Pressure due to two liquids in a U-tube

In summary, the conversation discusses different situations where pressures on either side of a dashed line must be equal for equilibrium. In situation 2, it is not possible for the pressure due to the red liquid to be equal to the pressure due to the same length of air column. In situation 1, the pressure of the red liquid must be greater than the pressure of the grey liquid. In situation 3, the pressures are equal since the liquid columns have the same heights. In situation 4, the higher level of the grey liquid column indicates it has a lower density than the red liquid. The person asking the question wanted to confirm if their understanding was correct.
  • #1
brotherbobby
671
159
Homework Statement
The figure below shows four situations in which a red liquid and a gray liquid are in a U-tube. In one situation the liquids cannot be in static equilibrium. (a) Which situation is that? (b) For the other three situations, assume static equilibrium. For each of them, is the density of the red liquid greater than, less than, or equal to the density of the gray liquid?
Relevant Equations
(Gauge) pressure due to a liquid at a depth ##h## below the surface : ##P_G = \rho_L gh## where ##\rho_L## is the density of the liquid.
1579161497137.png


(a) Situation (2) is my answer (which cannot be in static equilibrium). The pressures in the lower dashed line (##P_2##) has to be the same on either side (for equilibrium). That means the pressure due to the red liquid is equal to the pressure due to the same length of air column (length), which is not possible. I explain a bit in the cropped diagram shown below. ##P_A## = Pressure due to air and ##\color{red}{P_{LR}}## = Pressure due to red liquid.
1579171720632.png


(b) In situation 1, I begin by drawing on it.
1579171987195.png

We can see that the pressure ##P_2 = P_{\text{atm}} + \color{red}{P_{LR}} = P_{\text{atm}} + \color{grey}{ P_{LG}}\Rightarrow \color{red}{P_{LR}} =\color{grey}{ P_{LG}}##. Clearly since the height of the grey column is less than the red, equal pressures mean ##\boxed{\color{red} {\rho_{LR}} < \color{grey}{\rho_{LG}}}##. (You will note that I have ignored the little column of air by which the grey liquid is below the red liquid. I am assuming the pressure of air to be small compared to those due to the liquids. Or, we could add more grey liquid on the right side so that it's height is infinitesimally less than that of the red liquid which would cancel the pressure due to air on both sides exactly.)

Situation 3 is shown alongside.
1579172687442.png

Arguments are the same as before, focussing on equality of pressure ##P_2## on both sides which imply again that ##\color{red}{P_{LR}} = \color{grey}{ P_{LG}}##. But in this case since the liquid colums have the same heights, we conclude that ##\boxed{\color{red} {\rho_{LR}} = \color{grey}{\rho_{LG}}}##.Situation 4 is shown alongside.
1579173536034.png

Agin the arguments are same as above. Focussing on pressure ##P_2##, we find that the higher level of the grey liquid column must imply that it has a lower density than the red liquid, or ##\boxed{\color{red} {\rho_{LR}} > \color{grey}{\rho_{LG}}}##.
 

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  • #2
Your solution looks fine, did you have a question?
 
  • #3
I wanted to ask if I was right.
 

Related to Pressure due to two liquids in a U-tube

What is the principle behind the pressure due to two liquids in a U-tube?

The principle behind the pressure due to two liquids in a U-tube is Pascal's law, which states that pressure in a closed container is transmitted equally in all directions.

How does the density of the liquids affect the pressure in a U-tube?

The density of the liquids plays a critical role in determining the pressure in a U-tube. The higher the density of a liquid, the higher the pressure it exerts at a given depth. This is because pressure is directly proportional to the density of the liquid.

What happens to the pressure in a U-tube when the heights of the liquids are unequal?

When the heights of the liquids in a U-tube are unequal, the pressure at the bottom of the U-tube will be different on each side. The side with the higher liquid level will have a higher pressure, while the side with the lower liquid level will have a lower pressure.

Can the pressure in a U-tube be used to measure the density of a liquid?

Yes, the pressure in a U-tube can be used to indirectly measure the density of a liquid. By measuring the difference in height of the two liquids and the pressure difference between the two sides, the density of the liquid can be calculated using the formula P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the height difference.

Is there any practical application of the pressure due to two liquids in a U-tube?

Yes, the principle of pressure due to two liquids in a U-tube has many practical applications, such as in barometers for measuring atmospheric pressure, in manometers for measuring pressure in pipes, and in blood pressure cuffs for measuring blood pressure in the human body.

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