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`

Explanation

From the two figures we can see that

`2x+5 = 44.5 + 5 + 46`
`:." "x = 45.25 cm`
Let be the area of cross-section of the tube. Process is given isothermal. Hence, apply pV = constant in two sides of mercury column.
`pAx = P_1 A (44.5)" ....(i)"`
Similarly, `PAx = p_a A(46)" ...(ii)"`
or `P (45.25)=P_1 (46)" ...(iii)"`
From figure (b).
`p_2 = p_1 + 5 sin 30^@ " ....(iii)"`
Solving these three equations, we get
`p = 75.4" cm of Hg"`