The work done in blowing a soap bubble of radius $0.2 \, m$,given that the surface tension of the soap solution is $60 \times 10^{-3} \, N/m$,is:

$A$ mercury drop of radius $1 \ cm$ is sprayed into $10^6$ drops of equal size. The energy expended in joules is (surface tension of mercury is $460 \times 10^{-3} \ N/m$).

Fill in the blanks:
$(i)$ As the surface area is large,the potential energy of the molecules is ...... (more / less).
$(ii)$ As the surface area is small,the potential energy of its molecules is ...... (more / less).

$A$ 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:

The surface tension of the soap water solution is $\frac{1}{10 \pi} \text{ N m}^{-1}$. The free energy of the surface layer of a soap bubble of diameter $5 \text{ mm}$ will be:

The surface tension of a liquid is $5 \,N/m$. If a film is held on a ring of area $0.02 \,m^2$,its surface energy is about ........... $J$.