$A$ liquid drop of diameter $2 \text{ mm}$ breaks into $512$ droplets. The change in surface energy is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . (Take surface tension of liquid = $0.08 \text{ N/m}$)

The surface tension of a 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 volume $V$ is $W$. The work done in blowing the bubble of volume $2V$ from the same soap solution is

$A$ 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 $T$ is the surface tension of a soap solution, the amount of work done in blowing a soap bubble from a diameter $D$ to $2D$ is (in $\pi D^2 T$)

The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is ($T=$ surface tension of solution).