The radius of a circular plate is increasing at the rate of 0.01 $\mathrm{cm} / \mathrm{sec}$, when the…

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
  1. $60 \pi$ sq. cm $/ \mathrm{sec}$
  2. $0.24 \pi$ sq. cm $/ \mathrm{sec}$
  3. $1.2 \pi$ sq. cm $/ \mathrm{sec}$
  4. $24 \pi$ sq. cm $/ \mathrm{sec}$

Solution

We have $\frac{\mathrm{dr}}{\mathrm{dt}}=0.01$ $\begin{aligned} & \mathrm{A}=\pi \mathrm{r}^2 \\ & \therefore \quad \frac{\mathrm{dA}}{\mathrm{dt}}=\pi(2 \mathrm{r}) \frac{\mathrm{dr}}{\mathrm{dt}}=(2 \pi)(12)(0.01)=0.24 \pi \mathrm{sq} \cdot \mathrm{cm} / \mathrm{sec} \end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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