 Solution

For a long solenoid, solenoid inductance obeys $L={\mu }_0{N}^{2}A/\ell$, 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 givesWith unchanged wire material and gauge, halving the wire length also halves its electrical resistance: $R^{\prime }=R/2$. With the total capacitance $C_T$ fixed, the new angular frequency satisfiesThusNeglecting the small resistive correction gives the estimate $\sqrt{2}$.

There is a documented alternative interpretation: if half the turns are distributed over the original axial length, then $L^{\prime }=L/4$ while $R^{\prime }=R/2$, givingThis 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.^[ref]