JEE Main201910 Apr 2019Evening ShiftPhysicsRotational MotionActual
A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7 M 8 and is converted into uniform disc of radius 2 R . The second part is converted into a uniform solid sphere. Let I 1 be the moment of inertia of the disc about its axis and I 2 be the moment of inertia of the new sphere about its axis. The ratio I 1 / I 2 is given by:
Options
- A140
- B185
- C65
- D285
Correct answer
A. 140
Step-by-step solution
Moment of inertia of disc about the axis passing its centre and perpendicular to its surface, I 1 = 7 M 8 2 R 2 2 = 7 M R 2 4 . Radius of small sphere r is related as M 4 3 π R 3 = M 8 4 3 π r 3 r = R 2 Moment of inertia of sphere about its axis passing through its centre ∴ I 2 = 2 5 M 8 R 2 2 = M R 2 80 I 1 I 2 = 7 4 × 80 1 = 140