A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\sigma$ and surface tension $7.5\times10^{-4}\ \text{N cm}^{-1}$. The radius of the drop in cm will be: (Take $g = 10\ \text{m s}^{-2}$)
Answer: (A) $\dfrac{15}{\sqrt{2\rho - \sigma}}$
Weight = buoyancy on the immersed half + surface tension acting along the waterline circle ($2\pi r$):
$$\frac43\pi r^3\rho g = \frac23\pi r^3\sigma g + 2\pi rT\ \Rightarrow\ r^2 = \frac{3T}{g(2\rho - \sigma)}$$
With $T = 7.5\times10^{-2}\ \text{N m}^{-1}$ and $g = 10\ \text{m s}^{-2}$ (densities in SI):
$$r = \sqrt{\frac{0.0225}{2\rho - \sigma}}\ \text{m} = \frac{0.15}{\sqrt{2\rho - \sigma}}\ \text{m} = \frac{15}{\sqrt{2\rho - \sigma}}\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics