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Question

A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle 5cm containing mercury and the two equal ends containing air at the same pressure P. When the tube is held at an angle 60o with the vertical direction, the length of the air column above and below the mercury column are 46cm and 44.5 cm respectively. Calculate the pressure P in cm of mercury.(The temperature of the system is kept at 30oC)

A
85.4 cm of Hg column.
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B
75.4 cm of Hg column.
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C
65.4 cm of Hg column.
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D
95.4 cm of Hg column.
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Solution

The correct option is B 75.4 cm of Hg column.Solution:-Let the length of the air column initially in the tube be lcm.Since the total length of the tube remains the same before and after tilting,∴2l+5=46+44.5⇒l=45.25cmWhen the tube is held vertically at an angle of 60º, the Hg column will be displaced to the lower end. If the pressure of top and bottom ends be PA and PB respectively, thenPB=PA+5cos60°⇒PB−PA=5×12=2.5.....(1)As PV=constant holds for each air column.Therefore, for end A,PA=llAP0⇒PA=45.2546P0.....(2)Now, for end B,PB=llBP0PB=45.2544.5P0.....(3)Substituting the values of PA and PB in eqn(1), we have45.2544.5P0−45.2546P0=2.5⇒P0=2.5×46×44.545.25=75.4 cm of HgHence the pressure P is 75.4 cm of Hg.

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