A $U$ -tube of small and uniform cross section contains water of total length $4H$. The height difference between the water columns on the left and on the right is $H$ when the valve $K$ is closed. The valve is suddenly open, and water is flowing from left to right. Ignore friction. The speed of water when the heights of the left and the right water columns are the same, is :-
Diffcult
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$\mathrm{K}=\mathrm{U}_{i}-\mathrm{U}_{f}$
$\mathrm{U}_{i}=\Delta \mathrm{mg} \frac{\mathrm{H}}{2}-\mathrm{U}_{0}$
$\mathrm{U}_{f}=2\left[\frac{\Delta \mathrm{m}}{2} \mathrm{g} \frac{\mathrm{H}}{4}\right]-\mathrm{U}_{0}$
$\frac{1}{2} m v^{2}=K=\Delta m g \frac{H}{4}$
$\frac{1}{2}(\rho 4 \mathrm{HA}) \mathrm{v}^{2}=\rho \mathrm{HAg} \frac{\mathrm{H}}{4}$
$v=\sqrt{\frac{g H}{8}}$
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