Question

If a 5 cm long capillary tube with $ 0.1 $ mm internal diameter, open at both ends, is slightly dipped in water having surface tension 75 dyne/cm, would the water overflow out of the upper end of the capillary? Explain your answer.

Answer 1

Hint
To check whether the liquid will overflow, compare the height to which the liquid will rise in the given conditions with the height of the tube.The radius of curvature of the meniscus adjusts itself according to the height of rise of the liquid.
$\Rightarrow h = \dfrac{{2T\cos \theta }}{{\rho Rg}} $
where ‘ $ {{\rho }} $ ’ is the density of water, ‘ $ {{\theta }} $ ’ is the angle of contact made by water with the capillary’s surface which is $ {{0}}^\circ $ and ‘ $ {{g}} $ ’ is the gravitational acceleration given as $ 981 cm/{s^2} $ .

Complete step by step answer
We are provided with the following data:
Length of the tube: $ L = 5cm $
Surface tension: $ T = 75dyne/cm $
Since, we are provided with the diameter of the tube, we need to convert it into radius for our calculation purposes:
Radius of the tube: $ R = \dfrac{{0.1}}{2} = 0.05mm = 0.005cm $ $ [\because 1cm = 10mm] $
It is important to remember to use all the units from the same family.
We know that the height up to which water rises in a capillary tube is given by the formula:
 $\Rightarrow h = \dfrac{{2T\cos \theta }}{{\rho Rg}} $
Putting the given values of $ T $ , $ \theta $ , R, g and $ \rho $ in this equation and solving it, we get
 $\Rightarrow h = \dfrac{{2 \times 75 \times \cos 0^\circ }}{{1 \times 0.005 \times 981}}$
 $\Rightarrow h = \dfrac{{150}}{{4.905}} $
 $\Rightarrow h = 30.58cm $
We can see that the length of the tube is less than the height up to which the water is supposed to rise. But this does not mean the water will overflow because in the case of a capillary, a meniscus layer with a radius of curvature R’ is formed to keep the water from flowing out of the tube.
Since the question asks us to explain this further, we will now try to find the radius of the curvature of the meniscus formed.
We know that the product of height of the tube and radius of curvature of the meniscus always remains constant, hence:
 $\Rightarrow R'L = rh $
where ‘ $ r $ ’ is the ideal radius of curvature for the liquid when the length of the tube is sufficient enough. We also know that
 $\Rightarrow r = \dfrac{R}{{\cos \theta ^\circ }} = \dfrac{R}{{\cos 0^\circ }} = R $
To find R’ we put the known values in the equation:
 $\Rightarrow R' = \dfrac{{rh}}{L} = \dfrac{{Rh}}{L} $ $ [\because r = R'] $
Solving for R’, we get
 $\Rightarrow R = \dfrac{{0.005 \times 30.58}}{5} $
 $\Rightarrow R = \dfrac{{0.1529}}{5} = 0.03058cm $
 $ \therefore $ The radius of curvature of the meniscus is approximately $ 0.0306cm $ .

Note
It is wrong to assume that water will overflow out of the tube since the height is smaller than the actual length of the tube. One should remember that water molecules adhere to the surface of the tube and rise higher. As soon as the tube length is at its max, water will form a meniscus layer to keep the adhesion forces intact.