Question

A spherical balloon is being inflated so that its volume increase uniformly at the rate of $40c{m}^3/minute$. The rate of increase in its surface area when the radius is $8cm$ is

• A. $10c{m}^2/minute$
• B. $20c{m}^2/minute$
• C. $40c{m}^2/minute$
• D. none of these

Answer 1

The correct option is A $10c{m}^2/minute$

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/minute
Surface area of sphere $S=4\pi r^2$
$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$
$\Rightarrow \frac{dS}{dt}=10c{m}^2/minute$