Step 1: Concept
Torque $\tau = I\alpha$, where $I$ is moment of inertia and $\alpha$ is angular acceleration.
Step 2: Analysis
- $I_{\text{ring}} = MR^2$
- $I_{\text{disc}} = \frac{1}{2}MR^2$
Since $I_{\text{ring}} > I_{\text{disc}}$, for the same torque, $\alpha_{\text{disc}} > \alpha_{\text{ring}}$.
Step 3: Calculation
Higher angular acceleration leads to a higher change in angular velocity and frequency in the same time interval.
Step 4: Conclusion
Hence, the disc will rotate with greater angular frequency.
Final Answer: (C)