Question

A vessel in the shape of an inverted cone of height $10ft$ 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 $=10ft$
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,
$\frac{dl}{dt}=\frac{1}{\sqrt{3}}$
$\Rightarrow l^2=h^2+r^2$
$=h^2+\frac{h^2}{3}=\frac{4h^2}{3}$
$l=\frac{2}{\sqrt{3}}h$
$\Rightarrow \frac{dl}{dt}=\frac{2}{\sqrt{3}}\frac{an}{dt}$
$\therefore \frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}\frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{1}{2}$
Now, $V=\frac{\pi }{9}{h}^3$
$\frac{dv}{dt}=\frac{3\pi h^2}{9}\frac{dh}{dt}$
$=\frac{1}{3}\pi (2\sqrt{3}{)}^2\times \frac{1}{2}$
$\Rightarrow \frac{dv}{dt}=2\pi$