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(D) The volume of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$. Given the initial volume $V_0 = 4500 \pi \ m^3$. The rate of change of vo

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$\frac{2}{9}$

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(D) The volume of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$. Given the initial volume $V_0 = 4500 \pi \ m^3$. The rate of change of volume is $\frac{dV}{dt} = -72 \pi \ m^3/	ext{min}$. After $t = 49$ minutes,the volume $V$ is $V = V_0 + (\frac{dV}{dt}) 	imes t = 4500 \pi - 72 \pi 	imes 49$. $V = 4500 \pi - 3528 \pi = 972 \pi \ m^3$. Using $V = \frac{4}{3} \pi r^3$,we find the radius $r$ at $t = 49$: $972 \pi = \frac{4}{3} \pi r^3 \implies r^3 = 972 	imes \frac{3}{4} = 729$. $r = \sqrt[3]{729} = 9 \ m$. Now,differentiate $V = \frac{4}{3} \pi r^3$ with respect to $t$: $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$. Substitute the known values: $-72 \pi = 4 \pi (9)^2 \frac{dr}{dt}$. $-72 \pi = 4 \pi (81) \frac{dr}{dt} \implies -72 \pi = 324 \pi \frac{dr}{dt}$. $\frac{dr}{dt} = -\frac{72}{324} = -\frac{2}{9} \ m/	ext{min}$. The rate at which the radius decreases is $\frac{2}{9} \ m/	ext{min}$.

$A$ 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 has begun,is

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