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In a U-tube manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure 40 kPa. The height of the fluid column in the atmospheric side is 60 cm, and that on the gas side is 30 cm. The manometic fluid used is: (Take g = 9.8 m/s2).
- a)Water
- b)Liquid ammonia
- c)Oil
- d)Mercury
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In a U-tube manometer, one end is open to the atmosphere, the other e...
Gauge pressure = 40000 Pa. Height difference = 60 – 30 = 30 cm = 0.3 m. ρ*g*(h2 – h1) = 40000. We get, ρ = 13605 kg/m3 = Density of mercury.
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In a U-tube manometer, one end is open to the atmosphere, the other e...
Given data:
- The gauge pressure of the gas is 40 kPa.
- The height of the fluid column on the atmospheric side is 60 cm.
- The height of the fluid column on the gas side is 30 cm.
- Acceleration due to gravity, g = 9.8 m/s^2.
To find: The manometric fluid used.
Solution:
We know that the pressure difference between the two arms of the U-tube manometer is given by the difference in the heights of the fluid columns multiplied by the density of the manometric fluid and acceleration due to gravity. Mathematically,
ΔP = ρgh
where,
ΔP = pressure difference
ρ = density of manometric fluid
g = acceleration due to gravity
h = height difference between the fluid columns
Let's calculate the pressure difference in this case:
ΔP = (ρgh)gas - (ρgh)atm
where,
(ρgh)gas = pressure due to the gas
(ρgh)atm = pressure due to the atmosphere
Substituting the given values, we get:
ΔP = (ρgh)gas - (ρgh)atm
ΔP = (40 × 10^3 Pa) - (1 × 10^5 Pa)
ΔP = -60 × 10^3 Pa
Since the pressure difference is negative, the fluid column on the atmospheric side is higher than that on the gas side. This means that the density of the manometric fluid must be higher than that of the gas.
Now, let's check which of the given options satisfies this condition:
a) Water: Density of water = 1000 kg/m^3. Density of gas (air) = 1.2 kg/m^3. Hence, water cannot be the manometric fluid as its density is less than that of the gas.
b) Liquid ammonia: Density of liquid ammonia = 682 kg/m^3. This is a possible option as its density is higher than that of the gas.
c) Oil: The density of oil varies widely depending on the type of oil. However, most oils have a density lower than that of the gas. Hence, oil cannot be the manometric fluid.
d) Mercury: Density of mercury = 13600 kg/m^3. This is a possible option as its density is much higher than that of the gas.
Hence, the manometric fluid used in this U-tube manometer is mercury (option D).
- The gauge pressure of the gas is 40 kPa.
- The height of the fluid column on the atmospheric side is 60 cm.
- The height of the fluid column on the gas side is 30 cm.
- Acceleration due to gravity, g = 9.8 m/s^2.
To find: The manometric fluid used.
Solution:
We know that the pressure difference between the two arms of the U-tube manometer is given by the difference in the heights of the fluid columns multiplied by the density of the manometric fluid and acceleration due to gravity. Mathematically,
ΔP = ρgh
where,
ΔP = pressure difference
ρ = density of manometric fluid
g = acceleration due to gravity
h = height difference between the fluid columns
Let's calculate the pressure difference in this case:
ΔP = (ρgh)gas - (ρgh)atm
where,
(ρgh)gas = pressure due to the gas
(ρgh)atm = pressure due to the atmosphere
Substituting the given values, we get:
ΔP = (ρgh)gas - (ρgh)atm
ΔP = (40 × 10^3 Pa) - (1 × 10^5 Pa)
ΔP = -60 × 10^3 Pa
Since the pressure difference is negative, the fluid column on the atmospheric side is higher than that on the gas side. This means that the density of the manometric fluid must be higher than that of the gas.
Now, let's check which of the given options satisfies this condition:
a) Water: Density of water = 1000 kg/m^3. Density of gas (air) = 1.2 kg/m^3. Hence, water cannot be the manometric fluid as its density is less than that of the gas.
b) Liquid ammonia: Density of liquid ammonia = 682 kg/m^3. This is a possible option as its density is higher than that of the gas.
c) Oil: The density of oil varies widely depending on the type of oil. However, most oils have a density lower than that of the gas. Hence, oil cannot be the manometric fluid.
d) Mercury: Density of mercury = 13600 kg/m^3. This is a possible option as its density is much higher than that of the gas.
Hence, the manometric fluid used in this U-tube manometer is mercury (option D).
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In a U-tube manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure 40 kPa. The height of the fluid column in the atmospheric side is 60 cm, and that on the gas side is 30 cm. The manometic fluid used is: (Take g = 9.8 m/s2).a)Waterb)Liquid ammoniac)Oild)MercuryCorrect answer is option 'D'. Can you explain this answer?
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In a U-tube manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure 40 kPa. The height of the fluid column in the atmospheric side is 60 cm, and that on the gas side is 30 cm. The manometic fluid used is: (Take g = 9.8 m/s2).a)Waterb)Liquid ammoniac)Oild)MercuryCorrect answer is option 'D'. 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 manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure 40 kPa. The height of the fluid column in the atmospheric side is 60 cm, and that on the gas side is 30 cm. The manometic fluid used is: (Take g = 9.8 m/s2).a)Waterb)Liquid ammoniac)Oild)MercuryCorrect answer is option 'D'. 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 manometer, one end is open to the atmosphere, the other end attached to a pressurized gas of gauge pressure 40 kPa. The height of the fluid column in the atmospheric side is 60 cm, and that on the gas side is 30 cm. The manometic fluid used is: (Take g = 9.8 m/s2).a)Waterb)Liquid ammoniac)Oild)MercuryCorrect answer is option 'D'. Can you explain this answer?.
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