A balloon, which always remains spherical on inflation,

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 time (t) is given by,

$\frac{d V}{d t}=\frac{d V}{d r} \cdot \frac{d r}{d t}$ [By chain rule]

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

$=4 \pi r^{2} \cdot \frac{d r}{d t}$

It is given that $\frac{d V}{d t}=900 \mathrm{~cm}^{3} / \mathrm{s}$.

$\therefore 900=4 \pi r^{2} \cdot \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{900}{4 \pi r^{2}}=\frac{225}{\pi r^{2}}$

Therefore, when radius = 15 cm,

$\frac{d r}{d t}=\frac{225}{\pi(15)^{2}}=\frac{1}{\pi}$

Hence, the rate at which the radius of the balloon increases when the radius is $15 \mathrm{~cm}$ is $\frac{1}{\pi} \mathrm{cm} / \mathrm{s}$.