The excess pressure inside a soap bubble is three times that of another soap bubble. The ratio of their volumes is:

What should be the radius of a water drop so that the excess pressure inside it is $72 \ Nm^{-2}$ (in $mm$)? (The surface tension of water is $7.2 \times 10^{-2} \ Nm^{-1}$)

Under isothermal condition, two soap bubbles of radii $r_1$ and $r_2$ combine to form a single soap bubble of radius $R$. If $P$ is the external pressure and $T$ is the surface tension of the soap solution, find the expression for $T$.

$A$ 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 soap bubbles of radii $x$ and $y$ coalesce to form a single bubble of radius $z$. Then $z$ is equal to

Two soap bubbles of radii $2 \, cm$ and $4 \, cm$ join to form a double bubble in air. The radius of curvature of the interface is .......... $cm$.