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, work required to pull the hanging part on to the table is

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

Answer 1

Answer: (D)

$\frac{MgL}{18}$

Answer 2

The weight of hanging part $\left(\frac{L}{3}\right)$ of chain is $\left(\frac{1}{3}\,Mg\right)$. This weight acts at centre of gravity of the hanging part, which is at a distance of $\left(\frac{L}{6}\right)$ from the table.
As work done $=$ force $\times$ distance
$\therefore W = \frac{Mg}{3} \times \frac{L}{6} = \frac{MgL}{18}$