Question

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

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

Answer 1

Answer: (A)

Volume of sphere $V=\frac{4}{3} \pi r^{3}$
$\frac{dv}{dt}=\frac{4}{3} \pi 3r^{2} \cdot\frac{d r}{d t}$
$35=4\pi r^{2} . \frac{d r}{d t} or \, \frac{d r}{d t}=\frac{35}{4\pi r^{2}}$
Surface area of sphere=S=$4\pi r^{2} $
$\frac{d s}{d t}=4\pi\times2r\times\frac{d r}{d t}=8\pi r. \frac{d r}{d t}$
$\frac{d s}{d t}=\frac{70}{r} \quad\left(By \, using \left(1\right)\right)$
Now, diameter = $14\, cm, r = 7$
$\therefore \quad\frac{d s}{d t}=10 \quad$