JEE Advanced · 2006 · 09 Apr 2006 · Rotational Mechanics

Taking torque about the hinge side (one edge of the plate):

The balls strike the shaded half region (area $\frac{ab}{2}$) with n balls per unit area per unit time. Each ball undergoes elastic collision, so the momentum change per ball is $2mv$.

The force from the balls acts at the center of the shaded region, which is at distance $\frac{3b}{4}$ from the hinge.

Torque due to balls: ${\tau }_{balls}=\left(\frac{ab}{2}\cdot n\cdot 2mv\right)\times \frac{3b}{4}$

Torque due to weight of plate: ${\tau }_{weight}=Mg\times \frac{b}{2}$

For equilibrium:

$\frac{ab}{2}\cdot n\cdot 2mv\cdot \frac{3b}{4}=Mg\cdot \frac{b}{2}$

Simplifying:

$v=\frac{2Mg}{3abnm}=\frac{2\times 3\times 10}{3\times 1\times 2\times 100\times 0.01}=\frac{60}{6}=10$ m/s

Answer: 10 m/s