What should be the radius of a water drop so that the excess pressure inside it is $72 \ Nm^{-2}$ (in $mm$)? (The surface tension of water is $7.2 \times 10^{-2} \ Nm^{-1}$)

$A$ water drop of radius '$r$' and volume '$V$' is kept between two identical glass plates such that it forms a thin layer of area '$A$' between the plates. $A$ force '$F$' is applied such that the two plates separate from each other. The surface tension '$T$' of the liquid is:

The excess pressure inside a soap bubble $A$ in air is half the excess pressure inside another soap bubble $B$ in air. If the volume of the bubble $A$ is $n$ times the volume of the bubble $B$,then the value of $n$ is . . . . . . .

$8$ mercury drops coalesce to form one mercury drop. The energy changes by a factor of:

Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to -

$A$ certain number of spherical liquid drops of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid,which one of the following statements is true for the energy $(E)$ in the process?