A spherical balloon is being inflated at the rate of  $35\raisebox{1ex}{$cc$}\!\left/ \!\raisebox{-1ex}{$\min $}\right.$. The rate of increase in the surface area of the balloon when its diameter is 14 cm is

$\begin{array}{l}VolumeofballoonV=\frac{4}{3}\Pi r^3\\ diameter=14radiusr=7\\ V=\frac{4}{3}\Pi r^3\frac{dv}{dt}=35\raisebox{1ex}{$cu$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.\\ derivativewithrespecttot\\ \frac{dv}{dt}=\frac{4}{3}.\Pi 3r^2\frac{dr}{dt}\\ 35=\frac{4}{3}\Pi \times 3\times 49\frac{dr}{dt}\\ \frac{dr}{dt}=\frac{5}{28\Pi }\\ SurfaceareaS=4\Pi r^2\\ derivativewithrespecttot\\ \frac{ds}{dt}=4\Pi \left(2r\right)\frac{dr}{dt}\\ =4\Pi \times 2\times 7\times \frac{5}{28\Pi }\\ =10\end{array}$