$A$ water drop of radius $1\,cm$ is broken into $729$ equal droplets. If the surface tension of water is $75\,dyne/cm$,then the gain in surface energy up to the first decimal place will be $...\times 10^{-4}\,J$.

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

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

The amount of work done in blowing a soap bubble such that its diameter increases from $d_1$ to $d_2$ is ($T=$ surface tension of soap solution).

$A$ 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:

$A$ 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 $\mu J$)