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Q.
If air core is replaced by an iron core in an inductor, its self-inductance is increased from $0.02\, mH$ to $40\, mH$. The relative permeability of iron is
Inductance of a solenoid coil of length $L$, total number of turns $N$ and cross-sectional area $A$, is given as
$L=\frac{\mu_0 N^2 A}{l} \ldots \text { (i) }$
where, $\mu_0$ is the permeability in vacuum (or air).
If air core is replaced by iron core, inductance will be given as
$L^{\prime}=\frac{\mu N^2 A}{l} \ldots \text {.(ii) }$
On dividing Eq. (ii) by Eq (i). we get
$\frac{L^{\prime}}{L}=\frac{\mu N^2 A}{l} / \frac{\mu_0 N^2 A}{l}$
$ =\frac{\mu}{\mu_0}=\mu_r$
where, $\mu_r$ is the relative permeability of iron. Given,
$ L=0.02\, mH , L^{\prime}=40\, mH $
$ \because \mu_r=\frac{40}{0.02}=2000$