KCET2015PhysicsMotion in One Dimension
Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is ( I ). If the same rod is bent into a ring and its moment of inertia about its diameter is ( I^ ) then the ratio ( I I^ ) is
Options
- A( 3 2 ² )
- B( 8 3 ² )
- C( 2 3 ² )
- D( 5 3 ² )
Correct answer
C. ( 2 3 ² )
Step-by-step solution
Moment of inertia of a thin uniform rod is given as [ I= M L² 12 ] Moment of inertia of a ring is given as [ I^ = M R² 2 ] Therefore ( I I = ( M L² 12 ) ( M R² 2 ) = L² 6 R² ) Now, ( L=2 R ) Thus ( I I = (2 )² R² 6 R² = 4 ² 6 = 2 ² 3 )