A) \[10\sqrt{8}m/s\]
B) \[\frac{40}{3}m/s\]
C) \[\pi \]
D) \[{{\tan }^{-1}}(\sin )\]
Correct Answer: C
Solution :
Given, \[v=\frac{W}{q}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[AT]}\] \[=[M{{L}^{2}}{{T}^{-3}}{{A}^{-1}}]\] \[=[M{{L}^{2}}{{T}^{-3}}{{A}^{-1}}]\] \[\left[ M{{L}^{-2}}{{A}^{-2}} \right]\] \[\left[ M{{L}^{2}}{{T}^{-2}}{{A}^{-2}} \right]\]
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