The rate of change of volume of a sphere with respect to its surface area when the radius is \( 4 \)
\( \mathrm{cm} \) is

Given that, radius of the sphere is
\[
r=4
\]
We now that, volume sphere is given by
\[
V=\frac{4}{3} \pi r^{2} \rightarrow(1)
\]
and surface area of sphere is given by
\[
S=4 \pi r^{2} \rightarrow(2)
\]
Differentiating Eqs. (1) and (2) with respect to \( r \), we get
\[
\frac{d V}{d r}=\frac{4}{3} \times 3 \pi r^{2}=4 \pi r^{2}
\]
\[
\begin{array}{l}
\text { And } \frac{d S}{d r}=4 \times 2 \pi r=8 \pi r \\
\text { So, } \frac{d V}{d S}=\frac{d V / d r}{d S / d r}=\left[\frac{4 m r^{2}}{8 \pi r}\right]=\frac{r}{2} \\
\text { At } r=4 \mathrm{~cm}, \text { we get } \\
\left(\frac{r}{2}\right)_{r=4}=\frac{4}{2}=2 \mathrm{~cm}^{3} \mathrm{~cm}^{-2}
\end{array}
\]