# A uniform tube closed at one end, contains a pellet of mercury 10cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20cm. Find the length of the air column trapped when the tube is inverted so that the colsed end goes down. Atmospheric pressure = 75cm of mercury.

Vikas TU
14149 Points
4 years ago
Dear student
Let the curved surface area of the tube be A
Initial volume of air, V1=20×A cm = 0.2A m
Length of mercury, h= 0.1 m
The pressure of the trapped air when the tube is inverted and vertical be P1
The pressure of the mercury and trapped air balances the atmospheric pressure
Thus, P1+0.1ρg=0.75ρgor, P1=0.65ρg
When the tube is inverted with the closed end down, the pressure acting upon theair is given by
Atomospheric pressure + Mercury column pressure
Pressure of trapped air = Atmospheric pressure +  Mercury column pressure
P2=0.75ρg+0.1g=0.85ρg
Applying the Boyle's law when the temperature remains constant,
we get P1V1=P2V2 or, 0.65ρg×0.2A=0.85ρghA
or h = 0.15 m = 15 cm