Question

A uniform chain of mass $M$ and length $L$ is held on a horizontal frictionless table with $\frac{1}{n^{th}}$ of its length hanging over the edge of the table. The work done is pulling the chain up on the table is

• A. $\frac{MgL}{n}$
• B. $\frac{MgL}{2n}$
• C. $\frac{MgL}{n^2}$
• D. $\frac{MgL}{2n^2}$

Answer 1

The correct option is D

$\frac{MgL}{2n^2}$

The mass per unit length of the chain $m=\frac{M}{L}$. The mass of the hanging portion of the chain is $m^{\prime }=\frac{M}{n}$. This mass can be assumed to be concentrated at the centre of the hanging portion of the chain which is a distance of $x=\frac{L}{2n}$ from the edge of the table. Therefore, the work done in pulling the hanging portion of the chain on to the table top is $W=m^{\prime }gx=\frac{M}{n}\times g\times \frac{L}{2n}=\frac{MgL}{2n^2}$.