Question

A big water drop is formed by the combination of “n” small water drops of equal radii. The ratio of surface energy of “n” drops to the surface energy of big drop is
$\left( A \right){n^2}:1$
$\left( B \right)n:1$
$\left( C \right)\sqrt n :1$
$\left( D \right)\sqrt[3]{n}:1$

Answer 1

Hint – In this question use the concept that the volume of the larger water drop will eventually be the equal to n times the volume of the smaller drops that is$\dfrac{4}{3}\pi {R^3} = n\left( {\dfrac{4}{3}\pi {r^3}} \right)$, where R is the radius of the larger water drop and r is the radius of the smaller water drop. The water drops will be in the form of a sphere thus use volume of sphere formula. Then use the surface energy of the n small drops that is $nT \times 4\pi {r^2}$ where T is the surface tension working on the small water drop. This will help approach the solution of this problem.

Complete step-by-step answer:
Let the radius of the big water drop be R.
Now this big water drop is formed by the combination of n small equal radius water drops.
Let the radius of the small water drop be r.
Now as we know that the water drops are always in spherical shape, so the volume of the n small water drops is equal to the volume of the big water drops (according to volume conservation).
As we know that the volume of the sphere is $\dfrac{4}{3}\pi {R^3}$
$ \Rightarrow \dfrac{4}{3}\pi {R^3} = n\left( {\dfrac{4}{3}\pi {r^3}} \right)$
Now cancel out the common terms we have,
$ \Rightarrow {R^3} = n\left( {{r^3}} \right)$
Now take cube root on both sides we have,
$ \Rightarrow \sqrt[3]{{{R^3}}} = \sqrt[3]{{n\left( {{r^3}} \right)}}$
$ \Rightarrow R = r\sqrt[3]{n}$...................... (1)
Now the surface energy of n small drops is $nT \times 4\pi {r^2}$
Where, $4\pi {r^2}$ is the area of the small water drop and T is the surface tension working on the small water drop.
Now the surface energy on big water drop is $T \times 4\pi {R^2}$
Where T is the same surface tension as above as both the water drops are in the same medium.
Now the ratio of surface energy of n small drops to big drops is
$\dfrac{{nT \times 4\pi {r^2}}}{{T \times 4\pi {R^2}}}$
Now cancel out the common terms we have,
$ \Rightarrow \dfrac{{n \times {r^2}}}{{{R^2}}}$
Now from equation (1) we have,
$ \Rightarrow \dfrac{{n \times {r^2}}}{{{{\left( {r\sqrt[3]{n}} \right)}^2}}}$
Now simplify this we have,
$ \Rightarrow \dfrac{n}{{{{\left( {\sqrt[3]{n}} \right)}^2}}} = \dfrac{{{n^{1 - \dfrac{2}{3}}}}}{1} = \dfrac{{{n^{\dfrac{{3 - 2}}{3}}}}}{1} = \dfrac{{{n^{\dfrac{1}{3}}}}}{1} = \dfrac{{\sqrt[3]{n}}}{1}$
So this is the required ratio.
Hence option (D) is the correct answer.

Note – There will always be an excess energy at the surface of the bulk rather than at the bulk of any shape. So this excess energy at the surface respective to the bulk refers to the surface energy. There is often a confusion between surface energy and the surface tension. The net intermolecular force present on the molecules at the surface is the surface tension whereas the net energy of the bonds that are between the molecules at the surface is the surface energy.