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).

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

The radius of a soap bubble is increased from $\frac{1}{\sqrt{\pi}} \text{ cm}$ to $\frac{2}{\sqrt{\pi}} \text{ cm}$. If the surface tension of water is $30 \text{ dynes/cm}$,then the work done will be ....... $ergs$.

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

Energy needed in breaking a liquid drop of radius $R$ into $n$ smaller drops,each of radius $r$,is [where $T$ is the surface tension of the liquid].

$A$ film of soap solution is formed between two straight parallel wires of length $10 \,cm$ each, separated by $0.5 \,cm$. If their separation is increased by $1 \,mm$ while still maintaining their parallelism, how much work will have to be done? (Surface tension of solution $= 65 \times 10^{-3} \,N/m$)