Question

If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8c{m}^2/s$, then the rate of change of its volume is :

• A. constant
• B. proportional to $\sqrt{r}$
• C. proportional to $r^2$
• D. proportional to $r$

Answer 1

Answer: (D)

$V=\frac{4}{3}\pi r^3\Rightarrow \frac{dV}{dt}=4\pi r^2.\frac{dr}{dt}\dots \left(i\right)$
$S=4\pi r^2\Rightarrow \frac{dS}{dt}=8\pi r.\frac{dr}{dt}$
$\Rightarrow 8=8\pi r\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{1}{\pi r}$
Putting the value of $\frac{dr}{dt}$ in $\left(i\right)$, we get
$\frac{dV}{dt}=4\pi r^2\times \frac{1}{\pi r}=4r$
$\Rightarrow \frac{dV}{dt}$ is proportional to $r.$