Question

Two mercury drops of radius $r$ merge to form a bigger drop. The surface energy of the bigger drop will be ($T$ is surface tension of Hg)?

Answer 1

Hint: This problem can be solved by realizing the fact that the volume of the bigger drop will be equal to the sum of the volumes of the two smaller drops and the relation for the radius of the bigger drop with that of the smaller drops can be found out from that. The surface energy can be found out using the direct formula in terms of the surface tension and surface area.

Formula Used:
$V=\dfrac{4}{3}\pi {{R}^{3}}$
$U=TA$
$A=4\pi {{R}^{2}}$

Complete step by step answer:
We will use the direct formula for the surface energy of a body in terms of the surface tension and the surface area.
The surface energy $U$ of a body with surface tension $T$ and surface area $A$ is given by
$U=TA$ --(1)
Also, the volume $V$ of a sphere of radius $R$ is given by
$V=\dfrac{4}{3}\pi {{R}^{3}}$ --(2)
The surface area $A$ of a sphere of radius $R$ is given by
$A=4\pi {{R}^{2}}$ ---(3)
Now, let us analyze the question.
The radius of the smaller drops is $r$.
Let the radius of the bigger drop formed by the joining of two smaller drops be $R$.
Let the volume of each smaller drop be $v$.
Let the volume of the bigger drop be $V$.
The surface tension of mercury is $T$.
Let the surface area of the bigger drop be $A$ and the surface energy be $U$.
Now, using (2), we get
$v=\dfrac{4}{3}\pi {{r}^{3}}$ --(4)
$V=\dfrac{4}{3}\pi {{R}^{3}}$ --(5)
Now, since the larger drop is formed by the joining of two smaller drops, the volume of the large drop will be the sum of the volumes of the two smaller drops.
$\therefore V=v+v=2v$ --(6)
Putting (4) and (5) in (6), we get
$\dfrac{4}{3}\pi {{R}^{3}}=2\times \dfrac{4}{3}\pi {{r}^{3}}$
$\therefore {{R}^{3}}=2{{r}^{3}}$
Cube rooting both sides we get
$\sqrt[3]{{{R}^{3}}}=\sqrt[3]{2{{r}^{3}}}$
$\therefore R={{\left( 2 \right)}^{\dfrac{1}{3}}}r$ --(7)
Now, using (3), we get
$A=4\pi {{R}^{2}}$ --(8)
Putting (7) in (8), we get
$A=4\pi {{\left( {{\left( 2 \right)}^{\dfrac{1}{3}}}r \right)}^{2}}=4\pi {{\left( 2 \right)}^{\dfrac{2}{3}}}{{r}^{2}}$ --(9)
Now, using (1), we get
$U=TA$ --(10)
Putting (9) in (10), we get
$U=T\times 4\pi {{\left( 2 \right)}^{\dfrac{2}{3}}}{{r}^{2}}=4\pi {{\left( 2 \right)}^{\dfrac{2}{3}}}T{{r}^{2}}$
Hence, we have got the required expression for the surface energy of the larger drop.

Note:
Students must realize that the sum of the surface energies of the smaller drops will be $2\times 4\pi {{r}^{2}}$ which is greater than the sum of the surface energy of the larger drop. This is because a part of the surface energy is consumed to change the shape of the larger drop when the smaller drops join to form the larger drop by breaking the intermolecular bonds on the surface of the smaller drops.