Question

A spherical balloon is being inflated at the rate of 35 cc/min. The rate of increase in surface area (in sq.cm/min) of the balloon when its diameter is 14 cm, is

• A. $10\sqrt{10}$
• B. $\sqrt{10}$
• C. $100$
• D. $10$

Answer 1

The correct option is D $10$

Let V – volume, S – surface area, r – radius of spherical balloon
Given $\frac{dv}{dt}=35cc/min,2r=14cm\Rightarrow r=7cm$
$\therefore$ $V=\frac{4}{3}\pi r^3\Rightarrow \frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$
$\Rightarrow 35=4\pi .\left(49\right).\frac{dr}{dt}$
$\Rightarrow \frac{dr}{dt}=\frac{5}{28\pi }cm/min$
Now, $S=4\pi r^2$
$\frac{ds}{dt}=8\pi r\frac{dr}{dt}=8\pi .7.\frac{5}{28\pi }=10sq.cm/min$