Question

A vessel in the shape of an inverted cone of height $10\, ft$ 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/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

• A. $\frac{2 \pi}{\sqrt{3}}$
• B. $2 \pi$
• C. $2 \pi \sqrt{3}$
• D. $\pi \sqrt{3}$

Answer 1

Answer: (B)

Given height of cone $=10 ft$
vertical angle $=30^{\circ}$

$\therefore \tan 30=\frac{r}{h} $
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{r}{h}$
$ \Rightarrow r=\frac{h}{\sqrt{3}}$
$\because V=\frac{\pi}{3} \times \frac{h^{2}}{3} \times h$
$ \Rightarrow \frac{8 \pi}{\sqrt{3}}=\frac{\pi}{9} h^{3}$
$\left[\because V=\frac{8 \pi}{\sqrt{3}}\right]$
$h=2 \sqrt{3}$
$\therefore $ 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 $
$\Rightarrow \frac{d l}{d t}=\frac{2}{\sqrt{3}} \frac{a n}{d t}$
$\therefore \frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}} \frac{d h}{d t} $
$\Rightarrow \frac{d h}{d t}=\frac{1}{2}$
Now, $V=\frac{\pi}{9} h^{3}$
$\frac{d v}{d t}=\frac{3 \pi h^{2}}{9} \frac{d h}{d t}$
$=\frac{1}{3} \pi(2 \sqrt{3})^{2} \times \frac{1}{2} $
$\Rightarrow \frac{d v}{d t}=2 \pi$