MHT CET20242 May 2024Evening ShiftPhysicsRotational MotionActual
A solid sphere of mass ' m ', radius ' R ', having moment of inertia about an axis passing through center of mass as ' I ' is recast into a disc of thickness ' t ' whose moment of inertia about an axis passing through the rim (edge) & perpendicular to plane remains ' I '. Then the radius of disc is
Options
- A2 R 15
- B( 2 15 ) R
- C4 R 15
- DR 4
Correct answer
A. 2 R 15
Step-by-step solution
The moment of inertia of a solid sphere = 2 5 MR ^2 Moment of inertia of a disc through its rim, = 1 2 MR ^2+ MR ^2= 3 2 MR ^2 Since both the moment of inertias are equal, 2 5 MR ^2= 3 2 Mr ^2 , where r is the radius of the disc r= 2 R 15