A tank full of water has a small hole at its bottom. If one fourth of the tank is emptied in t1 seconds and the remaining three-fourths of the tank is emptied in t2 seconds. Then the ratio (t1t2) is

$t \propto \sqrt{h_{1}} - \sqrt{h_{2}}$

$t_{1} \propto \sqrt{h} - \sqrt{\frac{3 h}{4}}$

$t_{2} \propto \sqrt{\frac{3 h}{4}} - 0$

$\frac{t_{1}}{t_{2}} = \frac{\sqrt{h} - \sqrt{\frac{3 h}{4}}}{\sqrt{\frac{3 h}{4}}} = \frac{\sqrt{h} - \frac{1}{2} \sqrt{3 h}}{\frac{\sqrt{3 h}}{2}}$

$= \frac{\sqrt{h}}{\sqrt{h}} [\frac{2 - \sqrt{3}}{\sqrt{3}}]$ $= [\frac{2}{\sqrt{3}} - 1]$

A tank full of water has a small hole at its bottom. If one fourth of the tank is emptied in $t_{1}$ seconds and the remaining three-fourths of the tank is emptied in $t_{2}$ seconds. Then the ratio $(\frac{t_{1}}{t_{2}})$ is