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 radius of the drop will be

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

Answer 1

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


For equilibrium of drop
$F_B+F_T=mg$
$\frac{2}{3}\pi R^3\rho g+2\pi RT=\frac{4}{3}\pi R^{3}dg$

$2RT=\frac{2}{3}{R}^{3}g(2d-\rho )$

$R=\sqrt{\frac{3T}{(2d-\rho )g}}$

$\therefore \left(a\right)$