MHTCET 15 May 2024 Shift 2 Paper

Concept:Self-inductance depends on the square of the number of turns and the permeability of the core material.Explanation:For the air-cored coil, the initial self-inductance is:$$L_1 = \frac{\mu_0 N^2 A}{l} = 0.1\ \mathrm{H}$$Here, $\mu_0$ is the permeability of free space, $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the coil length.After inserting the soft iron core, the permeability becomes $\mu_0 \mu_r$, where $\mu_r = 1000$.The number of turns is reduced to $N' = \frac{N}{10}$.So the new self-inductance is:$$L_2 = \frac{\mu_0 \mu_r (N')^2 A}{l} = \frac{\mu_0 \mu_r \left(\frac{N}{10}\right)^2 A}{l}$$Taking the ratio of $L_2$ to $L_1$:$$\frac{L_2}{L_1} = \mu_r \left(\frac{N'}{N}\right)^2 = 1000 \times \left(\frac{1}{10}\right)^2 = 1000 \times \frac{1}{100} = 10$$Hence:$$L_2 = 10 \times L_1 = 10 \times 0.1\ \mathrm{H} = 1\ \mathrm{H}$$Answer:The new self-inductance is $1\ \mathrm{H}$.Correct option: C. 1 H.