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 acceleration due to gravity, the work required to pull the hanging part on to the table is [IIT 1985]

• A. $MgL$
• B. $\frac{MgL}{3}$
• C. $\frac{MgL}{9}$
• D. $\frac{MgL}{18}$

Answer 1

The correct option is D

$\frac{MgL}{18}$

If point mass m is pulled through a height h then work done W = mgh
Similarly for a chain we can consider its centre of mass at the middle point of the hanging part i.e. at a height of L/(2n) from the lower end and mass of the hanging part of chain $=\frac{M}{n}$
So work done to raise the centre of mass of the chain on the table is given by
$W=\frac{M}{n}\times g\times \frac{L}{2n}$ [As W = mgh]
$orW=\frac{MgL}{2n^2}$
As $\frac{1}{3}$ part of the chain is hanging from the edge of the table. So by substituting n = 3 in standard expression

$W=\frac{MgL}{2n^2}=\frac{MgL}{2(3{)}^2}=\frac{MgL}{18}$