Question

A cylindrical vessel of height $h$ and base area $S$ is filled with water. An orifice of area $s\ll S$ is opened in the bottom of the vessel. Neglecting the viscosity of water, time taken by all the water to pour out of the vessel.

Answer 1

Let at any moment of time, water level in the vessel be $H$ then speed of flow of water through the orifice, at the moment will be
$v=\sqrt{2gH}$ (1)
In the time interval $dt$, the volume of water ejected through orifice,
$dV=svdt$ (2)
On the other hand, the volume of water in the vessel at time $t$ equals
$V=SH$ (3)
Differentiating (3) with respect to time,
$\frac{dV}{dt}=S\frac{dH}{dt}$ or $dV=SdH$ (4)
Equations (2) and (4)
$SdH=svdt$ or $dt=\frac{S}{s}\frac{dH}{\sqrt{2gH}}$, from (2)
Integrating, ${\int }_0^{t}dt=\frac{S}{s\sqrt{2h}}{\int }_h^0\frac{dh}{\sqrt{H}}$
Thus, $t=\frac{S}{s}\sqrt{\frac{2h}{g}}$