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 ends containing air at the same pressure `P`. When the tuve is held at an angle of `60^(@)C` with the vertical 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

When the tube is kept inclined to the vertical, the length if the upper part is` l_(1)=46cm` and that of the lower part is `l_(2)=44.5cm`. When the tube lies horizontally, the length on each side is
`l_(0)=(l_(1)+l_(2))/(2)=(46cm+44.5cm)/(2)=45.25cm.`
` Let p_(1) and p_(2) `be the pressures in the upper and the lower parts when the tube is kept inclined, As the temperature is constant throughout, we can apply Boyle's law. For the upper part,
`p_(1)l_(1)A=pl_(0)A`
or, ``p_(1)=(pl_(0))/(l_(1))` ...(i)`
Similarly, for the lower part,
`p_(2)=(pl_(0))/(l_(2))` ...(ii)
Now consider the eqilibrium of the mercury pellet when the tube is kept in inclined pisition, Let m be the mass of the mercury. The forces along the length of the tube are
(a) `p_(1)`A down the tube
(b) `p_(2)`up the tube
and (c) mg cos `60^(@)` down the tube.
Thus ,`p_(2)=p_(1)+(mg)/(A) cos 60_(@).`
Putting from (i) and (ii),`
`(pl_(0))/(l_(2))=(pl_(0))/(l_(1))+(mg)/(2A)`
or,`pl_(0)((1)/(l_(2))-(1)/(l_(1)))=(mg)/(2A)`
`or, p=(mg)/((2Al_(0))((1)/(l_(2))-(1)/l_(1))`
If the pressure p is equal to a height h if mercury,
`p=hrhog.`
`also, m=(5cm)Arho`
`so that hrhog=((5cm)Arhog)/(2Al_(0)((1)/(l_(2))-(1)/(l_(1)))`
or, `h=(5cm)/(2(45.25cm)((1)/(44.5cm)-(1)/(46cm))`
'=75.39 cm.`
The pressure p is equal to 75.39cm of mercury.`
`