Question

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

Answer 1

moment of inertia of ring about axis passing through its centre and perpendicular to its plane is given by 
                  I = mr²
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 × (0.5)² = 0.25 × 0.25 = 0.0625 Kg.m²