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1. units,dimensions and measurement — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics1. units,dimensions and measurement1 Mark
MCQ
In terms of resistance $R$ and time $T$, the dimensions of ratio $\frac{\mu } {\varepsilon }$ of the permeability $\mu $ and permittivity $\varepsilon $  is
  • A
    $\left[ {R{T^{ - 2}}} \right]$
  • B
    $\left[ {{R^2}{T^{ - 1}}} \right]$
  • ✓
    $\left[ {{R^2}} \right]$
  • D
    $\left[ {{R^2}{T^2}} \right]$
✓

Answer

Correct option: C.
$\left[ {{R^2}} \right]$
c
$\begin{array}{l}
{\rm{Dimensions of}}\,\mu  = \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]\\
{\rm{Dimensions of}}\, \in \, = \,\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]\\
{\rm{Dimensions of}}\,R\, = \left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]\\
\therefore \frac{{{\rm{Dimensions of}}\,\mu }}{{{\rm{Dimensions of}}\, \in }} = \left[ {\frac{{ML{T^{ - 2}}{A^{ - 2}}}}{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}} \right]\\
 = \left[ {{M^2}{L^4}{T^6}{A^{ - 4}}} \right] = \left[ {{R^2}} \right]
\end{array}$

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