A thin ring has mass 0.25 kg and radius 0.5 m. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is _______.

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A thin ring has mass 0.25 kg and radius 0.5 m. Its M.I. about an axis passing through its centre and perpendicular to its plane is 0. 0625 kg m2. I = m$r^{2}$ where m is the mass of ring and r is the minimum seperation between axis of rotation and point of observation { e.g., radius of ring } here, m = 0.25 Kg r = 0.5 m now, I = 0.25 $\times$ (0.5$)^{2}$ = 0.25 $\times$ 0.25 = 0.0625 Kg.$m^{2}$

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