Question

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 $T$, value of $r$ just before bubbles detach is (density of water is ${\rho }_w)$

• A. $R^2\sqrt{\frac{{\rho }_{w}g}{T}}$
• B. $R^2\sqrt{\frac{3{\rho }_{w}g}{T}}$
• C. $R^2\sqrt{\frac{2{\rho }_{w}g}{3T}}$
• D. $R^2\sqrt{\frac{{\rho }_{w}g}{6T}}$

Answer 1

The correct option is C $R^2\sqrt{\frac{2{\rho }_{w}g}{3T}}$


The bubble will detach if,

$\text{Buoyant force}\ge \text{surface tension force}$

$\frac{4}{3}\pi R^3{\rho }_{w}g\ge \int T\left(dl\sin \theta \right)$

${\rho }_w\frac{4}{3}\pi R^{3}g\ge T\cdot 2\pi r\sin \theta$

Here, $\sin \theta =\frac{r}{R}$

Solving,

$r=\sqrt{\frac{2{\rho }_w{R}^{4}g}{3T}}=R^2\sqrt{\frac{2{\rho }_{w}g}{3T}}$

Hence, option $\left(\text{B}\right)$ is correct.