Two spherical soap bubble coalesce. If `V` is the consequent change in volume of the contained air and `S` the change in total surface area, show that
`3PV+4ST=0`
where `T` is the surface tension of soap bubble and `P` is
Atmospheric pressure

Two spherical soap bubble collapses. If V is the consequent change in volume of the contained air and S in the change in the total surface area and T is the surface tension of the soap solution, then it relation between P_(0), V, S and T are lambdaP_(0)V + 4ST = 0 , then find lambda ? (if P_(0) is atmospheric pressure) : Assume temperature of the air remain same in all the bubbles

If the rate of change in volume of spherical soap bubble is uniform then the rate of change of surface area varies as

If R is the radius of a soap bubble and S its surface tension, then the excess pressure inside is

An isolated and charged spherical soap bubble has a radius 'r' and the pressure inside is atmospheric. If 'T' is the surface tension of soap solution, then charge on drop is:

Excess pressure inside a soap bubble of radius r and surface tension T is

Air is pushed inot a soap bubble of radius r to duble its radius. If the surface tension of the soap solution is S, the work done in the process is