A spherical soap bubble of radius 3cm is formed inside another...

Pressure inside a bubble is given by, ${P}_{0} + \dfrac{4T}{r}$.

Pressure inside the bubble of smaller radius will be,

${P}_{2} - {P}_{1} = \left[\left({P}_{0} + \dfrac{4T}{6}\right) + \dfrac{4T}{3}\right]$

Pressure inside the bubble with radius $r$, ${P}_{0} + \dfrac{4T}{r}$

Now,

$⇒ {P}_{0} + \dfrac{4T}{6} + \dfrac{4T}{3} = {P}_{0} + \dfrac{4T}{r}$

$⇒ \dfrac{1}{r} = \dfrac{1}{6} + \dfrac{1}{3}$

$⇒ r = 2cm$