To find the angular velocity ($\omega$) of the wheel after $10$ seconds, we first need to understand the relationship between the force applied through the string, the torque produced by this force, and how this torque affects the wheel's angular acceleration ($\alpha$).

The torque ($\tau$) produced by the force ($F$) is given by the product of the force and the radius ($r$) of the wheel through which the force is applied:

$\tau = F \cdot r$

Given that $F = 40 \, \mathrm{N}$ and $r = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m}$, the torque can be calculated as:

$\tau = 40 \cdot 0.1 = 4 \, \mathrm{Nm}$

The torque is related to the angular acceleration ($\alpha$) and the moment of inertia ($I$) of the wheel by the equation:

$\tau = I \cdot \alpha$

Given that $I = 0.40 \, \mathrm{kg \cdot m}^2$, we can rearrange the above formula to solve for $\alpha$:

$\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10 \, \mathrm{rad/s}^2$

With the angular acceleration ($\alpha$), we can calculate the angular velocity ($\omega$) after a given time ($t$) using the formula:

$\omega = \omega_0 + \alpha \cdot t$

Where $\omega_0$ is the initial angular velocity. Since the wheel starts from rest, $\omega_0 = 0$. Thus, for $t = 10 \, \mathrm{s}$:

$\omega = 0 + 10 \cdot 10 = 100 \, \mathrm{rad/s}$

Therefore, the angular velocity ($\omega$) of the wheel after $10$ seconds is $100 \, \mathrm{rad/s}$, so $x = 100$.