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A mass m moves in a circle on a smooth horizontal plane with velocity v₀ at a radius R₀. The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R₀/2. The final value of the kinetic energy is:

1. 1/4 mv₀²
2. 2mv₀²
3. 1/2 mv₀²
4. mv₀²

Correct answer: 2mv₀²

Solution

As the radius decreases, angular momentum is conserved. Initial angular momentum is m*v₀*R₀, and final angular momentum is m*v_f*(R₀/2). Equating these gives v_f = 2v₀. Kinetic energy is (1/2)mv², so the final kinetic energy is (1/2)m*(2v₀)² = 2mv₀².

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