A) \[\frac{gml}{4}\]
B) \[\frac{gm{{l}^{2}}}{2}\]
C) \[\frac{gml}{\sqrt{2}}\]
D) \[\frac{gml}{18}\]
Correct Answer: D
Solution :
Initial weight of the hanging part of chain \[=\frac{mg}{3}\] Final weight of the hanging part of chain =0 So, mean weight of hanging part of chain \[\frac{\frac{mg}{3}+0}{2}=\frac{mg}{6}+0\] Work done is given by \[\frac{mg}{6}\times \frac{l}{3}=\frac{mgl}{18}\]You need to login to perform this action.
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