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 the acceleration due to gravity, then work required to pull the hanging part on the table will be:
1
\(\frac{MgL}{6}\)
2
\(\frac{MgL}{9}\)
3
\(\frac{MgL}{12}\)
4
\(\frac{MgL}{18}\)