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JEE Advanced 2018Paper 1
Exam
JEE Advanced
Year
2018
Paper
Paper 1
Subject
Physics
Chapter
Units and Dimensions

JEE Advanced 2018 · Paper 1 · Physics · Units and Dimensions

Paragraph: In electromagnetic theory, the electric and magnetic...

Paragraph: In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E]and[B]stand\left[E\right]and \left[B\right] stand for dimensions of electric and magnetic fields respectively, while[ {ε}_{0} ] and [ {\mu}_{0} ]stand for dimensionsofthepermittivityandpermeabilityoffreespacerespectively.[L]and[T]aredimensionsoflengthandtimerespectively.Allthequantitiesaregivenindimensions of the permittivity and permeability of free space respectively. \left[L\right]and \left[T\right] are dimensions of length and time respectively. All the quantities are given inSIunits<br/><br><strong>Question:</strong>Therelationbetween[ϵ0]and[μ0]isunits<br /><br> <strong> Question : </strong>The relation between \left[{ϵ}_{0}\right]and \left[{\mu}_{0}\right] is

Options

  1. A[μ0]=[ε0][L]2[T]−2\left[{\mu}_{0}\right] = \left[{ε}_{0}\right]{\left[L\right]}^{2}{\left[T\right]}^{- 2}
  2. B[μ0]=\left[{\mu}_{0}\right] =\left[{ε}_{0}\right]{\left[L\right]}^{- 2}{\left[T\right]}^{2}
  3. C[μ0]=[ε0]−1[L]2[T]−2\left[{\mu}_{0}\right] = {\left[{ε}_{0}\right]}^{- 1}{\left[L\right]}^{2}{\left[T\right]}^{- 2}
  4. D[μ0]=[ε0]\left[{\mu}_{0}\right] = {\left[{ε}_{0}\right]}^{- 1}{\left[L\right]}^{- 2}{\left[T\right]}^{2}

Correct answer

D. [μ0]=[ε0]\left[{\mu}_{0}\right] = {\left[{ε}_{0}\right]}^{- 1}{\left[L\right]}^{- 2}{\left[T\right]}^{2}

Step-by-step solution

We have,
C=1μ0ε0C = \dfrac{1}{\sqrt{{\mu}_{0}{ε}_{0}}}
∴[C2]=[1μ0ε0]∴ \left[{C}^{2}\right] = \left[\dfrac{1}{{\mu}_{0}{ε}_{0}}\right]
⇒L2T−2=1[μ0][ε0]⇒ {L}^{2}{T}^{- 2} = \dfrac{1}{\left[{\mu}_{0}\right]\left[{ε}_{0}\right]}
⇒[μ0]=[ε0]−1[L]−2[T]2⇒ \left[{\mu}_{0}\right] = {\left[{ε}_{0}\right]}^{- 1}{\left[L\right]}^{- 2}{\left[T\right]}^{2}
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