Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Let\[[{{\varepsilon }_{0}}]\]denote the dimensional formula of the permittivity of the vacuum and\[[{{\mu }_{0}}]\]at of the permittivity of the vacuum. If M = mass, L = length, T = time and I = electric current: (1) \[[{{\varepsilon }_{0}}]=[{{M}^{-1}}{{L}^{-3}}{{T}^{2}}I]\] (2) \[[{{\varepsilon }_{0}}]=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{I}^{2}}]\] (3) \[[{{\mu }_{0}}]=[ML{{T}^{-2}}{{I}^{-2}}]\] (4) \[[{{\mu }_{0}}]=[M{{L}^{2}}{{T}^{-1}}I]\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 and correct
D) 1, 2 and 4 are correct
Correct Answer: B
Solution :
\[F=\frac{1}{4\pi \,{{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[[{{\varepsilon }_{0}}]=\frac{[{{q}_{1}}][{{q}_{2}}]}{[F][{{r}^{2}}]}=\frac{{{[IT]}^{2}}}{[ML{{T}^{-2}}][{{L}^{2}}]}\] \[=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{I}^{2}}]\] Speed of light, \[c=\frac{1}{\sqrt{{{\varepsilon }_{0}}{{\mu }_{0}}}}\] \[\therefore \] \[[{{\mu }_{0}}]=\frac{1}{\sqrt{{{\varepsilon }_{0}}{{\mu }_{0}}}}\] \[=\frac{1}{[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{I}^{2}}]{{[L{{T}^{-1}}]}^{2}}}\] \[=[ML{{T}^{-2}}{{I}^{-2}}]\]
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