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A thin ring of mass M and radius R is spinning about its symmetry axis with angular speed ω. Two identical small masses, each of mass m, are then lightly fixed at the two opposite ends of a diameter of the ring. The decrease in kinetic energy is

1. (a) (M+2m)/M × ω² R²
2. (b) Mm/(M+2m) × ω² R²
3. (c) Mm/(M+m) × ω² R²
4. (d) (M+m)/(M+2m) × ω² R²

Correct answer: (b) Mm/(M+2m) × ω² R²

Solution

The correct option reflects the change in the moment of inertia of the system when the two small masses are added to the ring, which affects the total kinetic energy. The decrease in kinetic energy is calculated by considering the initial and final moments of inertia and their corresponding angular velocities, leading to the expression given in option (b).

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