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 $7.2 \times 10^{-2} \ Nm^{-1}$)

An ice cube of edge $1 \ cm$ melts in a gravity-free container. The approximate surface area of the water formed is (water is in the form of a spherical drop)

The excess pressure inside the first soap bubble is three times that inside the second soap bubble. The ratio of the volume of the first soap bubble to the volume of the second soap bubble is

$A$ spherical drop of water has a radius of $1\, mm$. If the surface tension of water is $70 \times 10^{-3} \, N/m$,then the difference of pressure between the inside and outside of the spherical drop is:

Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 \ N/m^2$. The radii of bubbles $A$ and $B$ are $2 \ cm$ and $4 \ cm$,respectively. The surface tension of the soap solution is $0.04 \ N/m$. Find the ratio $n_B / n_A$,where $n_A$ and $n_B$ are the number of moles of air in bubbles $A$ and $B$,respectively. [Neglect the effect of gravity.]

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is