The work done in breaking a drop of liquid of radius $R$ (Surface tension $T$) into $64$ equal drops is

Two spherical soap bubbles formed in vacuum have diameters $3.0 \, mm$ and $4.0 \, mm$. They coalesce to form a single spherical bubble. If the temperature remains unchanged,find the diameter of the bubble so formed in $mm$.

$A$ glass capillary tube of inner diameter $0.28 \ mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as that in the vessel (in $N/m^2$) is:
Surface tension of water $= 0.07 \ N/m$
Atmospheric pressure $= 10^5 \ N/m^2$

$A$ large number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given,surface tension of liquid $T$,density $\rho$):

An air bubble of radius $1 \ mm$ is at a depth of $8 \ cm$ below the free surface of a liquid column. If the surface tension and density of the liquid are $0.1 \ N \ m^{-1}$ and $2000 \ kg \ m^{-3}$ respectively, by what amount is the pressure inside the bubble greater than the atmospheric pressure (in $N \ m^{-2}$)? (Take $g = 10 \ m \ s^{-2}$)

Consider a water droplet of diameter $0.2 \,mm$ where the outside pressure is $1.5 \,N / cm^2$ at $25^{\circ} C$. The pressure inside the droplet, when the surface tension at $25^{\circ} C$ is $0.08 \,N / m$ is