Question

A uniform chain of length $l$ is placed on a smooth horizontal table, such that half of its length hangsover one edge. It is released from rest, the velocity with which it leaves the table is:

• A. $\sqrt{\frac{3gl}{4}}$
• B. $\sqrt{\frac{3gl}{2}}$
• C. $\sqrt{\frac{2gl}{3}}$
• D. $\sqrt{\frac{gl}{3}}$

Answer 1

Answer: (A)

$\sqrt{\frac{3gl}{4}}$

Keeping the horizontal level of table as reference of gravitational potential, decrease in the gravitational potential of the chain=$-\frac{m}{2}g\frac{l}{4}-(-mg\frac{l}{2})$

$=\frac{3}{8}mgl$

This loss in gravitational potential increases the kinetic energy of chain.

$⟹\frac{3}{8}mgl=\frac{1}{2}m{v}^2$

$⟹v=\sqrt{\frac{3gl}{4}}$