Q90 / 150 · MHT CET 2021 (24 Sep Shift 1) · Mathematics
The radius of a circular plate is increasing at the rate of 0.01 $\mathrm{cm} / \mathrm{sec}$, when the radius is $12 \mathrm{~cm}$. Then the rate at which the area increases is
- A$60 \pi$ sq. cm $/ \mathrm{sec}$
- Correct answerB$0.24 \pi$ sq. cm $/ \mathrm{sec}$
- C$1.2 \pi$ sq. cm $/ \mathrm{sec}$
- D$24 \pi$ sq. cm $/ \mathrm{sec}$
Solution
We have $\frac{\mathrm{dr}}{\mathrm{dt}}=0.01$
$$
\begin{aligned}
& \mathrm{A}=\pi \mathrm{r}^2 \\
& \therefore \quad \frac{\mathrm{dA}}{\mathrm{dt}}=\pi(2 \mathrm{r}) \frac{\mathrm{dr}}{\mathrm{dt}}=(2 \pi)(12)(0.01)=0.24 \pi \mathrm{sq} \cdot \mathrm{cm} / \mathrm{sec}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.