A liquid drop of density $\rho$ is floating half immersed in a liquid of surface tension $S$ and density…

A liquid drop of density $\rho$ is floating half immersed in a liquid of surface tension $S$ and density $\frac{\rho}{2}$. If the surface tension $S$ of the liquid is numerically equal to 10 times of acceleration due to gravity, then the diameter of the drop is:
  1. $\sqrt{\frac{20}{\rho}}$
  2. $\sqrt{\frac{80}{\rho}}$
  3. $\sqrt{\frac{60}{\rho}}$
  4. $\sqrt{\frac{40}{\rho}}$

Solution

A figure given here, shows a liquid drop half immersed in a liquid, where density of drop, $\rho_d=\rho$, density of liquid, $\rho_L=\frac{\rho}{2}$ and surface tension, $T=10 \mathrm{~g}$. Force acting on the drop due to surface tension,

where, $V=$ volume of the drop $=\frac{4}{3} \pi r^3$ Since, the ball is in equilibrium. $ \begin{gathered} \quad F=w \\ \log \pi D=\frac{3}{4} \rho g \frac{\pi D^3}{6} \\ D=\sqrt{\frac{80}{\rho}} \mathrm{m} \quad \text { [From Eqs. (i) and (ii)] } \end{gathered} $ Hence, the correct option is (b)

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

Practice more Mechanical Properties of Fluids questions on Aicharya