665. A rectangular reservoir with a horizontal cross-sectional area of $S=6 \mathrm{~m}^{2}$ is filled with water to a height of $H=5$ m. Determine the time it takes for all the water to flow out of the reservoir through a small hole in its bottom with an area of $s=0.01 \mu^{2}$, assuming that the velocity of the water flow is $0.6 \sqrt{2 \text { g } h}$, where $h$ is the height of the water level above the hole, and $g$ is the acceleration due to gravity.
Solution. According to the general scheme (1), divide the desired time $T$ into a large number $n$ of small intervals $\Delta t_{1}, \Delta t_{2}, \ldots, \Delta t_{n}$, and let the water level in the reservoir decrease by an amount $\Delta x=\frac{H}{n}$ during each such interval (Fig. 124).[^20]

If we assume that during each small time interval $\Delta t_{i}$ the speed of water flowing out through the hole at the bottom remains constant, equal to its value at the beginning of the interval $0.6 \sqrt{2 \mathrm{~g}\left(H-x_{i}\right)}$, then, equating the volume of water that flows out at this speed through the hole at the bottom over the interval $\Delta t_{i}$ to the volume of the part of the reservoir that has emptied over the same interval, we obtain the approximate equality

$$
0.6 s \sqrt{2 g\left(H-x_{i}\right)} \Delta t_{i} \approx S \Delta x
$$

from which

$$
\Delta t_{i} \approx \frac{S \Delta x}{0.6 s \sqrt{2 g\left(H-x_{i}\right)}}
$$

The approximate value of the total desired time $T$ will be equal to the sum

$$
T=\sum_{i=1}^{n} \Delta t_{i} \approx \sum_{i=1}^{n} \frac{S \Delta x}{0.6 s \sqrt{2 g\left(H-x_{i}\right)}}
$$

where, according to the problem statement, the points $x_{i}$ are contained in the interval $[0, H]$.

Having verified that the error in the obtained approximate value of $T$ tends to zero as $n$ increases, we find the exact value of $T$ as the limit of the integral sum (*) as $n \rightarrow+\infty$, i.e., as the corresponding definite integral

$$
T=\frac{S}{0.6 s \sqrt{2 g}} \int_{0}^{H}(H-x)^{-\frac{1}{2}} d x=\left.\frac{2 S}{0.6 s \sqrt{2 g}}(H-x)^{\frac{1}{2}}\right|_{H} ^{0}=\frac{S}{0.6 s} \sqrt{\frac{2 \overline{I}}{G}}
$$

Substituting the numerical values of the parameters, we get $T \approx$ $\approx 1010$ sec $\approx 16.83$ min.

If the decrease in water in the reservoir were constantly replenished, i.e., if the water level in it remained constant, then the speed of water flowing out would also be constant, equal to $0.6 \sqrt{2 \mathrm{~g} H}$. In this case, in each second, the volume of water flowing out through the hole at the bottom of the reservoir would be $0.6 s \sqrt{2 g H}$, equal to the volume of a right cylinder with base area $s$ and height $0.6 \sqrt{2 g H}$. Therefore, under the given assumption, the volume of water contained in the reservoir would flow out of it in time

$$
T_{1}=\frac{S H}{0.6 s \sqrt{2 g H}}=\frac{1}{2} \frac{S}{0.6 s} \sqrt{\frac{2 H}{g}} .
$$

Comparing this result with the previous one shows that the time of outflow $T$, without replenishing the decrease in water in the reservoir, is twice the time of outflow $T_{1}$, when the decrease in water is constantly replenished; $T=2 T_{1}$.