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Q. A thin tube of uniform cross section is sealed at both ends. It lies horizontally. The middle 5cm containing Hg and the two equal parts containing air at the pressure P0 . When the tube is held at an angle 60 with the vertical, the length of air column above and below the mercury are 46 and 44.5cm respectively. Calculate the pressure P0 in cm of Hg (the temperature of the system is kept at 30C ). Report answer after rounding off to the nearest integer value.

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
Let a is cross-sectional area I, II, I' , II' are sections of the tube.
Let P0 is initial pressure in both sections
PA and PB are pressures in section I' and II'
PBPA=5 cm×cos 60=25 cm ...(I)
PA×46×aRT=P0×4525×aRT=nI or nI
PA=P0452546 ...(II)
PB×445×aRT=P0×4525×aRTnII or nII
PB=P04525445 ...(III)
IIIIIPBPA=P04525[14451460]=25
\text{P}_{0} \frac{4 5 \cdot 2 5 \times 1 \cdot 5}{4 4 \cdot 5 \times 4 6 \cdot 0} = 2 \cdot 5
\text{P}_{0}=\frac{2 \cdot 5 \times 4 4 \cdot 5 \times 4 6 \cdot 0}{4 5 \cdot 2 5 \times 1 \cdot 5}=75\text{ cm}=\text{P}_{0}