Question

A spherical balloon is being inflated at the rate of $35cc/min$. The rate of increase in surface area (in $c{m}^2/min$.) of the balloon when its diameter is $14cm$, is

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

Answer 1

Answer: (D)

$V=\frac{4}{3}\pi r^3\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\left(\frac{35}{4\pi r^2}\right)...(i)$
Also, $S=4\pi r^2\Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt}=(8\pi r)\left(\frac{35}{4\pi r^2}\right)=\frac{70}{r}$
${\therefore \frac{dS}{dt}]}_{r=7}=\frac{70}{7}=10$