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 ........ $dyne/cm$.

When $10^6$ small drops coalesce to make a new larger drop,then the drop:

Consider an air bubble of radius $2 \,mm$ in a liquid at a depth of $5 \,cm$ below the free surface. The density of the liquid is $1000 \,kg/m^3$ and the surface tension is $0.1 \,N/m$. Find the pressure inside the air bubble relative to the pressure at the free surface of the liquid. (Take $g = 10 \,m/s^2$) (in $\,Pa$)

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

When $n$ identical mercury drops combine to form a single big drop,

The excess pressure inside a soap bubble of radius $2 \ cm$ is $50 \ dyne/cm^2$. The surface tension is