The surface tension of a soap solution is $T$. Work done in blowing a soap bubble of diameter $2d$ is (in $\pi d^2 T$)

$A$ water film is formed between two straight parallel wires,each of length $10 \text{ cm}$,kept at a separation of $0.5 \text{ cm}$. Now,the separation between them is increased by $1 \text{ mm}$ without breaking the water film. The work done for this is (surface tension of water $= 7.2 \times 10^{-2} \text{ N/m}$)

If the work done in blowing a soap bubble of volume $V$ is $W$,then the work done in blowing a soap bubble of volume $2V$ will be

$A$ drop of liquid of radius $R=10^{-2} \,m$ having surface tension $S=\frac{0.1}{4 \pi} \,Nm^{-1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $\Delta U=10^{-3} \,J$. If $K=10^\alpha$, then the value of $\alpha$ is:

$A$ wire of length $10 \, cm$ is used in a soap film experiment. What is the work done to move the wire by $1 \, mm$? (Surface tension $T = 7.2 \times 10^{-2} \, N/m$)

$A$ liquid drop having surface energy $E$ is sprayed into $512$ droplets of the same size. Then the final surface energy is