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.

Answer 1

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 P_1.

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