Mercury & Water Flow in U-Tube: Determining the Length & Height

[ronald29]

Mercury is poured into a U-tube. The left arm of the tube has cross-sectional area A1 of 10.0 cm2, and the right arm has a cross-sectional area A2 of 5.00 cm2. One hundred grams of water are then poured into the right arm.(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/cm3, what distance h does the mercury rise in the left arm?

 

[tiny-tim]

Welcome to PF!

Hi ronald29! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[nlightner]

(a) To determine the length of the water column in the right arm, we can use the equation for pressure in a fluid: P = ρgh, where P is pressure, ρ is density, g is the acceleration due to gravity, and h is the height of the fluid column.

First, we need to find the pressure at the bottom of the right arm, where the water is poured in. Since the water is poured in from the top, the pressure at the bottom is atmospheric pressure, which is typically 1 atm or 101,325 Pa.

Next, we need to find the pressure at the top of the water column. Since the water and mercury are in the same U-tube, they are in communication with each other and therefore have the same pressure at the same horizontal level. This means that the pressure at the top of the water column is also 1 atm.

Now, we can set up the equation: 1 atm = ρgh. We know the density of water (ρ = 1 g/cm3) and the acceleration due to gravity (g = 9.8 m/s2). We can convert the units to be consistent: 1 atm = (1 g/cm3)(9.8 m/s2)(100 cm/1 m)(1 Pa/1 N)h. Solving for h, we get h = 1.02 m.

Therefore, the length of the water column in the right arm is 1.02 m.

(b) To determine the distance h that the mercury rises in the left arm of the U-tube, we can use the equation for pressure again: P = ρgh. However, this time we need to find the pressure difference between the two arms of the U-tube.

The pressure at the top of the mercury column is atmospheric pressure (1 atm), and the pressure at the bottom of the mercury column is the pressure due to the weight of the water column (ρgh). So, the pressure difference is 1 atm - ρgh.

We know the density of mercury (ρ = 13.6 g/cm3) and the height of the water column (h = 1.02 m). We can use the same unit conversions as before to get 1 atm - (13.6 g/cm3)(9.8 m/s2)(100 cm/1 m)(1 Pa/1 N)(1.02 m) = 1