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

The amount of energy required to form a soap bubble of radius $2\,cm$ from a soap solution is nearly $..........\,\times 10^{-4}\,J$: (surface tension of soap solution $=0.03\,N\,m^{-1}$)

$A$ big water drop is formed by the combination of $n$ small water droplets of equal radii. The ratio of the surface energy of $n$ droplets to the surface energy of the big drop is

$A$ thin film of water is formed between two straight parallel wires each of length $8 \ cm$ separated by a distance of $0.6 \ cm$. The work done to increase the distance between the wires to $0.8 \ cm$ is (Surface tension of water $= 0.07 \ N/m$) (in $\mu J$)

Two small drops of mercury each of radius $R$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is -

The amount of work done in forming a soap film of size $10\,cm \times 10\,cm$ is (Surface tension $T = 3 \times 10^{-2}\,N/m$).