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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 a...
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 a...
Solution:
Given: Gauge pressure of gas = 40 kPa, ρmercury = 13600 kg/m3, ρwater = 1000 kg/m3
We know that the pressure difference between the two arms of a manometer is given by:
ΔP = ρgh
where ΔP is the pressure difference, ρ is the density of the manometric fluid, g is the acceleration due to gravity, and h is the height difference between the two arms of the manometer.
Let us assume that the height difference between the two arms of the manometer when using mercury is h1. Therefore, we can write:
ΔPmercury = ρmercury gh1
Similarly, when using water as the manometric fluid, let the height difference between the two arms be h2. Therefore, we can write:
ΔPwater = ρwater gh2
We know that the gauge pressure of the gas remains the same in both cases, which means that the pressure difference between the two arms of the manometer must remain the same. Therefore, we can write:
ΔPmercury = ΔPwater
Substituting the values of ΔPmercury and ΔPwater, we get:
ρmercury gh1 = ρwater gh2
Simplifying the above equation, we get:
h2/h1 = ρmercury/ρwater
Substituting the values of ρmercury and ρwater, we get:
h2/h1 = 13600/1000 = 13.6
Therefore, the height difference between the two arms of the manometer when using water as the manometric fluid is 13.6 times the height difference when using mercury. This can also be expressed as a percentage change, which is given by:
% Change = [(New Value - Old Value)/Old Value] x 100
Substituting the values, we get:
% Change = [(13.6h1 - h1)/h1] x 100 = 1260%
Therefore, the height difference changes by 1260% when the manometric fluid is changed from mercury to water. Hence, the correct answer is option A.
Given: Gauge pressure of gas = 40 kPa, ρmercury = 13600 kg/m3, ρwater = 1000 kg/m3
We know that the pressure difference between the two arms of a manometer is given by:
ΔP = ρgh
where ΔP is the pressure difference, ρ is the density of the manometric fluid, g is the acceleration due to gravity, and h is the height difference between the two arms of the manometer.
Let us assume that the height difference between the two arms of the manometer when using mercury is h1. Therefore, we can write:
ΔPmercury = ρmercury gh1
Similarly, when using water as the manometric fluid, let the height difference between the two arms be h2. Therefore, we can write:
ΔPwater = ρwater gh2
We know that the gauge pressure of the gas remains the same in both cases, which means that the pressure difference between the two arms of the manometer must remain the same. Therefore, we can write:
ΔPmercury = ΔPwater
Substituting the values of ΔPmercury and ΔPwater, we get:
ρmercury gh1 = ρwater gh2
Simplifying the above equation, we get:
h2/h1 = ρmercury/ρwater
Substituting the values of ρmercury and ρwater, we get:
h2/h1 = 13600/1000 = 13.6
Therefore, the height difference between the two arms of the manometer when using water as the manometric fluid is 13.6 times the height difference when using mercury. This can also be expressed as a percentage change, which is given by:
% Change = [(New Value - Old Value)/Old Value] x 100
Substituting the values, we get:
% Change = [(13.6h1 - h1)/h1] x 100 = 1260%
Therefore, the height difference changes by 1260% when the manometric fluid is changed from mercury to water. Hence, the correct answer is option A.
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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 GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE 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 GATE 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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