Fluid mechanics barometer problem

In summary, when constructing a barometer, a tube of length 1m is filled completely with mercury and inverted in a mercury cup. The barometer reading on a particular day is 76cm. However, when a 1m tube is filled with mercury up to 76cm and inverted, the height of the mercury column in the tube is <76cm. This is because in the latter case, there is still air remaining in the tube at atmospheric pressure, causing the overall weight of the mercury to be less and resulting in a lower mercury level.
  • #1
anmolnanda
20
0
to construct a barometer,a tube of length 1m is filled completely with mercury and is inverted in a mercury cup.the barometer reading on a particular day is 76cm.suppose a 1m tube is filled with mercury upto 76cm and then closed by a cork.It is inverted in a mercury cup and the cork is removed.Height of mercury column in the tube over the surface is
<76cm
why??
i want to know why it is <76cm as the books did not explain this
 
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  • #2
If it is filled completely, when you invert it the weight of the mercury is counteracted by the atmospheric pressure pushing down on the fluid in the cup (I assume you know how a barometer works).

So the fluid level drops until the weight of the fluid in the tube = the force of the atmospheric pressure. As the fluid level drops, it leaves a vacuum in above it in the tube.

Now, when you only fill it to 76cm there is air remaining in the tube at atmospheric pressure. This means that you have the weight of the mercury acting plus the force applied by this air at atmospheric pressure.

You still only have the atmospheric pressure acting on the fluid in the cup as above, so this force now has to equal the weight of the mercury plus the force applied by the air inside the tube. Because the force from the internal air is constant, the mercury level has to be lower. (Mercury level is lower which means overall weight of mercury is less. This weight plus the atmospheric air = external air pressure.)
 

FAQ: Fluid mechanics barometer problem

1. What is a fluid mechanics barometer problem?

A fluid mechanics barometer problem is a mathematical problem that involves using the principles of fluid mechanics to determine the pressure at a specific point in a fluid column. It typically involves a column of fluid (such as water or mercury) that is subjected to different pressures at different points, and the goal is to calculate the pressure at a specific height in the column.

2. How do you solve a fluid mechanics barometer problem?

To solve a fluid mechanics barometer problem, you need to use the fundamental principles of fluid mechanics, such as Pascal's law and Archimedes' principle. You also need to use equations such as Bernoulli's equation and the hydrostatic equation. These equations allow you to relate pressure, density, height, and other variables in the fluid column to solve for the unknown pressure.

3. What are some real-world applications of fluid mechanics barometer problems?

Fluid mechanics barometer problems have many practical applications, including weather forecasting, determining the pressure in pipelines and hydraulic systems, and measuring the depth of bodies of water. They are also used in the design of buildings and structures to ensure they can withstand the pressure of fluids inside or outside the structure.

4. What are the units of measurement for pressure in a fluid mechanics barometer problem?

The most common units of measurement for pressure in a fluid mechanics barometer problem are Pascals (Pa) or atmospheres (atm). In some cases, other units such as pounds per square inch (psi) or millimeters of mercury (mmHg) may be used.

5. Are there any common mistakes to avoid when solving a fluid mechanics barometer problem?

One common mistake to avoid when solving a fluid mechanics barometer problem is forgetting to convert units consistently. Make sure all the values in the problem are in the same units before plugging them into equations. It's also essential to pay attention to the direction of pressure changes and to use the correct equation for the specific situation.

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