Question

A uniform chain of mass m and length l overhangs a table with its two third parts on the table. The minimum work to be done by a person to put the hanging part back on the table is

• A. $\frac{mgl}{18}$
• B. $\frac{mgl}{6}$
• C. $\frac{mgl}{9}$
• D. $\frac{mgl}{3}$

Answer 1

The correct option is A $\frac{mgl}{18}$

$\text{Mass of}\frac{1}{3}\text{part of chain}=\frac{m}{3}$

$\text{Work done = Change in potential energy}$

$=\frac{m}{3}\times g\times \frac{l}{6}=\frac{mgl}{18}$