Surface tension of a soap solution is $2 \times 10^{-2} \, N/m$. The work done in producing a soap bubble of radius $2 \, cm$ is:

$A$ soap bubble of surface tension $0.04 \ N/m$ is blown to a diameter of $7 \ cm$. If $(15000 - x) \ \mu J$ of work is done in blowing it further to make its diameter $14 \ cm$, then the value of $x$ is . . . . . . . (Take $\pi = 22/7$)

$A$ drop of liquid of radius $R=10^{-2} \,m$ having surface tension $S=\frac{0.1}{4 \pi} \,Nm^{-1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $\Delta U=10^{-3} \,J$. If $K=10^\alpha$, then the value of $\alpha$ is:

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to total initial surface energy is

$A$ frame made of metallic wire enclosing a surface area $A$ is covered with a soap film. If the area of the frame of metallic wire is reduced by $50 \%$,the energy of the soap film will be changed by (in $\%$)

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