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
- A \(0.1 \mathrm{~cm}^3 / \mathrm{cm}^2\)
- B \(\frac{1}{2} \mathrm{~cm}^3 / \mathrm{cm}^2\)
- C \(1 \mathrm{~cm}^3 / \mathrm{cm}^2\)
- D \(2 \mathrm{~cm}^3 / \mathrm{cm}^2\)
Answer & Solution
Correct Answer
(C) \(1 \mathrm{~cm}^3 / \mathrm{cm}^2\)
Step-by-step Solution
Detailed explanation
Volume of sphere \((V)=\frac{4}{3} \pi r^3\)
Surface area of sphere \((A)=4 \pi r^2\)
\(\frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi \mathrm{r}^2 \text { and } \frac{\mathrm{dA}}{\mathrm{dr}}=8 \pi \mathrm{r}\)
\(\begin{aligned} & \therefore \quad\left(\frac{d V}{d A}\right)=\frac{\left(\frac{d V}{d r}\right)}{\left(\frac{d A}{d r}\right)}=\frac{4 \pi r^2}{8 \pi r}=\frac{r}{2} \\ & \therefore \quad\left(\frac{d V}{d A}\right)_{r=2}=1 \mathrm{~cm}^3 / \mathrm{cm}^2\end{aligned}\)
Surface area of sphere \((A)=4 \pi r^2\)
\(\frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi \mathrm{r}^2 \text { and } \frac{\mathrm{dA}}{\mathrm{dr}}=8 \pi \mathrm{r}\)
\(\begin{aligned} & \therefore \quad\left(\frac{d V}{d A}\right)=\frac{\left(\frac{d V}{d r}\right)}{\left(\frac{d A}{d r}\right)}=\frac{4 \pi r^2}{8 \pi r}=\frac{r}{2} \\ & \therefore \quad\left(\frac{d V}{d A}\right)_{r=2}=1 \mathrm{~cm}^3 / \mathrm{cm}^2\end{aligned}\)
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