Question

On heating water, bubbles beings formed at the bottom of the vessel detatch and rise. Take the bubbles to the 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 detatch is : (density of water is $\rho$).

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

Answer 1

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

The bubble will detach if p

Buoyant force $\ge$ surface tension

$\therefore \left(2\pi rT\right)\sin \theta =\frac{4}{3}\pi R^{3}pwg$

$T\times \frac{r}{R}\times 2\pi r=\frac{4}{3}\pi R^{3}pwg$

$r^2=\frac{2}{3}\frac{R^{4}pwg}{T}$

$\therefore r=R^2\sqrt{\frac{pwg}{3T}}$

Answer is A