A) \[[M{{L}^{5}}{{T}^{-2}}]\]
B) \[[M{{L}^{-1}}{{T}^{-2}}]\]
C) \[[M{{L}^{3}}T]\]
D) \[[M{{L}^{6}}T]\]
Correct Answer: A
Solution :
From the law of homogeneity of dimensions, the a dimensions of \[\frac{a}{{{V}^{2}}}\] must be that of pressure. \[\frac{[ML{{T}^{-2}}]}{[{{L}^{2}}]}=\frac{a}{[{{L}^{6}}]}\] \[a=[M{{L}^{5}}{{T}^{-2}}]\]You need to login to perform this action.
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