A large cylindrical tank has a horizontal cross-sectional area 20 times that of a small circular hole in its side wall. The hole is 0.20 m above the…

Let $h$ be the height of the surface above the hole, initially $h_0=1.8\ \mathrm{m}$ (the hole is $0.20\ \mathrm{m}$ above the base, so the base height is irrelevant). Torricelli's theorem gives the efflux speed $\sqrt{2gh}$ (the speed of the falling surface is neglected, as instructed). Continuity requires $A\left(-\dfrac{dh}{dt}\right)=a\sqrt{2gh}$, i.e. $\dfrac{dh}{\sqrt h}=-\dfrac{a}{A}\sqrt{2g}\,dt$. Integrating from $h_0$ to $h_0/4$ gives $2\left(\sqrt{h_0}-\sqrt{h_0/4}\right)=\sqrt{h_0}=\dfrac{a}{A}\sqrt{2g}\,t$. Hence $t=\dfrac{A}{a}\cdot\dfrac12\sqrt{\dfrac{2h_0}{g}}=20\times\dfrac12\times0.6=6\ \mathrm{s}$. A numerical integration of the differential equation confirms $6.00\ \mathrm{s}$. Physically, a shallow tank with a small hole drains slowly, and the time grows with the area ratio and with the square root of the head, which is why the head above the hole, not above the base, matters.