Question

A spherical balloon is filled with $4500\pi$ cubic meters of helium gas. If a leak in the ballon 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 ballon 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 $\frac{2}{9}$

volume of spherical ballon $=V=\frac{4}{3}\pi r^3\Rightarrow 4500\pi =\frac{4\pi r^3}{3}(\because Given,volume=4500\pi m^3)$
Differentiating both the sides, w.r.t we get,
$\frac{dV}{dt}=4\pi r^2\left(\frac{dr}{dt}\right)$
Now, it is given that $\frac{d\vee }{dt}=72\pi$
$\because$after 49min, Volume $=(4500-49\times 72)\pi$
$=(4500-3528)\pi$
$=972\pi m^3$
$\Rightarrow V=972\pi m^3\because 972\pi =\frac{4}{3}\pi r^3$
$\Rightarrow r^3=3\times 243=3\times 3^5=3^6=(3^2{)}^3=9$
Also, we have $\frac{d\vee }{dt}=72\pi$
$\because 72\pi =4\pi \times 9\times 9\left(\frac{dr}{dt}\right)\Rightarrow \frac{dr}{dt}=\left(\frac{2}{9}\right)$