A thin tube of uniform cross section has trapped air columns of lengths 46 and 44 cm above and below a middle pellet of mercury 4 cm in length when held at an angle of 60° to the vertical. When placed

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A thin tube of uniform cross section has trapped air columns of lengths 46 and 44 cm above and below a middle pellet of mercury 4 cm in length when held at an angle of 60° to the vertical. When placed horizontally, the columns are equal in length. The temperature is 27°C. The pressure of air columns, in cm Hg, when the tube is horizontal, is

TestHub · Accepted Answer

Tilted, the pellet's vertical height is 4cos 60° = 2.0 cm, so P lower = P upper + 2.0 cm Hg. Horizontal, each column is (46+44)/2 = 45 cm at common pressure P. Boyle's law: P u · 46 = 45P, Pₗ · 44 = 45P. Subtracting: 45P(1/44 - 1/46) = 2.0 ⇒ 0.04447P = 2.0 ⇒ P ≈ 45.0 cm Hg. Matches (a).

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