$A$ drop of mercury of radius $2\, mm$ is split into $8$ identical droplets. Find the increase in surface energy in $\mu J$. (Surface tension of mercury is $0.465\;J/m^2$)

The radius of a soap bubble is $r$, and the surface tension of the soap solution is $T$. Without increasing the temperature, how much energy is required to double its radius (in $\pi r^2 T$)?

Two small drops of mercury each of radius $R$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is -

The work done in blowing a soap bubble of volume $V$ is $W$. The work required to blow a soap bubble of volume $2V$ is [where $T$ is the surface tension of the soap solution].

The amount of work done in forming a soap film of size $10\,cm \times 10\,cm$ is (Surface tension $T = 3 \times 10^{-2}\,N/m$).

Fill in the blanks:
$(i)$ As the surface area is large,the potential energy of the molecules is ...... (more / less).
$(ii)$ As the surface area is small,the potential energy of its molecules is ...... (more / less).