The rate of change of volume of a sphere with respect to its surface…

The correct answer is Option (2) → 2 cm³/cm²

Both the volume V and the surface area S depend on the radius r, so the chain rule links them:

    \( \displaystyle V=\frac{4}{3}\pi r^3\ \Rightarrow\ \frac{dV}{dr}=4\pi r^2 \)

    \( \displaystyle S=4\pi r^2\ \Rightarrow\ \frac{dS}{dr}=8\pi r \)

    \( \displaystyle \frac{dV}{dS}=\frac{dV/dr}{dS/dr} \)

    \( \displaystyle =\frac{4\pi r^2}{8\pi r}=\frac{r}{2} \)

At r = 4 cm:

    \( \displaystyle \frac{dV}{dS}=\frac{4}{2}=2\ \text{cm}^3/\text{cm}^2 \)

Tip: For a sphere \( \frac{dV}{dS}=\frac{r}{2} \) at every radius, a result worth remembering.