A) \[[M{{L}^{-2}}{{T}^{-1}}{{I}^{0}}]\]
B) \[[{{M}^{0}}{{L}^{0}}{{T}^{3}}{{I}^{0}}]\]
C) \[[{{M}^{-1}}{{L}^{-2}}{{T}^{6}}{{I}^{2}}]\]
D) \[[{{M}^{0}}{{L}^{0}}{{T}^{2}}{{I}^{0}}]\]
Correct Answer: B
Solution :
\[{{C}^{2}}LR=[{{C}^{2}}{{L}^{2}}]\times \left[ \frac{R}{L} \right]=[{{T}^{4}}]\times \left[ \frac{1}{T} \right]=[{{T}^{3}}]\] As \[\left[ \frac{L}{R} \right]=T\] and \[\sqrt{LC}=T\]
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