A drop of liquid of density ' \(\rho\) ' is floating half immersed in a liquid of density ' \(d\) '. If ' \(T\) ' is the surface tension, then the diameter of the drop of the liquids is

The drop is in equilibrium under the action of the following forces:
Weight of the liquid, \(\mathrm{W}=\mathrm{Mg}=\frac{4}{3} \pi \mathrm{r}^3 \rho \mathrm{g}\) (downwards)
Upthrust \(=\) weight of the liquid displaced
\(
\therefore \mathrm{F}_{\mathrm{t}}=\frac{2}{3} \pi \mathrm{r}^3 \mathrm{dg}
\)
(upwards)
Force due to surface tension, \(\mathrm{F}=2 \pi \mathrm{rT}\) (upwards)
\(
\begin{aligned}
& \therefore \mathrm{Mg}=\mathrm{F}+\mathrm{F}_{\mathrm{t}} \\
& \therefore \mathrm{F}=\mathrm{Mg}-\mathrm{F}_{\mathrm{t}} \\
& \therefore 2 \pi \mathrm{r} T=\frac{4}{3} \pi \mathrm{r}^3 \rho g-\frac{2}{3} \pi \mathrm{r}^3 \mathrm{dg} \\
& \therefore \mathrm{T}=\frac{2}{3} \mathrm{r}^2 \rho g-\frac{1}{3} \mathrm{r}^2 \mathrm{dg} \\
& =\mathrm{r}^2 g\left(\frac{2}{3} \rho-\frac{1}{3} d\right) \\
& =\mathrm{r}^2 g\left(\frac{2 \rho-d}{3}\right) \\
& \therefore \mathrm{r}=\sqrt{\frac{3 \mathrm{~T}}{\mathrm{~g}(2 \rho-\mathrm{d})}} \\
& \therefore \mathrm{D}=\sqrt{\frac{12 \mathrm{~T}}{\mathrm{~g}(2 \rho-d)}}
\end{aligned}
\)