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

$\begin{array}{l}\text{Let the CSA of the tube be A.}\\ {\text{Initial volume of air, V}}_1\text{= 20A cm = 0}\text{.2A}\\ \text{Length of mercury, h = 0}\text{.1 m}\\ \\ \mathrm{Let\; the}\text{pressure of the trapped}\mathrm{air}\text{w}\mathrm{hen\; the\; tube\; is\; inverted\; and\; vertical}{\text{be P}}_1\text{.}\\ \text{Now, pressure of the mercury and trapped air}\mathrm{balances}\mathrm{the}\mathrm{atmospheric\; pressure}.\mathrm{Thus},\\ \\ P_1+0.1\rho g=0.75\rho g\\ \Rightarrow P_1=0.65\rho g\\ \text{When the tube is inverted with the closed end down, the pressure acting upon the trapped air is given by}\\ \\ \text{Atmospheric pressure + Mercury column pressure}\\ \mathrm{Now},\\ \text{Pressure of trapped air = Atmospheric pressure + Mercury column pressure}\left[\text{In equilibrium}\right]\\ P_2=0.75\rho g+0.1\rho g=0.85\rho g\\ \text{Applying the Boyle's law when the t}\mathrm{emperature\; remains}constant,\mathrm{we}\mathrm{get}\\ P_1{V}_1=P_2{V}_2\\ \text{Let the new height of the trapped air be x.}\\ \Rightarrow 0.65\rho g\text{0}\text{.2A =}0.85\rho gxA\\ \Rightarrow x=0.15\text{m = 15 cm}\end{array}$