Question

A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, work required to pull the hanging part on to the table is :

• A. MgL
• B. $\frac { \mathrm { MgL } } { 3 }$
• C.
• D. $\frac { \mathrm { MgL } } { 18 }$

Answer 1

Answer: (D)

$\frac { \mathrm { MgL } } { 18 }$

Answer 2

The weight of hanging part of chain is . This weight acts at centre of gravity of the hanging part, which is at a distance of from the table.

As work done = force × distance

$\therefore \mathrm { W } = \frac { \mathrm { Mg } } { 3 } \times \frac { \mathrm { L } } { 6 } = \frac { \mathrm { MgL } } { 18 }$