$A$ liquid drop having surface energy $E$ is spread into $729$ droplets of same size. The final surface energy of the droplets is

The area of a liquid film is $60\, cm^2$ and its surface tension is $T = 20\, dyne/cm$. What is the work done to increase its area to $120\, cm^2$ (in $, erg$)?

$A$ 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 $460 \times 10^{-3} \,N/m$)

$A$ 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 work done in blowing a soap bubble of radius $0.2\, m$ is (the surface tension of soap solution being $0.06\, N/m$).

The work done in blowing a soap bubble of volume $V$ is $W$. The work done in blowing the bubble of volume $2V$ from the same soap solution is