From equation of continuity,
$A v_1=a_2$
$\mathrm{A}\left(-\frac{\mathrm{dh}}{\mathrm{dt}}\right)=\mathrm{a} \sqrt{2 \mathrm{gh}}$
$\int_{\mathrm{h}_1}^{\mathrm{h}_2} \frac{-\mathrm{dh}}{\sqrt{\mathrm{h}}}=\int_0^{\mathrm{t}} \frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{~g}} \mathrm{dt}$
$-[2 \sqrt{\mathrm{h}}]_{\mathrm{h}_1}^{\mathrm{h}_2}=\frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{~g}} \mathrm{t}$
$2\left(\sqrt{\mathrm{h}_2}-\sqrt{\mathrm{h}_1}\right)=\frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{gt}}$
$\therefore \quad t=\frac{2 A}{a} \sqrt{\frac{H}{2 g}}=\frac{A}{a} \frac{\sqrt{2 H}}{\sqrt{g}}$
For height $h$, the time is $t_1=\frac{A}{A_0} \sqrt{\frac{2 h}{g}}$
For height $4 \mathrm{~h}$, the time is $t_2=\frac{A}{A_0} \sqrt{\frac{8 h}{g}}$
$\therefore \quad \frac{t_2}{t_1}=2$
$\therefore \quad \mathrm{t}_2=2 \mathrm{t}_1$
$\therefore \quad \mathrm{t}_2=2 \mathrm{t}$
$\ldots\left(\because \mathrm{t}_1=\mathrm{t}\right)$