A drop of liquid of density rho is floating h | vedclass

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Question

(A) For the drop to be in equilibrium,the upward forces (buoyant force and surface tension force) must balance the downward force (weight of the drop)

Vedclass · Accepted Answer

$\frac{15}{\sqrt{2\rho - \sigma}}$

Vedclass · Answer

(A) For the drop to be in equilibrium,the upward forces (buoyant force and surface tension force) must balance the downward force (weight of the drop).Let $R$ be the radius of the drop. The volume of the drop is $V = \frac{4}{3}\pi R^3$.Buoyant force $F_b = 	ext{Volume immersed} 	imes \sigma 	imes g = \frac{V}{2} \sigma g = \frac{2}{3}\pi R^3 \sigma g$.Surface tension force $F_T = T 	imes (2\pi R) = 2\pi RT$.Weight of the drop $W = mg = V ho g = \frac{4}{3}\pi R^3 ho g$.Equating forces: $F_b + F_T = W$$\frac{2}{3}\pi R^3 \sigma g + 2\pi RT = \frac{4}{3}\pi R^3 ho g$$2\pi RT = \frac{4}{3}\pi R^3 ho g - \frac{2}{3}\pi R^3 \sigma g = \frac{2}{3}\pi R^3 g (2ho - \sigma)$$T = \frac{R^2 g (2ho - \sigma)}{3} \Rightarrow R^2 = \frac{3T}{g(2ho - \sigma)}$Given $T = 7.5 	imes 10^{-4} \, N/cm = 7.5 	imes 10^{-2} \, N/m$ and $g = 10 \, m/s^2$:$R = \sqrt{\frac{3 	imes 7.5 	imes 10^{-2}}{10(2ho - \sigma)}} = \sqrt{\frac{22.5 	imes 10^{-2}}{10(2ho - \sigma)}} = \sqrt{\frac{2.25 	imes 10^{-2}}{2ho - \sigma}} = \frac{0.15}{\sqrt{2ho - \sigma}} \, m$Converting to $cm$ $(1 \, m = 100 \, cm)$:$R = \frac{0.15 	imes 100}{\sqrt{2ho - \sigma}} \, cm = \frac{15}{\sqrt{2ho - \sigma}} \, cm$.

$A$ drop of liquid of density $\rho$ is floating half-immersed in a liquid of density $\sigma$ and surface tension $T = 7.5 \times 10^{-4} \, N \, cm^{-1}$. The radius of the drop in $cm$ will be: (Take: $g = 10 \, m/s^2$)

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