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Question

# A thin tube of uniform cross-section is sealed at both ends. When it lies horizontally, the middle 5 cm contains mercury and the two equal ends contain air at the same pressure P. When the tube is held at an angle of 60∘ with the vertical, then the length of the air columns above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure P in centimetres of mercury. (The temperature of the system is kept at 30 ∘C)

A
45.8
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B
75.4
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C
67.5
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D
79.3
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Solution

## The correct option is B 75.4Let a be the area of cross-section of the tube. When the tube is exactly horizontal 5 cm Hg is in the middle. Length L of the air column in tube will be, 2L+5=44.5+5+46 ∴L=45.25 cm Now making the tube at 60∘ with the vertical, the pressure difference across mercury will be, ΔP=P1−P2=5cos60∘ ∴ΔP=2.5 cm of Hg Using Boyle's Law, PV=P1V1=P2V2 P1=PVV1 & P2=PVV2 ∴P×45.25a44.5a−P×45.25a46a=2.5 ∴P=75.4 cm of Hg Hence, option (B) is correct.

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