A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle 5cm containing mercury and the parts on its two sides containing air at the same pressure p. when the tube is held at and angle of `60_(@)` with the vertical, the length of the air column above and belw the mercury pellet are 46cm and 44.5cm respectively. Calculate the ressure - in centimeters of mercury, The timperature 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.`
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