Question

If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8\, cm^2/s$, 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)

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