Two spherical soap bubbles of radii r₁ and r₂ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

Step-by-step solution

Excess of pressure, inside the first bubble p₁= 4 T r₁ Similarly, p₂= 4 T r₂ Let the radius of the large bubble be R . Then, excess of pressure inside the large bubble, p= 4 T R Under isothermal condition, temperature remains constant. So, P V=p₁ V₁+p₂ V₂ 4 T R ( 4 3 R^3 )= 4 T r₁ ( 4 3 r₁^3 )+ 4 T r₂ ( 4 3 r₂^3 )R^2=r₁^2+r₂^2 R= r₁^2+r₂^2