Question:
A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $\rho$ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):
Correct Option: , 4
Solution:
(4) For the drops to be in equilibrium upward force on
drop $=$ downward force on drop
$T .2 \pi R=\frac{4}{3} \pi R^{3} d g-\frac{2}{3} \pi R^{3} \rho g$
$\Rightarrow T(2 \pi R)=\frac{2}{3} \pi R^{3}(2 d-\rho) g$
$\Rightarrow T=\frac{R^{2}}{3}(2 d-\rho) g \Rightarrow R=\sqrt{\frac{3 T}{(2 d-\rho) g}}$
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