If the rate of change in volume of spherical soap bubble is uniform, then the rate of change of surface area varies as :

Solution :
$$\eqalign{ & {\text{Let volume}} = V = \frac{4}{3}\pi {r^3}......\left( 1 \right) \cr & {\text{and surface area}} = S = 4\pi {r^2}......\left( 2 \right) \cr & {\text{Now,}} \cr & \left( 1 \right) \Rightarrow \frac{{dv}}{{dt}} = \frac{4}{3} \times 3\pi {r^2} \times \frac{{dr}}{{dt}} \cr & = 4\pi {r^2}\frac{{dr}}{{dt}}......\left( 3 \right) \cr & \left( 2 \right) \Rightarrow \frac{{ds}}{{dt}} = 4\pi \times 2 \times r\frac{{dr}}{{dt}} \cr & = \frac{{8\pi {r^2}}}{r}\frac{{dr}}{{dt}} \cr & = \frac{2}{r}\left[ {4\pi {r^2}\frac{{dr}}{{dt}}} \right]\, \cr & = \frac{2}{r}\frac{{dv}}{{dt}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{from 3}}} \right) \cr} $$