Two small droplets combine to form a single large drop. What is the ratio of the surface energy of the small droplets to the surface energy of the large drop?

$A$ spherical drop of liquid splits into $1000$ identical spherical drops. If $E_1$ is the surface energy of the original drop and $E_2$ is the total surface energy of the resulting drops,then $\frac{E_1}{E_2} = \frac{x}{10}$. The value of $x$ is:

Surface tension of a soap solution is $2 \times 10^{-2} \, N/m$. The work done in producing a soap bubble of radius $2 \, cm$ is:

$A$ water drop of radius $1\,cm$ is broken into $729$ equal droplets. If the surface tension of water is $75\,dyne/cm$,then the gain in surface energy up to the first decimal place will be $...\times 10^{-4}\,J$.

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 $T = 0.072 \,N/m$)

The radius of a soap bubble is $r$. The surface tension of the soap solution is $T$. Keeping the temperature constant,the radius of the soap bubble is doubled. The energy necessary for this process will be: (in $\pi r^2 T$)