Force due to Surface Tension question from MHT CET 2024 (3 May, Shift 2)

Answer: (c)

1. Weight = buoyancy + surface tension force along the waterline: $\frac{4}{3}\pi r^3\rho g = \frac{2}{3}\pi r^3 d g + 2\pi r T$.
2. $\frac{2}{3} r^2 g(2\rho - d) = 2T \Rightarrow r^2 = \dfrac{3T}{g(2\rho - d)}$.
3. Diameter $D = 2r \Rightarrow D^2 = \dfrac{12T}{g(2\rho - d)}$.
4. So the answer is (c) $\left[\frac{12T}{g(2\rho - d)}\right]^{1/2}$.