Find the time required for a cylindrical tank of radius 2.5 m and height 3 m to empty through a round hole of 2.5 cm with a velocity `2. 5sqrt(h)` m/s, `h` being the depth of the water in the tank.

Explanation

Let us allow the water to flow for time dt.
We suppose that in this time, the height of the water level reduces by dh. Therefore,
`pi(2.5)^(2)dh=-2.5sqrt(h)pi(0.025)^(2)dt`
or `(dh)/(dt)=-2.5 xx 10^(-4)sqrt(h)` (negative sings denotes that the rate of flow will decrease as t increases)
`int(dh)/sqrt(h) = -2.5 xx 10^(-4)int dt` or `2sqrt(h)=-2.5 xx 10^(-4)t+c`
At t=0, h=3, or `c=2sqrt(3)`
Hence, for `h=0, t=(2sqrt(3))/(2.5 xx 10^(-4))=8000sqrt(3)`s.