Question

A chain of length $l$ is lying in a smooth horizontal tube such that a fraction of its length $h$ hangs freely and the end touches the ground. At a certain moment the other end of chain is set free. The speed of this end of chain when it slips out of the tube is:

• A. ${\left[\right(2gh\left)\frac{dl}{dh}\right]}^{1/2}$
• B. $\sqrt{gh}$
• C. $\sqrt{2gl}$
• D. ${\left[2gh{\log }_e\frac{l}{h}\right]}^{\frac{1}{2}}$

Answer 1

The correct option is D ${\left[2gh{\log }_e\frac{l}{h}\right]}^{\frac{1}{2}}$

Let $m$ be the mass per unit length.
For hanging part, $mha=mhg-T....\left(1\right)$
For part in tube, $T=mxa....\left(2\right)$
solving (1) and (2), $mha=mhg-mxa\Rightarrow a=\frac{hg}{h+x}$
or, $\frac{dv}{dt}=\frac{hg}{h+x}$
or, $\frac{dv}{dx}.\frac{dx}{dt}=\frac{hg}{h+x}$
or, $vdv=\frac{hg}{h+x}dx$ $(\frac{dx}{dt}=v)$
or, ${\int }_0^{v}vdv={\int }_0^{l-h}\frac{hg}{h+x}dx$
or, $\frac{v^2}{2}=ghlo{g}_e\left(\frac{l}{h}\right)$
or, $v={\left[2ghlo{g}_e\right(\frac{l}{h}\left)\right]}^{1/2}$