A uniform chain of length L and mass m is lying on a smooth table. One third of its length is hanging verti cally down over the edge of the table. How much work need to be done to pull the hanging part back to the table ?

Options

A. mgl/25
B. mgl/50
C. mgl/5
D. mgl/10

Answer

b

Explanation

If `1/n`th part of chain is hanging over the edge of table then work done to pull the hanging part is `W=(mgl)/(2n^(2))`
Here `1/5` th mean n=5
thus `W=(mgl)/(2xx5^(2))=(mgl)/(50)`