Find out the work done (in $\times 10^{-3} \; J$) to expand the soap bubble to radius $R = 5 \; cm$ (Surface tension of water $= 0.1 \; N/m$).

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 $T = 0.072 \,N/m$)

$A$ big water drop is formed by the combination of $n$ small water droplets of equal radii. The ratio of the surface energy of $n$ droplets to the surface energy of the big drop is

The surface tension of soap solution is $0.03 \,N/m$. The work done in blowing to form a soap bubble of surface area $40 \,cm^2$ (in $J$) is:

The work done in increasing the diameter of a soap bubble from $2 \ cm$ to $4 \ cm$ is (Surface tension of soap solution $= 3.5 \times 10^{-2} \ N/m$)

The surface tension of a soap solution is $T$. Work done in blowing a soap bubble of diameter $2d$ is (in $\pi d^2 T$)