In $SI $ units the dimensions of $\sqrt {\frac{{{ \varepsilon _0}}}{{{\mu _0}}}} $ is
JEE MAIN 2019, Medium
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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 $,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} = {\left[ {\frac{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}{{ML{T^{ - 2}}{A^{ - 2}}}}} \right]^{\frac{1}{2}}}$
$= {\left[ {{M^{ - 2}}{L^{ - 4}}{T^6}{A^4}} \right]^{1/2}}$
$= \left[ {{M^{ - 1}}{L^{ - 2}}{T^3}{A^2}} \right]$
$\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$
$\left[ {{\mu _0}} \right] = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]$
Dimension of $,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} = {\left[ {\frac{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}{{ML{T^{ - 2}}{A^{ - 2}}}}} \right]^{\frac{1}{2}}}$
$= {\left[ {{M^{ - 2}}{L^{ - 4}}{T^6}{A^4}} \right]^{1/2}}$
$= \left[ {{M^{ - 1}}{L^{ - 2}}{T^3}{A^2}} \right]$
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