A chain AB of length `l` is located in a smooth horizontal tube so that its fraction of length `h` hangs freely and touches the surface of the table with its end B(figure). At a certain moment the end A of the chain is set free. With what velocity will this end of the chain slip out of the tube?

Explanation

Let the length of the chain inside the smooth horizontal tube at an arbitrary instant is x.
From the equation,
`mvecw=vecF+vecu(dm)/(dt)`
as `vecu=0`, `vecFuarruarrvecw`, for the chain inside the tube
`lambdaxw=T` where `lambda=m/l` (1)
Similarly for the overhanging part,
`vecu=0`
Thus `mw=F`
or `lambdahw=lambdahg-T` (2)
From (1) and (2),
`lambda(x+h)w=lambdahg` or, `(x+h)v(dv)/(ds)=hg`
or, `(x+h)v(dv)/((-dx))=gh`,
[As the length of the chain inside the tube decreases with time, `ds=-dx`.]
or, `vdv=-gh(dx)/(x+h)`
Integrating, `underset(0)overset(v)int vdv=-gh underset((l-h))overset(0)int(dx)/(x+h)`
or `v^2/2=gh1n(l/h)` or `v=sqrt(2gh1n(l/h))`