Explanation

(c) $Time\ taken\ t = \sqrt{\frac{2h}{g}}$
Time taken for the level of water to fall from$\ h\ $to$\ h/2$
$t_{1} = \sqrt{\frac{2}{g}}\left\lbrack \sqrt{h_{1}} - \sqrt{h_{2}}\ \right\rbrack = \sqrt{\frac{2}{g}}\left\lbrack \sqrt{h} - \sqrt{\frac{h}{2}} \right\rbrack$
$= \sqrt{\frac{2h}{g}}\left( 1 - \frac{1}{\sqrt{2}} \right)$
Similarly, time taken for the level of water to fall from$h/2$ to 0.
$t_{2} = \sqrt{\frac{2}{g}}\ \left\lbrack \sqrt{\frac{h}{2}} - 0 \right\rbrack$
$\Longrightarrow \ t_{2} = \sqrt{\frac{2h}{g}} \bullet \frac{1}{\sqrt{2}}$
$\therefore\ \ \frac{t_{1}}{t_{2}} = \frac{\sqrt{\frac{2h}{g}}\left( 1 - \frac{1}{\sqrt{2}} \right)}{\sqrt{\frac{2h}{g}}\frac{1}{\sqrt{2}}}$
$= \frac{\left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right)}{1/\sqrt{2}} = \sqrt{2} - 1$