a
(a) Capacitance of wire

\(C = 0.014 \times {10^{ - 6}} \times 200 = 2.8 \times {10^{ - 6}}F = 2.8\mu F\)

For impedance of the circuit to be minimum \({X_L} = {X_C}\)

==> \(2\pi \nu L = \frac{1}{{2\pi \nu C}}\)

==>\(L = \frac{1}{{4{\pi ^2}{\nu ^2}C}} = \frac{1}{{4{{(3.14)}^2} \times {{(5 \times {{10}^3})}^2} \times 2.8 \times {{10}^{ - 6}}}}\)

\( = 0.35 \times {10^{ - 3}}H = 0.35\,mH\)