A heavy, flexible uniform chain of length $\mathrm{\pi }r$ and mass $\lambda \mathrm{\pi }r$lies in a smooth semicircular tube $\mathrm{AB}$of the radius $r$. Assuming a slight disturbance to start the chain in motion, find the velocity $v$ with which it will emerge from the end $\mathrm{B}$ of the tube.

• A

  $\sqrt{4gr\left(\frac{3}{\text{π}}+\frac{\text{π}}{2}\right)}$
• B

  $\sqrt{3gr\left(\frac{2}{\text{π}}+\frac{\text{π}}{5}\right)}$
• Correct answer
  C

  $\sqrt{2\text{g}r\left(\frac{2}{\text{π}}+\frac{\text{π}}{2}\right)}$
• D

  $\sqrt{5gr\left(\frac{4}{\text{π}}+\frac{\text{π}}{3}\right)}$

Solution

Since friction absent, we can apply the law of conservation of energy. Centre of gravity of a semicircular arc is at a distance $\left(\frac{2\mathrm{\pi }}{r}\right)$ from the centre.

$\text{Initial potential energy}=\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)g\left(\frac{2\mathit{\text{r}}}{\text{π}}\right)$

$\text{Final potential energy}=\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)g\left(\frac{-\text{π}\mathit{\text{r}}}{2}\right)$

When the chain is completely slipped off the tube, all the links of the chain have the same velocity $v$.



$\text{Kinetic energy of chain}=\frac{1}{2}\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)v^2$

$\text{From COE,}\mathit{\text{λ}}\text{π}\mathit{\text{rg}}\left(\frac{2\text{π}}{\text{r}}\right)=\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)\mathit{\text{g}}\left(\frac{-\text{π}\mathit{\text{r}}}{2}\right)+\frac{1}{2}\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)v^2$

$\text{From which we find,}v=\sqrt{2gr\left(\frac{2}{\text{π}}+\frac{\text{π}}{2}\right)}$

Hence, the correct answer is option: 'C'.