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Q.
If S is the surface tension of a liquid, the energy needed to break a liquid drop of radius R into 64 drops is
Mechanical Properties of Fluids
Solution:
Let $r$ be the radius of each small drop. Then
Volume of 64 small drops = Volume of big drop
or $64 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$ or $r=\frac{R}{4}\,\,\,\,\,\dots(i)$
Surface area of big drop $=4 \pi R^{2}$
Surface area of 64 small drops $=64 \times 4 \pi r^{2}$
$\therefore $ Increase in surface area $=64 \times 4 \pi r^{2}-4 \pi R^{2}=4 \pi\left[64 r^{2}-R^{2}\right]$
$=4 \pi\left[4 R^{2}-R^{2}\right]=12 \pi R^{2}$ (Using (i))
Energy needed surface tension $\times$ increase in surface area
$=S \times 12 \pi R^{2}=12 \pi R^{2} S $