Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is . . . . . . . ($S$ is the surface tension of mercury). (in $\pi r^2 S$)

$A$ spherical drop of oil of radius $1\, cm$ is broken into $1000$ droplets of equal radii. If the surface tension of oil is $50\, dynes/cm$,the work done is

$A$ glass tube of uniform internal radius has a valve separating two identical ends. Initially,the valve is in a tightly closed position. End $1$ has a hemispherical soap bubble of radius $r$. End $2$ has a soap bubble with a radius of curvature $R$ $(R > r)$ as shown in the figure. Just after opening the valve,

The excess pressure due to surface tension in a spherical liquid drop of radius $r$ is directly proportional to

$A$ 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 $\rho$):

An air bubble doubles in radius when it rises from the bottom of the sea to the surface. If the atmospheric pressure is equal to the pressure exerted by a $10 \, m$ column of water,then the depth of the sea is $... \, m$. (Assume surface tension is negligible.)