The work done in splitting a drop of water of $1\, mm$ radius into $10^6$ droplets is (surface tension of water $T = 72 \times 10^{-3}\, N/m$):

When two soap bubbles of radius $r_1$ and $r_2$ $(r_2 > r_1)$ coalesce,the radius of curvature of the common surface is:

The excess pressure inside an air bubble of radius $r$ just below the surface of water is $p_1$. The excess pressure inside a drop of the same radius just outside the surface is $p_2$. If $T$ is surface tension,then

$A$ large number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given,surface tension of liquid $T$,density $\rho$):

$A$ spherical liquid drop of radius $R$ is divided into eight equal droplets. If surface tension is $T$,then the work done in this process will be

Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is