Answer
$f=1-\frac{V}{V_0}$.
Work Step by Step
Assume the tank is empty. The entire wire of length $\ell$ is non-superconducting. Suppose its resistivity is $\rho$.
The resistance of the wire, with an empty tank, is $R_0=\rho\frac{\ell}{A}=\frac{V_0}{I}$.
Now, let a length x of the wire become superconducting. That part of the wire has zero resistance. Only the length $\ell-x$ has resistance. The wire’s overall resistance is $R=\rho\frac{\ell-x}{A}=\rho\frac{\ell}{A}\frac{\ell-x}{\ell}=R_0\frac{\ell-x}{\ell}$.
The current is held steady, so we calculate the voltage using Ohm’s Law.
$$V=IR=\frac{V_0}{R_0}R_0\frac{\ell-x}{\ell}=V_0(1-\frac{x}{\ell})=V_0(1-f)$$
We solve for the relationship between f and V.
$$f=1-\frac{V}{V_0}$$
We see that measuring the voltage V reveals the fraction of the tank that is filled with liquid helium.
