Question

A glass capillary sealed at the upper end is of length 0.11 m and internal diameter$2\times {10}^{-5}m$. The tube is immersed vertically into a liquid of surface tension $5.06x{10}^{-2}N/m$.To what length in cm has the capillary to be immersed so that the liquid levels inside and outside the capillary become the same ? What will happen to the water levels inside the capillary if the seal is now broken?

Answer 1

Let $A$ be the area of the cross section of the tube and $l$ be the length of the tube. Then initial volume of air inside tube is $V_1=Al$ and initial pressure is $P_o$ (atmospheric pressure).
When the tube is immersed in liquid to a distance $x$, then level of liquid inside and outside are same as shown in figure. Now volume of air inside the tube is $V_2=A(l-x)$ and pressure inside the tube is $P_2=P_o+\frac{2T}{r}$
According to Boyle's law, we have
$P_{o}Al=(P_o+\frac{2T}{r})A(l-x)\Rightarrow x=\frac{l}{(1+\frac{r{P}_o}{2T})}$

on substituting the values, we have
$x=\frac{0.11}{(1+\frac{{10}^{-5}\times 1.012\times {10}^5}{2\times 5.06\times {10}^{-2}})}=0.01m=1cm$

When the seal is broken, pressure inside the tube is equal to atmospheric and rise of liquid through $h$ in the tube is given by, $h\rho g=\frac{2T}{r}$
$\Rightarrow h=\frac{2T}{r\rho g}$, on substituting the values, we have
$\Rightarrow h=\frac{2\times 5.06\times {10}^{-2}}{{10}^{-5}\times 1000\times 9.8}=1.03m$
but length of the tube outside liquid is $(l-x)=(0.11-0.01)m=0.1m$, so the tube will be of insufficient length. So the liquid will rise to the top.