• # question_answer 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 radius of the drop will be A)  $\sqrt{\frac{3T}{g(2d-\rho )}}$        B)  $\sqrt{\frac{3T}{g(d-\rho )}}$ C)  $\sqrt{\frac{6T}{g(2d-\rho )}}$        D)  $\sqrt{\frac{6T}{g(d-\rho )}}$

For equilibrium of drop, $\frac{2}{3}\pi {{R}^{3}}\,\rho g+2\pi RT=\frac{4}{3}\pi {{R}^{3}}dg$ $\Rightarrow$            $2RT=\frac{2}{3}{{R}^{3}}g\,(2d-\rho )$ $\therefore$    $R=\sqrt{\frac{3T}{(2d-\rho )g}}$