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Some drops, each of radius ‘r’ coalesce to form a large drop of radius ‘R’. The surface tension is T. Find the change in surface energy per unit volume.

(a) $T\left({1\over r}-{1\over R}\right)$

(b) $3T\left({1\over r}-{1\over R}\right)$

(c) $3T\left({1\over R}-{1\over r}\right)$

(d) $T\left({1\over R}-{1\over r}\right)$
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Answer: (b) $3T\left({1\over r}-{1\over R}\right)$

Solution:

Radius of small drop is r and radius of big drop is R.

When they coalesce, the volume will the same before and after coalesce.

${4\over3}\pi R^3=n.{4\over3}\pi r^3=\text{volume }(V)$

$R=n^{1\over3}r\quad\quad....(1)$

The surface area of large drop is $4πR^2$ and surface area of small drop is $4πr^2$

Let

$∆U\ -$ change in surface energy

$∆A\ -$ Change in surface area

$T\ -$ surface tension

Then, $∆U=T.∆A$

$∆U=T(4\pi R^2-4\pi r^2)$

$∆U={3T\over3}\left({4\pi R^3\over R}-{4\pi r^3\over r}\right)$

$∆U=3T\left({{4\pi R^3\over3}\over R}-{{4\pi r^3\over3}\over r}\right)$

$∆U=3T\left({V\over R}-{V\over r}\right)$

$\left|{∆U\over V}\right|=3T\left({1\over r}-{1\over R}\right)$

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