MHTCET 25 Apr 2025 Shift 2 Paper

Concept:For a coil of fixed radius, length, and core material, self-inductance depends on the square of the number of turns: $L \propto N^2$.Explanation:The first coil has $N_1 = 600$ turns and $L_1 = 108\ \mathrm{mH}$.The second coil has the same radius but $N_2 = 500$ turns, and its inductance $L_2$ is to be determined.Since $L \propto N^2$, the ratio of inductances equals the square of the ratio of the number of turns:$$\frac{L_2}{L_1} = \left(\frac{N_2}{N_1}\right)^2$$Rearranging gives:$$L_2 = L_1 \left(\frac{N_2}{N_1}\right)^2$$Substitute $L_1 = 108\ \mathrm{mH}$, $N_2 = 500$, and $N_1 = 600$:$$L_2 = 108 \times \left(\frac{500}{600}\right)^2 = 108 \times \left(\frac{5}{6}\right)^2$$$$L_2 = 108 \times \frac{25}{36} = 3 \times 25 = 75\ \mathrm{mH}$$Answer:The self-inductance of the coil with 500 turns is $75\ \mathrm{mH}$.Hence, the correct option is B. $75\ \mathrm{mH}$.