Units & Dimensions

JEE Main level, single correct, medium

121 of 149

Question 121 of 149

Which of the following combinations has the dimension of electrical resistance (ϵ0({ϵ}_{0} is the permittivity of vacuum and μ0{\mu }_{0} is the permeability of vacuum)?

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Answer: (D) μ0ϵ0\sqrt{\dfrac{{\mu }_{0}}{{ϵ}_{0}}}

[Resistance]=[Volt][Ampere]=[ML2T−3A−1][A]=[ML2T−3A−2]\left[\text{Resistance}\right]\text{=}\dfrac{\left[\text{Volt}\right]}{\left[\text{Ampere}\right]}=\dfrac{\left[M{L}^{2}{T}^{-3}{A}^{-1}\right]}{\left[A\right]}=\left[M{L}^{2}{T}^{-3}{A}^{-2}\right]
We know, μ0=[MLT−2A−2]ϵ0=M−1L−3T4A2{\mu }_{0}=\left[ML{T}^{-2}{A}^{-2}\right] {\epsilon }_{0}={M}^{-1}{L}^{-3}{T}^{4}{A}^{2}
Hence,
μ0ϵ0=[μ0][ϵ0]=[[MLT−2A−2][M−1L−3T4A2]]1/2\sqrt{\dfrac{{\mu }_{0}}{{\epsilon }_{0}}}=\sqrt{\dfrac{\left[{\mu }_{0}\right]}{\left[{\epsilon }_{0}\right]}}={\left[\dfrac{\left[ML{T}^{-2}{A}^{-2}\right]}{\left[{M}^{-1}{L}^{-3}{T}^{4}{A}^{2}\right]}\right]}^{1/2}
=[M2L4T−6A−4]1/2=ML2T−3A−2={\left[{M}^{2}{L}^{4}{T}^{-6}{A}^{-4}\right]}^{1/2}=M{L}^{2}{T}^{-3}{A}^{-2}

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