= Solution
Differentiate $b=Q/\sqrt M$ and use both balances:
$$
\frac{db}{dz}=\frac{Q'}{\sqrt M}-\frac{QM'}{2M^{3/2}}=2\alpha-\frac{B_0Q^2}{2M^{5/2}}.
$$
Substituting $Q^2=8\alpha M^{5/2}/(5B_0)+cQ_0^2$ gives
$$
\boxed{\frac{db}{dz}=\frac{6\alpha}{5}-\frac{cQ_0^2B_0}{2M^{5/2}}}.
$$
For $c<0$, the radius grows faster than the pure-plume slope $6\alpha/5$ and its growth rate decreases toward that slope as $M$ increases. For $c=0$, it grows at that slope exactly. For $0<c<1$, it grows more slowly, possibly initially contracting, and its slope increases toward $6\alpha/5$. Since $M'=B_0Q/M>0$ and $M\to\infty$, all branches satisfy $db/dz\to6\alpha/5$ as $z\to\infty$.
Initial necking requires
$$
2\alpha-\frac{B_0Q_0^2}{2M_0^{5/2}}<0\quad\Longleftrightarrow\quad \boxed{B_0Q_0^2>4\alpha M_0^{5/2}\quad\Longleftrightarrow\quad c>\frac35}.
$$
Thus a sufficiently <lazy plume> initially narrows as buoyancy accelerates its fluid faster than <fluid entrainment> supplies additional volume. It still has increasing volume and <momentum fluxes>; shrinking radius does not imply volume loss. At equality the initial slope is zero, and for $c>3/5$ the slope later becomes positive once $M^{5/2}=5cB_0Q_0^2/(12\alpha)$. This gives a finite minimum radius before the eventual <pure plume> regime.
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