For a lossless transmission line the characteristic impedance depends only on the ratio of its distributed inductance to its distributed capacitance :

\(Z_{0}=\sqrt{\dfrac{L}{C}}\qquad\Rightarrow\qquad L=Z_{0}^{2}C\)

Substitute the given values.

\(L=(75)^{2}\times70\times10^{-12}=5625\times70\times10^{-12}\)

\(L=3.9375\times10^{-7}\ \text{H/m}=0.39375\ \mu\text{H/m}\)

which is option 3. As in the previous question the distractors are the same digits with the decimal point shifted, so the powers of ten must be tracked carefully: \(10^{-12}\) from the picofarad and \(10^{6}\) from the microhenry differ by six orders, and 5625 × 70 = 393,750 supplies the rest.

A physical cross-check on the answer. The velocity of propagation on the line is

\(v=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{3.94\times10^{-7}\times70\times10^{-12}}}=\dfrac{1}{1.66\times10^{-9}}=2.02\times10^{8}\ \text{m/s}\)

which is about 0.67 c — exactly the velocity factor of a polyethylene-filled coaxial cable, whose relative permittivity is 2.25 and whose velocity factor is \(1/\sqrt{2.25}=0.667\). The numbers in the question are therefore those of a real RG-59-type cable, which confirms the result.

Why Z₀ is a resistance despite involving only reactive elements. The square root of henries per farad has the units of ohms, and physically the line absorbs energy from the source at a steady rate as the wave travels away — so an infinite line looks purely resistive. That is also why terminating a line in \(Z_{0}\) makes it appear infinitely long and produces no reflection.

For a coaxial geometry both L and C follow from the conductor radii:

\(Z_{0}=\dfrac{60}{\sqrt{\varepsilon_{r}}}\ln\dfrac{b}{a}\)

so the 75 Ω and 50 Ω standards correspond to particular diameter ratios — 75 Ω chosen historically for lowest attenuation and 50 Ω as a compromise between attenuation and power handling.

Hence, the inductance is 0.3937 μH/m.