To find the net impedance of the circuit, we need to calculate the impedance of each component.$\\$Given:$\\$• Inductance $L = \frac{50}{\pi}$ mH$= \frac{50 \times 10^{-3}}{\pi}$ H$\\$• Capacitance $C = \frac{10^3}{\pi}$ $\mu$F$= \frac{10^{-3}}{\pi}$ F$\\$• Resistance $R = 10$ $\Omega \\$• Frequency $f = 50$ Hz$\\$The angular frequency $\omega$ is given by:$\\ \omega = 2\pi f = 2\pi \times 50 = 100\pi$ rad/s$\\$The impedance of the inductor $Z_L$ is:$\\ Z_L = j\omega L = j(100\pi) \left(\frac{50 \times 10^{-3}}{\pi}\right) = j5$ $\Omega \\$The impedance of the capacitor $Z_C$ is:$\\ Z_C = \frac{1}{j\omega C} = \frac{1}{j(100\pi) \left(\frac{10^{-3}}{\pi}\right)} = \frac{1}{j100} = -j0.01$ $\Omega \\$The total impedance $Z$ of the series circuit is:$\\ Z = R + Z_L + Z_C = 10 + j5 - j0.01 \\ Z = 10 + j(5 - 0.01) \\ Z = 10 + j4.99 \\$The magnitude of the impedance $|Z|$ is:$\\ |Z| = \sqrt{R^2 + (X_L - X_C)^2} \\ |Z| = \sqrt{10^2 + 4.99^2} \\ |Z| = \sqrt{100 + 24.9001} \\ |Z| = \sqrt{124.9001} \\ |Z| \approx 11.17$ $\Omega \\$This corresponds to $5\sqrt{5}$ $\Omega$ (Option 3).