Question

A uniform chain of length $l$ and mass $m$ lies on a smooth table. A very small part of this chain hangs from the table. It begins to fall under the weight of the hanging end. Find the velocity of the chain when the length of the hanging part becomes $y$.

• A. $y\sqrt{\frac{g}{l}}$
• B. $2y\sqrt{\frac{g}{l}}$
• C. $y\sqrt{\frac{2g}{l}}$
• D. $y\sqrt{\frac{g}{2l}}$

Answer 1

The correct option is A $y\sqrt{\frac{g}{l}}$

Let, mass of chain $=m$
When length of the hanging part $=y$,
mass of hanging part of chain $=\frac{my}{l}$
According to principle of energy conservation :
Loss in gravitational potential energy of hanging part = Gain in $KE$ of the whole chain
$\Rightarrow \left(\frac{my}{l}\right)g\left(\frac{y}{2}\right)=\frac{1}{2}m{v}^2$
(because COM of hanging part has fallen by a distance of $\frac{y}{2}$)
$\Rightarrow v^2=\frac{g{y}^2}{l}\Rightarrow v=y\sqrt{\frac{g}{l}}$