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 << R and the surface tension of water is T, then value of r just before bubbles detach is
(density of water is $σ$ )

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$R^{2} \sqrt{\frac{2 ρ_{w} g}{3 T}}$

$R^{2} \sqrt{\frac{2 ρ_{w} g}{6 T}}$

$R^{2} \sqrt{\frac{ρ_{w} g}{T}}$

$R^{2} \sqrt{\frac{3 ρ_{w} g}{T}}$

answer is A.

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Detailed Solution

The bubble will detach, if buoyant force $\geq$ surface tension force
$\Rightarrow \frac{4}{3} {πR}^{3} ρ_{w} g \geq \int T \times dlsin θ; \Rightarrow (ρ_{w}) (\frac{4}{3} {πR}^{3}) g \geq (T) (2 πr) \sin θ here, \sin θ = \frac{r}{R} After Solving, r = \sqrt{\frac{2 ρ_{w} R^{4} g}{3 T}} \Rightarrow r = R^{2} \sqrt{\frac{2 ρ_{w} g}{3 T}}$