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, the work required to pull the hanging part on to the table is

Options

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

Answer

D

Explanation

(d) Mass`(M)/(3)` has its centre of mass at `(L)/(6)` below the table surface.

`:. W=mgh=((M)/(3))(g)((L)/(6))=(MgL)/(18)`
`W_(1)=+W_(g)+W_(T)=0, W_(T)=0`