Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 4 / 4A / Solution

Let $z$ measure height above the release point, and use $\mathcal{V}=4\pi a^3/3$ for the bubble volume to avoid confusing it with a speed. For constant radius, Newton's second law, the specified buoyancy and the given Stokes law giveWith positive $\alpha ,\rho ,\mu ,a$, defineThe differential equation becomes $\dot{u}+(u/\tau )=U/\tau$. Its solution with $u(0)=0$ is $u(t)=U(1-e^{-t/\tau })$. Integration using $z(0)=0$ yieldsAn arbitrary initial height is added to this expression. The effective inertia changes the relaxation time $\tau$, but not the terminal velocity obtained by balancing buoyancy and drag.

For the dissolving bubble assume $\beta >0$ and use the specified instantaneous terminal velocity, rather than the constant-radius transient formula. The bubble disappears at $t_{\ast }=a_0^2/\beta$, andIts total rise is therefore $\boxed{z(t_{\ast })=\rho g{a}_0^4/(9\mu \beta )}$. This second model neglects the initial acceleration and uses exactly the terminal-speed approximation stipulated for the shrinking bubble.