Question

A container with liquid filled inside it till depth $H$ from the top is as shown in the diagram. Find the maximum angular velocity with which the cylinder can be rotated about the axis shown, such that the liquid doesn’t spill out.

• A. $\frac{\sqrt{Hg}}{R}$
• B. $\frac{\sqrt{2gH}}{R}$
• C. $\frac{2\sqrt{gH}}{R}$
• D. $\frac{\sqrt{gH}}{2R}$

Answer 1

The correct option is C $\frac{2\sqrt{gH}}{R}$


Condition: When the water is just spilling out, the depth of the lowest point on the water surface will $2$ times the depth of the water surface before rotation.

The equation of parabola (surface of the water) when the container is rotating is given by,

$y=\frac{{\omega }^2}{2g}{x}^2$

Here, when it is just spiling out, $y=2H,x=R$

$\Rightarrow \frac{{\omega }^2{R}^2}{2g}=2H$

$\omega =2\sqrt{\frac{gH}{R^2}}=\frac{2}{R}\sqrt{Hg}$

Hence, option $\left(c\right)$ is correct.