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

The correct answer is Option (4) → 2 cm³/cm²

Let r be the radius. The volume and the surface area are

    \(\displaystyle V = \frac{4}{3}\pi r^3,\quad S = 4\pi r^2\)

Both depend on r, so by the chain rule

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

    \(\displaystyle \frac{dV}{dr} = 4\pi r^2,\quad \frac{dS}{dr} = 8\pi r\)

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

At r = 4 cm:

    \(\displaystyle \frac{dV}{dS} = \frac{4}{2} = 2\) cm³/cm²

Why the other options are incorrect

• Option (1) \(64\pi\) cm³/cm² — Incorrect. This is \(\frac{dV}{dr} = 4\pi(4)^2 = 64\pi\), the rate of change of volume with respect to the radius.

• Option (2) \(32\pi\) cm³/cm² — Incorrect. This is \(\frac{dS}{dr} = 8\pi(4) = 32\pi\), the rate of change of surface area with respect to the radius.

• Option (3) \(\frac{1}{2}\) cm³/cm² — Incorrect. This is \(\frac{dS}{dV} = \frac{2}{r} = \frac{1}{2}\), the rate taken the other way round.

NCERT Exemplar:
Application of Derivatives