A closed water tank has a cross-sectional area $A$. It has a small hole at a depth of $h$ from the free surface of water. The radius of the hole is $r$, so that $r \ll \sqrt{\frac{A}{\pi}}$. If $P_0$ is the pressure inside the tank above the water level, and $P_{a}$ is the atmospheric pressure, the rate of flow of the water coming out of the hole is:

Question

$\pi r^2 \sqrt{2gh}$

$\pi r^2 \sqrt{gh + \frac{2(P_0 - P_a)}{\rho}}$

$\pi r^2 \sqrt{2gh}$

$\pi r^2 \sqrt{2gh + \frac{2(P_0 - P_a)}{\rho}}$

Solution

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