The work done in blowing a soap bubble of radius $0.2 \, m$,given that the surface tension of the soap solution is $60 \times 10^{-3} \, N/m$,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 $\pi = \frac{22}{7}$.

Consider a soap film on a rectangular frame of wire of area $4 \times 4 \text{ cm}^2$. If the area of the soap film is increased to $4 \times 5 \text{ cm}^2$,the work done in the process will be (The surface tension of the soap film is $3 \times 10^{-2} \text{ N/m}$).

If $8$ identical droplets are formed from a single drop of radius $2\,mm$,what is the change in energy in $\mu J$? (Surface tension $T = 0.465\,J/m^2$)

The radius $R$ of the soap bubble is doubled under isothermal conditions. If $T$ is the surface tension of the soap bubble, the work done in doing so is given by (in $\pi R^2 T$)

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