Question

A spherical balloon is being inflated so that its volume increases uniformly at the rate of $40c{m}^3/min.$ At $r=8cm$, its surface area increases at the rate

• A. $8c{m}^2/min.$
• B. $10c{m}^2/min.$
• C. $20c{m}^2/min.$
• D. none of these

Answer 1

The correct option is B $10c{m}^2/min.$

Volume of sphere $V=\frac{4}{3}\pi r^3$
$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
$\Rightarrow \frac{dr}{dt}=\frac{5}{32\pi }cm/min$
Surface area of sphere $S=4\pi r^2$
$\Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt}$
$\Rightarrow \frac{dS}{dt}=10c{m}^2/min$