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Question

# A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal ends containing air at the same pressure Po. When the tube is held at an angle 60o with the vertical, the lengths of the air column above and below the mercury are 46 and 44.5 cm respectively. The pressure Po in cm of Hg is: (The temperature of the system is left at 30 K.)

A
80 cm of Hg
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B
75.4 cm of Hg
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C
70 cm of Hg
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D
90 cm of Hg
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Solution

## The correct option is B 75.4 cm of HgSince the total length of the tube remains same before and after tilting,2l+5=46+5+44.5When the tube is tilted,(Pbottom=Ptop+5cos60∘)cm of HgAlso PV=constant holds for each air column.Therefore, Pbottom=llbottomP0And Ptop=lltopP0Solving above four equations gives P0=75.4 cm of Hg

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