Four particles each of mass ' m ' are lying symmetrically on the rim of disc of mass ' $M$ ' and radius ' $R$ '. Moment of inertia of the system about an axis passing through one of the particle and perpendicular to plane of disc is

Moment of inertia of the system

The axis passes through particle P1 and is perpendicular to the disc plane. The disc’s moment of inertia about this axis, via the parallel axis theorem, is $I_{\text{disc}} = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.

For the four particles on the rim: P1 lies on the axis ($r_1 = 0$), P2 and P4 are each at distance $R\sqrt{2}$ ($I = 2mR^2$), and P3 is diametrically opposite ($r_3 = 2R$, $I = 4mR^2$). Their total moment of inertia is $I_{\text{particles}} = 0 + 2mR^2 + 4mR^2 + 2mR^2 = 8mR^2$.

The system’s total moment of inertia is therefore

$I_{\text{total}} = \frac{3}{2}MR^2 + 8mR^2 = \frac{(3M + 16m)R^2}{2}$

This corresponds to option B.