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 end containing air at the same pressure P. When the tube is held at an angle of `60^@` with the vetical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure P in centimeters of mercury. (The temperature of the system is kept at `30^@C`).

Answer

`75.4` cm of Hg

Explanation

`P'' = P' + 5 "cos"60^(@) = (P' + 2.5) "cm of Hg`
For constant temperature process: `P xx ((46 + 44.5)/(2)) = P' xx 46 = (P' + 2.5)(44.5)`
`implies P' = (44.5 xx 2.5)/(1.5)` & `P = 75.4 "cm og Hg"`