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, $\mathrm{V}=75 \pi$
$$
\begin{array}{ll}
\therefore & \pi \mathrm{t}=75 \pi \\
\therefore & \mathrm{t}=75 \text { seconds. }
\end{array}
$$