Question

If \[\sigma \] is surface tension the work done in breaking a big drop of radius, \[R\] into \[n\] drops of equal radius is:
A. \[R{{n}^{\dfrac{2}{3}}}\sigma \]
B. \[\left( {{n}^{\dfrac{2}{3}}}-1 \right)\sigma {{R}^{2}}\]
C. \[4\pi {{R}^{2}}\left( {{n}^{\dfrac{1}{3}}}-1 \right)\sigma \]
D. \[\pi {{R}^{2}}\left( n\dfrac{1}{3}-1 \right)\sigma \]

Answer 1

Hint: Here surface tension and radius of big droplet are given. We need to find the work done in breaking it into n smaller drops. While breaking a drop into n smaller drops surface area is increased. Hence work done will be against surface tension to increase the area. We have an equation which relates surface tension, area and work.
And if there is no addition or loss of quantity, volume will be constant.
Formula used:
Conservation of volume,
\[Volume\text{ }of\text{ }big\text{ }drop\text{ }=\text{ }Volume\text{ }of\text{ }n\text{ }no\text{ }of\text{ }smaller\text{ }drop\]
\[Volume\text{ }of\text{ }sphere=\dfrac{4}{3}\pi {{R}^{3}}\]
\[\text{Surface tension, T= }\dfrac{\text{work done}}{\text{change in area}}\]

Complete answer:
Even though the big drop is broken into smaller drops, the volume is conserved here as there is no addition or loss of quantity.
Then,
\[Volume\text{ }of\text{ }big\text{ }drop\text{ }=\text{ }Volume\text{ }of\text{ }n\text{ }no\text{ }of\text{ }smaller\text{ }drop\]
\[Volume\text{ }of\text{ }big\text{ }drop=\dfrac{4}{3}\pi {{R}^{3}}\]
\[Volume\text{ }of\text{ }n\text{ }no\text{ }of\text{ }smaller\text{ }drops\text{ }=n\times \dfrac{4}{3}\pi {{r}^{3}}\]
\[\dfrac{4}{3}\pi {{R}^{3}}=n\times \dfrac{4}{3}\pi {{r}^{3}}\]
\[{{R}^{3}}=n{{r}^{3}}\]
\[r=R{{n}^{\dfrac{-1}{3}}}\]
Since the Big drop of radius R is broken into n number of smaller drops the surface area of liquid is increased. Then the work done will be against the surface tension.
Then,
We know that,\[\text{Surface tension, T= }\dfrac{\text{work done}}{\text{change in area}}\]
\[T=\dfrac{W}{\Delta A}\]
Then,
\[\text{Workdone, W=T }\!\!\times\!\!\text{ }\Delta \text{A}\]
\[W=T\times \left( \text{Total surface area of n drops- total surface area of big drop} \right)\]
\[\left( Surface\text{ }area\text{ }of\text{ }a\text{ }sphere\text{ }=4\pi {{R}^{2}} \right)\]
\[W=T\times \left( 4n\pi {{r}^{2}}-4\pi {{R}^{2}} \right)\]
Given that, \[Surface\text{ }tension,\text{ }T=\sigma \]
\[W=\sigma \times \left( 4n\pi {{r}^{2}}-4\pi {{R}^{2}} \right)\]
Substitute \[r=R{{n}^{\dfrac{-1}{3}}}\]
Then,
\[W=\sigma \times \left( 4n\pi {{\left( R{{n}^{\dfrac{-1}{3}}} \right)}^{2}}-4\pi {{R}^{2}} \right)\]
\[W=\sigma \times \left( 4n\pi {{R}^{2}}{{n}^{\dfrac{-2}{3}}}-4\pi {{R}^{2}} \right)\]
\[W=4\pi {{R}^{2}}\left( {{n}^{\dfrac{1}{3}}}-1 \right)\sigma \]

So, the correct answer is “Option C”.

Additional Information:
Surface tension is an impact among a liquid’s surface layer. It causes the layer to possess characteristics like elastic and it's the result of the cohesive forces that attract liquid molecules to one another. This surface force keeps dense objects from sinking into the liquid. Molecules within a volume of liquid pull one another equally in all directions. However molecules below and to its sides pull a molecule on the liquid’s surface. Hence, the net force on this surface molecule is inward. So the surface behaves as if it were beneath tension.
As a result of inward pull, the surface of the liquid tends to be within the smallest possible space for a given volume of the liquid. This leads to the lowest energy level of the liquid. For a given volume, a sphere has a minimum area.

Note:
The work done to increase the surface area is stored in the droplet as potential energy. This work done to extend the area by \[1c{{m}^{2}}\] is known as surface energy. The unit of surface energy is, \[\text{Joules/}{{\text{m}}^{2}}\] in the S.I. system.