A) \[\frac{2}{9}mgl\]
B) \[\frac{4}{9}mgl\]
C) \[\frac{3}{4}mgl\]
D) \[\frac{5}{18}mgl\]
Correct Answer: B
Solution :
Suppose ground level is of the zero potential level, then, total mechanical energy at two instant can be given as. \[TM{{E}_{initially}}=TM{{E}_{finally}}\] \[\Rightarrow \]\[mg\frac{1}{2}+\frac{1}{2}m{{v}^{2}}=\left( \frac{m}{3} \right)g(1-1)+\frac{2m}{3}gl\] \[=\frac{5mgl}{18}+\frac{2}{3}mgl\] (always consider the height of centre of mass) \[\Rightarrow \]\[\frac{1}{2}m{{v}^{2}}=KE=\frac{4}{9}mgl\]You need to login to perform this action.
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