Question

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

When radius $r=8\text{cm}$, then the increase in radius in the next $1/2\text{min}$ is

• A. $0.025\text{cm}$
• B. $0.050\text{cm}$
• C. $0.075\text{cm}$
• D. $0.01\text{cm}$

Answer 1

The correct option is A $0.025\text{cm}$

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, $S=\frac{80}{8}=10{\text{cm}}^2/\text{min.}$
​​​​
Therefore, increase of $S$ in $\frac{1}{2}\text{min}=10\times \left(\frac{1}{2}\right)=5{\text{cm}}^2/\text{min.}$

If $r_1$ is the radius of the balloon after $\frac{1}{2}\text{min}$, then $4\pi r_1^2=4\pi (8{)}^2+5$
$\Rightarrow r_1^2-8^2=\frac{5}{4\pi }=0.397$ nearly.
$\therefore$ Required increase in the radius $=r_1=8.025-8$
$=0.025\text{cm}$