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Air is discharging from a large spherical balloon at the rate of 4 cubic meters per minute. Then, the rate at which the surface area is shrinking when the radius of the balloon is $8 \mathrm{~m}$, is
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The correct answer is:
$1 \mathrm{~m}^2 / \mathrm{min}$
Let air is discharging from a large spherical balloon, at the rate $\frac{d v}{d t}=4 \mathrm{~m}^3 / \mathrm{min}$ $\because V=\frac{4}{3} \pi r^3$, where $r$ is the radius of spherical balloon.
So,
$$
\begin{aligned}
& \text { So, } \begin{aligned}
\frac{d v}{d t} & =4 \pi r^2 \frac{d r}{d t} \text { at } r=8 \mathrm{~m} \\
4 & =4 \pi(64) \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{64 \pi} \mathrm{m} / \mathrm{min} \\
\because \text { Area } s=4 \pi r^2 & \Rightarrow \frac{d s}{d t}=4 \pi(2 r) \frac{d r}{d t} \text { at } r=8 \mathrm{~m} \\
\frac{d s}{d t} & =64 \pi \times \frac{1}{64 \pi}=1 \mathrm{~m}^2 / \mathrm{min} .
\end{aligned}
\end{aligned}
$$
So,
$$
\begin{aligned}
& \text { So, } \begin{aligned}
\frac{d v}{d t} & =4 \pi r^2 \frac{d r}{d t} \text { at } r=8 \mathrm{~m} \\
4 & =4 \pi(64) \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{64 \pi} \mathrm{m} / \mathrm{min} \\
\because \text { Area } s=4 \pi r^2 & \Rightarrow \frac{d s}{d t}=4 \pi(2 r) \frac{d r}{d t} \text { at } r=8 \mathrm{~m} \\
\frac{d s}{d t} & =64 \pi \times \frac{1}{64 \pi}=1 \mathrm{~m}^2 / \mathrm{min} .
\end{aligned}
\end{aligned}
$$
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