When two soap bubbles of radii $4 \, cm$ and $5 \, cm$ coalesce,what is the radius of the common interface surface in $cm$?

Consider an air bubble of radius $2 \,mm$ in a liquid at a depth of $5 \,cm$ below the free surface. The density of the liquid is $1000 \,kg/m^3$ and the surface tension is $0.1 \,N/m$. Find the pressure inside the air bubble relative to the pressure at the free surface of the liquid. (Take $g = 10 \,m/s^2$) (in $\,Pa$)

$A$ 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 $= 10^5 \ N/m^2$

Two bubbles $A$ and $B$ $(r_A > r_B)$ are joined through a narrow tube. Then

In a water tank,an air bubble rises from the bottom to the top surface of the water. If the depth of the water in the tank is $7.28 \ m$ and atmospheric pressure is $10 \ m$ of water,then the ratio of the radii of the bubble at the bottom of the tank and at the top surface of the water is (Temperature of the water in the tank is constant).

$A$ number of small water droplets of surface tension '$T$',each of radius '$r$',are combined to form a single drop of radius '$R$'. If the released energy is converted into kinetic energy,then the velocity acquired by the bigger drop is . . . . . . ($\rho$ - density of water)