1. Identify the force needed for circular motion

For the beaker to rotate with the disc without slipping, it must have centripetal acceleration toward the center:

$a_c = \omega^2 R$

So the required centripetal force is

$F_c = m\omega^2 R$

2. What provides this centripetal force?

The only horizontal force available is static friction between the beaker and the disc.

Maximum static friction is

$f_{\max} = \mu N$

Since the disc is horizontal,

$N = mg$

Hence,

$f_{\max} = \mu mg$

3. Condition for no slipping

For the beaker to revolve with the disc, the required centripetal force must not exceed maximum static friction:

$m\omega^2 R \le \mu mg$

Cancel $m$:

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

Therefore,

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

4. Match with the options

This corresponds to:

Option B:

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

5. Comparison with stored correct answer

Stored correct answer is B, which matches the derived result.