A uniform chain of length I 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, the work required to pull the hanging part on to the table is

The situation is shown in fig. (9).Given that the length of hanging part is L/3,  so its mass would be M/3.Hence the weight (Mg/3) will act at the center of gravity G of hanging part. It will act at a distance (L/6) ) below the surface of table. Work done in pulling the hanging part on the table = increase in P.E.=m×g×h=M3×g×L6=mgL18