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 $, erg$)?

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:

Twenty-seven droplets of water, each of radius $0.1 \,mm$, merge to form a single drop. Calculate the energy released during this process. (Take surface tension of water $T = 0.072 \,N/m$)

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

$A$ 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 work done in blowing a soap bubble of radius $R$ is $W_1$ and that of radius $2R$ is $W_2$. The ratio of $W_1$ to $W_2$ is