Pressure measurement in U tube with mercury and water

[sluo]

Homework Statement

Mercury is poured into a U tube. The left arm of the tube has a cross sectional area A_1 of 10 cm^2 and the right arm has a cross sectional area A_2 of 5 cm^2. One hundred grams of water are then poured into the right arm of the tube.

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 of the U tube?

Homework Equations

ρ = mass/volume
P = P_0 + ρgh

The Attempt at a Solution

For part a, let h_w denote the height of the water column. I figured that since the density of water is 1g/cm^3, we can just do

1 * 5 * h_w = 100

Since they told us that the mass of the water is 100g. This gives us 20cm, which seems reasonable.

For part b, I'm totally stumped. The picture in the book has the original mercury level marked somewhere within the column of water (i.e. the column of water is higher than the mercury in the left arm). I don't know how to figure out how far down the water pushed the mercury in the right arm.

 

[LawrenceC]

Hint: When the U tube balances out, the pressure at the lowest point of bend due to the left column of Hg must equal the pressure due to the right column of Hg plus H2O. Pressure is density times depth.

 

[sluo]

I didn't think of that... but I'm wondering how it helps because we don't know the height of the U tube or the original height of the mercury in each arm.

 

[haruspex]

First, calculate the difference in mercury levels after the water is added.

 

[Nickie]

The pressure measurement in a U tube with mercury and water involves understanding the relationship between pressure, density, and height. In this scenario, the pressure in the U tube is equal at both arms due to the continuity of fluid flow. Therefore, we can use the equation P = P_0 + ρgh to calculate the height of the water column in the right arm and the distance the mercury rises in the left arm.

For part a, we can use the given information to calculate the height of the water column. As you correctly stated, the density of water is 1 g/cm^3 and the mass is 100g. Therefore, we can rearrange the equation ρ = mass/volume to solve for volume, which is equal to the cross-sectional area (A_2) multiplied by the height of the water column (h_w). This gives us A_2 * h_w = 100 cm^3. Plugging in the given value for A_2 (5 cm^2), we get h_w = 20 cm.

For part b, we can use the same equation, P = P_0 + ρgh, to calculate the distance the mercury rises in the left arm. The pressure at the bottom of the left arm is equal to the pressure at the bottom of the right arm (since they are connected), so we can set the two equations equal to each other:

P_0 + ρgh = P_0 + ρgh_w

We know the density of mercury (13.6 g/cm^3) and the height of the water column (20 cm) from part a. We are trying to solve for the height of the mercury column (h). Rearranging the equation, we get:

h = (ρgh_w - ρgh)/ρg

Plugging in the values, we get h = (13.6 g/cm^3 * 20 cm - 1 g/cm^3 * 20 cm)/13.6 g/cm^3 * 9.8 m/s^2 = 1.47 m.

Therefore, the mercury rises approximately 1.47 m in the left arm of the U tube. This is because the pressure exerted by the column of water is equal to the pressure exerted by the column of mercury, so the heights must be in proportion to their densities.