JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2019
- Medium
- MCQ
Which of the following combinations has the dimension of electrical resistance ( \({ \varepsilon _0}\) is the permittivity of vacuum and \({\mu _0}\) is the permeability of vacuum) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\sqrt {\frac{{{\mu _0}}}{{{\varepsilon_0}}}} \)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\left[ {{\varepsilon _0}} \right] = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}\\
\left[ {{\mu _0}} \right] = M\,L\,{T^{ - 2}}{A^{ - 2}}\\
\left[ R \right] = \,M\,{L^2}\,{T^{ - 3}}{A^{ - 2}}\\
\left[ R \right] = \left[ {\sqrt {\frac{{{\mu _0}}}{{{\varepsilon _0}}}} } \right]
\end{array}\)
\left[ {{\varepsilon _0}} \right] = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}\\
\left[ {{\mu _0}} \right] = M\,L\,{T^{ - 2}}{A^{ - 2}}\\
\left[ R \right] = \,M\,{L^2}\,{T^{ - 3}}{A^{ - 2}}\\
\left[ R \right] = \left[ {\sqrt {\frac{{{\mu _0}}}{{{\varepsilon _0}}}} } \right]
\end{array}\)
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