Question

A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $\rho$ and surface tension $T$. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

• A. $r=\sqrt{\frac{2T}{3(d+\rho )g}}$
• B. $r=\sqrt{\frac{T}{(d-\rho )g}}$
• C. $r=\sqrt{\frac{T}{3(d+\rho )g}}$
• D. $r=\sqrt{\frac{3T}{(2d-\rho )g}}$

Answer 1

The correct option is D $r=\sqrt{\frac{3T}{(2d-\rho )g}}$

Free body diagram of drop,

For the drop to be in equilibrium,

$\text{Total upward force}=\text{Total downward force}$

$\Rightarrow F_S+F_B=mg$

$\Rightarrow F_S=mg-F_B$

$\Rightarrow T\left(2\pi R\right)=\frac{4}{3}\pi R^{3}dg-\frac{2}{3}\pi R^3\rho g$

$\Rightarrow T\left(2\pi R\right)=\frac{2}{3}\pi R^3(2d-\rho )g$

$\Rightarrow T=\frac{R^2}{3}(2d-\rho )g$

$\Rightarrow R=\sqrt{\frac{3T}{(2d-\rho )g}}$

Hence, $\left(D\right)$ is the correct answer.