The self-inductance of a solenoid of 600 turns is 108 mH. The self-inductance of a coil having 500 turns with the same length, the same radius and the same medium will be

$\text{ Given, }\mathrm{N}=600,\mathrm{L}=108\mathrm{mH}=108\times {10}^{-3}\mathrm{H}$
Let l and A be length and area of cross section of coil.
So, $\mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2\mathrm{A}}{\mathrm{l}}$
$\begin{array}{l}108\times {10}^{-3}=\frac{{\mathrm{\mu }}_0(600{)}^2\mathrm{A}}{\mathrm{l}}\\ \frac{{\mathrm{\mu }}_0\mathrm{A}}{\mathrm{l}}=\frac{108\times {10}^{-3}}{36\times {10}^4}\\ \frac{{\mathrm{\mu }}_0\mathrm{A}}{\mathrm{l}}=3\times {10}^{-7}\dots \dots 1\end{array}$
Now self-inductance of coil with 500 turns ${\mathrm{L}}^{\mathrm{\prime }}=\frac{{\mathrm{\mu }}_0(500{)}^2\mathrm{A}}{\mathrm{l}}=\left(\frac{{\mathrm{\mu }}_0\mathrm{A}}{\mathrm{l}}\right)(500{)}^2$
$\begin{array}{l}=3\times {10}^{-7}\times (500{)}^2\\ =3\times {10}^{-7}\times 25\times {10}^4\\ =75\mathrm{mH}\end{array} ........\mathrm{from}\left(1\right)$