A U-tube of small and uniform cross-section c | vedclass

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(B) Let the cross-sectional area be $A$ and the density of water be $\rho$. The total length of the water column is $L = 4H$. The mass of the water is

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$\sqrt{\frac{gH}{8}}$

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(B) Let the cross-sectional area be $A$ and the density of water be $\rho$. The total length of the water column is $L = 4H$. The mass of the water is $m = \rho A L = 4\rho AH$.Initially,the left column is at height $H/2$ above the mean level and the right column is at $H/2$ below the mean level. The potential energy of the system relative to the mean level is $U_i = \rho A (H/2) g (H/4) + \rho A (H/2) g (-H/4) = 0$ (considering the center of mass of each column).More simply,we can consider the potential energy change of the displaced water. $A$ volume $V = A(H/2)$ of water moves from the left side to the right side. The center of mass of this volume drops by a distance $H/2$. The change in potential energy is $\Delta U = -(\rho A (H/2)) g (H/2) = -\rho A g H^2 / 4$.By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy:$|\Delta U| = \frac{1}{2} m v^2$$\rho A g H^2 / 4 = \frac{1}{2} (4 \rho A H) v^2$$\rho A g H^2 / 4 = 2 \rho A H v^2$$v^2 = \frac{gH}{8}$$v = \sqrt{\frac{gH}{8}}$

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