A large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The…
A large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from $h$ to $h / 2$ and $h / 2$ to zero is
$\sqrt{2}$
$\frac{1}{\sqrt{2}}$
$\sqrt{2}-1$
$\frac{1}{\sqrt{2-1}}$
Solution
Time taken, $t=\sqrt{\frac{2 h}{g}}$
Time taken for the level of water to fall from $h$ to $h / 2$ to 0 ,
$t_1=\sqrt{\frac{2}{g}}\left(\sqrt{h_1}-\sqrt{h_2}\right)$
$=\sqrt{\frac{2}{g}}\left(\sqrt{h}-\sqrt{\frac{h}{2}}\right)$
$=\sqrt{\frac{2 h}{g}}\left(1-\frac{1}{\sqrt{2}}\right)$
Similarly, time taken for the level of water to fall from $h / 2$ to 0 .
$t_2=\sqrt{\frac{2}{g}}\left(\sqrt{\frac{h}{2}}-0\right)$
$\Rightarrow \quad t_2=\sqrt{\frac{2 h}{g}} \cdot \frac{1}{\sqrt{2}}$
$\therefore \quad \frac{t_1}{t_2}=\frac{\sqrt{\frac{2 h}{g}}\left(1-\frac{1}{\sqrt{2}}\right)}{\sqrt{\frac{2 h}{g}} \frac{1}{\sqrt{2}}}$
$=\frac{\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}{1 / \sqrt{2}}$
$=\sqrt{2}-1$