A balloon, which always remains spherical has a variable radius.

The volume of a sphere $(V)$ with radius $(r)$ is given by $V=\frac{4}{3} \pi r^{3}$.

Rate of change of volume (V) with respect to its radius (r) is given by,

$\frac{d V}{d r}=\frac{d}{d r}\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi\left(3 r^{2}\right)=4 \pi r^{2}$

Therefore, when radius = 10 cm,

$\frac{d V}{d r}=4 \pi(10)^{2}=400 \pi$

Hence, the volume of the balloon is increasing at the rate of $400 \pi \mathrm{cm}^{2}$.