= Solution
For a long <solenoid>, <solenoid inductance> obeys $L=\mu_0N^2A/\ell$, where $N$ is the number of turns, $A$ the cross-sectional <area> and $\ell$ the axial winding length. The natural reading of half as many turns occupying half the axial length gives
$$
N'=N/2,\qquad\ell'=\ell/2,\qquad L'=L/2.
$$
With unchanged wire material and gauge, halving the wire length also halves its <electrical resistance>: $R'=R/2$. With the total <capacitance> $C_T$ fixed, the new <angular frequency> satisfies
$$
\omega'^2=\frac2{LC_T}-\frac{R^2}{L^2}=2\omega_*^2+\frac{R^2}{L^2}.
$$
Thus
$$
\boxed{\frac{f'}{50\,\mathrm{Hz}}=\sqrt{2+\left[\frac{R}{L\omega_*}\right]^2}=1.43005,\qquad f'=71.50\,\mathrm{Hz}.}
$$
Neglecting the small resistive correction gives the estimate $\sqrt2$.
There is a documented alternative interpretation: if half the turns are distributed over the original axial length, then $L'=L/4$ while $R'=R/2$, giving
$$
\omega'^2=\frac4{LC_T}-\frac{4R^2}{L^2}=4\omega_*^2,\qquad\boxed{f'=2f_*=100\,\mathrm{Hz}.}
$$
This corresponds to treating “length of the winding” as wire length while preserving axial geometry. The examiner's report explicitly accepted either inductance scaling, provided the resistance change was included. The first interpretation follows the statement that the axial winding length is halved; the second records the accepted ambiguity rather than silently choosing a different geometry.https://www.whatdotheyknow.com/request/natural_science_tripos_physics_e/response/2127325/attach/3/FOI%202022%20639%20Sullivan%20materials.pdf?cookie_passthrough=1#page=63[Examiner commentary, B8(d), released PDF page 63: both inductance scalings accepted, with resistance halved]{ref}
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