NEETPhysicsRotational Motion
A uniform circular ring and a uniform circular disc, both having the same original mass M and radius R , have a sector corresponding to an angle of 90^ removed from them. Match the objects in List-I with their respective moments of inertia about an axis passing through their centre and perpendicular to their plane in List-II. List-I List-II (A) Remaining part of the ring (I) 3 4 M R^2 (B) Remaining part of the disc (
Options
- A(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
- B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- C(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
- D(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Correct answer
D. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Step-by-step solution
When a 90^ sector is removed, the fraction of mass removed is 90^ 360^ = 1 4 . The fraction of mass remaining is 3 4 . For a ring, the moment of inertia of any sector about the central axis is I = m R^2 , where m is the mass of that specific sector. (A) Remaining part of the ring: mass m = 3 4 M , so I = 3 4 M R^2 (Matches I). (C) Removed sector of the ring: mass m = 1 4 M , so I = 1 4 M R^2 (Matches III). For a disc, the moment of inertia of any sector about the central axis is I = 1 2 m R^2 . (B) Remaining part o