Let $R_1$ and $R_2$ be the radii of two mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

An air bubble of radius $r$ in water is at depth $h$ below the water surface. If $P$ is the atmospheric pressure, and $d$ and $T$ are the density and surface tension of water respectively, then the pressure inside the bubble 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?

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

$1000$ droplets of water, each having a diameter of $2 \ mm$, coalesce to form a single drop. Given the surface tension of water is $0.072 \ N/m$. The energy loss in the process is

If two soap bubbles of different radii are connected by a tube,then