The rate of change of volume of a sphere with respect to its surface area,

Let r be the radius of the sphere at any time t.

Volume of the sphere, $V=\frac{4}{3} \pi r^{3}$

$V=\frac{4}{3} \pi r^{3}$

Differentiating both sides with respect to $r$, we get

$\frac{d V}{d r}=\frac{4}{3} \pi \times \frac{d}{d r} r^{3}$

$\Rightarrow \frac{d V}{d r}=\frac{4}{3} \pi \times 3 r^{2}$

$\Rightarrow \frac{d V}{d r}=4 \pi r^{2}$

Surface area of the sphere, $S=4 \pi r^{2}$

$S=4 \pi r^{2}$

Differentiating both sides with respect to $r$, we get

$\frac{d S}{d r}=4 \pi \times \frac{d}{d r} r^{2}$

$\Rightarrow \frac{d S}{d r}=4 \pi \times 2 r$

$\Rightarrow \frac{d S}{d r}=8 \pi r$

Now, rate of change of volume of a sphere with respect to its surface area

$=\frac{d V}{d S}$

$=\frac{\frac{d V}{d r}}{\frac{d S}{d r}}$

$=\frac{4 \pi r^{2}}{8 \pi r}$

$=\frac{r}{2}$

When $r=2 \mathrm{~cm}$, we get

$\left(\frac{d V}{d S}\right)_{r=2 \mathrm{~cm}}=\frac{2 \mathrm{~cm}}{2}=1 \mathrm{~cm}$

Thus, the rate of change of volume of a sphere with respect to its surface area when the radius is $2 \mathrm{~cm}$ is $1 \mathrm{~cm}$.

The rate of change of volume of a sphere with respect to its surface area, when the radius is 2 cm, is ____1 cm____.