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

$A$ drop of mercury of radius $2\, mm$ is split into $8$ identical droplets. Find the increase in surface energy in $\mu J$. (Surface tension of mercury is $0.465\;J/m^2$)

$A$ water film is formed between two straight parallel wires,each of length $10 \text{ cm}$,kept at a separation of $0.5 \text{ cm}$. Now,the separation between them is increased by $1 \text{ mm}$ without breaking the water film. The work done for this is (surface tension of water $= 7.2 \times 10^{-2} \text{ N/m}$)

If the surface tension of a soap solution is $3 \times 10^{-2} \,N/m$, then the work done in forming a soap film of $20 \,cm \times 5 \,cm$ will be

$A$ 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 $\pi R^2 T$)

$A$ 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 = $0.08 \text{ N/m}$)