= Solution
For the <focal length of a rotating liquid mirror>, consider steady rigid rotation at <angular frequency> $\Omega=2\pi\nu$. The free surface is an <equipotential> of $gz-\Omega^2r^2/2$. Hence
$$
z-z_0=\frac{\Omega^2r^2}{2g}.
$$
A paraboloidal mirror has $z-z_0=r^2/(4f)$, giving
$$
\boxed{f=\frac{g}{2\Omega^2}=\frac{g}{8\pi^2\nu^2}\simeq3.1\,\mathrm m.}
$$
The numerical estimate uses the conventional standard acceleration $g_n=9.80665\,\mathrm{m\,s^{-2}}$, rather than an asserted local measurement or a recovered handbook entry.<NIST Guide to the SI, Footnotes by National Institute of Standards and Technology>[Footnote 23: conventional standard acceleration of free fall]{ref} The mercury <density> cancels; the ideal surface shape follows from gravity and rotation.
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