An air bubble of radius $1.0 \ mm$ is observed at a depth of $20 \ cm$ below the free surface of a liquid having surface tension $0.095 \ J/m^2$ and density $10^3 \ kg/m^3$. The difference between pressure inside the bubble and atmospheric pressure is . . . . . . $N/m^2$. (Take $g = 10 \ m/s^2$)

In air,a charged soap bubble of radius $R$ breaks into $27$ small soap bubbles of equal radius $r$. Then,the ratio of mechanical force acting per unit area of the big soap bubble to that of a small soap bubble is

Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops of the same liquid combine, then the excess pressure inside the bigger drop is (in $units$)

If two soap bubbles of equal radii $r$ coalesce,then the radius of curvature of the interface between the two bubbles will be:

$27$ drops of mercury coalesce to form a bigger drop. What is the relative increase in surface energy?

If two glass plates have water between them and are separated by a very small distance (see figure),it is very difficult to pull them apart. This is because the water in between forms a cylindrical surface on the side that gives rise to a lower pressure in the water in comparison to the atmosphere. If the radius of the cylindrical surface is $R$ and the surface tension of water is $T$,then the pressure in the water between the plates is lower by: