A uniform solid disc of mass 4 kg and radius 0.5 m is pivoted at...

The moment of inertia of a uniform solid disc about its central axis is:
$I = \dfrac{1}{2}MR^2 = \dfrac{1}{2} \times 4 \times (0.5)^2 = 2 \times 0.25 = 0.5 \text{ kg m}^2$

The torque produced by the tangential force applied at the rim is:
$\tau = F \times R = 10 \times 0.5 = 5 \text{ N m}$

Using Newton's second law for rotation, the angular acceleration $\alpha$ is:
$\alpha = \dfrac{\tau}{I} = \dfrac{5}{0.5} = 10 \text{ rad/s}^2$

Since the disc starts from rest ($\omega_i = 0$), the final angular velocity after $t = 3 \text{ s}$ is:
$\omega_f = \omega_i + \alpha t = 0 + (10 \times 3) = 30 \text{ rad/s}$

Answer: $30 \text{ rad/s}$