A) \[MgL\]
B) \[\frac{MgL}{3}\]
C) \[\frac{MgL}{9}\]
D) \[\frac{MgL}{18}\]
Correct Answer: D
Solution :
Since length of hanging part is\[\frac{L}{3}\], so its mass would be \[\frac{M}{3}\] and the weight \[\frac{Mg}{3}\] will act at the centre of gravity \[G\] of hanging part, which is at distance\[\frac{L}{6}\]. Now, work done in pulling the hanging part on the table\[=\]increase in \[P.E.\] \[\Rightarrow \] \[W=mgh=\frac{M}{3}\times g\times \frac{L}{6}=\frac{\mathbf{MgL}}{\mathbf{18}}\]You need to login to perform this action.
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