A chain is held on a frictionless table with `1//n` th of its length hanging over the edge. If the chain has a length L and a mass M, how much work is required to pull the hanging part back on the table?

Options

A. `MgL//2n^2`
B. `2MgL//-2n^2`
C. `MgL//4n^3`
D. `MgL//3n^(-1)`

Answer

A

Explanation

If m is the mass per unit length of the chain, the mass of the chain of length y will be ym and the force acting on it due to gravity will be mgy (assuming that y is the length of the chain hanging over the edge). So, the work done in pulling the dy length of the chain on the table
`dW = F(-dy)` [as y is decreasing]
İ.e., `dW = (mgy)(-dy) `[as F = mgy]
So, the work done in pulling the hanging portion on the table :
`W = - int_(L//n)^(0) mgydy = (MgL^2)/(2n^2)`
or `W = MgL//2n^2 `[as M= mL]