$$I_{CM}$$ is moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of the disc. $$I_{AB}$$ is its moment of inertia about an axis AB perpendicular to the plane and parallel to the axis CM at a distance $$\frac{2}{3}R$$ from the center, where R is the radius of the disc. The ratio of $$I_{AB}$$ and $$I_{CM}$$ is $$x : 9$$. The value of $$x$$ is _____.

For a circular disc of mass $$M$$ and radius $$R$$, the moment of inertia about the central axis perpendicular to its plane ($$I_{CM}$$) is $$I_{CM} = \frac{1}{2}MR^2$$

According to the Parallel Axis Theorem, the moment of inertia about any parallel axis $$AB$$ at a distance $$d$$ is $$I_{AB} = I_{CM} + Md^2$$

Substituting the given distance $$d = \frac{2}{3}R$$:

$$I_{AB} = \frac{1}{2}MR^2 + M\left(\frac{2}{3}R\right)^2$$

$$I_{AB} = \left(\frac{9 + 8}{18}\right)MR^2 = \frac{17}{18}MR^2$$

$$\frac{I_{AB}}{I_{CM}} = \frac{\frac{17}{18}MR^2}{\frac{1}{2}MR^2} = \frac{17}{18} \times \frac{2}{1} = \frac{17}{9}$$

$$x = 17$$