Question

A chain AB of length $l$ is lying in a smooth horizontal tube so that a fraction $h$ of its length $l$ , 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. The velocity of end A of the chain, when it slips out of tube , is:

• A. $h\sqrt{\frac{2g}{lh}}$
• B. $\sqrt{2gh\times lo{g}_e\left(\frac{l}{h}\right)}$
• C. $\sqrt{2gl\times lo{g}_e\left(\frac{l}{h}\right)}$
• D. $\frac{1}{hl}\sqrt{2g}$

Answer 1

The correct option is B $\sqrt{2gh\times lo{g}_e\left(\frac{l}{h}\right)}$

Initially the end part $\left(B\right)$ of chain just touches the table. After some time, a part of the chain $\left(BP\right)$ lies on the table and the length of the chain in the tube $\left(AO\right)$ is $x$.

Force of gravitation acts only on the hanging part of chain i.e $OP$

As the $BP$ part of chain rests on the table, so only the $AP$ part of chain will accelerate.

Let the mass of the whole chain be $M$.

$\therefore$ Mass of chain $OP$: $m=\frac{M}{l}h$

Force of gravitation, $F=mg=\frac{M}{l}hg$

Mass of chain $AP$: $M^{\prime }=\frac{M}{l}(h+x)$

$\therefore$ $M^{\prime }a=mg$

$\frac{M}{l}(h+x)a=\frac{M}{l}hg$ $⟹a=\frac{hg}{h+x}$

$\therefore$ $-v\frac{dv}{dx}=\frac{hg}{h+x}$

$⟹$ ${\int }_0^{v}vdv=-hg{\int }_{(l-h)}^0\frac{1}{h+x}dx$

$\therefore$ $\frac{v^2}{2}=-hg$ $lo{g}_e(h+x){|}_{(l-h)}^0$

$\frac{v^2}{2}=-hg$ $lo{g}_e\frac{h}{l}$

$⟹$ $v=\sqrt{2gh\times lo{g}_e(\frac{l}{h})}$