A) \[[{{\varepsilon }_{0}}]=[{{M}^{-1}}{{L}^{-3}}{{T}^{2}}A]\]
B) \[[{{\varepsilon }_{0}}]=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\]
C) \[[{{\varepsilon }_{0}}]=[{{M}^{-1}}{{L}^{2}}{{T}^{-1}}{{A}^{-2}}]\]
D) \[[{{\varepsilon }_{0}}]=[{{M}^{-1}}{{L}^{2}}{{T}^{-1}}A]\]
Correct Answer: B
Solution :
\[[{{\varepsilon }_{0}}]=\left[ \frac{{{C}^{2}}}{N-{{m}_{2}}} \right]=\frac{{{A}^{2}}{{T}^{2}}}{ML{{T}^{-2}}{{L}^{2}}}\] \[[{{\varepsilon }_{0}}]={{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}\]
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