Suppose there is a uniform circular disc of mass $$M$$ kg and radius $$r$$ m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $$A$$ of the disc is given by $$ \frac{x}{256}Mr^{2} $$. The value of $$x$$ is______.

Mass of removed parts (m):

For a hole with radius $$r/4$$:  $$m = \frac{\pi (r/4)^2}{\pi r^2} \times M = \frac{1}{16} M$$

Moment of Inertia of the Original Disc ($$I_{disc}$$):

$$I_{disc} = \frac{1}{2} Mr^2 = \frac{128}{256} Mr^2$$

Moment of Inertia of One Removed Part ($$I_{hole}$$):

$$I_{hole} = I_{cm\_hole} + m d^2$$

$$I_{hole} = \left[ \frac{1}{2} m \left( \frac{r}{4} \right)^2 \right] + m \left( \frac{3r}{4} \right)^2$$

$$I_{hole} = \left[ \frac{1}{2} \cdot \frac{M}{16} \cdot \frac{r^2}{16} \right] + \left[ \frac{M}{16} \cdot \frac{9r^2}{16} \right]$$

$$I_{hole} = \frac{Mr^2}{512} + \frac{18Mr^2}{512} = \frac{19}{512} Mr^2$$

Final Moment of Inertia:

$$I_{remainder} = I_{disc} - 2 \cdot I_{hole}$$

$$I_{remainder} = \frac{128}{256} Mr^2 - 2 \left( \frac{19}{512} Mr^2 \right)$$

$$I_{remainder} = \frac{128}{256} Mr^2 - \frac{19}{256} Mr^2 = \frac{109}{256} Mr^2$$

$$x = 109$$