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 applied an upward force on the droplet).

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

Answer 1

The correct option is D

$r=\sqrt{\frac{3T}{\left(2\rho -d\right)g}}$

Step 1. Given Data,

The density of a small spherical droplet$=d$,

The density of liquid$=\rho$,

Surface tension$=T$,

Let the radius of the droplet is $r$.

Step 2. We have to calculate the radius of droplet $r$,

From the Free body diagram of the droplet, we can say,

In equilibrium, the net external force acting on the sphere is zero.

Hence, $mg=B+F_r$ (where, $m$ is mass, $g$ is gravity, $B$ is Normal Force, and $F_r$ upward force due to surface tension.)

$m=\rho V(V=\frac{4}{3}{\mathrm{\pi r}}^3,$is the volume of the sphere and is $\rho$ density of liquid$)$

Since, the Normal Force is acting on the semisphere, $B=d\left(\frac{2}{3}{\mathrm{\pi r}}^3\right)g$

Upward Force due to surface tension $F_r=\left(2\mathrm{\pi r}\right)T$,

$$\begin{gathered} \Rightarrow mg=B+F_r \\ \Rightarrow \rho \left(\frac{4}{3}{\mathrm{\pi r}}^3\right)g=d\left(\frac{2}{3}{\mathrm{\pi r}}^3\right)g+\left(2\mathrm{\pi r}\right)T \\ \Rightarrow \left(2\mathrm{\pi r}\right)T=\rho \left(\frac{4}{3}{\mathrm{\pi r}}^3\right)g-d\left(\frac{2}{3}{\mathrm{\pi r}}^3\right)g \\ \Rightarrow \left(2\mathrm{\pi r}\right)T=\left(2\rho -d\right)\left(\frac{2}{3}{\mathrm{\pi r}}^3\right)g \\ \Rightarrow T=\left(2\rho -d\right)\left(\frac{1}{3}{r}^2\right)g \\ \Rightarrow r^2=\frac{3T}{\left(2\rho -d\right)g} \\ \Rightarrow r=\sqrt{\frac{3T}{\left(2\rho -d\right)g}} \end{gathered}$$

Hence, option D is correct.