Question

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

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