The excess pressure due to surface tension in a spherical liquid drop of radius $r$ is directly proportional to

The pressure inside two soap bubbles, $A$ and $B$, is $1.01 \,atm$ and $1.02 \,atm$ respectively. The ratio of their respective radii $(r_A : r_B)$ is (outside pressure $= 1 \,atm$).

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.]

If the shape of the liquid surface is curved, then

If two soap bubbles of different radii are connected by a tube,then

Three small identical bubbles of water having the same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is: