On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If $$r \ll R$$, and the surface tension of water is T, value of r just before bubbles detach is: (density of water is $$\rho_w$$)

$$F_B = \text{Volume} \times \rho_w \times g = \frac{4}{3}\pi R^3 \rho_w g$$ (Buoyancy Force)

The surface tension of water pulls downward along the circular contact perimeter of radius $$r$$. From the geometry of the contact region, let $$\theta$$ be the angle that the bubble's surface makes with the vertical at the contact rim. For a spherical shape where $$r \ll R$$: $$\sin\theta \approx \frac{r}{R}$$

The downward component of the surface tension force acting along the circular perimeter of radius $$r$$ ($$2\pi r$$) is $$F_{ST} = T \cdot (2\pi r) \cdot \sin\theta$$

$$F_{ST} = 2\pi r T \left(\frac{r}{R}\right) = \frac{2\pi r^2 T}{R}$$

Just before detachment, the upward buoyancy force is exactly counterbalanced by the downward surface tension force holding the bubble to the bottom:

$$F_{ST} = F_B$$

$$\frac{2\pi r^2 T}{R} = \frac{4}{3}\pi R^3 \rho_w g$$

$$r^2 = \frac{4 R^4 \rho_w g}{6 T} = \frac{2 \rho_w g}{3T} R^4$$

$$r = R^2 \sqrt{\frac{2\rho_w g}{3T}}$$