Question

A hemispherical tank of radius 2 metres is initially full of water and has an outlet of $12c{m}^2$ cross sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law $V\left(t\right)=0.6\sqrt{2gh\left(t\right)}$, where V(t) and h(t) are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time t, and g is the acceleration due to gravity. If $T$ the time it takes to empty the tank, the value of $\frac{27T\sqrt{g}}{\pi }{10}^{-3}$ is equal to:

Answer 1

Let the height of the water level above the base at time $t$ be $h\left(t\right)$ cm and let it fall $dh$ in time $dt$.

Let the change in volume be $dv$.

Let $r$ be the radius of the surface of water at time $t$.

$r^2=(200{)}^2-(200-h{)}^2$
$=400h-h^{2}cm$ and $dv=-\pi r^{2}dh$
$\Rightarrow \frac{dv}{dt}=-\pi r^2\frac{dh}{dt}$
We are given velocity of water flow is $V=\frac{3}{5}\sqrt{2gh}$.

Rate of flow of water $=12V$. Thus $\frac{dv}{dt}=12V=\frac{36}{5}\sqrt{2gh}$
Thus $-\pi r^2\frac{dh}{dt}=\frac{36}{5}\sqrt{2g}\sqrt{h}$.
$\Rightarrow \frac{dh}{dt}=-\frac{36}{5\pi }\sqrt{2g}\frac{\sqrt{h}}{400h-h^2}$
$=-\frac{36}{5\pi }\sqrt{2g}\frac{1}{400\sqrt{h}-h^{3/2}}$
Separating variables, we have
$(400\sqrt{h}-h^{3/2})dh=-\frac{36}{5\pi }\sqrt{2g}dt$
$\Rightarrow \frac{800}{3}{h}^{3/2}-\frac{2}{5}{h}^{5/2}=-\frac{36}{5\pi }\sqrt{2g}t+C$
At t = 0, h = 200 cm
$C=\frac{800}{3}(200{)}^{3/2}-\frac{2}{5}(200{)}^{5/2}=\sqrt{2}\frac{56}{15}\times {10}^{5}cm$
We have to find t, so that h = 0. Let t = T when h = 0, then
$0=-\frac{36}{5\pi }\sqrt{2g}T+\sqrt{2}\frac{56}{15}\times {10}^5$
$\Rightarrow T=\frac{\sqrt{2}\times 56\times 5\pi }{15\times 36\times \sqrt{2g}}{10}^5=\frac{14}{27}\frac{\pi }{\sqrt{8}}{10}^5$ units
Thus $\frac{27T\sqrt{g}}{\pi }{10}^{-3}=1400$