Question

The rate of the change of the surface area of a sphere of radius $\mathrm{r}$ when the radius is increasing at the rate of $2\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$ is proportional to

• A. $\frac{1}{\mathrm{r}}$
• B. $\frac{1}{{\mathrm{r}}^2}$
• C. $\mathrm{r}$
• D. ${\mathrm{r}}^2$

Answer 1

The correct option is C

$\mathrm{r}$

Explanation for correct option

Given: Radius$=\mathrm{r}$ and $\frac{\mathrm{dr}}{\mathrm{dt}}=2$

Now surface area of sphere $\mathrm{S}=4{\mathrm{\pi r}}^2$

Rate of change of surface area of sphere $=\frac{\mathrm{dS}}{\mathrm{dt}}$

So, $\frac{\mathrm{dS}}{\mathrm{dt}}=4\mathrm{\pi }\frac{\mathrm{d}\left({\mathrm{r}}^2\right)}{\mathrm{dt}}$

$\Rightarrow$$\frac{\mathrm{dS}}{\mathrm{dt}}=4\mathrm{\pi }\left(2\mathrm{r}\right)\frac{\mathrm{d}\left(\mathrm{r}\right)}{\mathrm{dt}}$ $\left[\because \frac{\mathrm{d}\left({\mathrm{m}}^{\mathrm{n}}\right)}{\mathrm{dx}}=\left({\mathrm{nm}}^{\mathrm{n}-1}\right)\frac{\mathrm{dm}}{\mathrm{dx}}\right]$

$\Rightarrow$$\frac{\mathrm{dS}}{\mathrm{dt}}=8\mathrm{\pi r}\frac{\mathrm{d}\left(\mathrm{r}\right)}{\mathrm{dt}}$

$\Rightarrow$$\frac{\mathrm{dS}}{\mathrm{dt}}=8\mathrm{\pi r}\left(2\right)$ $\left[\because \frac{\mathrm{dr}}{\mathrm{dt}}=2\right]$

$\Rightarrow$$\frac{\mathrm{dS}}{\mathrm{dt}}=16\mathrm{\pi r}$

$\Rightarrow$$\frac{\mathrm{dS}}{\mathrm{dt}}\propto \mathrm{r}$

Hence, the correct option is option(C) i.e. $\mathrm{r}$