Force due to Surface Tension question from JEE Main 2014

Answer: (d)

1. Upward buoyancy on the bubble $\approx \frac{4}{3}\pi R^3 \rho_w g$ (since $r \ll R$).
2. Surface tension acts along the contact circle $2\pi r$; its vertical (downward) part is $2\pi r T\sin\theta$ with $\sin\theta = \frac{r}{R}$.
3. Just before detaching: $\dfrac{2\pi r^2 T}{R} = \frac{4}{3}\pi R^3\rho_w g \Rightarrow r^2 = \dfrac{2R^4\rho_w g}{3T}$.
4. $r = R^2\sqrt{\dfrac{2\rho_w g}{3T}}$. So the answer is (d).