First slide

Surface tension and surface energy

Question

A large number of droplets, each of radius a, coalesce, to form a bigger drop of radius b. Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is ( $σ$ surface tension, $ρ$ density)

Moderate

A

${[\frac{σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2}$
B

${[\frac{2 σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2}$
C

${[\frac{6 σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2}$
D

${[\frac{3 σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2}$

Solution

Let say number of droplets are n.
$Energy released = [n \times 4 π a^{2} - 4 π b^{2}] σ$
$Now, n \times \frac{4}{3} π a^{3} = \frac{4}{3} π b^{3} or n = \frac{b^{3}}{a^{3}}$
Therefore, energy released is
$\begin{array}{l} = [\frac{b^{3}}{a^{3}} \times 4 π a^{2} - 4 π b^{2}] σ = 4 π b^{2} [\frac{b}{a} - 1] σ \\ Now, K.E of big drop=energy released \\ \Rightarrow \frac{1}{2} (\frac{4}{3} π b^{3}) ρ v^{2} = 4 π b^{2} [\frac{b}{a} - 1] σ \\ ⇒ v = {[\frac{6 σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2} \end{array}$

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