Two soap bubbles of radii $R$ and $r$ come in contact. $R$ is more than $r$. The radius of curvature of the common surface is

The radius of a soap bubble is $r$ and the surface tension of the soap solution is $S$. The electric potential to which the soap bubble must be raised by charging it so that the pressure inside the bubble becomes equal to the pressure outside the bubble is $(\varepsilon_0 = \text{permittivity of the free space})$

$A$ water drop of $0.01 \ cm^3$ is squeezed between two glass plates and spreads into an area of $10 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,then the normal force required to separate the glass plates from each other will be: (in $N$)

$1000$ spherical drops of water each $10^{-8} \ m$ in diameter coalesce to form one large spherical drop. The amount of energy liberated in this process (in joule) is (surface tension of water is $0.075 \ N/m$).

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

$A$ 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 = . . . . . . $.