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In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m3, ρwater = 1000 kg/m3).
- a)1260%
- b)92.64 %
- c)Remains unchanged (0%)
- d)13.6%
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
In a U-tube mercury manometer, one end is exposed to the atmosphere an...
Assuming the same vertical distance between the two ends of the manometer and the same diameter of the tube, the height difference between the two fluids will change due to the difference in their densities. The density of mercury is approximately 13.6 times greater than the density of water.
Initially, the gauge pressure of the gas was balanced by the weight of the mercury column in the manometer. Let's assume that the height difference between the two mercury levels was h_m. Then, the pressure difference between the two ends of the manometer was:
ΔP_m = ρ_mgh_m
where ρ_m is the density of mercury, g is the acceleration due to gravity, and h_m is the height difference between the two mercury levels.
The gauge pressure of the gas was 40 kPa, which is equivalent to:
ΔP_gas = 40,000 Pa
When we change the manometric fluid to water, the pressure difference between the two ends of the manometer will be balanced by the weight of the water column. Let's assume that the height difference between the two water levels is h_w. Then, the pressure difference between the two ends of the manometer will be:
ΔP_w = ρ_wgh_w
where ρ_w is the density of water and h_w is the height difference between the two water levels.
The pressure difference between the two ends of the manometer must remain the same, regardless of the manometric fluid used. Therefore:
ΔP_m = ΔP_w
ρ_mgh_m = ρ_wgh_w
Solving for h_w, we get:
h_w = h_m(ρ_m/ρ_w)
Substituting the densities of mercury and water, we get:
h_w = h_m(13.6/1)
h_w = 13.6h_m
Therefore, the height difference between the two water levels will be 13.6 times greater than the height difference between the two mercury levels. If the height difference between the two mercury levels was, for example, 10 cm, then the height difference between the two water levels would be:
h_w = 13.6(10 cm) = 136 cm
So, the height difference changes by a factor of 13.6.
Initially, the gauge pressure of the gas was balanced by the weight of the mercury column in the manometer. Let's assume that the height difference between the two mercury levels was h_m. Then, the pressure difference between the two ends of the manometer was:
ΔP_m = ρ_mgh_m
where ρ_m is the density of mercury, g is the acceleration due to gravity, and h_m is the height difference between the two mercury levels.
The gauge pressure of the gas was 40 kPa, which is equivalent to:
ΔP_gas = 40,000 Pa
When we change the manometric fluid to water, the pressure difference between the two ends of the manometer will be balanced by the weight of the water column. Let's assume that the height difference between the two water levels is h_w. Then, the pressure difference between the two ends of the manometer will be:
ΔP_w = ρ_wgh_w
where ρ_w is the density of water and h_w is the height difference between the two water levels.
The pressure difference between the two ends of the manometer must remain the same, regardless of the manometric fluid used. Therefore:
ΔP_m = ΔP_w
ρ_mgh_m = ρ_wgh_w
Solving for h_w, we get:
h_w = h_m(ρ_m/ρ_w)
Substituting the densities of mercury and water, we get:
h_w = h_m(13.6/1)
h_w = 13.6h_m
Therefore, the height difference between the two water levels will be 13.6 times greater than the height difference between the two mercury levels. If the height difference between the two mercury levels was, for example, 10 cm, then the height difference between the two water levels would be:
h_w = 13.6(10 cm) = 136 cm
So, the height difference changes by a factor of 13.6.
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Community Answer
In a U-tube mercury manometer, one end is exposed to the atmosphere an...
Since the gauge pressure remains the same ρ*(h2 – h1) = constant. The height difference in mercury manometer is 0.30 m and that in a water manometer is 4.08 m. Percent change is thus, 1260%. Be careful about the denominator used for computing percent change.
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In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m3, ρwater = 1000 kg/m3).a)1260%b)92.64 %c)Remains unchanged (0%)d)13.6%Correct answer is option 'A'. Can you explain this answer?
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In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m3, ρwater = 1000 kg/m3).a)1260%b)92.64 %c)Remains unchanged (0%)d)13.6%Correct answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m3, ρwater = 1000 kg/m3).a)1260%b)92.64 %c)Remains unchanged (0%)d)13.6%Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a U-tube mercury manometer, one end is exposed to the atmosphere and the other end is connected to a pressurized gas. The gauge pressure of the gas is found to be 40 kPa. Now, we change the manometric fluid to water. The height difference changes by: (ρmercury = 13600 kg/m3, ρwater = 1000 kg/m3).a)1260%b)92.64 %c)Remains unchanged (0%)d)13.6%Correct answer is option 'A'. Can you explain this answer?.
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