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Q.
Two small drops of mercury, each of radius $R$, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is
MHT CETMHT CET 2019Mechanical Properties of Fluids
Solution:
Let $r$ be radius of common drop
$
\begin{array}{l}
\frac{4}{3} \pi r^3=2 \times \frac{4}{3} \pi R^3 \\
r=2^{1 / 3} R
\end{array}
$
surface tension of 2 drops $2 \times 4 \pi R^2 T$
surface tension of common drop $4 \pi r^2 T$
ratio $\frac{2 \times 4 \pi R^2 T }{4 \pi r ^2 T }=\frac{2 R ^2}{2^{2 / 3} R ^2}=\frac{2^{1 / 3}}{1}$