AP EAPCET 25-Aug-2021 Shift 2 Paper

From the perpendicular axis theorem, we have $I_{2}=I_{x}+I_{y}$ where, $I_{2}$ is moment of inertia about Z-axis, $I_{x}$ is moment of inertia about X-axis and $I_{y}$ is moment of inertia about Y-axis. Moment of inertia of a disc, $I_{x}=\frac{1}{2} M R^{2}$ Now, $I_{x}=I_{y}$, because two perpendicular diameters are equivalent. So, $I_{\mathrm{diameter}}=\frac{1}{2} \times \frac{1}{2} M R^{2} = \frac{M R^{2}}{4}$ ⇒ $\frac{M R^{2}}{4}=25$ (given) ⇒ $M=\frac{100}{R^{2}}\ \mathrm{kg}$ Also every diameter passes through centre of mass. So, $I_{\mathrm{COM}}=\frac{M R^{2}}{4}=25\ \mathrm{kg}\cdot\mathrm{m}^{2}$ Using parallel axis theorem, $I=I_{\mathrm{COM}}+M x^{2}$ (where, $.x=\frac{R}{2}$) $=25+M\frac{R^{2}}{4}$ $=25+\frac{100}{R^{2}} \times \frac{R^{2}}{4}$ $=50\ \mathrm{kg}\cdot\mathrm{m}^{2}$