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Physics Question on mechanical properties of fluid

A cylindrical vessel contains a liquid of density ρ\rho 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 with which the liquid comes out of the vessel is

A

2gh\sqrt{2gh}

B

2(gh+mgρA)\sqrt{2\left(gh+\frac{mg}{\rho A}\right)}

C

2(gh+mgA)\sqrt{2\left(gh+\frac{mg}{A}\right)}

D

2gh+mgA\sqrt{2gh+\frac{mg}{A}}

Answer

2(gh+mgρA)\sqrt{2\left(gh+\frac{mg}{\rho A}\right)}

Explanation

Solution

Difference in pressure energy = K.E of the liquid ρgh+mgA=12ρν2;V=2(gh+mgρA)\rho gh + \frac{mg}{A} = \frac{1}{2} \rho \nu^2 ; V = \sqrt{2 \left( gh + \frac{mg}{\rho A} \right) }