Question

A spherical balloon is being inflated so that its volume increases uniformly at the rate of $40{\text{cm}}^3\text{/min.}$

At radius $r=8\text{cm}$, its surface area increases at the rate

• A. $8{\text{cm}}^2\text{/min}$
• B. $10{\text{cm}}^2\text{/min}$
• C. $20{\text{cm}}^2\text{/min}$
• D. None of these

Answer 1

The correct option is B $10{\text{cm}}^2\text{/min}$

Let $V$ be the volume and $r$ the radius of the balloon at any time.
Then, $V=\left(\frac{4}{3}\right)\pi r^3$
$\therefore \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
$\Rightarrow 40=4\pi r^2\frac{dr}{dt}$ (Given)
$\Rightarrow \frac{dr}{dt}=\frac{10}{\pi r^2}...\left(1\right)$

Now, let $S$ be the surface area of the balloon when its radius is $r$. Then
$S=4\pi r^2$
$\therefore \frac{dS}{dt}=8\pi r\frac{dr}{dt}$

From $\left(1\right)$ and $\left(2\right),$
$\frac{dS}{dt}=8\pi r\frac{10}{\pi r^2}=\frac{80}{r}$

When $r=8,$ the rate of increase of surface area is, $S=\frac{80}{8}=10{\text{cm}}^2/\text{min.}$