A liquid drop of density $\mathrm{\rho }$ is floating half immersed in a liquid of surface tension S and density $\mathrm{\rho }/2$. If the surface tension S of liquid is numerically equal to 10 times of acceleration due to gravity, then the diameter of the drop is

A drop of liquid is depicted here half submerged in another liquid, with its density, 

${\mathrm{\rho }}_{\mathrm{d}}=\mathrm{\rho }$

density of liquid ${\mathrm{\rho }}_{\mathrm{L}}=\frac{\mathrm{\rho }}{2}$

surface tension, T=10g.

Surface tension's force acting on the drop,

$\mathrm{F}=\mathrm{T}l=10\mathrm{g}\left(\mathrm{\pi D}\right)\mathrm{\_\_\_\_\_\_\_\_\_}\left(1\right)$

The measured drop's weight,

$$\begin{gathered} \mathrm{w}={\mathrm{w}}_{\mathrm{d}}-{\mathrm{w}}_{\mathrm{L}}=\mathrm{\rho Vg}-\frac{\mathrm{\rho }}{2}\frac{\mathrm{V}}{2}\mathrm{g} \\ =\frac{3}{4}\mathrm{\rho Vg\_\_\_\_\_\_}2 \end{gathered}$$

The ball is in equilibrium as a result.
$$\begin{gathered} \therefore \mathrm{F}=\mathrm{w} \\ 10\mathrm{g\pi D}=\frac{3}{4}\mathrm{\rho g}\frac{{\mathrm{\pi D}}^3}{6} \\ \mathrm{D}=\sqrt{\frac{80}{\mathrm{\rho }}} \end{gathered}$$

Hence the correct answer is $\sqrt{\frac{80}{\mathrm{\rho }}}.$