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Physics Question on mechanical properties of fluid

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

A

r=2T3(d+ρ)gr=\sqrt{\frac{2T}{3\left(d+\rho\right)g}}

B

r=T(d+ρ)gr=\sqrt{\frac{T}{\left(d+\rho\right)g}}

C

r=T(d−ρ)gr=\sqrt{\frac{T}{\left(d-\rho\right)g}}

D

r=3T(2d−ρ)gr=\sqrt{\frac{3T}{\left(2d-\rho\right)g}}

Answer

r=3T(2d−ρ)gr=\sqrt{\frac{3T}{\left(2d-\rho\right)g}}

Explanation

Solution

FBD of droplet
B+F=mgB + F = mg
B=(23πR3)ρgB = \left(\frac{2}{3}\pi R^{3}\right)\rho g
F=T(2πR)F = T\left(2\pi R\right)
m=d(43πR3)m =d \left(\frac{4}{3}\pi R^{3}\right)
(23πR3)ρg+T(2πR)=(43πR3)g\left(\frac{2}{3}\pi R^{3}\right)\rho g + T\left(2\pi R\right) = \left(\frac{4}{3}\pi R^{3}\right)g
T(2πR)=(23πR3)g[2d−ρ]T\left(2\pi R\right) = \left(\frac{2}{3}\pi R^{3}\right)g \left[2d-\rho\right]
R=3T(2d−ρ)gR = \sqrt{\frac{3T}{\left(2d-\rho\right)g}}