A) \[M{{L}^{2}}{{T}^{-2}}\]
B) \[{{L}^{-1/2}}{{T}^{2}}\]
C) \[{{L}^{-1/2}}{{T}^{-1}}\]
D) \[L{{T}^{-2}}\]
Correct Answer: B
Solution :
[b] \[[F]=[A]\times {{[x]}^{1/2}}=[B]{{[t]}^{2}}\] \[\Rightarrow \frac{[A]}{[B]}=\frac{[t{{[}^{2}}}{{{[x]}^{1/2}}}=[{{L}^{-1/2}}{{T}^{2}}]\]
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