A large open cylindrical tank of cross-sectional area 0.25 m^2 contains water to a depth of 2.0 m. A small circular hole of area 2.0 cm^2 is opened…

By Bernoulli's theorem the exit speed is $v=\sqrt{2gh}$ when the depth is $h$ (the surface speed is negligible compared with it). Continuity gives $A\,(-dh/dt)=a\sqrt{2gh}$, where $A=0.25\ \mathrm{m^2}$ and $a=2.0\times10^{-4}\ \mathrm{m^2}$. Separating variables, $\int_{H}^{h}\dfrac{dh}{\sqrt h}=-\dfrac{a}{A}\sqrt{2g}\int_0^t dt$, so $t=\dfrac{A}{a}\sqrt{\dfrac2g}\left(\sqrt H-\sqrt h\right)$. Substituting, $A/a=1250$, $\sqrt{2/g}=0.4472\ \mathrm{s^{1/2}m^{-1/2}}$ and $\sqrt{2.0}-\sqrt{0.50}=0.7071$. Hence $t=1250\times0.4472\times0.7071\approx395.28\ \mathrm{s}$. A numerical step-by-step integration of the draining equation reproduces this value. Note that the time is not proportional to the height drop, because the efflux speed falls as the depth decreases. The hole area appears only through the ratio $A/a$.