To find the net impedance, Z, of the circuit, we need to calculate the inductive reactance (XL), the capacitive reactance (XC), and use the resistance (R) in the circuit. Since R is given as 10 $ \Omega $, we calculate the reactances as follows:

Inductive reactance, $ X_L = 2\pi fL $, where $ f $ is the frequency and $ L $ is the inductance. With $ f = 50 $ Hz and $ L $ given as $ \frac{50}{\pi} $ mH, or $ \frac{50 \times 10^{-3}}{\pi} $ H, we find $ X_L = 2 \pi \times 50 \times \frac{50 \times 10^{-3}}{\pi} = 5 $ $ \Omega $.

Capacitive reactance, $ X_C = \frac{1}{2\pi fC} $, where $ C $ is the capacitance. Given $ C = \frac{10^3}{\pi} $ μF, or $ \frac{10^{-3}}{\pi} $ F, results in $ X_C = \frac{1}{2 \pi \times 50 \times \frac{10^{-3}}{\pi}} = 10 \Omega $.

With these values, the net impedance $ Z $ of the circuit can be calculated using the formula:

$ Z = \sqrt{R^2 + (X_L - X_C)^2} $

Substituting $ R = 10 \Omega $, $ X_L = 5 \Omega $, and $ X_C = 10 \Omega $, we find:

$ Z = \sqrt{10^2 + (5 - 10)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \Omega $

Therefore, the correct option is $ 5\sqrt{5} \Omega $.