Question

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 << \(\sqrt{\frac{A}{\pi}}\). If P_o is the pressure inside the tank above water level, and P_a is the atmospheric pressure, the rate of flow of the water coming out of the hole is [ ρ is the density of water]

• A. \(\pi r^2\sqrt{2gh+\frac{2(P_o-P_a)}{\rho}}\)
• B. \(\pi r^2\sqrt{2gH}\)
• C. \(\pi r^2\sqrt{gh+\frac{2(P_o-P_a)}{\rho}}\)
• D. \(\pi r^2\sqrt{2gh}\)

Answer 1

The Correct Option is A

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