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A small ring of mass m can slide on a smooth rod. The ring is attached to a particle of mass m by a string of length l. A horizontal velocity $v_{0} = \sqrt{2 g l}$ is given to the ring

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The maximum angle that the string will make in subsequent motion is $30 º$ with vertical

The maximum angle that the string will make in subsequent motion is $60 º$ with vertical

Velocity of ring when string makes maximum angle with vertical is $\frac{\sqrt{g l}}{2}$

Velocity of particle when string makes maximum angle with vertical is $\sqrt{\frac{g l}{2}}$

answer is A, D.

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Detailed Solution

At maximum angle of string with vertical, velocity of ring and particle will be equal using conservation of momentum $m v_{0} = (m + m) v \Rightarrow v = \frac{v_{0}}{2} = \frac{\sqrt{2 g l}}{2} = \sqrt{\frac{g l}{2}}$

Using conservation of energy

$\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v^{2} + \frac{1}{3} m v^{2} + m g l (1 - \cos θ)$

$\cos θ = \frac{1}{2} \Rightarrow θ = 60 º$

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