# Not That Simple!

A cylinder contains a fluid of some density. At a distance $$R$$ from the centre of the base of the cylinder, a string of length $$L$$ is connected, other end of which is connected to a light ball. Initially there is tension in the string. Now this whole assembly is rotated with angular velocity $$\omega$$ about an axis passing through the centre of the cylinder. (As shown in the figure).

When rotated, It is found that the vertical depression of the light ball is $$H$$.

Find $$\omega$$ (Angular velocity of rotation)

Given:

$$g= 10 \text{ m/s}^{2}$$.

$$L= 1\text{ m}$$.

$$H= 0.5\text{ m}$$.

$$R= 2 \text{ m}$$.

The interesting fact is that the answer doesn't depend on the densities of the fluid and the ball, but the direction in which the ball moves depends on the respective densities.

Hint: If you think the ball moves away from the axis, you might be wrong!

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