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

Options

A. `MgL`
B. `(MgL)/(3)`
C. `(MgL)/(9)`
D. `(MgL)/(18)`

Answer

D

Explanation

The weight of hanging part `((L)/(3))` from the table
As work done = force `xx` distance
`:. W = (Mg)/(3) xx (L)/(6) = (MgL)/(18)`