Question

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 liquid is numerically equal to $10$ times of acceleration due to gravity, then the diameter of the drop is

• A. $\sqrt{\frac{20}{\rho}}$
• B. $\sqrt{\frac{80}{\rho}}$
• C. $\sqrt{\frac{60}{\rho}}$
• D. $\sqrt{\frac{40}{\rho}}$

Answer 1

Answer: (B)

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, $T=10\, g$.
Force acting on the drop due to surface tension,
$F=T l=10 \,g (\pi D) \dots$(i)
The observed weight of the drop,
$w=w_{d}-w_{L}=\rho V g-\frac{\rho}{2} \frac{V}{2} g$
$=\frac{3}{4} \rho V g\dots$(ii)
where, $V=$ volume of the drop $=\frac{4}{3} \pi r^{3}$
Since, the ball is in equilibrium.
$\therefore F=w$
$10 \,g\, \pi D=\frac{3}{4} \rho g \frac{\pi D^{3}}{6}$
$D=\sqrt{\frac{80}{\rho}} m $ [From Eqs. (i) and (ii)]