Two soap bubbles, $A$ and $B$, have radii in the ratio $3 : 2$. The ratio of the excess pressure inside bubble $A$ to that of bubble $B$ is:

The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times the volume of the second,where $n$ is:

When a mercury drop of radius $R$ breaks into $n$ droplets of equal size,the radius $r$ of each droplet is

The adjoining diagram shows three soap bubbles $A, B$ and $C$ prepared by blowing the capillary tube fitted with stop cocks $S_1, S_2$ and $S_3$. With stop cock $S$ closed and stop cocks $S_1, S_2$ and $S_3$ opened,what will happen?

$A$ soap bubble,having a radius of $1\; mm$,is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2}\; N/m$. The pressure inside the bubble is equal to the pressure at a point $Z_{0}$ below the free surface of water in a container. Taking $g=10\; m/s^{2}$ and the density of water $\rho = 10^{3}\; kg/m^{3}$,the value of $Z_{0}$ is......$cm$.

Small drops of liquid of the same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is: