A uniform chain has a mass M and length L. It is placed on a frictionless table with length l0 hanging over the edge. The chain begins to slide down. Then, the speed ν with which the end leaves the edge is given by

Decrease in gravitational potential energy = increase in kinetic energy12mv2 +-mgL2=0+-ml0Lgl02 ⇒v=gLL2-l02

A uniform chain has a mass M and length L. It is placed on a frictionless table with length l0 hanging over the edge. The chain begins to slide down. Then, the speed ν with which the end leaves the edge is given by

Decrease in gravitational potential energy = increase in kinetic energy12mv2 +-mgL2=0+-ml0Lgl02 ⇒v=gLL2-l02

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A uniform chain has a mass M and length L. It is placed on a frictionless table with length l₀ hanging over the edge. The chain begins to slide down. Then, the speed $ν$ with which the end leaves the edge is given by

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$v = \sqrt{\frac{g}{L} (L + l_{0})}$

$v = \sqrt{\frac{g}{L} (L - l_{0})}$

$v = \sqrt{2 g (L - l_{0})}$

$v = \sqrt{\frac{g}{L} (L^{2} - l_{0}^{2})}$

answer is C.

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Decrease in gravitational potential energy = increase in kinetic energy

$\frac{1}{2} m v^{2} + (- m g \frac{L}{2}) = 0 + (- \frac{m l_{0}}{L} g \frac{l_{0}}{2}) \Rightarrow v = \sqrt{\frac{g}{L} (L^{2} - l_{0}^{2})}$

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A uniform chain has a mass M and length L. It is placed on a frictionless table with length l0 hanging over the edge. The chain begins to slide down. Then, the speed ν with which the end leaves the edge is given by

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A uniform chain has a mass M and length L. It is placed on a frictionless table with length l₀ hanging over the edge. The chain begins to slide down. Then, the speed $ν$ with which the end leaves the edge is given by