In $S.I.$ system,the total energy of the free surface of a liquid drop is $2 \pi$ times the surface tension of the liquid. The diameter of the drop is . . . . . . . (in $m$)

If $T$ is the surface tension of a soap solution, then the amount of work done in increasing the diameter of a soap bubble from $D$ to $2D$ is: (in $\pi D^2 T$)

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

The radius $R$ of the soap bubble is doubled under isothermal conditions. If $T$ is the surface tension of the soap bubble, the work done in doing so is given by (in $\pi R^2 T$)

$A$ liquid drop of diameter $2 \text{ mm}$ breaks into $512$ droplets. The change in surface energy is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . (Take surface tension of liquid = $0.08 \text{ N/m}$)

The amount of work done to break a big water drop of radius $R$ into $27$ small drops of equal radius is $10 \ J$. The work done required to break the same big drop into $64$ small drops of equal radius will be (in $J$)