A uniform capillary tube closed at one end contained dry air trapped by a thread of mercury 8.5×10^-2m long.When the tube was held horizontally the length of the air column was 5×10^-2m,when it was held vertically with the closed end downwards the length was 4.5×10^-2m.Determine the value of the atmospheric pressure.g=10m/s, density of mercury 1.36×10^4kg/m^-3
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Answer is 0.08
Answer Is 1.04*10^5 pa
To determine the value of the atmospheric pressure, we can make use of two conditions: when the tube is held horizontally and when it is held vertically.
Let's analyze each condition separately:
1. When the tube is held horizontally:
In this case, the length of the air column is equal to the distance between the thread of mercury and the closed end of the tube, which is given as 5×10^-2m.
2. When the tube is held vertically:
When the tube is held vertically with the closed end downwards, the mercury in the tube exerts a pressure on the air column, causing it to compress. The height of the mercury column can be calculated as the difference between the lengths of the thread of mercury and the air column, which is 8.5×10^-2m - 4.5×10^-2m = 4×10^-2m.
Now, let's derive the equations to find the atmospheric pressure (P) using these conditions:
1. When the tube is held horizontally:
The pressure at the closed end of the tube (P1) is equal to the atmospheric pressure (P0) plus the pressure due to the column of air (P_a). Since the tube is open to the atmosphere, P_a can be considered as zero (P_a = 0). Therefore, P1 = P0.
2. When the tube is held vertically:
The pressure at the closed end of the tube (P1) is equal to the atmospheric pressure (P0) plus the pressure due to the column of air (P_a) plus the pressure due to the column of mercury (P_m). We can consider P1 as the pressure at the bottom of the mercury column, which can be calculated using the equation P = P0 + ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.
Therefore, P1 = P0 + ρ_mgh_m, where ρ_m is the density of mercury, g is the acceleration due to gravity, and h_m is the height of the mercury column.
Since P1 is equal to P0 in both conditions, we can equate the two equations:
P0 = P0 + ρ_mgh_m
Simplifying this equation, we obtain:
0 = ρ_mgh_m
Now we can solve for the atmospheric pressure (P0):
P0 = 0 / (ρ_mgh_m)
Using the given values:
ρ_m = 1.36×10^4 kg/m^3
g = 10 m/s^2
h_m = 4×10^-2 m
P0 = 0 / (1.36×10^4 kg/m^3*10 m/s^2*4×10^-2 m)
P0 = 0
Therefore, the value of the atmospheric pressure is zero.