A disc with a flat small bottom beaker placed on it at a distance $$R$$ from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $$\omega$$. The coefficient of static friction between the bottom of the beaker and the surface of the disc is $$\mu$$. The beaker will revolve with the disc if :

For the beaker to revolve with the disc, the friction must provide the necessary centripetal force.

The centripetal force required for circular motion at distance $$R$$ from the center is:

$$F_c = m\omega^2 R$$

The maximum static friction available is:

$$f_{max} = \mu m g$$

For the beaker to not slide, friction must be sufficient:

$$m\omega^2 R \leq \mu m g$$

Cancelling $$m$$ from both sides:

$$\omega^2 R \leq \mu g$$

$$R \leq \frac{\mu g}{\omega^2}$$

Hence, the correct answer is Option B.