Work done in increasing the size of a soap bubble from a radius of $3 \ cm$ to $5 \ cm$ is nearly (Surface tension of soap solution $= 0.03 \ Nm^{-1}$) (in $\pi \ mJ$)

How much work is required to form a bubble of radius $10 \ cm$ from a liquid with surface tension $\frac{3}{100} \ N/m$?

Twenty-seven droplets of water, each of radius $0.1 \,mm$, merge to form a single drop. Calculate the energy released during this process. (Take surface tension of water $T = 0.072 \,N/m$)

$A$ water film is formed between two parallel wires of $10 \text{ cm}$ length. The distance of $0.5 \text{ cm}$ between the wires is increased by $1 \text{ mm}$. The work done in the process is (surface tension of water $= 72 \text{ mN/m}$).

The surface tension of the soap water solution is $\frac{1}{10 \pi} \text{ N m}^{-1}$. The free energy of the surface layer of a soap bubble of diameter $5 \text{ mm}$ will be:

$A$ 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 $\mu J$)