# Mercury Manometer

#### nasi112

##### Junior Member

I don't understand why my calculations are wrong. Can anyone help me?

#### HallsofIvy

##### Elite Member
It is interesting that you title this "mercury manometer" but your picture shows water, not mercury!

#### Dr.Peterson

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

View attachment 26044

I don't understand why my calculations are wrong. Can anyone help me?
The image is quite faded, and I don't want to strain my eyes. Can you explain your reasoning in words?

#### nasi112

##### Junior Member
Thanks a lot HallsofIvy and Dr.Peterson for giving me some of your valuable time by looking at this problem which perhaps is not for mathematicians.

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 $$\displaystyle = 0.85 \cdot 1000$$