A) \[M{{L}^{5}}{{T}^{-2}}\]
B) \[M{{L}^{-1}}{{T}^{-2}}\]
C) \[{{M}^{0}}{{L}^{3}}{{T}^{0}}\]
D) \[{{M}^{0}}{{L}^{6}}{{T}^{0}}\]
Correct Answer: A
Solution :
By principle of dimensional homogenity \[\left[ \frac{a}{{{V}^{2}}} \right]=\left[ P \right]\] \[\therefore \] \[[a]=[P]\ [{{V}^{2}}]=[M{{L}^{-1}}{{T}^{-2}}]\ \times [{{L}^{6}}]\] = \[[M{{L}^{5}}{{T}^{-2}}]\]
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