Question

Two identical drops of $\mathrm{Hg}$ coalesce to form a bigger drop. Find the ratio of the surface energy of a bigger drop to a smaller drop.

• A. $2^{2/3}$
• B. $2^{3/2}$
• C. $5^{2/3}$
• D. $3^{3/2}$

Answer 1

The correct option is A

$2^{2/3}$

Step 1: Given data

Two identical drops of $\mathrm{Hg}$ coalesce to form a bigger drop

Step 2: Formula used

Volume, $V=\frac{4}{3}{\mathrm{\pi r}}^3$

$S=T.A$

$\left(S\right)$ =surface energy, $\left(T\right)$ =surface tension, $\left(A\right)$ = surface area

Step 3: Find the relation between the radius of bigger and smaller drop

Let two drops of radii $r$ combines together to form a drop of radii $R$.

Therefore,

$$\begin{gathered} 2\times \frac{4}{3}\times {\mathrm{\pi r}}^3=\frac{4}{3}\times {\mathrm{\pi R}}^3 \\ \Rightarrow 2r^3=R^3 \\ \Rightarrow R=2^{\frac{1}{3}}r \\ \Rightarrow \frac{R}{r}=\frac{2^{\frac{1}{3}}}{1}---------\left(1\right) \end{gathered}$$

Step 4: Calculate the ratio of surface tensions of both drops

Now surface energy $\left(S\right)$ = surface tension $\left(T\right)\times$ surface area$\left(A\right)$

Ratio of surface energy for the bigger and smaller drop:

$\frac{S\text{'}}{S}=\frac{T\times 4{\mathrm{\pi R}}^2}{T\times 4{\mathrm{\pi r}}^2}$, As tension is same in both cases and surface area $=4{\mathrm{\pi r}}^2$,

Hence, the ratio of the surface energy of bigger drop to smaller drop becomes,

$$\begin{gathered} \frac{S\text{'}}{S}=\frac{R^2}{r^2} \\ \Rightarrow \frac{S\text{'}}{S}={\left(\frac{2^{{\displaystyle \frac{1}{3}}}}{1}\right)}^2.by\left(1\right) \\ \Rightarrow \frac{S\text{'}}{S}=\frac{2^{{\displaystyle \frac{2}{3}}}}{1} \\ \Rightarrow \frac{S\text{'}}{S}=2^{\frac{2}{3}} \end{gathered}$$

$\frac{S\text{'}}{S}=\frac{2^{2/3}}{1}=2^{2/3}:1$

Hence, option A is correct.