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

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