A cylindrical vessel contains a liquid of density ρ up to 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:

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2gh

2(gh+mgρA)

2(gh+mgA)

2gh+mgA

answer is B.

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Detailed Solution

Applying Bernoulli’s theorem at point 1 and 2, Difference in pressure energy between 1 and 2 = difference in kinetic energy between 1 and 2or ρhg+mgA=12ρv2or v=2gh+2mgρA=2(gh+mgρA)

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