Question

Two small drops of mercury, each of radius $R$, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

• A. $1 : 2^{1/3}$
• B. $2^{1/3}: 1$
• C. $2 : 1$
• D. $1 : 2$

Answer 1

The Correct Option is B

Radius of one drop of mercury is $R$.
$\therefore$ The volume of one drop$ =\frac{4}{3}\pi R^{3}$
$\therefore$; Total volume of the two drops,
$V=2\times\frac{4}{3}\pi R^{3}=\frac{8}{3}\pi R^{3}$
Let the radius of the large drop formed be $R '$ The volume of the large drop is also V.
$\therefore \frac{4}{3}\pi R'^{3}=\frac{8}{3}\pi R^{3} \Rightarrow R '^{3}=2R^{3}\Rightarrow R'=2^{1/3}R.$
Now the surface area of the two drops is
$S_{1}=2\times4\pi R^{2}=8\pi R^{2}$
and the surface area of the resultant drop is
$S_{1}=2\times4\pi R'^{2}=4\pi2^{2/3}R^{2}$
Let T be the surface tension of mercury. Therefore the surface energy of the two drops before coalescing is
$U_{1}=S_{1}T=8\pi R^{2}T$
and the surface energy after coalescing,
$U_{1}=S_{2}T=2^{2/3}\times4\pi R^{2}T$
$\therefore \frac{U_{1}}{U_{2}}=\frac{8\pi R^{2}T}{2^{2 / 3}\times4\pi R^{2}T}=\frac{2}{2^{2 / 3}}=2^{1 / 3}$