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MHT CET · Maths · Application of Derivatives

The rate of change of the volume of a sphere with respect to its surface area, when its radius is 2 cm, is _______ \(\mathrm{cm}^3 / \mathrm{cm}^2\).

  1. A 0.1
  2. B 0.5
  3. C 1
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(C) 1

Step-by-step Solution

Detailed explanation

Volume of sphere \((V)=\frac{4}{3} \pi r^3\)
Surface area of sphere \((A)=4 \pi r^2\)
\(\begin{aligned}
& \therefore \quad \frac{d V}{d r}=4 \pi r^2 \text { and } \frac{d A}{d r}=8 \pi r \\
& \therefore \quad \frac{d V}{d A}=\frac{\frac{d V}{\frac{d r}{d A}}}{\frac{d r}{d r}}=\frac{4 \pi r^2}{8 \pi r}=\frac{r}{2} \\
& \therefore \quad\left(\frac{d V}{d A}\right)_{r=2}=\frac{2}{2}=1 \mathrm{~cm}^3 / \mathrm{cm}^2
\end{aligned}\)