Question

A balloon is pumped at the rate of $a$ cm³/minute. The rate of increase of its surface area when the radius is $b$ cm is

• A. $\frac{2a^2}{b^4}$ ${\mathrm{cm}}^2/\min$
• B. $\frac{a^2}{2b^2}$${\mathrm{cm}}^2/\min$
• C. $\frac{2a}{b}$ ${\mathrm{cm}}^2/\min$
• D. None of these

Answer 1

The correct option is C

$\frac{2a}{b}$ ${\mathrm{cm}}^2/\min$

Explanation for the correct option:

Step- 1: Find the increasing rate of surface area of balloon:

Given, the balloon is pumped at a rate $=a{\text{cm}}^3/\min$

$\Rightarrow$ $\frac{dV}{dt}=a$

Put the value of $V$(volume of sphere)

$V=\frac{4}{3}{\mathrm{\pi r}}^3$

$\Rightarrow$$\frac{dV}{dt}=\frac{4}{3}\times 3{\mathrm{\pi r}}^2\frac{d\mathrm{r}}{d\mathrm{t}}$

$\Rightarrow$ $a=4{\mathrm{\pi r}}^2\frac{d\mathrm{r}}{d\mathrm{t}}$

$\Rightarrow$ $\frac{dr}{dt}=\frac{a}{4{\mathrm{\pi r}}^2}$ — $\left(1\right)$

Step- 2 : Rate of incerease of surface areas:

Now, surface area of balloon will be $S=4{\mathrm{\pi r}}^2$

Differentiate it with respect to $r$

$\Rightarrow$$\frac{dS}{dt}=4\mathrm{\pi }\left(2\mathrm{r}\right)\frac{d\mathrm{r}}{d\mathrm{t}}$

$=8\mathrm{\pi r}\frac{dr}{dt}$

$=8\mathrm{\pi r}\left(\frac{\mathrm{a}}{4{\mathrm{\pi r}}^2}\right)$ [From equation $\left(1\right)$]

$=\frac{2a}{r}$

Put the value of $r=b$

$\Rightarrow \frac{dS}{dt}=\frac{2\mathrm{a}}{\mathrm{b}}$

Thus, the rate of increase of surface area of balloon is $\frac{2\mathrm{a}}{\mathrm{b}}$ ${\mathrm{cm}}^2/\min$

Hence, option ‘C’ is correct.