The rate of change of volume of a sphere with respect to its surface area, when radius is 4 cm, is equal to
Answer & explanation
Correct answer: option 4
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