If $T$ is the surface tension of a soap solution, the amount of work done in blowing a soap bubble from a diameter $D$ to $2D$ is (in $\pi D^2 T$)

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 -

$A$ soap bubble is blown to a diameter of $7 \ cm$. $36960 \ erg$ of work is done in blowing it further. If the surface tension of the soap solution is $40 \ dyne/cm$,then the new radius is . . . . . . $cm$. Take $\pi = \frac{22}{7}$.

The surface tension of a soap solution is $0.03 \,N/m$. The work done in blowing to form a soap bubble of surface area $40 \,cm^2$ (in $J$) is:

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).

$A$ big water drop is formed by the combination of $n$ small water droplets of equal radii. The ratio of the surface energy of $n$ droplets to the surface energy of the big drop is