A large tank filled with water to a height ‘h’ is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to $\frac{h}{2}$ and from $\frac{h}{2}$ to zero is

Time taken for the level to fall from H to h/2, $t=\frac{A}{A_0}\sqrt{\frac{2}{g}}\left[\sqrt{H}-\sqrt{H^{\prime }}\right]$

According to problem- the time taken for the level to fall from h to h/2  $t_1=\frac{A}{A_0}\sqrt{\frac{2}{g}}\left[\sqrt{h}-\sqrt{\frac{h}{2}}\right]$

similarly time taken for the level to fall from h/2 to zero  

$t_2=\frac{A}{A_0}\sqrt{\frac{2}{g}}\left[\sqrt{\frac{h}{2}}-0\right]\therefore \frac{t_1}{t_2}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1.$