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Physics Question on mechanical properties of fluid

A closed water tank has cross-sectional area A. It has small hole at a depth of h from the free surface of water. The radius of the hole is r so that r << Aπ\sqrt{\frac{A}{\pi}}. If Po is the pressure inside the tank above water level, and Pa is the atmospheric pressure, the rate of flow of the water coming out of the hole is [ ρ is the density of water]
A closed water tank

A

πr22gh+2(Po−Pa)ρ\pi r^2\sqrt{2gh+\frac{2(P_o-P_a)}{\rho}}

B

πr22gH\pi r^2\sqrt{2gH}

C

πr2gh+2(Po−Pa)ρ\pi r^2\sqrt{gh+\frac{2(P_o-P_a)}{\rho}}

D

πr22gh\pi r^2\sqrt{2gh}

Answer

πr22gh+2(Po−Pa)ρ\pi r^2\sqrt{2gh+\frac{2(P_o-P_a)}{\rho}}

Explanation

Solution

The correct answer is (A) : πr22gh+2(Po−Pa)ρ\pi r^2\sqrt{2gh+\frac{2(P_o-P_a)}{\rho}}.