A disc of radius R and thickness R 6 has a moment of inertia I about an axis passing through its centre and perpendicular to its plane. The disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is

Step-by-step solution

Volume of disc is A.d = π . R 2 × R 6 = R 3 × π 6 Moment of inertia of disc is I = 1 2 M R 2 When the disk is remolded in solid sphere of volume V having radius r, then π R 3 6 = 4 3 × π r 3 ⇒ R 3 6 × 3 4 = r 3 ∴ r 3 = R 3 8 ⇒ r = R 2 Moment of inertia of sphere is given by 2 5 M . r 2 = 2 5 × M . R 2 4 = M R 2 10 = M R 2 2 × 1 5 = I 5