= Solution
Take $z$ positive downwards. The leading axial <velocity> of a <slender viscous thread> is independent of radius. <Incompressible flow> then gives $u_r=-rw'/2$, while steady <mass conservation> gives $(a^2w)'=0$. The <Newtonian fluid stress tensor> has $\sigma_{rr}=-p-\mu w'$ and $\sigma_{zz}=-p+2\mu w'$. The <Young–Laplace equation>, neglecting axial curvature at slender order, gives $\sigma_{rr}=-p_e-\gamma/a$ and hence $p=p_e+\gamma/a-\mu w'$.
The axial cut transmits both the integrated <Newtonian fluid stress tensor> and the circumferential <surface tension>. After subtracting the ambient <pressure>, its tensile <force> is
$$
T=\pi a^2(3\mu w'-\gamma/a)+2\pi\gamma a
=3\pi\mu a^2w'+\pi\gamma a.
$$
The coefficient three is the <Trouton ratio> for axial extension. The last term is essential: the <capillary tensile force of a slender thread> is $\pi\gamma a$, not $2\pi\gamma a$, because the <capillary pressure> subtracts half the interfacial line contribution. Axial <force> balance is $T'+\pi\rho ga^2=0$. Together the governing equations are
$$
\boxed{(a^2w)'=0,\qquad 3\mu(a^2w')'+\rho ga^2+\gamma a'=0.}
$$
Define the <dimensionless variables>
$$
\ell=\sqrt{\frac{\mu w_0}{\rho g}},\quad Z=\frac z\ell,\quad W=\frac w{w_0},\quad
\Gamma=\frac{\gamma\ell}{\mu a_0w_0}=\frac{\gamma}{\rho g a_0\ell}.
$$
Then $a=a_0W^{-1/2}$ and the axial equation becomes
$$
3\frac d{dZ}\left(\frac{W_Z}{W}\right)+\frac1W+\Gamma\frac d{dZ}W^{-1/2}=0.
$$
For the increasing power-law branch, all three terms have the same dependence when $\alpha=2$. Substitution leaves $1-6k^2-\Gamma k=0$. Thus
$$
\boxed{W=(1+kZ)^2,\quad \alpha=2,\quad k=\frac{\sqrt{\Gamma^2+24}-\Gamma}{12}.}
$$
This remains valid at $\Gamma=0$, when $k=1/\sqrt6$. The other root is negative and describes a decreasing local branch up to its singular endpoint, not a thread accelerating downwards indefinitely. Positive <surface tension> decreases $k$, so it slows the fall at a given height on this branch. Thinning reduces the capillary tensile <force> $\pi\gamma a$. Its negative axial derivative supplies an upward capillary <force> on a thread segment, balancing part of the weight. The viscous tensile <force> therefore supplies a smaller remaining balance; on the quadratic family this reduces the stretching rate.
The condition $W(0)=1$ alone does not select a unique solution of this second-order <ordinary differential equation>. The boxed power law is the requested verified branch, with tensile <force> tending to zero at large height; an additional nozzle tension or downstream condition is required to select it physically. The slender, negligible-inertia approximation must remain valid over the height range considered.
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