Surface tension of a soap solution is $1.9 \times 10^{-2} \text{ N/m}$. The work done in blowing a soap bubble of $2.0 \text{ cm}$ diameter will be:

In $S.I.$ system,the total energy of the free surface of a liquid drop is $2 \pi$ times the surface tension of the liquid. The diameter of the drop is . . . . . . . (in $m$)

$A$ liquid drop having surface energy $E$ is sprayed into $512$ droplets of same size. The final surface energy is (in $E$)

$A$ 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 $= 7.2 \times 10^{-2} \ N/m$)

Energy required to form a soap bubble of diameter $20 \, cm$ will be (Surface tension for soap solution is $30 \, dynes/cm$)

$1000$ small water drops of equal size combine to form a big drop. The ratio of final surface energy to the total initial surface energy is