In SI, units, the dimensions of sqrt {fr | vedclass

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Question

$\begin{array}{l}
Dimension\,of\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} \
\left[ {{\varepsilon _0}} ight] = \left[ {{M^{ - 1}}{L^{ - 3}}{T

Vedclass · Accepted Answer

${A^2}{T^3}{M^{ - 1}}{L^{ - 2}}$

Vedclass · Answer

$\begin{array}{l}
Dimension\,of\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}} \
\left[ {{\varepsilon _0}} ight] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} ight]\
\left[ {{\mu _0}} ight] = \left[ {ML{T^{ - 2}}{A^{ - 2}}} ight]\
Dimension\,of\,\,\sqrt {\frac{{{\varepsilon _0}}}{{{\mu _0}}}}  = {\left[ {\frac{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}{{ML{T^{ - 2}}{A^{ - 2}}}}} ight]^{\frac{1}{2}}}\
 = {\left[ {{M^{ - 2}}{L^{ - 4}}{T^6}{A^4}} ight]^{1/2}}\
 = \left[ {{M^{ - 1}}{L^{ - 2}}{T^3}{A^2}} ight]
\end{array}$

In $SI\, units$, the dimensions of $\sqrt {\frac{{{ \varepsilon _0}}}{{{\mu _0}}}} $ is

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