Let [tex]r_A[/tex] and [tex]r_B[/tex] be the respective radii of balloons A and B. If the fixed total volume is V, then
[tex]V = \dfrac{4\pi}3\left({r_A}^3 + {r_B}^3\right)[/tex]
and knowing [tex]r_A=10\,\rm cm[/tex] and [tex]r_B=9\,\rm cm[/tex] at the start, we have V = 6916π/3 cm³. Then when [tex]r_A=12\,\rm cm[/tex], the radius of the other sphere is [tex]r_B=1\,\rm cm[/tex].
Differentiating both sides with respect to time t gives a relation between the rates of change of the radii:
[tex]0 = 4\pi \left({r_A}^2 \dfrac{dr_A}{dt} + {r_B}^2 \dfrac{dr_B}{dt}\right) \implies \dfrac{dr_B}{dt} = -\left(\dfrac{r_A}{r_B}\right)^2 \dfrac{dr_A}{dt}[/tex]
We're given [tex]\frac{dr_A}{dt} = 0.1\frac{\rm cm}{\rm s}[/tex] the whole time. At the moment [tex]r_A=12\,\rm cm[/tex], the radius of balloon B is changing at a rate of
[tex]\dfrac{dr_B}{dt} = -\left(\dfrac{12\,\rm cm}{1\,\rm cm}\right)^2 \left(0.1\dfrac{\rm cm}{\rm s}\right) = \boxed{-14.4 \dfrac{\rm cm}{\rm s}}[/tex]
