A sphere increases its volume at the rate of \(\pi \mathrm{cc} / \mathrm{s}\). The rate at which its surface area increases when the radius is \(1 \mathrm{~cm}\) is

Given rate of increase volume of sphere is
\[
\frac{d V}{d t}=\pi
\]
We know that,
Volume of sphere; \(V=\frac{4}{3} \pi r^{3}\)
\(\Rightarrow \quad \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}\)
\(\Rightarrow \quad \pi=4 \pi r^{2} \frac{d r}{d t} ; \quad\) [from Eq. (i)]
\(\Rightarrow \quad \frac{d r}{d t}=\frac{1}{4 r^{2}}\)
5
Also, we know that, surface of sphere,
\[
\begin{aligned}
&S=4 \pi r^{2} \\
&\Rightarrow \quad \frac{d S}{d t}=8 \pi r \frac{d r}{d t}=8 \pi r \cdot \frac{1}{4 r^{2}} \quad[\text { from Eq. (ii)] } \\
&\Rightarrow \quad[\because \text { given } r=1] \\
&\Rightarrow \quad \frac{d S}{d t}=2 \pi
\end{aligned}
\]
So, rate of increase in surface \(=2 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}\)