$A$ big drop is formed by coalescing $1000$ small identical drops of water. If $E_1$ be the total surface energy of $1000$ small drops of water and $E_2$ be the surface energy of the single big drop of water,then the ratio $E_1 : E_2$ is $x : 1$,where $x = . . . . . . $.

$A$ spherical soap bubble of radius $3\,cm$ is formed inside another spherical soap bubble of radius $6\,cm$. If the internal pressure of the smaller bubble of radius $3\,cm$ in the above system is equal to the internal pressure of another single soap bubble of radius $r\,cm$,then the value of $r$ is:

The pressure inside a soap bubble $A$ is $1.01 \text{ atm}$ and that in a soap bubble $B$ is $1.02 \text{ atm}$. The ratio of the volume of $A$ to that of $B$ is

The pressure of air in a soap bubble of $0.7 \ cm$ diameter is $8 \ mm$ of water above the pressure outside. The surface tension of the soap solution is ........ $dyne/cm$.

$A$ drop of water of volume $0.05\, cm^3$ is pressed between two glass plates,as a consequence of which it spreads and occupies an area of $40\, cm^2$. If the surface tension of water is $70\, dyne/cm$,then the normal force required to separate out the two glass plates will be in Newton.

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