Question

A chain of mass $m$ and length $l$ is placed on a table with one-sixth of it hanging freely from the table edge. The amount of work done to pull the chain on to the table is

• A. $\frac{mgl}{4}$
• B. $\frac{mgl}{6}$
• C. $\frac{mgl}{72}$
• D. $\frac{mgl}{36}$

Answer 1

The correct option is C $\frac{mgl}{72}$

Given, the total mass of chain $=m$

Length of chain $=l$

So, mass per unit length is $(\lambda =m/l)$

If we consider a small element $dx$ at a distance '$x$' from the table, then the small mass element $dm$ is given as:

$dm=\frac{m}{l}dx$

So, work done against gravity in lifting the small mass element $dm$ through $x$ distance is :

$dW=dmgx=\left(m/l\right)gxdx$

Now, total work done is obtained by integrating $dW$ from limits $x=0$ to $x=\frac{l}{6}$ as one-sixth of the chain hangs from table.

${\int }_0^{W}dW=\int \frac{mg}{l}x\left(dx\right)=\frac{mg}{l}{\int }_0^{\frac{l}{6}}xdx$

$W=\frac{mg}{2l}{\left(\frac{l}{6}\right)}^2=\frac{mgl}{72}$