Question

A drop of liquid of density p is floating half-immersed in a liquid of density d. If p is the surface tension the diameter of the drop of the liquid is:

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

Answer 1

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

$2\pi r\sigma +\frac{1}{2}\times \frac{4}{3}\pi r^{3}dg=\frac{4}{3}\pi r^3\rho g$
Or $2\pi r\sigma =\frac{\pi r^{3}g}{3}[4\rho -2d]or{r}^2=\frac{3\times 2\pi \sigma }{\pi g(2\rho -2d)}$
Or $r^2=\frac{3\sigma }{g(2\rho -d)}orr=\sqrt{\frac{3\sigma }{g(2\rho -d)}}$
$Diameter=2r=\sqrt{\frac{12\sigma }{g(2\rho -d)}}$