Work done in increasing the size of a soap bubble from a radius of $3\, cm$ to $5\, cm$ is (Surface tension of soap solution $= 0.03\, Nm^{-1}$)

$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, the energy spent in the blowing is: (in $\pi T r^2$)

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 $3 \times 10^{-2} \text{ N/m}$).

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 $, erg$)?

Energy required to form a soap bubble of diameter $20 \, cm$ will be (Surface tension for soap solution is $30 \, dynes/cm$)

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is