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`).

Explanation

Length of pipe = `(46+44.5 +5) cm = 95.5 cm `
Suppose the pressure in both the columns of air in the horizontal the pressure in both the columns of air in the horizontal position be P .
` :. l = (46 +44.5)//2 = 45.25 cm `
When the tube is inclined , let `P_(1) and P_(2)` be the pressure and `l_(1) and l_(2)` be the lengths of lower and upper air columns respectively .
Using Boyle.s law ,
`PA l_(1) = P_(1) Al_(2) = P_(2)Al_(2)` (where A is area of cross - section of the tube )
Using Boyle.s law
`PA l = P_(2)Al_(2)`
`P_(1) = P_(2) ++rhoH_(g)g .l _(g) cos 60^(@)`
` (P_(2) + rho H_(g)g l _(Hg) cos 60^(@))l_(1) =P_(2)l_(2)` [ From equaton (1)]
`P_(2) = (l_(Hg)xx rho _(Hg) g xx l_(1))/(2(l_(2)-l_(1))`
`P = (l_(Hg)xxrho_(Hg) xx g xx l_(1)l_(2))/(2(l_(2)-l_(1))xxl) Pa = (l_(Hg)xxl_(1)xxl_(2))/(2(l_(2)-l_(1))xxl) ` cm of Hg column
` = 75.4 ` cm of Hg column