If the radius of a circle is increasing at a uniform rate of $2\mathrm{cm}{\mathrm{s}}^{-1}$. The area of increasing the area of a circle, at the instant when the radius is $20\mathrm{cm}$, is

The correct option is C

$80\mathrm{\pi }{\mathrm{cm}}^2{\mathrm{s}}^{-1}$

Explanation for Correct answer:

Option (C) $80\mathrm{\pi }{\mathrm{cm}}^2{\mathrm{s}}^{-1}$

Step 1 Given data

The radius($\mathrm{r}$) of a circular plate is increasing at the rate of $2\mathrm{cm}{\mathrm{s}}^{-1}$

That is, $\frac{\mathrm{dr}}{\mathrm{dt}}=2\mathrm{cm}{\mathrm{s}}^{-1}$

The radius of the circular plate is $20\mathrm{cm}$

Step 2 Formula

The area of the circle is $\mathrm{A}={\mathrm{\pi r}}^2$

Step 3 The rate at which the area increases

The radius ($\mathrm{r}$) of a circular plate is increasing at the rate $\frac{\mathrm{dr}}{\mathrm{dt}}=2\mathrm{cm}{\mathrm{s}}^{-1}$

The radius of the circular plate is $\mathrm{r}=20\mathrm{cm}$

The area of the circular plate is $\mathrm{A}={\mathrm{\pi r}}^2$

The rate at which the area is,

$$\begin{gathered} \frac{\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{d}\left({\mathrm{\pi r}}^2\right)}{\mathrm{dt}} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}.....\left(1\right) \end{gathered}$$

Substituting values in the equation (1)

$$\begin{gathered} \frac{\mathrm{dA}}{\mathrm{dt}}=2\times \mathrm{\pi }\times 20\times 2 \\ \frac{\mathrm{dA}}{\mathrm{dt}}=80\mathrm{\pi }{\mathrm{cm}}^2{\mathrm{s}}^{-1} \end{gathered}$$

Therefore, the rate at which the area increases is $80\mathrm{\pi }{\mathrm{cm}}^2{\mathrm{s}}^{-1}$.

Hence, option (C) is correct.