MHT CET202015 Oct 2020Evening ShiftPhysicsRotational MotionActual
From a disc of mass 'M' and radius 'R' a circular hole of diameter R is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is
Options
- A11 MR ² 32
- B7 MR ² 32
- C9 MR ² 32
- D13 MR ² 32
Correct answer
D. 13 MR ² 32
Step-by-step solution
( I _ T otal disc = MR ² 2 ) As mass is proportional to area, ( M _ Removed = M 4 ) Now, about the same perpendicular axis: (I_ Removed = M 4 (R / 2)² 2 + M 4 ( R 2 )²= 3 MR ² 32 ) ( I _ Remaining Disc = I _ Total - I _ Removed ) (= MR ² 2 - 3 MR ² 32 ) (= 13 MR ² 32 )