Question

Two small drops of mercury, each radius $R$ coalesce to form a single drop. The ratio of the total surface energies before and after the change is:
A. $1:{2^{\dfrac{1}{3}}}$
B. ${2^{\dfrac{1}{3}}}:1$
C. $2:1$
D. $1:2$

Answer 1

The interaction between the molecules of liquid is cohesive, this causes the tension on the surface of the liquid called surface tension. This phenomenon occurs only on the surface of the liquid as inside the liquid the force on each molecule is balanced because molecules from all directions apply the force. Surface energy is the force acting between the molecules of liquid at the surface.

Complete step by step solution:
Let the radius of small drops of mercury be $r$ and the radius of large drops be $R$.
Now according to the question, the two small drops of mercury combine to form a single large drop.
In this case, the volume of two small drops will be equal to the volume of a large drop.
The volume of a sphere is given by the following formula.
$\dfrac{4}{3}\pi {\left( {radius} \right)^3}$
So, we can write the following.
$\dfrac{4}{3}\pi \left( {{r^3} + {r^3}} \right) = \dfrac{4}{3}\pi {R^3}$
Let us simplify the expression to get the relation between the radius of the two.
$2{r^3} = {R^3} \Rightarrow R = \sqrt[3]{2}r$ ………………….(1)
Now, the surface area of the sphere can be calculated using the following formula.
$S = 4\pi {\left( {radius} \right)^2}$
Surface energy is the product of surface area and surface tension.
So, the surface energy of small drops is given below.
${E_r} = 2 \times 4\pi {r^2} \times T$………………………(2)
Similarly, we can write the surface energy of a large sphere as below.
${E_R} = 4\pi {R^2} \times T = 4\pi {\left( {\sqrt[3]{2}r} \right)^2}T$
We replaced the radius of the large sphere in terms of the radius of the small sphere. And on further simplifying.
${E_R} = 4\pi {\left( 2 \right)^{\dfrac{2}{3}}}{r^2}T$ ………………(3)
Now, let us the ratio of equations (2) and (3).
$\dfrac{{{E_r}}}{{{E_R}}} = \dfrac{{2 \times 4\pi {r^2}T}}{{4\pi {{\left( 2 \right)}^{\dfrac{2}{3}}}{r^2}T}} = {2^{1 - \dfrac{2}{3}}} = \dfrac{{{2^{\dfrac{1}{3}}}}}{1}$
Hence, the correct option is (B) $\dfrac{{{2^{\dfrac{1}{3}}}}}{1}$.

Note:
Whenever one object is transformed into the other their volume remains the same. Here, the volume of two small spheres will be equal to the volume of the large sphere.
Surface energy arises due to surface tension on the surface of the liquid.