An air bubble of radius $0.1\, cm$ is in a liquid having surface tension $0.06\, N/m$ and density $10^3\, kg/m^3$. The pressure inside the bubble is $1100\, N/m^2$ greater than the atmospheric pressure. At what depth $h$ (in $m$) is the bubble below the surface of the liquid? $(g = 9.8\, m/s^2)$

The pressures inside two soap bubbles $A$ and $B$ are $1.02 \text{ atm}$ and $1.04 \text{ atm}$ respectively. The ratio of the volume of bubble $A$ to that of bubble $B$ is (outside pressure = $1 \text{ atm}$)

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 the internal pressures of two soap bubbles are $1.01 \text{ atm}$ and $1.02 \text{ atm}$ respectively, find the ratio of their volumes.

Two soap bubbles of radii $r_1$ and $r_2$ coalesce in vacuum under isothermal conditions to form a bigger bubble of radius $R$. What is the radius of the bigger bubble?

$A$ glass capillary tube of inner diameter $0.28 \ mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as that in the vessel (in $N/m^2$) is:
Surface tension of water $= 0.07 \ N/m$
Atmospheric pressure $= 10^5 \ N/m^2$