Question

A drop of liquid of density $\rho$ is floating half-immersed in a liquid of density $d$. If $\sigma$ is the surface tension the diameter of the drop of liquid.

• A. $\sqrt{\frac{3\sigma }{g(2\rho -d)}}$
• B. $\sqrt{\frac{6\sigma }{g(2\rho -d)}}$
• C. $\sqrt{\frac{6\sigma }{g(\rho -d)}}$
• D. $\sqrt{\frac{12\sigma }{g(2\rho -d)}}$

Answer 1

The correct option is D $\sqrt{\frac{12\sigma }{g(2\rho -d)}}$

Upward force $F_u=2\pi r\sigma +\frac{1}{2}\times \frac{4\pi r^3}{3}dg$
Downward Force $F_d=\left(4\pi r^3/3\right)\rho g$
Equating, we get
$2\pi r\sigma +1/2\times \left(4\pi r^3/3\right)dg=\left(4\pi r^3/3\right)\rho g$
$\therefore r=\sqrt{\frac{3\sigma }{g(2\rho -d)}}$
or diameter = $\sqrt{\frac{12\sigma }{g(2\rho -d)}}$