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Question: A U-tube open at both ends is partially filled with water. Oil having a density of 840 kg/m^{3} is then poured into the right arm and forms a column *L* = 4.68 cm high (Figure b).

(a) Determine the difference *h* in the heights of the two liquid surfaces.

Answer: To show more clearly how to set up the locations for calculating and comparing pressures, please reference to the following figure:

In case B, the pressures at location 1 and 2 are equal ||P_1=P_2|| and therefore $$P_0+\rho_wg(L-h)=P_0+\rho_ogL$$

In case C, the fast traveling air above the left side of the U-tube will create smaller pressure. The right side of the U-tube will still have atmosphere pressure (P_{0}) and will push the liquid up to the left side. Again, the pressures at location 1 and 2 are always equal: ||P_1=P_2||. So we have $$P+\rho_wgL=P_0+\rho_ogL.$$ Therefore $$P=P_0-(\rho_w-\rho_o)gL$$

To find the speed needed for creating the smaller pressure, we can set up the Bernoulli along the dashed line shown in the figure below, which gives $$P+\frac12\rho_{air}v^2=P_0$$

Here the vertical position (y) is identical on both sides of the equation so it is canceled. The speed of air on the right hand side is zero, so only have the ||P_0|| term.

Then we have $$\frac12\rho_{air}v^2=P_0-P=(\rho_w-\rho_o)gL.$$