Drops of liquid of density $d$ are floating half immersed in a liquid of density $\rho$. If the surface…
Drops of liquid of density $d$ are floating half immersed in a liquid of density $\rho$. If the surface tension of the liquid is $T$, then the radius of the drop is
$\sqrt{\frac{3 T}{g(3 d-\rho)}}$
$\sqrt{\frac{6 T}{g(2 d-\rho)}}$
$\sqrt{\frac{3 T}{g(2 d-p)}}$
$\sqrt{\frac{3 T}{g(4 d-3 p)}}$
Solution
According to the question,
$\frac{4}{3} \pi r^3 d g=\frac{2}{3} \pi r^3 \rho g+T \times 2 \pi r$
or
$r=\sqrt{\frac{3 T}{g(2 d-\rho)}}$