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A small spherical droplet of density d is floating exactly half immersed in a liquid of density $ρ$ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

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$R = \sqrt{\frac{2T}{3 (d + ρ) g}}$

$R = \sqrt{\frac{T}{(d - ρ) g}}$

$R = \sqrt{\frac{T}{(d + ρ) g}}$

$R = \sqrt{\frac{3T}{(2d - ρ) g}}$

answer is D.

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Detailed Solution

For the drops to be in equilibrium upward force on drop = downward force on drop

$T. 2πR = \frac{4}{3} {πR}^{3} dg - \frac{2}{3} {πR}^{3} ρg$

$\Rightarrow T (2πR) = \frac{2}{3} {πR}^{3} (2d-ρ) g$

$\Rightarrow T= \frac{R^{2}}{3} (2d-ρ) g \Rightarrow R = \sqrt{\frac{3T}{(2d-ρ) g}}$

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