Question

A water tank as the shape of an inverted right circular cone, whose semi-vertical angle is ${\tan }^{-1}\left(\frac{1}{2}\right)$. Water is poured into it at a constant rage of $5$ cubic meter per minute. The rate (in m/min), at which the level of water is rising at the instant when the depth of water in the tank is $10m$ is

• A. $\frac{2}{\pi }$
• B. $\frac{1}{5\pi }$
• C. $\frac{1}{10\pi }$
• D. $\frac{1}{15\pi }$

Answer 1

The correct option is B $\frac{1}{5\pi }$

$\tan \theta =\frac{1}{2}=\frac{r}{h}$
$r=\frac{h}{2}$
$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \frac{h^3}{4}$
$\frac{dV}{dt}=\frac{\pi }{12}{\left(3h\right)}^2\left(\frac{dh}{dt}\right)$
$5=\frac{\pi }{4}\left(100\right)\frac{dh}{dt}\Rightarrow \frac{dh}{dt}=\frac{1}{5\pi }$