First slide

Surface tension and surface energy

Question

8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of

Moderate

A

1
B

2
C

4
D

6

Solution

As volume remains constant therefore $R = n^{1 / 3} r$
$\frac{Energy of big drop}{Energy of small drop} = \frac{4 {πR}^{2} T}{4 {πr}^{2} T} = \frac{R^{2}}{r^{2}} = (8)^{2 / 3} = 4$

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