First slide

Surface tension

Question

Two mercury drops, each of radius r, merge to form a bigger drop. If $σ$ is the surface tension of mercury then the surface energy released is

Moderate

A

$(8 {πr}^{2} - 4 \times 2^{\frac{2}{3}} {πr}^{2}) σ$
B

zero
C

negative
D

$(4 {πr}^{2} σ - 8 \times 2^{\frac{1}{3}} r^{2}) σ$

Solution

$2 \times \frac{4}{3} {πr}^{3} = \frac{4}{3} {πR}^{3} or R = 2^{\frac{1}{3}} r$
Final surface energy = $4 {πR}^{2} σ = 4 π 2^{\frac{2}{3}} r^{2} σ$
Initial surface energy = 2 $\times 4 {πr}^{2} σ$
$∴ Energy released = [8 {πr}^{2} - 4 \times 2^{\frac{2}{3}} {πr}^{2}] σ$

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