Properties of Matter

Concept:Fluid flow through a small hole under gravity, using Bernoulli's principle and continuity equation, leads to a differential equation for the height decrease.Explanation:Let the liquid heights in the left and right chambers be $h_1$ and $h_2$.Each chamber has floor area $A_0 = 1\ \mathrm{m}^2$.Total volume conservation gives $h_1 + h_2 = 2\ \mathrm{m}$ (since initial left height was $2\ \mathrm{m}$).Thus $h_2 = 2 - h_1$.The hole area $a = \sqrt{10} \times 10^{-4}\ \mathrm{m}^2$.Continuity: $A_0 v_1 = a v$, where $v_1$ is the downward speed of the left surface and $v$ is the speed through the hole.Bernoulli's equation between the left surface (at atmospheric pressure $P_0$) and the hole (where pressure is $P_0 + \rho g h_2$ on the right side) yields:$\rho g (h_1 - h_2) = \frac{1}{2}\rho (v^2 - v_1^2)$.Since $a \ll A_0$, $v_1^2$ is negligible, so $v = \sqrt{2g(h_1 - h_2)}$.Substitute $h_2$: $v = \sqrt{2g(2h_1 - 2)} = 2\sqrt{g(h_1 - 1)}$.The rate of height decrease: $-\frac{dh_1}{dt} = v_1 = \frac{a}{A_0}v = \frac{a}{A_0} \cdot 2\sqrt{g(h_1 - 1)}$.Rearrange: $\frac{dh_1}{\sqrt{h_1 - 1}} = -\frac{2a\sqrt{g}}{A_0} dt$.Integrate from $t=0$ ($h_1=2$) to $t=500\ \mathrm{s}$:$\int_2^{h_1} \frac{dh_1}{\sqrt{h_1-1}} = -\frac{2a\sqrt{g}}{A_0} \int_0^{500} dt$.Left integral: $2\sqrt{h_1-1} - 2\sqrt{2-1} = 2\sqrt{h_1-1} - 2$.Right side: $-\frac{2a\sqrt{g}}{A_0} \cdot 500$.Plug $a = \sqrt{10} \times 10^{-4}$, $g = 10$, $A_0 = 1$:$-\frac{2 \cdot \sqrt{10} \times 10^{-4} \cdot \sqrt{10}}{1} \cdot 500 = -2 \cdot 10 \times 10^{-4} \cdot 500 = -2 \times 10^{-3} \times 500 = -1$.Thus $2\sqrt{h_1-1} - 2 = -1$, so $2\sqrt{h_1-1} = 1$, $\sqrt{h_1-1} = 0.5$.Squaring: $h_1 - 1 = 0.25$, hence $h_1 = 1.25\ \mathrm{m}$.Answer:$h_1 = 1.25\ \mathrm{m}$, which matches option A.