A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle 10 cm containing mercury and the parts on its two sides containing air at the same pressure P. When the tube is held vertically such that, the length of the air column above and below the mercury pellet are 36 cm and 34 cm respectively. Calculate the pressure P, if the temperature of the system is kept at 20∘C.
When the tube is kept vertical, the length of the upper part is l1=36 cm, and that of the lower part is l2=34 cm. When the tube lies horizontally, the length on each side is -
l0=[l1+l22]=[36 +34 2]=35 cm.
Let P1 and P2 be the pressures in the upper and the lower parts when the tube is vertical.
As the temperature is constant throughout, we can apply Boyle's law.
For the upper part,
P1l1A=Pl0A
⇒P1=(Pl0l1)....(i).
Similarly, for the lower part,
P2=(Pl0l2)....(ii).
Thus,
P2=P1+mgA
Putting from (i) and (ii),
(Pl0l2)=(Pl0l1)+(mgA)
⇒Pl0(1l2−1l1)=mgA
⇒P=⎡⎢
⎢
⎢
⎢⎣mgAl0(1l2−1l1)⎤⎥
⎥
⎥
⎥⎦.
If the pressure P is equal to a height h of mercury,
P=ρgh.
Also mas of the mercury pellet is m=(10 cm)Aρ
⇒ρgh=⎡⎢
⎢
⎢
⎢⎣(10 cm)AρgAl0(1l2−1l1)⎤⎥
⎥
⎥
⎥⎦
⇒h=⎡⎢
⎢
⎢
⎢⎣(10)(35)×(134−136)⎤⎥
⎥
⎥
⎥⎦=174.8 cm.
The pressure P is equal to 174.8 cm of Hg.