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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2019 · 8 April · Morning Shift · Physics · Units and Dimensions

In SI units, the dimensions of ϵ _ 0 μ _ 0 is:

In SISI units, the dimensions of ϵ0μ0\sqrt{\dfrac{{ϵ}_{0}}{{\mu}_{0}}} is:

Options

  1. AAT2M−1L−1A{T}^{2}{M}^{- 1}{L}^{- 1}
  2. BA2T3M−1L−2{A}^{2}{T}^{3}{M}^{- 1}{L}^{- 2}
  3. CA−1TML3{A}^{- 1}TM{L}^{3}
  4. DAT−3ML3/2A{T}^{- 3}M{L}^{3 / 2}

Correct answer

B. A2T3M−1L−2{A}^{2}{T}^{3}{M}^{- 1}{L}^{- 2}

Step-by-step solution

As we know, c=1μ0ε0c = \dfrac{1}{\sqrt{{\mu}_{0}{ε}_{0}}}
∴ε0μ0=ε02μ0ε0=ε0c∴ \sqrt{\dfrac{{ε}_{0}}{{\mu}_{0}}} = \sqrt{\dfrac{{ε}_{0}^{2}}{{\mu}_{0}{ε}_{0}}} = {ε}_{0}c
∴[ε0μ0]=[ε0][c]=[M−1L−3T4A2][LT−1]∴ \left[\sqrt{\dfrac{{ε}_{0}}{{\mu}_{0}}}\right] = \left[{ε}_{0}\right]\left[c\right] = \left[{M}^{- 1}{L}^{- 3}{T}^{4}{A}^{2}\right] [ L{T}^{- 1} ]
=[M−1L−2T3A2]= [ {M}^{- 1}{L}^{- 2}{T}^{3}{A}^{2} ]
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