JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2019
- Medium
- MCQ
In \(SI\, units\), the dimensions of \(\sqrt {\frac{{{ \varepsilon _0}}}{{{\mu _0}}}} \) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \({A^2}{T^3}{M^{ - 1}}{L^{ - 2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Dimension\,of\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} \\
\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\\
\left[ {{\mu _0}} \right] = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]\\
Dimension\,of\,\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} = {\left[ {\frac{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}{{ML{T^{ - 2}}{A^{ - 2}}}}} \right]^{\frac{1}{2}}}\\
= {\left[ {{M^{ - 2}}{L^{ - 4}}{T^6}{A^4}} \right]^{1/2}}\\
= \left[ {{M^{ - 1}}{L^{ - 2}}{T^3}{A^2}} \right]
\end{array}\)
Dimension\,of\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} \\
\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\\
\left[ {{\mu _0}} \right] = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]\\
Dimension\,of\,\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} = {\left[ {\frac{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}{{ML{T^{ - 2}}{A^{ - 2}}}}} \right]^{\frac{1}{2}}}\\
= {\left[ {{M^{ - 2}}{L^{ - 4}}{T^6}{A^4}} \right]^{1/2}}\\
= \left[ {{M^{ - 1}}{L^{ - 2}}{T^3}{A^2}} \right]
\end{array}\)
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