= Solution
Let $h_0=h(0)$ and $S=\pi R^2$. The closed tank has constant mass, so
$$
S\{\rho_u h+\rho_l(H-h)\}
=S\{\rho_u(0)h_0+\rho_l(H-h_0)\}.
$$
Therefore $(\rho_l-\rho_u)h=(\rho_l-\rho_u(0))h_0$. Under the <Boussinesq approximation>, define $g'=g(\rho_l-\rho_u)/\rho_l$ and obtain
$$
\boxed{B=g'h=g'_0h_0=\text{constant}.}
$$
The gravitational <potential energy> is especially easy to evaluate with the stipulated downward depth coordinate: the gravitational potential per unit mass is $-gz$. Hence
$$
E_P=-gS\left\{\rho_u\frac{h^2}2+\rho_l\frac{H^2-h^2}2\right\}
=-\frac{g\rho_lSH^2}2+\frac{\rho_lSBh}2.
$$
This proves the <potential energy of two-layer entrainment> relation
$$
\boxed{\dot E_P=\frac12\rho_lSB\dot h.}
$$
The positive sign expresses the lifting and homogenization of dense entrained fluid within the upper layer. The fixed thin interface is neglected in this leading two-layer energy integral.
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