The rate of change of volume of a sphere with respect to its surface area, when the radius is 6 cm is:

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

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

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

Surface area of sphere: $S = 4\pi r^2$

Required: $\frac{dV}{dS}$

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

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

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

When $r = 6$ cm:

$\frac{dV}{dS} = \frac{6}{2} = 3$

Therefore, the rate of change of volume with respect to surface area is 3 cm.