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A thin tube of uniform cross-section is sealed at both ends. It lies horizontally. The middle \( 5 c m \) contains \( \mathrm{Hg} \) and two equal ends contain air at the same pressure \( P_{0} . \) When the tube is held at an angle of \( 60^{\circ} \) with the vertical, the length of the air column above and below the Hg are \( 46 \mathrm{cm} \) and \( 44.5 \mathrm{cm} \). Calculate pressure \( P_{0} \) in cm of Hg. Assume temperature of the system to be constant.

1. \( 55 \mathrm{cm} \) of \( \mathrm{Hg} \)
2. \( 65 \mathrm{cm} \) of \( \mathrm{Hg} \)
3. \( 70.4 \mathrm{cm} \) of \( \mathrm{Hg} \)
4. \( 75.4 \mathrm{cm} \) of \( \mathrm{Hg} \)

Correct answer: \( 70.4 \mathrm{cm} \) of \( \mathrm{Hg} \)

Solution

Each air pocket is compressed/expanded isothermally, so its pressure changes inversely with its length. The two pressures differ by the hydrostatic head of the mercury column along the vertical direction, which lets you solve for the original pressure.

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