A liquid drop of density rho is floating half | vedclass

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(C) For the drop to be in equilibrium,the downward force (weight) must be balanced by the upward buoyant force and the upward surface tension force ac

Vedclass · Accepted Answer

$\left[\frac{12T}{g(2\rho-d)}\right]^{1/2}$

Vedclass · Answer

(C) For the drop to be in equilibrium,the downward force (weight) must be balanced by the upward buoyant force and the upward surface tension force acting along the circumference of the contact circle.Weight of the drop = $W = \frac{4}{3}\pi r^3 ho g$Buoyant force = $F_B = 	ext{Volume immersed} 	imes d 	imes g = \frac{1}{2} \left(\frac{4}{3}\pi r^3ight) dg = \frac{2}{3}\pi r^3 dg$Surface tension force = $F_T = T 	imes 2\pi r$Equating forces: $W = F_B + F_T$$\frac{4}{3}\pi r^3 ho g = \frac{2}{3}\pi r^3 dg + 2\pi rT$Rearranging terms: $2\pi rT = \frac{4}{3}\pi r^3 ho g - \frac{2}{3}\pi r^3 dg$$2\pi rT = \frac{2}{3}\pi r^3 g (2ho - d)$$T = \frac{1}{3} r^2 g (2ho - d)$$r^2 = \frac{3T}{g(2ho - d)}$$r = \sqrt{\frac{3T}{g(2ho - d)}}$Diameter $D = 2r = 2\sqrt{\frac{3T}{g(2ho - d)}} = \sqrt{\frac{12T}{g(2ho - d)}}$.

$A$ liquid drop of density $\rho$ is floating half-immersed in a liquid of density $d$. If $T$ is the surface tension,then the diameter of the liquid drop is ($g$ = acceleration due to gravity).

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