A spherical balloon is being inflated so that its volume increase uniformly at the rate of 40 cm³ /minute. The rate of increase in its surface area when the radius is 8 cm, is 

Let V be the volume, S be the surface area and r be the radius of the balloon. Then,  

$V=\frac{4}{3}\pi r^3$and $S=4\pi r^2$

$\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ and $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$

When $r=8\mathrm{cm}$and $\frac{dV}{dt}=40{\mathrm{cm}}^3/\text{ minute }$ 

$\frac{dV}{dt}=4{\pi }^2\frac{dr}{dt}\Rightarrow 40=4\pi \times 8^2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{5}{32\pi }$

putting , $r=8,\frac{dr}{dt}=\frac{5}{32\pi }$ in $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$ 

$\frac{dS}{dt}=8\pi \times 8\times \frac{5}{32\pi }=10{\mathrm{cm}}^2/\text{ minute }$