$A$ water drop is divided into $8$ equal droplets. The pressure difference between the inner and outer side of the big drop will be

Two long parallel glass plates have water between them. The contact angle between glass and water is zero. If the separation between the plates is $d$ (where $d$ is small) and the surface tension of water is $T$,with atmospheric pressure $P_0$,then what is the pressure inside the water just below the air-water interface?

$A$ 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:

Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 \ N/m^2$. The radii of bubbles $A$ and $B$ are $2 \ cm$ and $4 \ cm$,respectively. The surface tension of the soap solution is $0.04 \ N/m$. Find the ratio $n_B / n_A$,where $n_A$ and $n_B$ are the number of moles of air in bubbles $A$ and $B$,respectively. [Neglect the effect of gravity.]

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ surface tension of water). (in $\pi TR^2$)

$A$ large number of water drops,each of radius $r$,combine to form a single drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$,the rise in heat energy per unit volume will be: