MHT CET · Maths · Application of Derivatives
A spherical snow ball is forming so that its volume is increasing at the rate of \(8 \mathrm{~cm}^3 / \mathrm{sec}\). Find the rate of increase of radius when radius is \(2 \mathrm{~cm}\)
- A \(\pi \mathrm{cm} / \mathrm{sec}\)
- B \(\frac{1}{8 \pi} \mathrm{cm} / \mathrm{sec}\)
- C \(2 \pi \mathrm{cm} / \mathrm{sec}\)
- D \(\frac{1}{2 \pi} \mathrm{cm} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2 \pi} \mathrm{cm} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3\)
\(\therefore \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \mathrm{r}^2\right) \frac{\mathrm{dr}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}\)
\(\therefore 8=4 \pi(2)^2 \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{2 \pi}\)
\(\therefore \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \mathrm{r}^2\right) \frac{\mathrm{dr}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}\)
\(\therefore 8=4 \pi(2)^2 \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{2 \pi}\)
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