A chain of mass M and length L is held on a frictionless table with a fraction 1/n of its length hanging over the edge. If the chain is released, the velocity of the chain when it just leaves the table is:

A. gL(1 - 1/n^2) — The loss in potential energy of the hanging part equals the gain in kinetic energy of the whole chain.

JEE Main · PhysicsHard

$A chain of mass and length is held on a frictionless table with a fraction of its length hanging over the edge. If the chain is released, the velocity of the chain when it just leaves the table is:$

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Correct answer: A — $g L (1 - 1/ n^{2})$

$Initial potential energy of the hanging part of mass M / n is U_{i} = - (M / n) g (L /2 n) .$
$Final potential energy when the whole chain hangs is U_{f} = - M g (L /2) .$
$By conservation of energy, Δ K = - Δ U, so \frac{1}{2} M v^{2} = - [- (M g L /2) - (- (M / n) g (L /2 n))] .$
$Simplifying gives v^{2} = g L (1 - 1/ n^{2}) .$
$Hence the answer is (A).$