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Question
A uniform tube closed at one end, contains a pellet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure = 75 cm of mercury.
Solution
Let the CSA of the tube be A .
Initial volume of air , V1 = 20A cm = 0.2A
Length of mercury , h = 0.1 m
Let the pressure of the trapped air when the tube is inverted and vertical be P1.
Now , Pressure of the mercury and trapped air balances the atmospheric pressure . Thus ,
P1 + `0.1rho g` = `0.75rho g`
⇒ P1 = `0.65rho g`
when the tube is inverted with the closed end down , the pressure acting upon the trapped air is
Atmospheric pressure + Mercury column pressure
Now ,
Pressure of trapped air = Atmospheric Pressure + Mercury column Pressure [In equilibrium]
P2 = `0.75 rho g` + `0.1rho g` = `0.85 rho g`
Applying the Boyle's law when the temperature remains constant , we get
P1 V1 = P2V2
Let the new height of the trapped air be x .
⇒ `0.65 rho g`0.2A = `0.85rho g`xA
⇒ x = 0.15 m = 15 cm
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