= Solution
Let $U(x)$ and $\Delta(x)$ denote the characteristic vertical speed and width. Integrating the result of part (a), with $b(x)=Bx^{-1/5}$, gives the scaling
$$
\rho U^2\Delta\sim Bx^{4/5}.
$$
The <boundary-layer scaling> in the momentum equation balances vertical inertia against transverse viscosity:
$$
\rho\frac{U^2}{x}\sim\mu\frac{U}{\Delta^2},
\qquad
U\sim\frac{\mu x}{\rho\Delta^2}.
$$
Substituting this into the momentum-flux relation gives
$$
\frac{\mu^2x^2}{\rho\Delta^3}\sim Bx^{4/5},
$$
and therefore
$$
\boxed{\Delta\sim
\left(\frac{\mu^2}{\rho B}\right)^{1/3}x^{2/5}}.
$$
Returning to the viscous-inertial balance then gives
$$
\boxed{U\sim
\left(\frac{B^2}{\rho\mu}\right)^{1/3}x^{1/5}}.
$$
These are the <similarity scaling of a two-dimensional laminar plume>.
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