}}{\mu_{0}}$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Units & Measurement

Which of the following combinations has the dimension of electrical resistance (

ϵ_{0}

is the permittivity of vacuum and

μ_{0}

is the permeability of vacuum)?

Answer: (D)

Write the dimensions of each quantity:

[ε_{0}] = M^{- 1} L^{- 3} T^{4} A^{2}

and

[μ_{0}] = M L T^{- 2} A^{- 2}

, while resistance is

[R] = M L^{2} T^{- 3} A^{- 2}

. Then

\frac{μ _{0}}{ε _{0}} = \frac{M L T ^{- 2} A ^{- 2}}{M ^{- 1} L ^{- 3} T ^{4} A ^{2}} = M^{2} L^{4} T^{- 6} A^{- 4} = M L^{2} T^{- 3} A^{- 2}

, which matches

[R]

exactly. This is just the impedance of free space

Z_{0} = μ_{0} / ε_{0}

, which indeed carries the units of resistance.

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option