Slow to rise

We denote the following terms first:
$\rho_2$ is the density of $L_2$ .
$\rho_1$ is the density of $L_1$ .
$V_0$ is the initial volume of the liquid.
$h_3$ is the height of liquid 1.
$h_2$ is the height of liquid 2.
$h$ is the helight of the cylinder below $h_3$ .
$v_0$ is $v_\text{cylinder}$ with respect to the ground.

Step 1:

We can write the volume of liquid in the beaker in terms of the liquid flow and also in terms of the beaker's geometry.

$V_0 + \pi r^2 (h + h_2) + t \left( \dfrac{dV}{dt} \right) = \pi R^2 (h_2 + h_3)$

Step 2:

We can write the mass of the cylinder in terms of the initial setup, and in terms of the system after the pouring begins. As the mass of the cylinder is constant, we have the following equality.

$\overbrace{h \pi r^2 \rho_1 g + h_2 \pi r^2 \rho_2 g}^{\textrm{During pour}} = \overbrace{\pi r^2 \ell \dfrac {\rho_1}2 g}^{\textrm{Initial}}$

Upon differentiating we find,

$\rho_1 \dfrac{dh}{dt} + \rho_2 \dfrac{dh_2}{dt} = 0 \Rightarrow \dfrac{dh_2}{dt} = \dfrac{-3dh}{2dt}$

Step 3:

We can also write the volume of liquid below the interface in two different ways.

$V_0 + \pi r^2 h = \pi R^2 h_3 \\ r^2 \dfrac{dh}{dt} = \dfrac{R^2 dh_3}{dt} \Rightarrow \dfrac{dh_3}{dt} = \frac{1}{16}\dfrac{dh}{dt}$

Differentiating the first equation gives

$\pi r^2 \left( \dfrac{dh}{dt} + \dfrac{dh_2}{dt} \right) + \dfrac{dV}{dt} = \pi R^2 \left( \dfrac{dh}{dt} + \dfrac{dh_3}{dt} \right)$

Substituting we find $\dfrac{dh}{dt} = -\dfrac1{90\pi} .$

Now, $v_0 -\dfrac{dh_3}{dt} = -\dfrac{dh}{dt} \qquad \qquad$

$\begin{aligned} v_0 &= \dfrac{dh}{dt} \left( \dfrac1{16} - 1 \right) \\ &= -\dfrac{15}{16} \times \dfrac{-1}{90\pi} \\ &= \dfrac1{96\pi}. \end{aligned}$

Original writeup