A small spherical droplet of density d is floating exactly half immersed in a iiquid 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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$r = \sqrt{\frac{2 T}{3 (d + ρ) g}}$

$r = \sqrt{\frac{T}{(d - ρ) g}}$

$r = \sqrt{\frac{T}{(d + ρ) g}}$

$r = \sqrt{\frac{3 T}{(2 d - ρ) g}}$

answer is D.

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Detailed Solution

For the drop to be in equilibrium
Upward force = down ward force $F_{s . t} + F_{b} = F_{g}$

$T \times 2 π R + \frac{2}{3} π R^{3} ρ g = \frac{4}{3} π R^{3} d g$ $T \times 2 π R = \frac{2}{3} π R^{3} g (2 d - ρ)$

$R = \sqrt{\frac{3 T}{(2 d - ρ) g}}$

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