On the surface of the liquid in equilibrium,molecules of the liquid possess

$A$ water drop of diameter $2\,cm$ is broken into $64$ equal droplets. The surface tension of water is $0.075\,N/m$. In this process,the gain in surface energy will be ...........$J$.

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T$ is the surface tension of water) (in $\pi R^2 T$)

The work done in blowing a soap bubble of radius $0.2 \, m$,given that the surface tension of the soap solution is $60 \times 10^{-3} \, N/m$,is:

$A$ 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 $= 7.2 \times 10^{-2} \text{ N/m}$)

$A$ 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 $\pi T r^2$)?