Question

A water tank has 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 rate of $5$ cubic meter per minute. Then 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 :

Answer 1

$\tan \theta =\frac{1}{2}\Rightarrow \frac{r}{h}=\frac{1}{2}\Rightarrow r=\frac{h}{2}$

Volume of the tank, $V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^2\times h=\frac{1}{12}\pi h^3$
Differentiating both sides w.r.t. $t$, we get
$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{4}{\pi h^2}\frac{dV}{dt}$
$\Rightarrow \frac{dh}{dt}{|}_{h=10}=\frac{4}{\pi (10{)}^2}\times 5$
$\Rightarrow \frac{dh}{dt}{|}_{h=10}=\frac{1}{5\pi }$