Question

Find the time required for a cylindrical tank of height $H$ and radius $r$ to drain half of the water through a circular hole of the area $a$ at the bottom. The flow-through hole is according to the law $v\left(t\right)=k\sqrt{2gh\left(t\right)}$, where $v\left(t\right)$ and $h\left(t\right)$ is the velocity of flow through hole and height above the hole respectively at the time $t$ and $g$ is the gravitational acceleration.

• A. $\frac{\pi r^2\sqrt{H}(\sqrt{2}-1)}{ak\sqrt{g}}$
• B. $\frac{\pi r^2\sqrt{2H}}{ak\sqrt{g}}$
• C. $\frac{\pi r^2\sqrt{H}}{ak\sqrt{g}}$
• D. $\frac{\pi r^2\sqrt{H}}{ak\sqrt{2g}}$

Answer 1

The correct option is A $\frac{\pi r^2\sqrt{H}(\sqrt{2}-1)}{ak\sqrt{g}}$


From above figure after $t$ time level of water is $h$
$dV=\pi r^{2}dha.v\left(t\right)=ak\sqrt{2gh\left(t\right)}\Rightarrow -\frac{dV}{dt}=ak\sqrt{2gh\left(t\right)}\Rightarrow -\pi r^{2}dh=ak\sqrt{2gh\left(t\right)}.dt\Rightarrow -\pi r^2\frac{dh}{\sqrt{h\left(t\right)}}=ak\sqrt{2g}.dt$
For required time
$\Rightarrow -\pi r^2{\int }_H^{\frac{H}{2}}\frac{dh}{\sqrt{h\left(t\right)}}=ak\sqrt{2g}{\int }_0^{t}dt\Rightarrow t=\frac{\pi r^2\sqrt{H}(\sqrt{2}-1)}{ak\sqrt{g}}$