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 upwards force on the droplet)

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

Answer 1

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

Given, density of the droplet, $=d$
Density of the liquid, $=\rho$
And surface tension, $=T$
Let the radius of the droplet be $r$.

We know that $F_B=V_{sub}\rho g$

From $FBD$ of the droplet, in equilibrium,

$F_B+F_{\text{surface}}=mg\dots \left(i\right)$
Here, $F_B=$buoyancy force $V_{sub}\rho g$
$F_B=\left(\frac{2}{3}\pi r^3\right)\rho g$
And force due to surface tension,
$F_{\text{surface}}=T\left(2\pi r\right)$
And mass of the spherical droplet,
$m=d\left(\frac{4}{3}\pi r^3\right)$

So, by substitute these values in equation $\left(i\right)$

$\left(\frac{2}{3}\pi r^3\right)\rho g+T\left(2\pi r\right)=d\left(\frac{4}{3}\pi r^3\right)g$

$T\left(2\pi r\right)=\left(\frac{2}{3}\pi r^3\right)g(2d-\rho )$
$r=\sqrt{\frac{3T}{(2d-\rho )g}}$
Final answer (a)