If the surface tension of a soap solution is $3 \times 10^{-2} \,N/m$, then the work done in forming a soap film of $20 \,cm \times 5 \,cm$ will be

$A$ film of water is formed between two straight parallel wires of length $10 \ cm$ each,separated by $0.5 \ cm$. If their separation is increased by $1 \ cm$ while still maintaining their parallelism,how much work will have to be done? (Surface tension of water $= 7.2 \times 10^{-2} \ N/m$)

$A$ soap bubble of radius $\frac{1}{\sqrt{\pi}} \text{ cm}$ is expanded to radius $\frac{3}{\sqrt{\pi}} \text{ cm}$. The surface tension of the soap solution is $25 \text{ dyne/cm}$. The work done during expansion in $\text{erg}$ is:

$A$ mercury drop of radius $1 \ cm$ is divided into $10^6$ droplets of equal size. If the surface tension of mercury is $35 \times 10^{-3} \ N/m$,then the change in surface energy in the process is:

The surface energy of a liquid drop is $V$. It is sprayed into $1000$ equal droplets. The surface energy of all the droplets is

If $T$ is the surface tension,the amount of work done in blowing a soap bubble from a diameter $d$ to diameter $D$ is