The potential energy of a molecule on the surface of a liquid compared to one inside the liquid is

Two mercury drops (each of radius $r$) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

What will be the work done in increasing the radius of a soap bubble from $\frac{r}{2}$ to $2r$, if the surface tension of the soap solution is $T$ (in $\pi r^2 T$)?

Surface tension of a soap solution is $2 \times 10^{-2} \, N/m$. The work done in producing a soap bubble of radius $2 \, cm$ is:

The work done in blowing a soap bubble of radius $0.2 \, m$,given that the surface tension of the soap solution is $60 \times 10^{-3} \, N/m$,is:

$A$ water drop breaks into $64$ identical droplets, each with a surface area of $10^{-7} \,m^2$. If the surface tension of water is $0.07 \,N/m$, what is the increase in surface energy during this process?