A chain AB of length l is lying in a smooth horizontal tube so that a fraction h of its length l, hangs freely and touches the surface of the table with its end B. At a certain moment, the end A of the chain is set free. The velocity of end A of the chain, when it slips out of tube, is

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h2glh

2ghloge(lh)

2glloge(lh)

2gh

answer is B.

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Detailed Solution

Let at any instant the length of the chain inside the tube be x. Thena=(MLh)gML(h+x) ⇒a=hgh+x⇒−vdvdx=ghh+x⇒∫0vv dv=−gh∫ℓ−h0dxh+x ⇒v22=−ghloge(h+x)|ℓ−h0 ⇒v2=−gh[logeh−logeℓ] ⇒v2=2ghloge(ℓh)⇒v=2ghloge(ℓh)

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