A liquid drop of density ' $Q$ ' is floating half immersed in a liquid of density ' $d$ '. Diameter of the…
A liquid drop of density ' $Q$ ' is floating half immersed in a liquid of density ' $d$ '. Diameter of the liquid drop is
( $\mathrm{Q}>\mathrm{d}, \mathrm{g}=$ acceleration due to gravity, $\mathrm{T}=$ surface tension)
The force balance invovles, surface-tension force and boyancy force balance the weight of the droplet, so the equation is:
$(2 \pi r) T+\left(\frac{4}{3} \pi r^3\right) \mathrm{dg}=\left(\frac{4}{3} \pi r^3\right) \mathrm{Qg}$
Where, $\mathrm{r}$ is the radius of the drop, $\mathrm{T}$ the interfacial tension between the liquids, $g$ is acceleration due to gravity.
Therefore,
$\mathrm{T}=\left(\frac{2}{3} \mathrm{r}^2\right)(\mathrm{Q}-\mathrm{d}) \mathrm{g}$
Diameter $=2 r=\left[\frac{6 T}{g(Q-d)}\right]^{\frac{1}{2}}$.
.