The surface energy of a liquid drop is $U$. It splits up into $512$ equal droplets. The surface energy becomes (in $U$)

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

$A$ soap bubble of initial radius $R$ is to be blown up. The surface tension of the soap film is $T$. The surface energy needed to double the diameter of the bubble is (in $\pi R^2 T$)

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 $m$)

$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, what is the energy spent in blowing the bubble (in $\pi T r^2$)?

The surface tension of a soap solution is $3.5 \times 10^{-2} \text{ N/m}$. The work required to increase the radius of a soap bubble from $1 \text{ cm}$ to $2 \text{ cm}$ is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . $(\pi = 22/7)$