A uniform chain of length l and mass m overhangs a smooth table with its one third part lying on the table. The kinetic energy of the chain as it completely slips off the table is

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$\frac{2}{9} mgl$

$\frac{4}{9} mgl$

$\frac{3}{4} mgl$

$\frac{5}{18} mgl$

answer is D.

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Detailed Solution

${(∆ K)}_{g a i n} = {(∆ U)}_{l o s s}$

$∴ K = U_{i} - U_{f} = - (\frac{2 m}{3}) (g) (\frac{1}{3}) - (- \frac{m g l}{2}) = \frac{5 m g l}{18}$

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