Question

Water is dripping out from a conical funnel of semi-vertical angle $\frac{\pi }{4}$ at the uniform rate of $2c{m}^2/s$ in the surface area through a tiny hole at the vertex of the bottom. When the slant height of cone is $4cm$, then rate of decrease of the slant height of water is

• A. $\frac{\sqrt{2}}{3\pi }cm/s$
• B. $\frac{\sqrt{2}}{\pi }cm/s$
• C. $\frac{\sqrt{2}}{4\pi }cm/s$
• D. None of these

Answer 1

Answer: (C)

If $s$ represents the surface area, then
$\frac{ds}{dt}=2c{m}^2/s$

$s=\pi r\cdot I=\pi l\cdot \sin \frac{\pi }{4}I=\frac{\pi }{\sqrt{2}}{I}^2\left(\because r=I\sin \frac{\pi }{4}\right)$
Therefore, $\frac{ds}{dt}=\frac{2\pi }{\sqrt{2}}l\cdot \frac{dl}{dt}=\sqrt{2}\pi l\cdot \frac{dl}{dt}$
When
$\frac{ds}{dt}=2c{m}^2/s,l=4cm$
$\frac{dl}{dt}=\frac{1}{\sqrt{2}\pi \cdot 4}\cdot 2=\frac{1}{2\sqrt{2}\pi }=\frac{\sqrt{2}}{4\pi }cm/s$