A light rod of length $l$ is free to rotate about one of its ends in vertical plane. A ring of mass m can slide freely without friction on the rod. Initially rod is kept horizontal and ring is situated at a distance $3 l / 5$ from fixed end. Now, rod is released choose correct statement(s).

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Angular momentum of the ring about point of suspension of rod when it looses contact with rod is $m \sqrt{\frac{72 g l^{3}}{125}}$

Acceleration of the ring is uniform

Angular velocity of the rod remains constant through out the motion.

Velocity of ring when it leaves contact with rod is $\sqrt{\frac{8 g l}{5}}$

answer is A, B, D.

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Detailed Solution

Three will be no normal reaction between rod and ring. Because rod is massless. The motion of ring will be free fall under gravity.
Question Image
Velocity of ring when it leaves contact with rod
$= \sqrt{2 \times g \times \frac{4 l}{5}} = \sqrt{\frac{8 g l}{5}}$
Angular momentum
$= m \frac{3 l}{5} \sqrt{\frac{8 g l}{5}} = m \sqrt{\frac{72 g l^{3}}{125}}$