A uniform chain of mass `M` and length `L` is lying on a frictionless table in such a way that its `1//3` parts is hanging vertically down. The work done in pulling the chain up the table is

Explanation

Let us visualise a state when length c of the string is lying on the table as shown in figure .

Force applied by the person to pull the string slowly is equal to the weight of hanging portion .
`F = m/l (l-x) g `
Let us pull the string by an amount dx so that applied force may be assumed to be constant . Wrok done by the person can be written as follows :
dW = Fdx
`rArr " " dW = (mg)/l (l-x) dx `
Initially two - third length is on the table and finally entire length is on teh table . So we can integrate work done by the person .
`W = (mg)/l int_(2l//3)^(l) (l-x) dx `
` rArr " " W = (mg)/l int_(2l//3)^(l) (l - x) dx `
`rArr " " W = (mg)/l [ {l^(2) - (l^(2))/2 } - {(2l^(2))/3 - (2l^(2))/9}] `
` rArr " " W = (mg)/l [{(l^(2))/2 }={(4l^(2))/9}] `
`rArr " " W = (mgl)/18 `