Two discs A and B of same material and thickness have radii R and 3 R respectively. Their moments of inertia about their axis will be in the ratio

Step-by-step solution

Key concept: The moment of inertia of a homogeneous disc about its central axis is I = 1 2 M R^2 , where mass scales proportionally to the square of the radius for constant thickness and density. Given R_B = 3R_A , it follows from M R^2 that M_B = 9M_A . Then for disc A: I_A = 1 2 M_A R_A^2 For disc B: I_B = 1 2 (9M_A) (3R_A)^2 = 1 2 (9M_A)(9R_A^2) = 81 2 M_A R_A^2 The ratio is I_A I_B = 1 2 M_A R_A^2 81 2 M_A R_A^2 = 1 81 Final answer: C