Question

Drops of liquid of density $d$ are floating half immersed in a liquid of density $\rho$. If the surface tension of liquid is $T$ then the radius of the drop will be $(d=$ density of liquid drop).

• A. $\sqrt{\frac{3T}{g(2d-\rho )}}$
• B. $\sqrt{\frac{6T}{g(2d-\rho )}}$
• C. $\sqrt{\frac{2T}{g(2d-\rho )}}$
• D. $\sqrt{\frac{3T}{g(4d-3\rho )}}$

Answer 1

The correct option is A $\sqrt{\frac{3T}{g(2d-\rho )}}$

Force due to surface tension$=T\times 2\pi r$

$T\times 2\pi r+\frac{2}{3}\pi r^3\times \rho g=\frac{4}{3}\pi r^3\times dg2T+\frac{2}{3}{r}^2\rho g=\frac{4}{3}{r}^{2}dg{r}^2(\frac{4}{3}dg-\frac{2}{3}\rho g)=2Tr=\sqrt{\frac{2T}{\frac{2}{3}g(2d-\rho )}}r=\sqrt{\frac{3T}{g(2d-\rho )}}$