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In terms of resistance $R$ and time $T$, the dimensions of ratio $\frac{\mu}{\varepsilon}$ of the permeability $\mu$ and permittivity $\varepsilon$ is:
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Verified Answer
The correct answer is:
$\left[\mathrm{R}^2\right]$
$\left[\mathrm{R}^2\right]$
Dimensions of $\mu=\left[\right.$ MLT $\left.^{-2} \mathrm{~A}^{-2}\right]$
Dimensions of $\in=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
Dimensions of $\mathrm{R}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$$
\begin{aligned}
&\therefore \frac{\text { Dimensions of } \mu}{\text { Dimensions of } e}=\frac{\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]} \\
&=\left[\mathrm{M}^2 \mathrm{~L}^4 \mathrm{~T}^{-6} \mathrm{~A}^{-4}\right]=\left[\mathrm{R}^2\right]
\end{aligned}
$$
Dimensions of $\in=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
Dimensions of $\mathrm{R}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$$
\begin{aligned}
&\therefore \frac{\text { Dimensions of } \mu}{\text { Dimensions of } e}=\frac{\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]} \\
&=\left[\mathrm{M}^2 \mathrm{~L}^4 \mathrm{~T}^{-6} \mathrm{~A}^{-4}\right]=\left[\mathrm{R}^2\right]
\end{aligned}
$$
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