One end of the pile of chain falls vertically through a hole in its support and pull the remaining links steadily. The links which are at rest acquire the velocity of the hanging position suddenly and without loving interaction with the remaining stationery links and with the support. If the acceleration a of the falling chain is `g//k`. Find `k`. Ignore any friction.

Explanation

Let the velocity of the end of the falling chain be `v_(0)` when a length `x` has already fallen. Taking mass per unit length of the chain as `lambda`, we can write,
`lambdaxg-((lambdavdt)v)/(dt)=lambdax*v(dv)/(dx)`
`gx-v^(2)=x(vdx)/(dx)`
or `v (dv)/(dx)+(v^(2))/(x)=g`
This equation may be solved by using Intergrating factor. The solution is
`v^(2)=(2gx)/(3)+(C )/(x^(2))`. where `C` is an arbitary constant.
At `t=0`, `x=0` and `v=0`
`:. C=0` and `v^(2)=(2gx)/(3)`