What should be the diameter of a soap bubble,in order that the excess pressure inside it is $25.6 \ Nm^{-2}$ (in $cm$)? [surface tension of soap solution $= 3.2 \times 10^{-2} \ Nm^{-1}$]

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 :

Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is (in $ : 1$)

If the excess pressure inside a soap bubble of radius $3 \,mm$ is equal to the pressure of a water column of height $0.8 \,cm$, then the surface tension of the soap solution is ( $\rho_{\text{water}} = 1000 \,kg/m^3, g = 9.8 \,m/s^2$ ).

The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal condition,they combine with each other to form a single bubble. The radius of the resultant bubble is

Derive the formula for the excess of pressure (pressure difference) inside a liquid drop and a soap bubble.