A uniform chain of length `L` and mass `M` is tying on a smoth table and one third of its length is banging vertically down table 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

`U_1=-(1)/(2)((M)/(L)xx(L)/(3))g(L)/(3)`
`=-(MgL)/(18)`
`U_2=0`
`W_(1rarr2)=U_2-U_1=(MgL)/(18)`