Question

An incompressible liquid is filled in a container. There is air above the liquid and a massless movable piston with a hole is also placed on the air. A thin capillary tube of inner radius 0.1 mm is dipped vertically into the liquid through the hole on the piston as shown in the figure. The air in the container is isothermally compressed from the original volume V0 to $\frac{100}{101}$V0 with the movable piston. Considering air as an ideal gas, the height (h) of the liquid column in the capillary above the liquid level in cm is:

(given Twater =0.075N/m, P0=105N/m2 , g=10m/sec2, $\rho l$=103 kg/m3, $\theta=0$)

Answer 1

T=constant $\Rightarrow$PV=constant

$P\propto\frac{1}{V}$, volume =$\frac{100}{101}$ times, so pressure = $\frac{101}{100}$times=1.01 times\Rightarrow$$P_f=1.01P_0

$P_0-\frac{2T}{R}=1.01P_0-\rho gh\,\,\,\,\Rightarrow\,\,\,\rho gh=0.01P_0+\frac{2T}{R}$

$(10)^3(10)h=(0.01)(10^5)+\frac{2\times 0.075}{0.1\times10^{-3}}\Rightarrow h=\frac{2500}{10^4}m=25\,cm$