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A closed U-tube manometer has a gas on one end with pressure 5 Pa and other end has vacuum, what is the approximate height difference of the liquid if the density of liquid is 10 Kg/m3?
  • a)
    5 mm
  • b)
    10 mm
  • c)
    50 mm
  • d)
    100 mm
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A closed U-tube manometer has a gas on one end with pressure 5 Pa and ...
Pressure of gas = Pressure due to height difference of liquid = d*g*h, ⇒ 5 = 10*10*h ⇒ h = 0.05 m = 50 mm.
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Community Answer
A closed U-tube manometer has a gas on one end with pressure 5 Pa and ...

Given data:
Pressure at one end of the U-tube manometer = 5 Pa
Density of the liquid = 10 Kg/m³

Assumption:
Acceleration due to gravity, g = 9.81 m/s²

Calculating height difference:
Let h be the height difference of the liquid in the U-tube manometer.

The pressure difference between the two ends of the manometer can be calculated using the formula:
ΔP = ρgh

Where:
ΔP = Pressure difference
ρ = Density of the liquid
g = Acceleration due to gravity
h = Height difference of the liquid

Given that the pressure at one end is 5 Pa and the other end is a vacuum, the pressure difference is:
ΔP = P₁ - P₂
ΔP = 5 - 0
ΔP = 5 Pa

Substitute the values in the formula:
5 = 10 * 9.81 * h
h = 5 / (10 * 9.81)
h ≈ 0.051 m

Converting the height from meters to millimeters:
h ≈ 0.051 m * 1000 mm/m
h ≈ 51 mm

Therefore, the approximate height difference of the liquid in the U-tube manometer is 50 mm, which corresponds to option C.
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A closed U-tube manometer has a gas on one end with pressure 5 Pa and other end has vacuum, what is the approximate height difference of the liquid if the density of liquid is 10 Kg/m3?a)5 mmb)10 mmc)50 mmd)100 mmCorrect answer is option 'C'. Can you explain this answer?
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