A cylindrical vessel of height h and base area S is filled with water. An orifice of area `s lt lt S` is opened in the bottom of the vessel. Neglecting the viscosity of water, determine how soon all the water will pour out of the vessel.

Explanation

Let at any moment of time, water level in the vessel be H then speed of flow of water through the orifice, at that 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`
Differentiating (3) with respect to time,
`(dV)/(dt)=S(dH)/(dt)` or `dV=SdH` (4)
Eqs. (2) and (4)
`SdH=svdt` or `dt=S/s(dH)/(sqrt(2gH))` from (2)
Integrating `underset(0)overset(t)intdt=(S)/(ssqrt(2h))underset(h)overset(0)int(dh)/(sqrtH)`
Thus, `t=S/ssqrt((2h)/(g))`