A uniform chain of mass m & length L is kept on a smooth horizontal table such that `(L)/(n) `portion of the chaing hangs from the table. The work dione required to slowly bringsthe chain completely on the table is

Explanation

Mass per unit length of chain `m=M//L`
Let y be the length of hanging part of the chain.
Force required to be applied `=` weigth of length y
`F=(my)g`
Small amount of work done in pulling the chain through a small distance dy is
`dW=F(-dy)` (Neg. sign for decreasing y)
`=-mgydy`
Total work done in pulling the hanging part on the table,
W`=int_(y=(L)/(n))^(y=0)-mgydy`
`W=-mg((y^(2))/(2))_((y=L)/(n))^(y=0)`
`W=(-mg)/(2)(0-(L^(2))/(n^(2)))`
`W=(mg)/(2)(L^(2))/(n^(2))=(M)/(L)xx(g)/(2)(L^(2))/(n^(2))=(MgL)/(2n^(2))`