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

Answer 1

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

$V=\frac{4}{3}\pi r^3$
$4500\pi =\frac{4}{3}\pi r_0^3\Rightarrow r_0=15m$
Now $\frac{dV}{dt}=\frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt}$
$-72\pi =4\pi r^2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{-18}{r^2}.......\left(i\right)$
$\int r^{2}dr=-\int 18dt\Rightarrow \frac{r^3}{3}=-18t+C$
At t = 0, r = 15 m
$So,\frac{(15{)}^3}{3}=-18\left(0\right)+C\Rightarrow C=1125$
$\Rightarrow r^3=-54t+\mathrm{3375.......}\left(ii\right)$
At time t = 49 min r = 9 m
from eq. (i) ${\left(\frac{dr}{dt}\right)}_{t=49}=\frac{-18}{(9{)}^2}=-\frac{2}{9}$
(Negative sign shows decrement in radii)