A string is wrapped around the rim of a wheel of moment of inertia $$0.40$$ kg m$$^2$$ and radius $$10$$ cm. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of $$40$$ N. The angular velocity of the wheel after $$10$$ s is $$x$$ rad/s, where $$x$$ is _______

Moment of inertia $$I = 0.40$$ kg m$$^2$$, radius $$r = 10$$ cm = 0.1 m, force $$F = 40$$ N, time $$t = 10$$ s, initial angular velocity = 0.

The torque is calculated as $$\tau = Fr = 40 \times 0.1 = 4$$ N·m.

The angular acceleration is $$\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10$$ rad/s$$^2$$.

After time $$t$$, the angular velocity is given by:

$$\omega = \omega_0 + \alpha t = 0 + 10 \times 10 = 100 \text{ rad/s}$$

The answer is $$x = \boxed{100}$$.