A right circular cone has height \(9 \mathrm{~cm}\) and radius of base \(5 \mathrm{~cm}\). It is inverted and water is poured into it. If at any instant, the water level rises at the rate \(\frac{\pi}{\mathrm{A}} \mathrm{cm} / \mathrm{sec}\). where \(\mathrm{A}\) is area of the water surface at that instant, then cone is completely filled in

For the conical vessel, \(\mathrm{h}=9 \mathrm{~cm}, \mathrm{r}=5 \mathrm{~cm}\)
\(\therefore \quad\) Full volume of the vessel,
\(
\begin{aligned}
\mathrm{V} & =\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\
& =\frac{1}{3} \pi \times 25 \times 9 \\
& =75 \pi \mathrm{cm}^3
\end{aligned}
\)
Now, \(\frac{\mathrm{h}}{\mathrm{r}}=\frac{9}{5}\)
\(
\begin{aligned}
& \therefore \quad \mathrm{r}=\frac{5 \mathrm{~h}}{9} \\
& \therefore \quad \mathrm{A}=\pi \mathrm{r}^2=\pi \frac{25 \mathrm{~h}^2}{81}
\end{aligned}
\)
According to the given condition, \(\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{\pi}{\mathrm{A}}=\pi \frac{81}{\pi 25 \mathrm{~h}^2}=\frac{81}{25 \mathrm{~h}^2}\)
\(\therefore \quad h^2 \mathrm{dh}=\frac{81}{25} \mathrm{dt}\)
Integrating on both sides, we get
\(
\begin{aligned}
& \frac{\mathrm{h}^3}{3}=\frac{81}{25} \mathrm{t}+\mathrm{c}_1 \\
\therefore \quad & \mathrm{h}^3=\frac{243}{25} \mathrm{t}+\mathrm{c}, \text { where } \mathrm{c}=3 \mathrm{c}_1
\end{aligned}
\)
Naturally, \(\mathrm{h}=0\), when \(\mathrm{t}=0\) and hence, \(\mathrm{c}=0\)
\(
\begin{aligned}
\therefore \quad \mathrm{h}^3 & =\frac{243}{25} \mathrm{t} \\
\therefore \quad \mathrm{V} & =\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\
& =\frac{1}{3} \pi \frac{25 \mathrm{~h}^2}{81} \mathrm{~h} \\
& =\frac{25}{243} \pi \mathrm{h}^3 \\
& =\frac{25}{243} \pi \frac{243}{25} \mathrm{t} \\
\therefore \quad \mathrm{V} & =\pi \mathrm{t}
\end{aligned}
\)
But volume of vessel, \(\mathrm{V}=75 \pi\)
\(
\begin{array}{ll}
\therefore & \pi \mathrm{t}=75 \pi \\
\therefore & \mathrm{t}=75 \text { seconds. }
\end{array}
\)