Given: Moment of inertia, $I = 0.10 \text{ kg-m}^2$ Radius, $R = 10 \text{ cm} = 0.1 \text{ m}$ Force, $F = 10 \text{ N}$ Time, $t = 2 \text{ s}$ Initial angular velocity, $\omega_0 = 0$ (since it starts from rest)

The torque acting on the wheel is given by: $\tau = F \times R = 10 \times 0.1 = 1 \text{ N m}$

Using Newton's second law for rotational motion, $\tau = I\alpha$, we can find the angular acceleration $\alpha$: $\alpha = \frac{\tau}{I} = \frac{1}{0.10} = 10 \text{ rad/s}^2$

Now, using the first equation of rotational kinematics to find the final angular velocity $\omega$: $\omega = \omega_0 + \alpha t$ $\omega = 0 + 10 \times 2 = 20 \text{ rad/s}$