Volume of the spherical balloon $=\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3$ $\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{4}{3} \pi \mathrm{r}^3\right) \cdot \frac{\mathrm{dr}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \cdot \frac{\mathrm{dr}}{\mathrm{dt}}$ $\because \frac{\mathrm{dv}}{\mathrm{dt}}=900 \mathrm{~cm}^3 / \mathrm{sec} . \quad \therefore 900=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}$ When radius is $15 \mathrm{~cm}$, putting $r=15$ $\therefore 900=4 \times \frac{22}{7} \times 15 \times 15 \times \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{7}{22}$ Hence, the radius of the balloon is increasing at the rate of $\frac{7}{22} \mathrm{~cm} / \mathrm{sec}$. i.e. $\frac{1}{\pi} \mathrm{cm} / \mathrm{sec}$