Question

When a capillary tube is dipped in a liquid, the liquid rises to a height $h$ in the tube. The free liquid surface inside the tube is hemispherical in shape. The tube is now pushed down so that the height of the tube outside the liquid is less than $h$. Then

• A. The liquid will come out of the tube like in a small fountain
• B. The liquid will ooze out of the tube slowly
• C. The liquid will fill the tube but not come out of its upper end
• D. The free liquid surface inside the tube will not be hemispherical

Answer 1

Answer: (C), (D)

The correct options are
C The liquid will fill the tube but not come out of its upper end
D The free liquid surface inside the tube will not be hemispherical

Consider a capillary tube of radius r immersed in a liquid of surface tension $T$ and density $\rho$. The height to which the liquid will rise in the capillary tube is given as

$h=\frac{2Tcos\theta }{R\rho g}=\frac{2T}{R\rho g}$

or

$hR=\frac{2T}{\rho g}$ which is a constant.

Thus we get,

$h_1{R}_1=h_2{R}_2$

When the tube is pushed down we are increasing the $h$, thereby reducing the radius $R$ of the of the liquid meniscus. Thus as $h$ increases, the level of the liquid becomes more and more flat but does not overflow.

The angle of contact at the free liquid surface inside the capillary tube will change such that the vertical component of the surface tension forces just balance the weight of the liquid column.