question_answer

A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, then work required to pull the hanging part on to the table is

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 \[P.E.\] \[\Rightarrow \]               \[W=mgh=\frac{M}{3}\times g\times \frac{L}{6}=\frac{\mathbf{MgL}}{\mathbf{18}}\]