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Q.
If 'n' of a liquid each with surface energy E join to form a single drop, in this process
Mechanical Properties of Fluids
Solution:
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
$
\begin{array}{l}
n \quad \times \frac{4}{3} \pi r ^{3}=\frac{4}{3} \pi R ^{3} \\
\Rightarrow R = n ^{\frac{1}{3}} r
\end{array}
$
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 $=\left(4 \pi R ^{2}\right) \times T$
$
\begin{array}{l}
=\left(4 \pi\left( n ^{\frac{1}{3}} r \right)^{2}\right) \times T \\
= n ^{\frac{2}{3}}\left(4 \pi r ^{2} \times T \right)= n ^{\frac{2}{3}} E \text { from (I) }
\end{array}
$
Clearly
$
n ^{\frac{2}{3}} E < nE
$
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
$
nE - n ^{\frac{2}{3}} E = E \left( n - n ^{\frac{2}{3}}\right)
$