Question

Two drops of equal radius coalesce to form a bigger drop. What is the ratio of surface energy of the bigger drop to the smaller one?
A. \[{2^{\dfrac{1}{2}}}:1\]
B. $1:1$
C. \[{2^{\dfrac{2}{3}}}:1\]
D. None of these

Answer 1

Hint: Surface energy is defined as the work done per unit area by the force which creates a new surface. The nature of this force is cohesive which means it attracts molecules of the same type.
The mathematical expression is similar to surface tension which is $\gamma = \dfrac{F}{{2L}}$ where $\gamma $ is the surface tension, F is the force acting and l is the length.
The surface energy of a spherical drop is given as $S.E = S(4\pi {r^2})$ where S is the surface tension and r is the radius of the drop.


Complete step by step solution:
Let the radius of the bigger drop be $R$ and the radii of the smaller drops be $r$ .
Since the two smaller drops combined to form a bigger drop, the initial volume must be equal to the final volume.
So, we have $V = {v_1} + {v_2}$ where $V$ denotes the final volume and ${v_{1\,}}\,,{v_2}$ denote the initial volumes of the drops.
Since it is given that the radii are equal for the smaller drops, ${v_1} = {v_2}$ .
And so, the expression becomes $V = 2{v_1}$
Using the formula of volume of a sphere which is $V = \dfrac{4}{3}\pi {r^3}$
$\dfrac{4}{3}\pi {R^3} = 2 \times \dfrac{4}{3}\pi {r^3}$
Cancelling out the common terms
\[{R^3} = 2{r^3}\]
\[ \Rightarrow R = {2^{\dfrac{1}{3}}}r\]
The surface energy of a spherical drop is given as $S.E = S(4\pi {r^2})$ where S is the surface tension and r is the radius of the drop.
From the expression we can see that the surface energy is directly proportional to the square of the radius. $S.E \propto {r^2}$.
Hence \[\dfrac{{S.{E_{l\arg e}}}}{{S.{E_{small}}}} = \dfrac{{{R^2}}}{{{r^2}}}\]
But we know that \[R = {2^{\dfrac{1}{3}}}r\]
Substituting in the equation we get,
\[\dfrac{{S.{E_{l\arg e}}}}{{S.{E_{small}}}} = \dfrac{{{{({2^{\dfrac{1}{3}}}r)}^2}}}{{{r^2}}}\]
Further solving,
\[\dfrac{{S.{E_{l\arg e}}}}{{S.{E_{small}}}} = \dfrac{{{2^{\dfrac{2}{3}}}{r^2}}}{{{r^2}}}\]
\[\dfrac{{S.{E_{l\arg e}}}}{{S.{E_{small}}}} = {2^{\dfrac{2}{3}}}\]
Hence option C is correct.

Note: Always keep a note of the object mentioned in the question. Two similar objects drop and bubble can be asked but we must carefully apply the required formula and prevent any confusion between the two. Also, the terms surface tension and surface energy might be perceived as similar. However, both are different and the difference must be carefully noted by the students.