$A$ big liquid drop splits into $n$ similar small drops under isothermal conditions, then in this process

The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:

Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is (in $ : 1$)

The work done in splitting a drop of water of $1\, mm$ radius into $10^6$ droplets is (surface tension of water $T = 72 \times 10^{-3}\, N/m$):

$A$ liquid drop of volume $V$ is placed on the surface of a glass plate. Then,another glass plate is placed on it such that the liquid forms a thin layer of area $A$ between the surfaces of the two plates. To separate the plates,a force $F$ has to be applied normal to the surfaces. The surface tension of the liquid is

An air bubble of radius $1.0 \ mm$ is observed at a depth of $20 \ cm$ below the free surface of a liquid having surface tension $0.095 \ J/m^2$ and density $10^3 \ kg/m^3$. The difference between pressure inside the bubble and atmospheric pressure is . . . . . . $N/m^2$. (Take $g = 10 \ m/s^2$)