NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
From a circular disc of radius R and mass 9 M , a small disc of radius R 3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is
Options
- A4 M R 2
- B40 9 M R 2
- C10 M R 2
- D37 9 M R 2
Correct answer
A. 4 M R 2
Step-by-step solution
I r e m a i n i n g = I w h o l e - I r e m o v e d or I = 1 2 9 M R 2 - 1 2 m R 3 2 + 1 2 m 2 R 3 2 ...(i) Here, m = 9 M π R 2 × π R 3 2 = M Substituting in Eq. (i), we have I = 4 M R 2