- Thread starter nasi112
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It is interesting that you title this "mercury manometer" but your picture shows water, not mercury!

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It also shows mercury; apparently A is the interface between water and mercury.It is interesting that you title this "mercury manometer" but your picture shows water, not mercury!

The image is quite faded, and I don't want to strain my eyes. Can you explain your reasoning in words?

Since the pressure at point \(\displaystyle A\) and \(\displaystyle B\) are equal, we can use this formula

\(\displaystyle P_o + \rho_w gh_w = P_o + \rho_m gh_m\)

where \(\displaystyle P_o\) is the atmospheric pressure, in this case it is the Air Pressure. It is not a big deal since it will be cancelled.

the equation will become

\(\displaystyle \rho_w h_w = \rho_m h_m\)

\(\displaystyle \rho_w (h + 0.61 + 0.61) = \rho_m h\)

All I have to do now is to plug the values of water density, mercury density, and solve for h.

\(\displaystyle 1000h + 1220 = 13600h\)

\(\displaystyle h = 0.0968 \ m\)

for (b) i am thinking of \(\displaystyle 2 \ psi = 13800 \ N/m^2\)

for (c) I have to repeat part (a) with a new density \(\displaystyle = 0.85 \cdot 1000\)