General Physics Ques 17
- Let $\left[\varepsilon_0\right]$ denote the dimensional formula of the permittivity of the vacuum and $\left[\mu_0\right]$ that of the permeability of the vacuum. If $M=$ mass, $L=$ length, $T=$ time and $I=$ electric current.
(1998, 2M)
(a) $\left[\varepsilon_0\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^2 \mathrm{I}\right]$
(b) $\left[\varepsilon_0\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right]$
(c) $\left[\mu_0\right]=\left[\mathrm{MLT}^{-2} \mathrm{I}^{-2}\right]$
(d) $\left[\mu_0\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1} \mathrm{I}\right]$
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Answer:
Correct Answer: 17.( b, c )
Solution:
$ \begin{aligned} F & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \\ {\left[\varepsilon_0\right] } & =\frac{\left[q_1\right]\left[q_2\right]}{[F]\left[r^2\right]}=\frac{[\mathrm{IT}]^2}{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right] \end{aligned} $
Speed of light, $c=\frac{1}{\sqrt{\varepsilon_0 \mu_0}}$
$ \begin{aligned} \therefore \quad\left[\mu_0\right] & =\frac{1}{\left[\varepsilon_0\right][c]^2}=\frac{1}{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right]\left[\mathrm{LT}^{-1}\right]^2} \\ & =\left[\mathrm{MLT}^{-2} \mathrm{I}^{-2}\right] \end{aligned} $
BETA
