It’s differentiation time!

This related rates problem can be expressed as:

$\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh}.$

If the water has radius $r$ and height $h$ at any given moment, then the following relationship holds:

$\frac{r}{h} = \frac{2.5}{17} \Rightarrow r = \frac{5h}{34}$

and the conical water volume at that same moment computes to:

$V(h) = \frac{\pi}{3} \cdot r^{2}h = \frac{\pi}{3} \cdot (\frac{5h}{34})^{2}h = \frac{25\pi}{3468} \cdot h^{3}$

and $\frac{dV}{dh} = \frac{25\pi}{1156} \cdot h^{2}.$

So now the water level rate (at $h = 4$ ) is computed as:

$\frac{dh}{dt} = \frac{2}{\frac{25\pi}{1156} \cdot 4^{2}} \approx \boxed{1.84}$ m/min.