Question

Mercury is poured into a U-tube as shown in Figure P14.21a. The left arm of the tube has cross-sectional area A1 of 10.0 cm^2 and the right arm has a cross-sectional area A2 of 5.00 cm^2. One hundred grams of water are then poured into the right arm as shown in Figure P14.21b. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm^3, what distance h does the mercury rise in the left arm?

Answer 1

Well, water in the tube forms a cylinder of height H and the base area:
\[base.area = A_2 \]
The density of water is uniform. From the definition of density, where the greek letter, Rho, represents density. The mass m of water is related to its volume V, which for a cylinder can be expressed in terms of its height and the area of it's base.

\[m = \rho V = \rho A_2H \]
Where H = height. Now, just rearrange using algebra and solve for H!
For part b, you are use:
\[P = P_0 + \rho_{H_2O}gH \]
The mercury above the considered level also determines this pressure
\[P = P_0 + \rho_{mercury}g(H_1+H_2 \]
The displacement of mercury by water has not significantly changed the pressure distribution and therefore we can assume that this process did
not affect its volume. Hence, we can assume:
\[A_1h_1=A_2h_2\]
Now you obtained three equations with three unknowns, combined to solve for the height, h:
\[h_1 = \frac{ \rho_{H_2O} \times H }{ \rho_{Hg} \times \left( 1+\frac{ A_1 }{ A_2 } \right) }\]

Answer 2

@abb0t I can understand where H in part a came from but where did h1, h2, H1, and H2 come from in part b?