Question

A glass tube has an inner diameter of \[1mm\], outer diameter of \[1.1mm\]. It is kept vertical and partially dipped in water. Calculate the downward pull due to surface tension. (Surface tension of water =$75 dynes/cm$)

Answer 1

Hint: Surface tension is the property of liquids. The liquids tend to minimize the surface area due to the difference in the attractive force between the liquid-liquid surface and the liquid-air surface. Here we must calculate the force due to the surface tension.
Formula: $T=\dfrac{F}{L}$

Complete step-by-step answer:
We know that the surface tension $T$ is also defined as the force $F$ experienced due to unit length $L$ of the tube it is filled with.
 Mathematically, it is given as
$T=\dfrac{F}{L}$
Here, given that the has inner diameter of \[d_{i}=1mm\], outer diameter of \[d_{o}=1.1mm\], then $r_{i}=0.5mm=0.05cm$ and $r_{o}=0.55m=0.055cm$
Then the maximum length of the tube is given as $L=2\pi(r_{i}+r_{o})=2\pi(0.05+0.055)=2\pi(0.105)$
Also given that the surface tension of water is $T=75 dynes/cm$
Then the force $F=TL$
Substituting the values, we get, $F=75\times 2\times \pi\times 0.105=49.49 dynes$
Hence the force or the downward pull experienced by the liquid is \[49.49dynes\]

Additional information:
Surface tension is the tendency of liquids to occupy minimum surface area. It is defined as the force acting per unit length of the liquid. It can also be defined as the energy needed by the liquid to reduce its surface area. This is observed at the liquid- air interface of the liquid, where the force of attraction among the liquid molecules, is greater than that of the liquid-air force. At the surface of the liquid there are two forces, one is the liquid-liquid interaction which draws the liquid molecule inwards, and the other is the tangential force parallel to the liquid surface. This is responsible for the spherical shape of drops and meniscus formation of liquids when filled in tubes. These are also responsible for the separation of oil and water in the oil-water mixtures.

Note:
Surface tension is expressed in different forms and units. It can be expressed in different forms like $\gamma=\dfrac{dyn}{cm}=\dfrac{erg}{cm^{2}}=\dfrac{10^{-3}N}{m}=\dfrac{10^{-3}J}{m^{2}}$. At the surface of the liquid there are two forces, one is the liquid-liquid interaction which draws the liquid molecule inwards, and the other is the tangential force parallel to the liquid surface.