91Ó°ÊÓ

• The diameter of the hole is d = 6 mm .
• The depth below the water level tank is h = 14 m.

In this problem, the speed of efflux is the speed of flowing water out of the hole, and it can be calculated by using the relation of speed from Bernoulli’s equation.

The relation of speed can be written as:

$v=\sqrt{2gh}$

Here, g is the gravitational acceleration.

Substitute$9.80m/s^2$ for g and 14 m for h in the above relation.

$$\begin{gathered} v=\sqrt{2\left(9.80m/s^2\right)\left(14m\right)} \\ v=16.56m/s \end{gathered}$$

Thus, the speed of efflux is 16.56 m/s .

The relation of volume discharged per unit time can be written as:

$$\begin{gathered} \frac{dV}{dt}=A×v \\ \frac{dV}{dt}=\frac{{\mathrm{Ï€»å}}^2}{4}×v \end{gathered}$$

Here, A is the area.

Substitute 6 nm for d and 15.56m/s for v in the above relation.

role="math" localid="1668141127354" $$\begin{gathered} \frac{dV}{dt}=\frac{\mathrm{π}{\left(6\mathrm{mm}×{\displaystyle \frac{1\mathrm{m}}{1000\mathrm{mm}}}\right)}^2}{4}×\left(16.56m/s\right) \\ \frac{dV}{dt}=4.67×{10}^{-4}{m}^3/s \end{gathered}$$

Thus, the discharged volume per second is$4.67×{10}^{-4}{m}^3/s$ .