Question

Water is dripping out at a steady rate of $1c{m}^3/s$ through a tiny hole at the vertex of the conical vessel, whose axis is vertical. If the slant height of water in the vessel is $4cm$, then the rate of decrease of slant height, where the vertical angle of the conical vessel is $\frac{\pi }{3}$, is

• A. $\frac{1}{2\sqrt{3}\pi }cm/s$
• B. $2\sqrt{3}\pi cm/s$
• C. $\sqrt{3}\pi cm/s$
• D. None of these

Answer 1

Answer: (A)

Given that, $\frac{dV}{dt}=-1c{m}^3/s$, where $V$ is the volume of water in the conical vessel.
From the figure, $I=4cm,h=I\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}l$ and $r=I\sin \frac{\pi }{6}=\frac{1}{2}$
Therefore, $V=\frac{1}{3}\pi r^{2}h=\frac{1\cdot \pi }{3}{\left(\frac{l}{2}\right)}^2\cdot \left(\frac{\sqrt{3}}{2}l\right)=\frac{\sqrt{3}}{24}{/}^3\pi$
$\frac{dV}{dt}=\frac{\sqrt{3}\pi }{8}{l}^2\frac{dl}{dt}$

Therefore, $-1=\frac{\sqrt{3}\pi }{8}16\cdot \frac{dl}{dt}$
$\Rightarrow \frac{dl}{dt}=\frac{-1}{2\sqrt{3}\pi }cm/s$
Therefore, the rate of decrease of slant height $=\frac{1}{2\sqrt{3}\pi }cm/s$.