In terms of resistance $R$ and time $T$, the dimensions of ratio $\frac{\mu } {\varepsilon }$ of the permeability $\mu $ and permittivity $\varepsilon $ is
JEE MAIN 2014, Diffcult
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${\rm{Dimensions of}}\,\mu = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]$
${\rm{Dimensions of}}\, \in \, = \,\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$
${\rm{Dimensions of}}\,R\, = \left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]$
$\therefore \frac{{{\rm{Dimensions of}}\,\mu }}{{{\rm{Dimensions of}}\, \in }} = \left[ {\frac{{ML{T^{ - 2}}{A^{ - 2}}}}{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}} \right]$
$= \left[ {{M^2}{L^4}{T^6}{A^{ - 4}}} \right] $
$= \left[ {{R^2}} \right]$
${\rm{Dimensions of}}\, \in \, = \,\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$
${\rm{Dimensions of}}\,R\, = \left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]$
$\therefore \frac{{{\rm{Dimensions of}}\,\mu }}{{{\rm{Dimensions of}}\, \in }} = \left[ {\frac{{ML{T^{ - 2}}{A^{ - 2}}}}{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}} \right]$
$= \left[ {{M^2}{L^4}{T^6}{A^{ - 4}}} \right] $
$= \left[ {{R^2}} \right]$
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