Question

A shpherical balloon is pumped, at the constant rate of $3m^3/min$. The rate of increasing of its surface area at certain instant is found to be $5m^2/min$. At this instant its radius is equal to:

• A. $\frac{1}{5}m$
• B. $\frac{3}{5}m$
• C. $\frac{6}{5}m$
• D. $\frac{2}{5}m$

Answer 1

The correct option is C $\frac{6}{5}m$

$V=\frac{4}{3}\pi r^3$

$\frac{dV}{dt}=\frac{4}{3}\pi \times 3r^2\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{3}{4\pi r^2}$

Surface area; $S=4\pi r^2$

$\frac{dS}{dt}=4\pi \times 2r\frac{dr}{dt}$

$5=4\pi \times 2r\times \frac{3}{4\pi r^2}$

$r=\frac{6}{5}m$