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A vessel in the shape of an inverted cone of height $10 \mathrm{ft}$ and semi vertical angle $30^{\circ}$ is full of water. Due to a hole at the vertex, the slant height of the water in the vessel is decreasing at a constant rate of $\frac{1}{\sqrt{3}}$ feet per minute. The rate (in cu. feet $/ \mathrm{min}$ ) at which the volume of water in the vessel is decreasing, when the volume of water is $\frac{8 \pi}{\sqrt{3}}$ cubic feet, is
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The correct answer is:
$2 \pi$
Given height of cone $=10 \mathrm{ft}$
vertical angle $=30^{\circ}$
$\begin{aligned} & \therefore \tan 30=\frac{r}{h} \Rightarrow \frac{1}{\sqrt{3}}=\frac{r}{h} \Rightarrow r=\frac{h}{\sqrt{3}} \\ & \because \quad V=\frac{\pi}{3} \times \frac{h^2}{3} \times h \Rightarrow \frac{8 \pi}{\sqrt{3}}=\frac{\pi}{9} h^3 \quad\left[\because V=\frac{8 \pi}{\sqrt{3}}\right] \\ & h=2 \sqrt{3}\end{aligned}$
$\begin{aligned}
\therefore \text { Given, } \frac{d l}{d t} & =\frac{1}{\sqrt{3}} \Rightarrow l^2=h^2+r^2=h^2+\frac{h^2}{3}=\frac{4 h^2}{3} \\
l & =\frac{2}{\sqrt{3}} h \Rightarrow \frac{d l}{d t}=\frac{2}{\sqrt{3}} \frac{d h}{d t} \\
\therefore \quad \frac{1}{\sqrt{3}} & =\frac{2}{\sqrt{3}} \frac{d h}{d t} \Rightarrow \frac{d h}{d t}=\frac{1}{2}
\end{aligned}$
Now,
$\begin{aligned}
V & =\frac{\pi}{9} h^3 \\
\frac{d v}{d t} & =\frac{3 \pi h^2}{9} \frac{d h}{d t}=\frac{1}{3} \pi(2 \sqrt{3})^2 \times \frac{1}{2} \Rightarrow \frac{d v}{d t}=2 \pi
\end{aligned}$
vertical angle $=30^{\circ}$
$\begin{aligned} & \therefore \tan 30=\frac{r}{h} \Rightarrow \frac{1}{\sqrt{3}}=\frac{r}{h} \Rightarrow r=\frac{h}{\sqrt{3}} \\ & \because \quad V=\frac{\pi}{3} \times \frac{h^2}{3} \times h \Rightarrow \frac{8 \pi}{\sqrt{3}}=\frac{\pi}{9} h^3 \quad\left[\because V=\frac{8 \pi}{\sqrt{3}}\right] \\ & h=2 \sqrt{3}\end{aligned}$
$\begin{aligned}
\therefore \text { Given, } \frac{d l}{d t} & =\frac{1}{\sqrt{3}} \Rightarrow l^2=h^2+r^2=h^2+\frac{h^2}{3}=\frac{4 h^2}{3} \\
l & =\frac{2}{\sqrt{3}} h \Rightarrow \frac{d l}{d t}=\frac{2}{\sqrt{3}} \frac{d h}{d t} \\
\therefore \quad \frac{1}{\sqrt{3}} & =\frac{2}{\sqrt{3}} \frac{d h}{d t} \Rightarrow \frac{d h}{d t}=\frac{1}{2}
\end{aligned}$
Now,
$\begin{aligned}
V & =\frac{\pi}{9} h^3 \\
\frac{d v}{d t} & =\frac{3 \pi h^2}{9} \frac{d h}{d t}=\frac{1}{3} \pi(2 \sqrt{3})^2 \times \frac{1}{2} \Rightarrow \frac{d v}{d t}=2 \pi
\end{aligned}$
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