Question

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

Answer 1

Correct Answer - D
Method 1:The hanging part of the chain which is to be pulled can be considered as a point mass situated at the centre of the hanging part. The equivalent diagram is drawn. The work dine in bringing the mass up will be equal to the change in potential energy of the mass.
`W=`change in potential energy
`=M/3xxgxxL/6=(MgL)/18`

Method 2: The per uni length of the chain `=M/L`. let us consider a finitesimally small length of the chain `dx` at a distance `x` from the bottom. To move `dx` to the top, a force equal to the weight of chain `x` will have to be applied upwards.

Weight of chain of length`x=(M/Lx)g`
Small amount of work done in moving `dx` to the top
`dW=vecF.vec(dx)=F dx=(M/Lx)g dx`
The total amount of work done in moving the one third length of the hanging chain on the table will be
`W=int_(0)^(L/3) ML x gdx=M/Lg`
`int_(0)^(L/3) x dx=M/Lg[x^(2)/2]_(0)^(L/3)=(MgL)/18`