Question

Water is being poured into a conical funnel at a rate of $4{\text{cm}}^3\text{/sec},$ whose semi vertical angle is $\frac{\pi }{6}.$ When slant height is $4\text{cm},$ which of the following(s) is correct

• A. Rate of change of height $=\frac{1}{\pi }\text{cm/sec}$
• B. Rate of change of slant height $=\frac{1}{\pi \sqrt{3}}\text{cm/sec}$
• C. Rate of change of height $=\frac{2}{\pi \sqrt{3}}\text{cm/sec}$
• D. Rate of change of slant height $=\frac{2}{\pi \sqrt{3}}\text{cm/sec}$

Answer 1

Answer: (A), (D)

Rate of change of slant height $=\frac{2}{\pi \sqrt{3}}\text{cm/sec}$

From figure :

$h=l\cos \theta =\frac{\sqrt{3}l}{2},r=l\sin \theta =\frac{l}{2}$
Now, $V=\frac{\pi r^{2}h}{3}=\frac{\pi l^3}{8\sqrt{3}}$
$\Rightarrow \frac{dV}{dt}=\frac{\sqrt{3}\pi l^2}{8}\cdot \frac{dl}{dt}=4$
At $l=4$
$\frac{16\sqrt{3}\pi }{8}\cdot \frac{dl}{dt}=4$
$\therefore \frac{dl}{dt}=\frac{2}{\pi \sqrt{3}}$
and $h=\frac{\sqrt{3}l}{2},$
$\Rightarrow \frac{dh}{dt}=\frac{\sqrt{3}}{2}\frac{dl}{dt}=\frac{1}{\pi }$