There are two liquid drops of different radii. The excess pressure inside over the outside is

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:

When two soap bubbles of radii $a$ and $b$ $(b > a)$ coalesce,the radius of curvature of the common surface is:

The radius of a soap bubble is increased from $R$ to $2R$. The work done in this process in terms of surface tension $S$ is: (in $\pi R^2 S$)

If the excess pressure inside a soap bubble of radius $3 \,mm$ is equal to the pressure of a water column of height $0.8 \,cm$, then the surface tension of the soap solution is ( $\rho_{\text{water}} = 1000 \,kg/m^3, g = 9.8 \,m/s^2$ ).

Pressure inside two soap bubbles are $1.01 \, atm$ and $1.02 \, atm$. The ratio between their volumes is: