When two soap bubbles of radius $r_1$ and $r_2$ $(r_2 > r_1)$ coalesce,the radius of curvature of the common surface is:

$A$ drop of water with a volume of $0.05 \ cm^3$ is pressed between two glass plates,causing it to spread between the plates. The area of contact with each plate is $40 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,the minimum normal force required to separate the two glass plates in newtons is approximately: (assuming the angle of contact is zero)

Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

$A$ large number of water droplets each of radius '$r$' combine to form a large drop of radius '$R$'. If the surface tension of water is '$T$' and the mechanical equivalent of heat is '$J$',then the rise in temperature due to this process is:

The lower end of a capillary tube of diameter $2.00 \; mm$ is dipped $8.00 \; cm$ below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at the temperature of the experiment is $7.30 \times 10^{-2} \; N m^{-1}$. Atmospheric pressure $= 1.01 \times 10^{5} \; Pa$,density of water $= 1000 \; kg m^{-3}$,$g = 9.80 \; m s^{-2}$. Also,calculate the excess pressure.

In air,a charged soap bubble of radius $R$ breaks into $64$ small soap bubbles of equal radius $r$. The ratio of mechanical force per unit area of the big soap bubble to that of a small bubble is