A, B and C are disc, solid sphere, and spherical shell respectively with same radii and masses. These masses are placed as shown in the figure. The moment of inertia of the given system about PQ is $$\frac{x}{15}I$$, where I is the moment of inertia of the disc about its diameter. The value of x is ______.

The moment of inertia of the disc about its diameter:

$$I = \frac{1}{4}MR^2 \implies MR^2 = 4I$$

Using the Parallel Axis Theorem ($$I = I_{cm} + Md^2$$), where $$d$$ is the perpendicular distance from the axis PQ to the center of each object:

Disc A: The axis passes through its center as a diameter ($$d = 0$$). $$I_A = \frac{1}{4}MR^2 = I$$

Solid Sphere B: The axis is at a distance $$d = R$$ from its center. $$I_B = \frac{2}{5}MR^2 + M(R)^2 = \frac{7}{5}MR^2$$

Spherical Shell C: The axis is at a distance $$d = R$$ from its center. $$I_C = \frac{2}{3}MR^2 + M(R)^2 = \frac{5}{3}MR^2$$

$$I_{sys} = I_A + I_B + I_C$$

$$I_{sys} = \frac{1}{4}MR^2 + \frac{7}{5}MR^2 + \frac{5}{3}MR^2$$

$$I_{sys} = \left( \frac{15 + 84 + 100}{60} \right) MR^2 = \frac{199}{60} MR^2$$

$$I_{sys} = \frac{199}{60} (4I)$$

$$I_{sys} = \frac{199}{15} I$$

$$x = 199$$