Water drains from a filled cylindrical vessel of base

Water drains from a filled cylindrical vessel of base area $S=420\ \text{cm}^2$ and height $H=40$ cm through a hole of area $s=2\ \text{cm}^2$ at the bottom. In what time will the vessel empty?

Step 1: Torricelli's law and the flow rate

\[v=\sqrt{2gh},\qquad Q=sv=s\sqrt{2gh}.\]

Step 2: Conservation of volume

As the level drops, $-S\,dh/dt$ equals the outflow rate:

\[S\frac{dh}{dt}=-s\sqrt{2gh}.\]

Step 3: Separate variables

\[h^{-1/2}\,dh=-\frac{s}{S}\sqrt{2g}\,dt.\]

Step 4: Integrate from full to empty

\[\int_H^0h^{-1/2}\,dh=-\frac{s}{S}\sqrt{2g}\int_0^Tdt \quad\Longrightarrow\quad -2\sqrt{H}=-\frac{s}{S}\sqrt{2g}\,T.\]

Step 5: Solve for $T$

\[T=\frac{2S\sqrt{H}}{s\sqrt{2g}}=\frac{S}{s}\sqrt{\frac{2H}{g}}.\]

\[\boxed{T=\frac{S}{s}\sqrt{\frac{2H}{g}}}\]

Step 6: Numerical value

With $S=0.042\ \text{m}^2$, $s=0.0002\ \text{m}^2$, $H=0.4$ m, $g=9.8\ \text{m/s}^2$:

\[T=\frac{0.042}{0.0002}\sqrt{\frac{0.8}{9.8}}=210\times0.2857\approx60.0\ \text{s}.\]

\[\boxed{T\approx60\ \text{s}=1\ \text{minute}}\]

Step 7: Physical interpretation

The drain time is proportional to $S/s$ (larger tank or smaller hole drains more slowly) and to $\sqrt{H}$, not $H$ itself — because the outflow speed drops as the level falls, doubling the initial depth only multiplies the drain time by $\sqrt2\approx1.41$, not by $2$.