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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2019 · 12 April · Morning Shift · Physics · Units and Dimensions

Which of the following combinations has the dimension of electrical...

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)?

Options

  1. Aϵ0μ0\dfrac{{ϵ}_{0}}{{\mu}_{0}}
  2. Bϵ0μ0\sqrt{\dfrac{{ϵ}_{0}}{{\mu}_{0}}}
  3. Cμ0ϵ0\dfrac{{\mu}_{0}}{{ϵ}_{0}}
  4. Dμ0ϵ0\sqrt{\dfrac{{\mu}_{0}}{{ϵ}_{0}}}

Correct answer

D. μ0ϵ0\sqrt{\dfrac{{\mu}_{0}}{{ϵ}_{0}}}

Step-by-step solution

\left[\text{Resistance}\right]\text{=}\dfrac{\left[\text{\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]{ε}_{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}}{{ε}_{0}}} = \sqrt{\dfrac{\left[{\mu}_{0}\right]}{\left[{ε}_{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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