QJEEM22Ph9ed7: JEE Main 2022 Physics - Class 11 JEEM22Ph9ed7 Solution: A disc with a flat small bottom beaker placed on it at a…

The force that prevents the beaker from sliding is the static friction force. For an object in uniform circular motion, the net force acting on the object (which is the friction force in this case) is equal to the centripetal force.

The static friction force is given by the normal force (which is equal to the weight of the object) multiplied by the coefficient of static friction, which is :

$F_{\text{friction}} = \mu mg.$

The centripetal force needed to keep an object moving in a circle of radius R at angular velocity ω is given by :

$F_{\text{centripetal}} = mR\omega^2.$

For the beaker not to slide off, the static friction force must be at least as large as the centripetal force. Therefore, we have :

$\mu mg \geq mR\omega^2.$

After canceling the mass m from both sides, we get :

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

So, the correct answer is Option B :

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