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

Options

A. `h sqrt((2g)/(Lh))`
B. `sqrt(2gh log_(e)((L)/(h)))`
C. `sqrt(2gl log_(e)((L)/(h)))`
D. `(1)/(hL) sqrt(2g)`

Answer

B

Explanation

Let 'x' be the length of chain inside the tube
`a = (((Mh)/(L))g)/((M)/(L)(h+x)) = (hg)/(h+x), -(vdv)/(dx) = (hg)/(h+x)`
on integrating both sides, `-int_(0)^(v) vdv = int_(l-h)^(0) (hg)/(h+x) dx`.