Question

While heating water, bubbles begin to form at the bottom of the vessel. After a while, they detach and rise. Consider a bubble to be a sphere of radius $R$ and making a circular contact of radius $r$ with the bottom of the vessel. Here, $r<<R$, the surface tension of water is $T$, and density of water is ${\rho }_w$. The value of $r$, just before the bubble detaches is,

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

Let us draw the situation of the bubble.

As the bubble just detaches from the surface,
Buoyant force $=$ surface tension force

$\frac{4}{3}\pi R^3{\rho }_{w}g=\int T\times dl\sin \theta$

$\left({\rho }_w\right)\left(\frac{4}{3}\pi R^3\right)g=\left(T\right)\left(2\pi r\right)\sin \theta$

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

$\left({\rho }_w\right)\left(\frac{4}{3}\pi R^4\right)g=\left(T\right)\left(2\pi r^2\right)$

Solving the above equation, we get,

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

Hence, $\left(A\right)$ is the correct answer.