Water is running in a hemispherical bowl of radius 180 cm at the rate of 108 cubic decimeters per minute. How fast the water level is rising when depth of the water level in the bowl is 120 cm ? ( 1 decimeter \(=10 \mathrm{~cm}\) )

Radius of hemispherical bowl \((\mathrm{r})=180 \mathrm{~cm}\)
Rate of flow
\(\begin{aligned}
\left(\frac{\mathrm{dV}}{\mathrm{dt}}\right) & =108 \mathrm{dm}^3 / \mathrm{min} \\
& =108000 \mathrm{~cm}^3 / \mathrm{min} \\
& =\frac{108000}{60} \mathrm{~cm}^3 / \mathrm{sec} \\
& =1800 \mathrm{~cm}^3 / \mathrm{sec}
\end{aligned}\)
Let depth of water in bowl be \(x\).
\(\therefore \quad\) Volume of water in hemispherical
\(\begin{aligned}
& \quad \text { bowl }(\mathrm{V})=\frac{\pi}{3} x^2(3 r-x) \\
& \quad \mathrm{V}=\frac{\pi}{3} x^2(3 \times 180-x) \\
& \therefore \quad \mathrm{V}=180 \pi x^2-\frac{\pi}{3} x^3
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{\mathrm{dV}}{\mathrm{dt}}=360 \pi x \frac{\mathrm{~d} x}{\mathrm{dt}}-\pi x^2 \frac{\mathrm{~d} x}{\mathrm{dt}} \\
& \left.\frac{\mathrm{dV}}{\mathrm{dt}}\right|_{x=120}=\frac{\mathrm{d} x}{\mathrm{dt}}\left(360 \pi \times 120-120^2 \pi\right)
\end{aligned}\)
\(\begin{array}{ll}\therefore & 1800=\frac{d x}{d t}\left(360 \pi \times 120-120^2 \pi\right) \\ \therefore & 15=\frac{d x}{d t}(360 \pi-120 \pi) \\ & \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{15}{240 \pi}=\frac{1}{16 \pi} \mathrm{~cm} / \mathrm{sec}\end{array}\)