If the volume of a spherical ball is increasing at the rate of 4$\mathrm{\pi }$cc / sec, then the rate of increase of its radius (in cm/sec), when the volume 288$\mathrm{\pi }$ cc is

$$\begin{gathered} \frac{\mathrm{dv}}{\mathrm{dt}}=4\mathrm{\pi } \mathrm{v}=288\mathrm{\pi } \\ \frac{4}{3}{\mathrm{\pi r}}^3=288\mathrm{\pi }\Rightarrow {\mathrm{r}}^3=216 \\ \Rightarrow \mathrm{r}=6 \\ \mathrm{v}=\frac{4}{3}{\mathrm{\pi r}}^3\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=4{\mathrm{\pi r}}^2\frac{\mathrm{dv}}{\mathrm{dt}} \\ \Rightarrow 4\mathrm{\pi }=4\mathrm{\pi }\times 36\times \frac{\mathrm{dv}}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{36} \end{gathered}$$