Given: Mass of the hollow cylinder, $M = 3\text{ kg}$ Radius of the hollow cylinder, $R = 40\text{ cm} = 0.4\text{ m}$ Force applied, $F = 30\text{ N}$

The moment of inertia of a hollow cylinder about its central axis is given by $I = MR^2$. $I = 3 \times (0.4)^2 = 3 \times 0.16 = 0.48\text{ kg m}^2$

The torque ($\tau$) acting on the cylinder due to the force is: $\tau = F \times R = 30 \times 0.4 = 12\text{ N m}$

From Newton's second law for rotational motion, we have $\tau = I\alpha$, where $\alpha$ is the angular acceleration. $\alpha = \frac{\tau}{I} = \frac{12}{0.48} = 25\text{ rad/s}^2$.