Question

A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowest. Its semi vertical angle is ${\tan }^{-1}\left(0.5\right)$. Water is powered into it at a constant rate of $5$ cubic meter per minute, then the rate at which the level of the water is rising at the instant when the depth of water in the tank is $10m$ is $\left(inmeter/min\right)$

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

Answer 1

The correct option is A $\frac{1}{2\pi }$

$ta{n}^{-1}x$ = 0.5

$\Rightarrow$ $r/h$ = 0.5

$\Rightarrow$ $r=h/2$ $-\left(i\right)$

Now, Volume of inverted cone

V = $\frac{\pi r^{2}h}{3}$ (where r is the radius and h is the height of cone)

V = $\frac{\pi h^3}{12}$ $from\left(i\right)$

$dV$ = $\frac{\pi h^2}{4}$ $dh$ (Taking derivative on both sides)

At $h=4m$,

$\frac{dV}{dt}$ = $4\pi$ $\frac{dh}{dt}$ (Dividing both sides by $dt$)

$\frac{dh}{dt}$ = $5/4\pi$

= $0.398m/hour$