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

Answer

A

Explanation

Let dx be the length of an element at distance x from the table of mass dx length `=(m/l)dx` work done to put back on the table
`w=(m/l)dxg(x)`
so, totla work done to put `1/3` part back on the table
`W=int_0^(1/3)(m/l)gxdx`
`rarr w=(m/l)g[x^2/2]_0^(1/3)`
`rarr =(mgl^2)/(18l)=(mgl)/18`