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 $\pi r^2 S$)

$A$ small soap bubble of radius $4 \ cm$ is trapped inside another bubble of radius $6 \ cm$ without any contact. Let $P_2$ be the pressure inside the inner bubble and $P_0$ be the pressure outside the outer bubble. The radius of another bubble with a pressure difference $P_2 - P_0$ between its inside and outside would be....... $cm$.

An ice cube of edge $1 \ cm$ melts in a gravity-free container. The approximate surface area of the water formed is (water is in the form of a spherical drop)

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change 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

Three small identical bubbles of water having the same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is: