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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 $49 \mathrm{~min}$ after the leakage began is
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The correct answer is:
$\frac{2}{9}$
$\frac{2}{9}$
$V=\frac{4}{3} \pi r^3 \Rightarrow 4500 \pi=\frac{4 \pi r^3}{3}$
$\Rightarrow \quad r=15 \mathrm{~m}$
After $49 \min =(4500-49.72) \pi$
$=972 \pi \mathrm{m}^3$
$\begin{aligned} \Rightarrow & & 972 \pi & =\frac{4}{3} \pi r^3 \\ \Rightarrow & & r^3 & =3 \times 243=3 \times 3^5\end{aligned}$
$\begin{aligned} r & =9 \\ \frac{d v}{d t} & =4 \pi r^2\left(\frac{d r}{d t}\right) \\ 72 \pi & =4 \pi \times 9 \times 9\left(\frac{d r}{d t}\right) \\ \frac{d r}{d t} & =\left(\frac{2}{9}\right)\end{aligned}$
$\Rightarrow \quad r=15 \mathrm{~m}$
After $49 \min =(4500-49.72) \pi$
$=972 \pi \mathrm{m}^3$
$\begin{aligned} \Rightarrow & & 972 \pi & =\frac{4}{3} \pi r^3 \\ \Rightarrow & & r^3 & =3 \times 243=3 \times 3^5\end{aligned}$
$\begin{aligned} r & =9 \\ \frac{d v}{d t} & =4 \pi r^2\left(\frac{d r}{d t}\right) \\ 72 \pi & =4 \pi \times 9 \times 9\left(\frac{d r}{d t}\right) \\ \frac{d r}{d t} & =\left(\frac{2}{9}\right)\end{aligned}$
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