If the surface tension of a soap solution is $3 \times 10^{-2} \,N/m$, then the work done in forming a soap film of $20 \,cm \times 5 \,cm$ will be

$A$ 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 $\pi = \frac{22}{7}$.

If the surface tension of a liquid is $T$,the gain in surface energy for an increase in liquid surface by $A$ is

$A$ mercury drop of radius $1 \ cm$ is sprayed into $10^6$ drops of equal size. The energy expended in joules is (surface tension of mercury is $460 \times 10^{-3} \ N/m$).

The area of a liquid film is $60\, cm^2$ and its surface tension is $T = 20\, dyne/cm$. What is the work done to increase its area to $120\, cm^2$ (in $, erg$)?

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