Let us consider two solenoids $A$ and $B$, made from same magnetic material of relative permeability $\mu_r$…
Let us consider two solenoids $A$ and $B$, made from same magnetic material of relative permeability $\mu_r$ and equal area of cross-section. Length of $A$ is twice that of $B$ and the number of turns per unit length in $A$ is half that of $B$. The ratio of self inductances of the two solenoids, $L_A: L_B$ is
$1: 2$
$2: 1$
$8: 1$
$1: 8$
Solution
$L=\mu_0 \mu_r \times n \times A \times N$
$L=\mu_0 \mu_r n \times A \times \frac{N}{l} \times l$
$L=\mu_0 \mu_r \times n^2 \times A \times I \quad \Rightarrow L \propto n^2 I$
$\Rightarrow \frac{L_A}{L_B}=\frac{n_A^2}{n_B^2} \times \frac{I_A}{I_B}$
$\Rightarrow \frac{L_A}{L_B}=\frac{1}{4} \times 2=\frac{1}{2}$