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 $\frac{\mathrm{h}}{2}$ and from$\frac{\mathrm{h}}{2}$ to zero is 

Time taken for the level to fall from H to H':  $\mathrm{t}=\frac{\mathrm{A}}{{\mathrm{A}}_0}\sqrt{\frac{2}{\mathrm{g}}}\left[\sqrt{\mathrm{H}}-\sqrt{\mathrm{H}\text{'}}\right]$
According to problem- the time taken for the level to fall from h to $\frac{\mathrm{h}}{2}: {\mathrm{t}}_1=\frac{\mathrm{A}}{{\mathrm{A}}_{\mathrm{o}}}\sqrt{\frac{2}{\mathrm{g}}}\left[\sqrt{\mathrm{h}}-\sqrt{\frac{\mathrm{h}}{2}}\right]$
and similarly, time taken for the level to fall from h to zero: ${\mathrm{t}}_2=\frac{\mathrm{A}}{{\mathrm{A}}_{\mathrm{o}}}\sqrt{\frac{2}{\mathrm{g}}}\left[\sqrt{\frac{\mathrm{h}}{2}}-0\right]$

$\therefore \frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}=\frac{1-{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}-0}=\sqrt{2}-1.$