Question

Two soap bubbles of radius $R$ each stick to each other in mid air due to which their common interface is in the shape of a circle. Area of the circle is

• A. $\frac{\pi R^2}{2}$
• B. $\frac{3}{4}\pi R^2$
• C. $\frac{\pi R^2}{4}$
• D. $\frac{16}{25}\pi R^2$

Answer 1

The correct option is B $\frac{3}{4}\pi R^2$


For the configuration to be stable (net force = 0 at the point P), $\theta ={120}^0$. Therefore angle made by force of surface tension with the vertical at the surface of the bubble is ${60}^0$.


Radius of the circle formed by the interface
$r=Rsin{60}^0=\frac{\sqrt{3}}{2}R$
$\therefore$ Area of the interface $A=\pi r^2=\frac{3}{4}\pi R^2$