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:

The 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 surface tension of the soap water solution is $\frac{1}{10 \pi} \text{ N m}^{-1}$. The free energy of the surface layer of a soap bubble of diameter $5 \text{ mm}$ will be:

The surface tension of a liquid is $0.5 \, N/m$. If a film is held on a ring of area $0.02 \, m^2$,its surface energy 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 $\pi = \frac{22}{7}$.

How much energy is required to double the radius of a soap bubble of radius $R$ and surface tension $T$ (in $\pi R^2 T$)?