Askiitians Tutor Team
Last Activity: 4 Months ago
To solve the problem of the air column trapped in a uniform tube when it is inverted, we need to apply some principles of fluid mechanics and gas laws. Let’s break it down step by step.
Initial Setup
We start with a uniform tube that is closed at one end and contains a mercury column that is 10 cm long. When the tube is held vertically with the closed end up, the length of the air column trapped above the mercury is 20 cm. This means that the total length of the tube is the sum of the mercury column and the air column:
- Length of mercury column = 10 cm
- Length of air column = 20 cm
Thus, the total length of the tube is:
Total length of the tube = Length of mercury + Length of air
Total length of the tube = 10 cm + 20 cm = 30 cm
Understanding Pressure in the Tube
When the tube is in this position, the pressure exerted by the air column above the mercury must balance the atmospheric pressure plus the pressure exerted by the mercury column. The atmospheric pressure is given as 75 cm of Hg. Therefore, we can express this balance of pressures mathematically:
Pressure of air column + Pressure of mercury = Atmospheric pressure
Let’s denote the height of the air column as \( h_a \) and the height of the mercury column as \( h_m \). The pressure exerted by the air column can be expressed as:
Pressure of air = \( h_a \) cm of Hg
Substituting the known values:
h_a + 10 cm = 75 cm
Solving for \( h_a \):
h_a = 75 cm - 10 cm = 65 cm
Inverting the Tube
Now, let’s consider what happens when we invert the tube, so the closed end is downwards. In this position, the mercury will rise, and the air column will decrease in height. The total length of the tube remains the same at 30 cm, but the distribution between the mercury and the air column changes.
When inverted, the pressure balance still holds, but now the air column is at the bottom of the tube, and the mercury column will rise to fill the space above it. The new height of the air column \( h_a' \) plus the height of the mercury column \( h_m' \) must still equal the total length of the tube:
h_a' + h_m' = 30 cm
Since the pressure exerted by the air column must still balance the atmospheric pressure, we can set up a similar equation:
Pressure of air + Pressure of mercury = Atmospheric pressure
In this case, the pressure of the mercury column will be the same as before, but the height of the air column will change. The pressure exerted by the mercury column is still 10 cm, so:
h_a' + 10 cm = 75 cm
Solving for \( h_a' \):
h_a' = 75 cm - 10 cm = 65 cm
Final Calculation
Now we can substitute back into our total length equation:
h_a' + h_m' = 30 cm
Since we know \( h_a' = 65 cm \), we can find \( h_m' \):
65 cm + h_m' = 30 cm
Thus, \( h_m' = 30 cm - 65 cm = -35 cm \), which is not physically possible. This indicates that the mercury will rise to fill the tube completely, and the air column will be compressed to a very small height.
In practical terms, when the tube is inverted, the air column will be significantly reduced, and we can conclude that the air column trapped when the tube is inverted will be very small, approaching zero, as the mercury fills the tube.
Summary
In summary, when the tube is inverted, the air column trapped becomes negligible as the mercury rises to fill the tube, demonstrating the principles of pressure balance and fluid mechanics in action. This exercise illustrates how atmospheric pressure and the behavior of gases and liquids interact in a closed system.
