Question

A balloon is pumped at the rate of a cm${}^3$/minute. The rate of increase of its surface area when the radius is b cm, is

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

Answer 1

The correct option is C $\frac{2a}{b}$ cm${}^2$/min

$V=\frac{4\pi }{3}.r^3$
$\frac{dV}{dt}=4\pi .r^2.\frac{dr}{dt}$
Or
$a=4\pi .b^2.\frac{dr}{dt}$
Or
$\frac{a}{4b^2\pi }=\frac{dr}{dt}$.
Now
$s=4\pi r^2$
Or
$\frac{ds}{dt}=8\pi .r\frac{dr}{dt}$
Or
$\frac{ds}{dt}=8\pi .b.\frac{a}{4\pi .b^2}$
Or
$\frac{ds}{dt}=\frac{2a}{b}$.