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