Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' I ₁ '. The same rod is bent into a ring and its moment of inertia about the diameter is ' I₂ '. Then I₁ / I₂ is

Step-by-step solution

The moment of inertia for a rod of mass M and length L about its center perpendicular axis is I₁ = 1 12 M L^2 . When bent into a ring, the circumference becomes L = 2 R , giving radius R = L 2 . The moment of inertia for a ring about its diameter is I₂ = 1 2 M R^2 = 1 2 M ( L 2 )^2 = M L^2 8 ^2 . The ratio is I₁ I₂ = 1 12 M L^2 1 8 ^2 M L^2 = 8 ^2 12 = 2 ^2 3 , matching option B .