Question

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

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

Answer 1

Answer: (A)

Volume of sphere $V=\frac{4}{3}\pi r^3$
$\frac{dv}{dt}=\frac{4}{3}\pi 3r^2\cdot \frac{dr}{dt}$
$35=4\pi r^2.\frac{dr}{dt}or\frac{dr}{dt}=\frac{35}{4\pi r^2}$
Surface area of sphere=S=$4\pi r^2$
$\frac{ds}{dt}=4\pi \times 2r\times \frac{dr}{dt}=8\pi r.\frac{dr}{dt}$
$\frac{ds}{dt}=\frac{70}{r}\left(Byusing\left(1\right)\right)$
Now, diameter = $14cm,r=7$
$\therefore \frac{ds}{dt}=10$