Question

Two mercury drops(each of radius r) merge to form a bigger drop. The surface energy of the bigger drop if T is the surface tensions

• A. $2^{5/3}\pi r^{2}T$
• B. $4\pi r^{2}T$
• C. $2\pi r^{2}T$
• D. $2^{8/3}\pi r^{2}T$

Answer 1

The correct option is D $2^{8/3}\pi r^{2}T$

Since final volume = 2 initial volume

$R^3=2r^3$

$R=2^{1/3}r$

So final surface area = $4\pi R^2$

= $4\pi 2^{2/3}{r}^2$

Area = $2^{8/3}\pi r^2$

So surface energy = T $\times$ area

= $2^{8/3}\pi r^{2}T$