What is the ratio of the number of moles of air in two soap bubbles of radii $R_1$ and $R_2$?

The pressure inside a soap bubble $A$ is $1.01 \text{ atm}$ and that in a soap bubble $B$ is $1.02 \text{ atm}$. The ratio of the volume of $A$ to that of $B$ is

$A$ 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......$cm$.

Pure water rises through a height $h$ in a capillary tube of internal radius $r$. Surface tension of water is $T$. The pressure difference between the water level in the container and the lowest point of the concave meniscus is

$A$ soap bubble is in the form of a circular tube having a radius of curvature $R$ and a radius of curvature perpendicular to it of $5R$. Find the excess pressure in the bubble.

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 $units$)