JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2014
- Hard
- MCQ
In terms of resistance \(R\) and time \(T\), the dimensions of ratio \(\frac{\mu } {\varepsilon }\) of the permeability \(\mu \) and permittivity \(\varepsilon \) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\left[ {{R^2}} \right]\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
{\rm{dimensions of}}\,\mu = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]\\
{\rm{dimensions of}}\, \in \, = \,\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\\
{\rm{dimensions of}}\,R\, = \left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]\\
\therefore \frac{{{\rm{dimensions of}}\,\mu }}{{{\rm{dimensions of}}\, \in }} = \left[ {\frac{{ML{T^{ - 2}}{A^{ - 2}}}}{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}} \right]\\
= \left[ {{M^2}{L^4}{T^{-6}}{A^{ - 4}}} \right] = \left[ {{R^2}} \right]
\end{array}\)
{\rm{dimensions of}}\,\mu = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]\\
{\rm{dimensions of}}\, \in \, = \,\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\\
{\rm{dimensions of}}\,R\, = \left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]\\
\therefore \frac{{{\rm{dimensions of}}\,\mu }}{{{\rm{dimensions of}}\, \in }} = \left[ {\frac{{ML{T^{ - 2}}{A^{ - 2}}}}{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}} \right]\\
= \left[ {{M^2}{L^4}{T^{-6}}{A^{ - 4}}} \right] = \left[ {{R^2}} \right]
\end{array}\)
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