Two separate soap bubbles of radii 3 × 10 - 3 m and 2 × 10 - 3 m respectively, formed of same liquid (surface tension 65 × 10 - 2 N m - 1 ) come together to form a double bubble. The radius of interface of double bubble is

Step-by-step solution

Excess pressure inside first bubble of radius r 1 p 1 = p 0 + 4 T r 1 Excess pressure inside second bubble of radius r 2 p 2 = p 0 + 4 T r 2 Excess pressure inside double bubble, p = p 2 - p 1 p = 4 T r 1 - r 2 r 1 r 2 If R be the radius of double bubble, then p = 4 T R or 4 T r 1 - r 2 r 1 r 2 = 4 T R ⇒ R = r 1 r 2 r 1 - r 2 = 3 × 10 - 3 × 2 × 10 - 3 3 - 2 × 10 - 3 = 6 × 10 - 3   m R = 6 × 10 - 3   m