Let r be the radius of the part of the cone filled with water and let h be the height. Then,
[tex]\begin{gathered} \frac{h}{7}=\frac{r}{4} \\ h=\frac{7r}{4} \end{gathered}[/tex]
In ten minutes the volume, V, of the part of the cone filled with water is given by
[tex]V=0.065\times10=0.65m^3[/tex]
Therefore,
[tex]\begin{gathered} \frac{1}{3}\pi\times r^2\times(\frac{7r}{4})=0.65 \\ r_{10m}=0.354688m \\ h_{10m}=0.620704m \end{gathered}[/tex]
We are given that, the rate at which the cone is being filled is given by
[tex]\frac{dV}{dt}=0.065m^3\text{ /min}[/tex][tex]\begin{gathered} V=\frac{1}{3}\pi r^2h \\ But\text{ }r=\frac{4h}{7} \\ \text{therefore} \\ V=\frac{16\pi}{147}h^3 \\ \frac{dV}{dt}=\frac{16\pi}{49}\frac{h^2dh}{\text{ dt}} \\ \frac{dh}{dt}=0.065\times\frac{49}{16\pi(0.620704)^2}=0.164464m\text{ /min} \end{gathered}[/tex]
Hence, the rate at which the water level is rising is 0.164 meters per minute
