The work done to blow a soap bubble to a volume V is W . If the same bubble is blown further to a volume 8V , the excess pressure inside the bubble in this new state will be :

Step-by-step solution

Let the initial radius of the soap bubble corresponding to volume V be R . Volume V = 4 3 R^3 4 R^3 = 3V A soap bubble has two free surfaces (inner and outer). The work done to blow it from zero volume to volume V is: W = T A = T 2 (4 R^2) = 8 R^2 T T = W 8 R^2 When the bubble is blown to a new volume V' = 8V , let its new radius be R' . Since volume is proportional to the cube of the radius, R'^3 = 8R^3 R' = 2R . The excess pressure inside a soap bubble of radius R' is given by: P' = 4T R' Substitute R' = 2R and t