Drops of liquid of density d are floating hal | vedclass

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(C) For a drop floating in equilibrium,the downward gravitational force must be balanced by the upward buoyant force and the upward surface tension fo

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

Vedclass · Answer

(C) For a drop floating in equilibrium,the downward gravitational force must be balanced by the upward buoyant force and the upward surface tension force.Weight of the drop = $\frac{4}{3} \pi r^3 d g$.Buoyant force (half immersed) = $\frac{2}{3} \pi r^3 ho g$.Surface tension force acting along the circumference of the contact circle = $T 	imes 2 \pi r$.Equating the forces: $\frac{4}{3} \pi r^3 d g = \frac{2}{3} \pi r^3 ho g + 2 \pi r T$.Dividing by $\pi r$: $\frac{4}{3} r^2 d g = \frac{2}{3} r^2 ho g + 2 T$.Rearranging terms: $r^2 g (\frac{4}{3} d - \frac{2}{3} ho) = 2 T$.$r^2 g (\frac{2}{3}) (2 d - ho) = 2 T$.$r^2 = \frac{3 T}{g(2 d - ho)}$.Therefore,$r = \sqrt{\frac{3 T}{g(2 d - ho)}}$.

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

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