Question bank/JEE Main/Physics

A small spherical droplet of density d is floating exactly half
immerse 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)

A$r = \sqrt{\frac{T}{(d + \rho)g}}$B$r = \sqrt{\frac{3T}{(2d + \rho)g}}$C$r = \sqrt{\frac{T}{(d + \rho)g}}$D$r = \sqrt{\frac{2T}{3(d + \rho)g}}$

Show answer and solution▾

$\Rightarrow d \cdot \frac{4}{3}\pi r^{3}g = \rho \cdot \frac{2}{3}\pi r^{3}g + 2\pi rT$
$\Rightarrow \frac{2}{3}r^{2}g(2d - \rho) = 2T$
$\Rightarrow r = \sqrt{\frac{3T}{(2d - \rho)}}$

Source: Grass-G / P-24 Physcis Paper 31 Dec.docx