A) \[[{{M}^{3}}\,{{L}^{2}}\,{{T}^{-\,4}}\,{{A}^{-\,4}}]\]
B) \[[{{M}^{-\,2}}\,{{L}^{-\,2}}\,{{T}^{3}}\,{{A}^{2}}]\]
C) \[[{{M}^{-\,3}}\,{{L}^{-\,2}}\,{{T}^{4}}\,{{A}^{4}}]\]
D) \[[{{M}^{-\,3}}\,{{L}^{-\,2}}\,{{T}^{2}}\,{{A}^{0}}]\]
Correct Answer: C
Solution :
\[\operatorname{X}\,\,=\,\,3Y{{Z}^{2}}\] \[Y=\frac{X}{3{{Z}^{2}}}=\frac{\dimension\,\,of\,\,capaci\operatorname{tance}}{{{(dimension\,\,of\,\,magnetic\,\,induction)}^{2}}}\] \[=\,\,\,\frac{[{{M}^{-\,1}}{{L}^{-\,2}}{{T}^{4}}{{A}^{2}}]}{{{[M{{T}^{-\,1}}{{A}^{-\,1}}]}^{2}}}=\frac{[{{M}^{-\,1}}{{L}^{-\,2}}{{T}^{4}}{{A}^{2}}]}{[{{M}^{2}}{{T}^{-\,2}}{{A}^{-\,2}}]}=[{{M}^{-\,3}}{{L}^{-\,2}}{{T}^{4}}{{A}^{4}}]\]
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