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 coalese. If V be the change in volume of the contained air, A is the change in total surface area then show that 3PV + 4AT = 0 Where T is the surface tension and P is atmospheric pressure.

Two spherical soap bubbles coalesce. If V is the consequent change in volume of the contained air S in the change in the total surface area. T is surface tension of the soap solution and P_0 is the atmospheric pressure, then ST//P_0 V = -n/4, where n = --

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

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