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 $= 0.03\, Nm^{-1}$)

Consider a soap film on a rectangular frame of wire of area $3 \times 3 \,cm^2$. If the area of the soap film is increased to $5 \times 5 \,cm^2$, the work done in the process will be (surface tension of soap solution is $2.5 \times 10^{-2} \,N/m$).

If $T$ is the surface tension of a liquid,the energy needed to break a liquid drop of radius $R$ into $64$ drops is:

The surface tension of 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 work done in blowing a soap bubble of radius $r$ from a solution of surface tension $T$ is:

The surface tension of a soap solution is $3.5 \times 10^{-2} \, N m^{-1}$. The amount of work done required to increase the radius of a soap bubble from $10 \, cm$ to $20 \, cm$ is $..... \times 10^{-4} \, J$.