First slide

Surface tension and surface energy

Question

If $σ$ be the surface tension, the work done in breaking a big drop of radius R into n drops of equal radius is

Moderate

A

$R n^{2 / 3} σ$
B

$(n^{2 / 3} - 1) R σ$
C

$(n^{1 / 3} - 1) R σ$
D

$4 π R^{2} (n^{1 / 3} - 1) σ$

Solution

Volume will be conserved
$\frac{4}{3} {πR}^{3} = n \frac{4}{3} {πr}^{3}$
R= $n^{1 / 3} r$
workdone = final energy - initial energy
= [ $n 4 π n^{- 2 / 3} R^{2}$ − $4 π R^{2}$ ]σ
= 4π $R^{2}$ ( $n^{1 / 3}$ −1)σ

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