A) \[M{{L}^{2}}T\]
B) \[{{M}^{-1}}{{L}^{2}}{{T}^{1}}\]
C) \[M{{L}^{2}}{{T}^{-2}}\]
D) \[ML{{T}^{-2}}\]
Correct Answer: B
Solution :
[b] Here, b and \[{{x}^{2}}={{L}^{2}}\]have same dimensions Also, \[a=\frac{{{x}^{2}}}{E\times t}=\frac{{{L}^{2}}}{(M{{L}^{2}}{{T}^{-2}})T}={{M}^{-1}}{{T}^{1}}\] \[a\times b=[{{M}^{-1}}{{L}^{2}}{{T}^{1}}]\]
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