The excess pressure inside a bubble in water is $P_1$. The excess pressure inside a drop of the same radius is $P_2$. Then:

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:

If a soap bubble expands, the pressure inside the bubble:

If the radius of a soap bubble is four times that of another,then the ratio of their excess pressures will be

The excess pressure inside a spherical drop of water $A$ is four times that of another drop $B$. Then the ratio of mass of drop $A$ to that of drop $B$ is

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :