Question

If $n$ drops of a liquid, each with surface energy $E$ join to form a single drop, then

• A. some energy will be released in the process
• B. some energy will be absorbed in the process
• C. the energy released will be $E(n^{2/3}-n)$
• D. the energy absorbed will be $E(n^{2/3}-n)$

Answer 1

Answer: (A), (C)

the energy released will be $E(n^{2/3}-n)$

Given,
Surface energy of smaller liquid drop$=E$
Number of smaller drops merged to form a single larger drop $=n$
Let us suppose, $T$ is surface tension of the liquid, $r$ is the radius of smaller drop and $R$ is the radius of larger drop.
As volume of liquid will be same,
Volume of $n$ smaller drops $=$ Volume of a single larger drop
$\Rightarrow n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\Rightarrow R=n^{1/3}r$ ........(1)
We know, surface energy $=$ surface tension $\times$ surface area
$\Rightarrow E=T\times A$
Now, Initial surface energy of $n$ smaller drops, $E_i=T\times n\times 4\pi r^2=nE$ .....(2)
$[E=4T\pi r^2]$
Final surface energy of larger drop,
$E_f=T\times 4\pi R^2=4\pi r^2{n}^{2/3}T$
$[$ from (1) $]$
$=n^{2/3}E$ ......(3)
Now, change in energy
$\Delta E=E_f-E_i=E(n^{2/3}-n)$
$[$ from (2) and (3) $]$
As, $n^{2/3}<n$ $\Rightarrow \Delta E$ is -ve.
Therefore, $E(n^{2/3}-n)$ amount energy will be released.
Option A, C are correct.