Q#34 (Past Exam Paper – November 2015 Paper 11 Q19)
A U-tube closed at one end contains mercury. Air at a pressure of 5.0×104 Pa is trapped at the closed end. The other end is open to the atmosphere and is fitted with a piston of mass 5.0 kg and cross-sectional area 5.0×10-4 m2.
The density of mercury is 13 600 kg m-3 and atmospheric pressure is 1.01×105 Pa.
What is the height h of the mercury column?
A 37 cm B 44 cm C 74 cm D 110 cm
Solution:
Answer: D.
Pressure of liquid = hρg
At the same level in the liquid mercury, the pressure on both sides of the U-tube should be the same.
Consider the pressure at the lower level indicated – where the piston is in contact with the liquid mercury.
Pressure at left end = Pressure at right end
The pressure at the left is due to the weight of the piston and due to atmospheric pressure above the piston.
The pressure at the right is due to the length h of the column of mercury and due to the trapped air above the mercury.
The pressure of the trapped air plus the pressure of the mercury column of height h must equal the total pressure supplied by the piston and the atmosphere.
Note:
Pressure due to piston = Force / Area = Weight / Area = mg / A
where m is the mass of the piston and A its area
Pressure at left end = Pressure at right end
Pressure due to piston and atmosphere = Pressure due to mercury column and trapped air
mg/A + Patm = hρg + Pair
[(5×10)/5.0×10-4] + 1.01×105 = (h × 13600 × 9.8) + 5.0×104
Solving gives
Height h = 110 cm
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