A small hole is cut in the bottom of a water storage tank. The depth of the water is $12.00 \mathrm{~m}$. If the diameter of the small hole in the bottom of the tank is $4.00 \mathrm{~mm}$, then what is the flow rate of the water leaving the tank?
A) $1.36 \times 10^{-4} \mathrm{~m} 3 / \mathrm{s}$
B) $1.93 \times 10^{-4} \mathrm{~m} 3 / \mathrm{s}$
C) $2.65 \times 10^{-4} \mathrm{~m} 3 / \mathrm{s}$
D) $2.53 \times 10^{-4} \mathrm{~m} 3 / \mathrm{s}$
E) $1.05 \times 10^{-4} \mathrm{~m} 3 / \mathrm{s}$

Question

Quizplus · Accepted Answer

The flow rate of water leaving the tank can be calculated using Torricelli's Law, which states that the speed of efflux, $v$, is given by $v = \sqrt{2gh}$, where $g$ is the acceleration due to gravity ($9.81 \, \text{m/s}^2$) and $h$ is the height of the water column above the hole ($12.00 \, \text{m}$). The flow rate, $Q$, is then $Q = Av$, where $A$ is the cross-sectional area of the hole. Given the diameter of the hole is $4.00 \, \text{mm}$ or $0.004 \, \text{m}$, the area $A = \pi r^2 = \pi (0.002)^2 = 1.25664 \times 10^{-5} \, \text{m}^2$. Calculating $v = \sqrt{2 \times 9.81 \times 12.00} \approx 15.34 \, \text{m/s}$, and thus $Q = 1.25664 \times 10^{-5} \times 15.34 \approx 1.93 \times 10^{-4} \, \text{m}^3/\text{s}$.

A Small Hole Is Cut in the Bottom of a Water