The edge of a cube is equal to the radius of a sphere. If the edge and the radius increase at the same rate, then the ratio of the increases in surface areas of the cube and sphere is 

Let the length of an edge of the cube be x units. Let $S_1$and $S_2$ be the surf ace areas of the cube and sphere respectively. Then, 

$S_1=6x^2$ and $S_2=4\pi x^2$

$\Rightarrow \frac{d{S}_1}{dt}=12x\frac{dx}{dt}$ and $S_2=8\pi x\frac{dx}{dt}\Rightarrow \frac{{\displaystyle \raisebox{1ex}{$d{s}_1$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}}{{\displaystyle \raisebox{1ex}{$d{S}_2$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}}=\frac{3}{2\pi }$