A uniform chain of mass `m` and length `l` is placed on a smooth table so that one`-`third length hanges freely as shown in the figure. Now the chain is released, with what velocity chain slips off the table?

Explanation

Let at some instant the hanging length be `x`.

pulling force `=(m)/(l)xg`, pulled mass `=m`
Acceleration of chain `a=(m/lxg)/(m)=g(x)/(l)`
`v(dv)/(dx)=g(x)/(l)`
`int_(0)^(v)vdv=(g)/(l)int_(l//3)^(l)xdx`
`|v^(2)/2|_(0)^(v)=(g)/(l)|x^(2)/2|_(l//3)^(l)`
`(v^(2))/(2)=(g)/(2l)xx{l^(2)-((l)/(3))^(3)}=(g)/(2l){t^(2)-(l^(2))/(9)}`
`v^(2)=(g)/(l)xx(8l^(2))/(9)=(8gl)/(9)`
`v=(2sqrt(2gl))/(3)`