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Question: Answered & Verified by Expert
Two small particles each of mass \( m \) are attached at the two ends of a semicircular ring of same mass as shown. \( R \) is radius of ring and \( O \) is the centre of the ring. Moment of inertia about an axis passing through \( O \) and normal to its plane is
PhysicsCenter of Mass Momentum and CollisionJEE Main
Options:
  • A \( 3 m R^{2} \)
  • B \( \frac{3 m R^{2}}{2} \)
  • C \( 2 m R^{2} \)
  • D \( \frac{m R^{2}}{2} \)
Solution:
1636 Upvotes Verified Answer
The correct answer is: \( 3 m R^{2} \)
Here ring is half but its mass is m so its moment of inertia will me mR2
I=mR2 + mR2 + mR2
Inet=mR2+2mR2
I=3mR2
 

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