The rate of change of volume of a sphere with respect to its surface area, when radius is 4 cm, is equal to

$2\, cm^3/cm^2$

The correct answer is Option (4) → $2\, cm^3/cm^2$

Volume of sphere: $V = \frac{4}{3}\pi r^{3}$

Surface area: $S = 4\pi r^{2}$

Required: $\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$

$\frac{dV}{dr} = 4\pi r^{2}$

$\frac{dS}{dr} = 8\pi r$

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

At $r = 4$:

$\frac{dV}{dS} = \frac{4}{2} = 2$

Required value = 2 cm