A water tank has a shape of inverted right circular cone whose semi-vertical angle is \(\tan ^{-1}\left(\frac{1}{2}\right)\). Water is poured into it at constant rate of 5 cubic meter/minute. The rate in meter/minute at which level of water is rising. at the instant when depth of water in the tank is. \(10 \mathrm{~m}\) is

Semi-vertical angle \(=\tan ^{-1}\left(\frac{1}{2}\right)\)
Let \(\alpha=\tan ^{-1}\left(\frac{1}{2}\right)\)
\(\tan \alpha=\frac{1}{2}\)
\(\frac{\mathrm{r}}{\mathrm{h}}=\frac{1}{2}\)
\(\mathrm{r}=\frac{\mathrm{h}}{2}\)
Given, \(\frac{d \mathrm{~V}}{\mathrm{dt}}=5 \mathrm{~m}^3 / \mathrm{min}\).
\(\mathrm{V}=\) Volume of cone
Volume of cone \(=\frac{1}{3} \pi r^2 \mathrm{~h}\)
\(\mathrm{V}=\frac{1}{3} \pi\left(\frac{\mathrm{h}}{2}\right)^2 \times \mathrm{h}\)
\(\mathrm{V}=\frac{1}{12} \pi \mathrm{h}^3\)
Differentiating w, r.t. t, we get
\(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{1}{12} \times \pi \times 3 \mathrm{~h}^2 \times \frac{\mathrm{dh}}{\mathrm{dt}}\)
\(5=\frac{1}{4} \pi h^2 \frac{\mathrm{dh}}{\mathrm{dt}}\)
\(\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{20}{\pi \mathrm{h}^2}\)
Now, \(\mathrm{h}=10\)... [Given]
\(\therefore \quad \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{20}{\pi \times(10)^2}\)
\(\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{1}{5 \pi}\)
\(\therefore \quad\) Rate of change of water level is \(\frac{1}{5 \pi} \mathrm{m} / \mathrm{min}\).