= Solution
To connect the prescribed interfacial <drag coefficient> to the energy supply, state the lumped <interfacial power model for entrainment> explicitly:
$$
P_I=S\tau u_I=c_D\rho_lS u_I^3,\qquad \dot E_P=C_WP_I.
$$
It estimates the power transmitted by lid-driven flow to the mixing interface. Constant geometric and transfer factors are included in $c_D$ or $C_W$. It is the local stress-power interpretation required for the depth laws in this question, not an identity equating arbitrary lid shaft power to interfacial power. With the <potential energy of two-layer entrainment> from (a),
$$
\dot h=\frac{2C_Wc_Du_I^3}{B}.
$$
In model S, $u_I=\Omega R$ and $B$ are constant, so
$$
\boxed{\dot h=\frac{2C_Wc_D\Omega R}{\mathrm{Ri}_B},\qquad
\mathrm{Ri}_B=\frac{B}{\Omega^2R^2}.}
$$
Thus the depth increases at a constant rate proportional to the inverse <bulk Richardson number>, with $h=h_0+2C_Wc_D\Omega R\,t/\mathrm{Ri}_B$, until the layer reaches $H$. Here the prescribed $u_I$ must be read as a characteristic speed. Literal pure <solid-body rotation> has zero radial velocity, so the PDF's parenthetical description of $u_I$ as radial cannot simultaneously be a strict component statement and the assumption $u_I=\Omega R$.
If the phrase about lid work is taken literally as shaft power $P_{\rm lid}=\Omega\mathcal T_{\rm lid}$, the interfacial stress alone does not specify the lid torque or its energy transfer to the interface. The preceding closure is therefore a visible model assumption, not a hidden consequence of <solid-body rotation>.
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