General Physics Ques 22
- The relation between $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ is
(a) $\left[\mu_0\right]=\left[\varepsilon_0\right][\mathrm{L}]^2[\mathrm{~T}]^{-2}$
(b) $\left[\mu_0\right]=\left[\varepsilon_0\right][\mathrm{L}]^{-2}[\mathrm{~T}]^2$
(c) $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[L]^2[T]^{-2}$
(d) $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[L]^{-2}[T]^2$
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Answer:
Correct Answer: 22.( d )
Solution:
$ \begin{aligned} c & =\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \\ c^2 & =\frac{1}{\mu_0 \varepsilon_0} \end{aligned} $
$\begin{aligned} \mu_0 & =\varepsilon_0^{-1} \cdot c^{-2} \\ {\left[\mu_0\right] } & =\left[\varepsilon_0\right]^{-1}\left[\mathrm{~L}^{-2} \mathrm{~T}^2\right]\end{aligned}$
BETA
