Question 12

For Tank 1, let $y$ be the instantaneous height of water. Velocity of efflux $v_1 = \sqrt{2gy}$. Rate of flow: $A \left(-\frac{dy}{dt}\right) = a\sqrt{2gy}$ $t_1 = \int_0^h \frac{A}{a\sqrt{2gy}} dy = \frac{A}{a} \sqrt{\frac{2h}{g}}$

For Tank 2, the pipe ends at ground level, so the pressure difference is determined by the total head $(y + H)$. Velocity of efflux $v_2 = \sqrt{2g(y+H)}$. Rate of flow: $A \left(-\frac{dy}{dt}\right) = a\sqrt{2g(y+H)}$ $t_2 = \frac{A}{a\sqrt{2g}} \int_0^h \frac{dy}{\sqrt{y+H}} = \frac{A}{a\sqrt{2g}} [2\sqrt{y+H}]_0^h = \frac{2A}{a\sqrt{2g}} (\sqrt{h+H} - \sqrt{H})$

The ratio $\frac{t_1}{t_2} = \frac{\frac{A}{a} \sqrt{\frac{2h}{g}}}{\frac{2A}{a\sqrt{2g}}(\sqrt{h+H} - \sqrt{H})} = \frac{\sqrt{h}}{\sqrt{h+H} - \sqrt{H}}$ Given $H = \frac{16}{9}h$: $\sqrt{h+H} = \sqrt{h + \frac{16}{9}h} = \sqrt{\frac{25}{9}h} = \frac{5}{3}\sqrt{h}$ $\sqrt{H} = \sqrt{\frac{16}{9}h} = \frac{4}{3}\sqrt{h}$ $\frac{t_1}{t_2} = \frac{\sqrt{h}}{\frac{5}{3}\sqrt{h} - \frac{4}{3}\sqrt{h}} = \frac{1}{1/3} = 3$ Final answer is 3.