Question

A uniform chain has a mass $m$ and length $l$. It is held on a frictionless table with one-sixth of its length hanging over the edge. The work done in just puling the hanging part back on the table is:

• A. $mg\frac{l}{72}$
• B. $\frac{mgl}{36}$
• C. $\frac{mgl}{12}$
• D. $\frac{mgl}{6}$

Answer 1

The correct option is A $mg\frac{l}{72}$

$\text{Given:}$ Mass $=m$ length $=l$

$\text{Formula used:}$ $\left(PE\right)=mgh$

$\text{Step 1: Initial \& final figure [Ref. Fig. 1]}$

$\text{Step 2: Workdone}$

Workdone by external force = Change in potential energy

Considering COM of hanging part at zero potential energy.

Therefore Initial P.E. $(PE{)}_i=0$

Mass of hanging part $m^{\prime }=\frac{m}{6}$ (Chain is uniform)

Hight from center of mass $h=\frac{l}{12}$

Final P.E. $(PE{)}_f=m^{\prime }gh=\frac{mg}{6}\left(\frac{l}{12}\right)$; When complete chain comes to table.

$\Rightarrow W=(PE{)}_f-(PE{)}_i=(\frac{mgl}{72}-0)$

$W=\frac{mgl}{72}$

Hence correct option is $A$.

$\text{ALTERNATE SOLUTION:}$

$\text{Step 1: Workdone due to dx length [Ref. Fig. 2]}$

Consider the situation as shown in the figure.

Let $F$ be the force required to lift the chain at the instant as shown by a small distance $dX$

$\Rightarrow dW=F.dX$ where; $F=\frac{mgx}{l}$

$\Rightarrow {\int }_0^{w}dW={\int }_0^{\frac{l}{6}}\frac{mgx}{l}dx$

$\Rightarrow W=\frac{mg}{l}{\left[\frac{x^2}{2}\right]}_0^{l/6}=\frac{mg{l}^2}{72l}$

$\Rightarrow W=\frac{mgl}{72}$

Hence correct option is $A$.