Physics - Fluid Question with Solution | TestHub

From equation of continuity, $A{v}_1=a_2$ $\mathrm{A}\left(-\frac{\mathrm{dh}}{\mathrm{dt}}\right)=\mathrm{a}\sqrt{2\mathrm{gh}}$ ${\int }_{{\mathrm{h}}_1}^{{\mathrm{h}}_2}\frac{-\mathrm{dh}}{\sqrt{\mathrm{h}}}={\int }_0^{\mathrm{t}}\frac{\mathrm{a}}{\mathrm{A}}\sqrt{2\mathrm{g}}\mathrm{dt}$ $-[2\sqrt{\mathrm{h}}{]}_{{\mathrm{h}}_1}^{{\mathrm{h}}_2}=\frac{\mathrm{a}}{\mathrm{A}}\sqrt{2\mathrm{g}}\mathrm{t}$ $2\left(\sqrt{{\mathrm{h}}_2}-\sqrt{{\mathrm{h}}_1}\right)=\frac{\mathrm{a}}{\mathrm{A}}\sqrt{2\mathrm{gt}}$ $\therefore t=\frac{2A}{a}\sqrt{\frac{H}{2g}}=\frac{A}{a}\frac{\sqrt{2H}}{\sqrt{g}}$ For height $h$, the time is $t_1=\frac{A}{A_0}\sqrt{\frac{2h}{g}}$ For height $4\mathrm{h}$, the time is $t_2=\frac{A}{A_0}\sqrt{\frac{8h}{g}}$ $\therefore \frac{t_2}{t_1}=2$ $\therefore {\mathrm{t}}_2=2{\mathrm{t}}_1$ $\therefore {\mathrm{t}}_2=2\mathrm{t}$ $\dots \left(\because {\mathrm{t}}_1=\mathrm{t}\right)$