A metallic ring of mass 1 kg has moment of inertia 1 kg m^{2 }when rotating about one of its diameters. It is molten and remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own axis

The MI of the thin ring about its diameter,

\(Ir = \frac{1}{2}MR^2 = 1 kg\ m^2\)

Since the ring is melted and recast into a thin disc of same radius R, the mass of the disc equals the mass of the ring = M. The MI of the thin disc about its own axis (i.e. transverse symmetry axis) is

\(I_d= \frac{1}{2}MR^2 = I_r\)

\(∴ I_d = 1 kg\ m^2\)