A cone whose height is always equal to its diameter is increasing in volume at the rate of $40 \mathrm{~cm}^{3} / \mathrm{sec}$. At what rate is the radius increasing when its circular base area is $1 \mathrm{~m}^{2}$ ?
(a) $1 \mathrm{~mm} / \mathrm{sec}$
(b) $0.001 \mathrm{~cm} / \mathrm{sec}$
(c) $2 \mathrm{~mm} / \mathrm{sec}$
(d) $0.002 \mathrm{~cm} / \mathrm{sec}$
(d) $0.002 \mathrm{~cm} / \mathrm{sec}$
Let $r$ be the radius, $h$ be the height and $V$ be the volume of the cone at any time $t .$ Then,
$V=\frac{1}{3} \pi r^{2} h$
$\Rightarrow V=\frac{2}{3} \pi r^{3} \quad[\because h=2 r]$
$\Rightarrow \frac{d V}{d t}=2 \times 10^{4} \frac{d r}{d t} \quad\left[\because \pi r^{2}=1 \mathrm{~m}^{2}\right.$ or $\left.10^{4} \mathrm{~cm}^{2}\right]$
$\Rightarrow \frac{d r}{d t}=\frac{1}{2 \times 10^{4}} \frac{d V}{d t}$
$\Rightarrow \frac{d r}{d t}=\frac{40}{2 \times 10^{4}}$
$\Rightarrow \frac{d r}{d t}=0.002 \mathrm{~cm} / \mathrm{sec}$
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