Updated with bugs fixed (see https://www.facebook.com/andy0130tw/posts/10203153062786174).
A cylinder is filled with incompressible liquid. The initial volume of the liquid is $V$, and the bottom area of the cylinder is $A$. Now we dig a hole at its bottom, which area is $a$. How much time it will take to drain all the liquid? The acceleration of gravity is $g$. (viscous force and friction are neglected)
Let the volume of the liquid in the cylinder $V$, varying with time. Its differential is (notice the sign)
$$ dV = A,dh. $$
The speed $v$ of flowed liquid can be calculated with Bernoulli's principle. Here we simply consider that energy is conserved,
$$ \begin{align}
v &= \sqrt{2gh}. \
\frac{dV}{dt} &= -av \
dV &= -a\sqrt{2gh},dt.
\end{align} $$
(Note that due to the assumption $A \gg a$, the kinetic energy of the small volume of the liquid on the top is ignored. As a result, the lost potential energy is all converted into kinetic energy of the liquid flowing out from the hole of equally mass.)
We combine these equations and thus link $h$ with $t$, solving the differential equation by separating variables,
$$ \begin{align}
A,dh &= -a\sqrt{2gh},dt \
\frac{dh}{\sqrt{h}} &= -\frac{a}{A}\sqrt{2g},dt \
\int \frac{dh}{\sqrt{h}} &= \int -\frac{a}{A}\sqrt{2g},dt \
&= \frac{a}{A}\sqrt{2g}\int -dt \
2\sqrt{h} &= \frac{a}{A}\sqrt{2g}(C-t),
\end{align} $$
it turns out that the solution is
$$ h = {\left( \frac{a}{A} \right)}^2 \frac{g{\left( C - t \right)}^2}{2}, $$
with $C$ constant.
Let the initial height of the liquid $h_0$, obviously,
$$ Ah_0 = V; $$
the time to drain the liquid $T$. The boundary conditions are therefore
$$h(0) = h_0,; h(T) = 0.$$
Substituting these conditions, we can get
$$ \begin{align}
C = T &= \frac{A}{a} \sqrt{\frac{2h_0}{g}} \
&= \frac{1}{a}\sqrt{\frac{2AV}{g}}.
\end{align} $$
And $T$ is the answer we are looking for.
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