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

Question

Aicharya · Accepted Answer

$v=\frac{4}{3} \pi r^2$ After $49$ minutes volume $=4500 \pi-49(72 \pi)=972 \pi$ $\frac{4}{3} \pi r^3=972 \pi \quad \Rightarrow r^3=729 \quad \Rightarrow r=9$ $v=\frac{4}{3} \pi r^3$ $\frac{d v}{d t}=\frac{4}{3} \pi 3 r^2 \frac{d r}{d t}$ $72 \pi=4 \pi r^2 \frac{d r}{d t}$ $\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{72}{4 \cdot 9 \cdot 9}=\frac{2}{9}$

A spherical balloon is filled with $4500 ~\pi$ cubic meters of helium gas. If a leak in the balloon causes…

Solution