Question

A chain AB of length l is loaded in a smooth horizobtal table so that its fraction of length h hangs freely and thouches 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 of the chain slip out of the table?

• A. $\sqrt{2ghin\left(\frac{l}{h}\right)}$
• B. $\sqrt{ghin\left(\frac{l}{h}\right)}$
• C. $\sqrt{2ghin\left(\frac{h}{l}\right)}$
• D. $\sqrt{3ghin\left(\frac{l}{h}\right)}$

Answer 1

The correct option is A $\sqrt{2ghin\left(\frac{l}{h}\right)}$

If we look at the hanging part of the chain, the forces acting on it are:

the weight of the hanging part, the tension (T) due to the part of chain in the tube and the normal reaction from the ground(N).

Let the length of chain remaining on table be x and mass of the chain be m.

So the mass of the chain hanging will be mh/l and that on table will be mx/l

If we balance forces on the hanging part of the chain, we get

$mgh/l-T-N=mha/l$ ..................(1)

If we balance forces for the part in the tube,

$T=mxa/l$ ....................(2)

As N = 0, from (1) and (2) we get, $\frac{mgh}{l}-\frac{mxa}{l}=\frac{mha}{l}$
mghl−mxal=mhal

i.e. $a=\frac{gh}{h+x}$
a=ghh+x

i.e. $\frac{-vdv}{dx}=\frac{gh}{h+x}$−vdvdx=ghh+x

Integrate the expression with limits of v from 0 to v and x from l-h to 0

${\int }_0^v-vdv={\int }_{L-h}^0\frac{gh}{h+x}dx$

We get,

$v=\sqrt{2gh.\ln \frac{l}{h}}$