A thin tube of uniform cross-section is sealed at both ends. It lies horizontally. The middle contains and two equal ends contain air at the same pressure When the tube is held at an angle of with the vertical, the length of the air column above and below the Hg are and . Calculate pressure in cm of Hg. Assume temperature of the system to be constant.

Physics · Pressure due to a fluid column, Pascal's law and its applications, effect of gravity on fluid pressure

A thin tube of uniform cross-section is sealed at both ends. It lies horizontall

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.

A. \( 55 \mathrm{cm} \) of \( \mathrm{Hg} \)
B. \( 65 \mathrm{cm} \) of \( \mathrm{Hg} \)
C. \( 70.4 \mathrm{cm} \) of \( \mathrm{Hg} \)
D. \( 75.4 \mathrm{cm} \) of \( \mathrm{Hg} \)

Step-by-step solution

Initial horizontal: total length = 46+5+44.5 = 95.5 cm, so each air column initially 45.25 cm. Using Boyle's law: P_upper * 46 = P0 * 45.25 and P_lower * 44.5 = P0 * 45.25. Pressure difference across Hg column due to vertical height = 5 cos60° = 2.5 cm Hg. So P_lower - P_upper = 2.5. Solve: P0 * 45.25*(1/44.5 - 1/46) = 2.5 => P0 ≈ 75.4 cm Hg.

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Step-by-step solution

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