A coil of wire of radius ' $r$ ' has 600 turns and a self-inductance of 108 mH . The self-inductance of a coil with same radius and 500 turns is

The self-inductance $L$ of a coil is proportional to the square of its number of turns: $L \propto N^2$.

Given initial values $L_1 = 108\text{ mH}$ and $N_1 = 600$, and final turns $N_2 = 500$, the inductance becomes:

$L_2 = L_1 \left(\frac{N_2}{N_1}\right)^2 = 108 \left(\frac{500}{600}\right)^2 = 108 \left(\frac{5}{6}\right)^2 = 108 \cdot \frac{25}{36}$

$L_2 = 3 \cdot 25 = 75\text{ mH}$

Final answer: $75\text{ mH}$