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`).

Options

A. 75.4 cm of Hg
B. 7.54 cm of Hg
C. 754 cm of Hg
D. 75.4 mm of Hg

Answer

A

Explanation

At horizontal position, let the length of air column in tube be L cm.
`:. 2L + 5 = 46 + 5 + 44.5 cm`
`L = 45.25 cm`
When the tube is held at `60^(@)` with the vertical, the mercury column will slip down.
`P_(Y) + 5 cos 60^(@) = P_(X)`
`P_(X) - P_(Y) = (5)/(2) = 2.5 cm Hg`.....(i)
From end X, `P_(0) xx 45.25 = P_(X) xx 44.5`
`P_(X) = (45.25)/(44.5) P_(0)`.....(ii)
From end Y, `P_(0) xx 45.25 = P_(Y) xx 46`
`P_(Y) = (45.25)/(46) P_(0)`.....(iii)
Substituting the values of `P_(X) " and " P_(Y)` in equation (i) we get
`P_(0) = 75.4`