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Question

# A cylindrical vessel contains a liquid of density ρ upto a height h. The liquid is closed by a piston of mass m and area of cross-section A. There is a small hole at the bottom of the vessel. The speed v with which the liquid comes out of the hole is

A
2gh
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B
2(gh+mgρA)
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C
2(gh+mgA)
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D
2gh+mgA
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Solution

## The correct option is B √2(gh+mgρA) Let atmospheric pressure be Po. Pressure at point 1=Po+mgA+ρgh Pressure at point 2= Atmospheric pressure =Po Applying Bernoulli's theorem at point 1 and 2, we get (Po+ρgh)+mgA+ρgh1=Po+12ρv2+ρgh2 (Since h1=h2) ⇒v=√2(gh+mgρA) m/s

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