A cylindrical water tank, $5m$ radius, $20m$ height is being drained at the bottom of the tank. The amount of water released with average speed $0.5\sqrt{h}$ ($h$ is the height of the water tank). Ask after how long the tank will run out?
Some answers are given:
a) 20 hours
b) 620 minutes
c) 400.862 minutes
d) 1404.962 minutes
My method is as follows: $(\pi*5^{2}*20)/(0.5\sqrt{20})=702.481$ minutes
I do not know where was wrong. Looking forward to your help. Thanks!
Let $V$ and $H$ be the volume and height of the tank, resp., and $T$ the time to empty it.
If $0.5\sqrt{H}$ was truly only the average rate of outflow $\frac{V}{T}$ (in $m^3min^{-1}$), your calculation would be correct.
But your answer does not occur in the multiple choice question. So, recheck the problem definition of your exercise.
If we have instead an instantaneous rate depending on the height $h$ of the water level we get $$\frac{dV(h(t))}{dt}=-0.5\sqrt{h(t)}$$
With $V(h) = \pi r^2 h$ being the volume of the water at water level $h$ we get
$$\frac{dh(t)}{dt} = -\frac 1{2\pi r^2}\sqrt h \text{ with } h(0) = H$$
Separating the variables and integrating gives
$$-2\pi r^2\int_H^0 \frac{dh}{\sqrt h} = \int_0^T dt$$
or $$4\pi r^2 \sqrt H = T \Rightarrow \boxed{T \approx 1404.96} (min)$$