Question

A string of mass $m$ and length $l$ resets over a frictionless table with ${\left(\frac{1}{4}\right)}^{th}$ of its length hanging from a side. The work done in bringing the hanging part back on the table is

• A. $mgl/4$
• B. $mgl/32$
• C. $mgl/16$
• D. None of these

Answer 1

The correct option is B $mgl/32$

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 $x=\frac{l}{4}$ as one-fourth of the chain hangs from table.

${\int }_0^{W}dW=\int \frac{mg}{l}x\left(dx\right)=\frac{mg}{l}{\int }_0^{\frac{l}{4}}xdx$

$W=\frac{mg}{2l}{\left(\frac{l}{4}\right)}^2=\frac{mgl}{32}$