Question

If 'n' of a liquid each with surface energy E join to form a single drop, in this process

• A. Some energy is absorbed
• B. Energy absorbed is $E(n-n^{2/3})$
• C. Energy released is $E(n-n^{2/3})$
• D. Energy released is $E(n^{2/3}-1)$

Answer 1

Answer: (C)

Let us say $r$ is the radius of each single drop before these coalesce and $R$ is the radius of the new bubble after these coalesce. Since volume of liquid remains same, we have
$<br/>\begin{array}{l}<br/>n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\\ <br/>\Rightarrow R=n^{\frac{1}{3}}r<br/>\end{array}<br/>$
Initial surface energy of single drop $=E=\left(4\pi r^2\right)\times T$
Initial surface energy of $n$ drops $=nE=n\left(4\pi r^2\right)\times T$
final surface energy of big single drop $=\left(4\pi R^2\right)\times T$
$<br/>\begin{array}{l}<br/>=\left(4\pi {\left(n^{\frac{1}{3}}r\right)}^2\right)\times T\\ <br/>=n^{\frac{2}{3}}\left(4\pi r^2\times T\right)=n^{\frac{2}{3}}E\text{ from (I) }<br/>\end{array}<br/>$
Clearly
$<br/>n^{\frac{2}{3}}E<nE<br/>$
i.e. final surface energy < initial surface energy that means some energy will be released in this process. And the energy released in this process is given by
$<br/>nE-n^{\frac{2}{3}}E=E\left(n-n^{\frac{2}{3}}\right)<br/>$