Question

A spherical balloon is filled with $4500\pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72\pi$ cubic meters per minute, then the rate $($in meters per minute$)$ at which the radius of the balloon decreases $49$ minutes after the leakage began is :

• A. $\frac{9}{7}$
• B. $\frac{7}{9}$
• C. $\frac{2}{9}$
• D. $\frac{9}{2}$

Answer 1

The correct option is C $\frac{2}{9}$

Let $r$ be the radius of the balloon and $V$ be its volume. It is given that
$\frac{dV}{dt}=-72\pi \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=-72\pi \Rightarrow \frac{4}{3}\pi r^3=-72\pi t+C$
When $t=0,\frac{4}{3}\pi r^3=C=4500\pi$
$\therefore \frac{4}{3}\pi r^3=4500\pi -72\pi t$
When $t=49,\frac{4}{3}\pi r^3=4500\pi -\left(72\pi \right)\left(49\right)=972\pi$
$\Rightarrow r^3=729$
$\Rightarrow r=9$
Also, $\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=-72\pi$
$\Rightarrow 4\pi r^2\frac{dr}{dt}=-72\pi$
$\Rightarrow \frac{dr}{dt}=-\frac{72\pi }{4\pi r^2}=-\frac{18}{r^2}$
So ${\frac{dr}{dt}|}_{r=9}=-\frac{2}{9}$