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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$r = \sqrt{\frac{T}{(d + ρ) g}}$

$r = \sqrt{\frac{2 T}{3 (d + ρ) g}}$

$r = \sqrt{\frac{T}{(d - ρ) g}}$

$r = \sqrt{\frac{3 T}{(2 d - ρ) g}}$

answer is D.

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Detailed Solution

Weight is balanced by Buoyancy force &surface tension forces $\Rightarrow m g = F_{B} + F_{S T}$

$\begin{array}{l} \Rightarrow m g = V_{i m} ρ g + T \times 2 π R \\ \Rightarrow V d g = \frac{V}{2} ρ g + T \times 2 π R \\ \Rightarrow V (d - \frac{ρ}{2}) g = T \times 2 π R \\ \Rightarrow \frac{4}{3} π R^{3} (\frac{2 d - ρ}{2}) g = T \times 2 π R \\ \Rightarrow R = \sqrt{\frac{3 T}{(2 d - ρ) g}} \end{array}$