A liquid drop of density ' \(\rho\) ' is floating half immersed in a liquid of density ' \(d\) '. If ' \(T\) ' is the surface tension then the diameter of the liquid drop is ( \(\mathrm{g}=\) acceleration due to gravity)

\(\begin{aligned} & \text { Surface Tension }=\mathrm{T} \times 2 \pi \mathrm{r} \\ & \frac{4}{3} \pi \mathrm{r}^3 \rho \mathrm{~g}=2 \pi \mathrm{rT}+\frac{1}{2} \times \frac{4}{3} \pi \mathrm{r}^3 \mathrm{dg}\end{aligned}\)
\(\begin{array}{ll}\therefore \quad & 2 \pi r T=\frac{4}{3} \pi r^3 \rho g-\left(\frac{4}{3} \pi r^3 d g\right) \times \frac{1}{2} \\ & 2 \pi T=\frac{4}{3} \pi r^2 g\left(\rho-\frac{d}{2}\right) \\ \therefore & r^2=\frac{2 \pi T}{\frac{4}{3} \pi g\left(\rho-\frac{d}{2}\right)} \\ \therefore & r^2=\frac{3 T}{g(2 \rho-d)} \Rightarrow r=\sqrt{\frac{3 T}{g(2 \rho-d)}} \\ & \text { Diameter }=2 r=\sqrt{\frac{12 T}{g(2 \rho-d)}}\end{array}\)