There are two long co-axial solenoids of same length $$l$$. The inner and outer coils have radii $$r_1$$ and $$r_2$$ and number of turns per unit length $$n_1$$ and $$n_2$$, respectively. The ratio of mutual inductance to the self-inductance of the inner-coil is:

We consider two very long co-axial solenoids, both of the same length $$l$$. The inner solenoid (which we shall call coil 1) has radius $$r_1$$ and number of turns per unit length $$n_1$$. The outer solenoid (coil 2) has radius $$r_2$$ and number of turns per unit length $$n_2$$. We wish to find the ratio of their mutual inductance $$M$$ to the self-inductance $$L_1$$ of the inner solenoid.

First, we recall the magnetic field inside a very long solenoid. The standard result is

$$B=\mu_0\,n\,I,$$

where $$n$$ is the number of turns per unit length and $$I$$ is the current through that solenoid. This field is uniform over the entire cross-section of the solenoid and directed along its axis.

Let a current $$I$$ flow only in the inner coil. Using the above formula, the magnetic field produced by this inner coil is

$$B_1=\mu_0 n_1 I.$$

The cross-sectional area of the inner solenoid is

$$A_1=\pi r_1^2.$$

Hence, the magnetic flux through each single turn of the inner solenoid is

$$\Phi_{\text{single (self)}} = B_1 A_1 = \mu_0 n_1 I \,\pi r_1^2.$$

The total number of turns in the inner solenoid is

$$N_1 = n_1 l.$$

Therefore, the total flux linked with all the turns of the inner solenoid is

$$\Phi_{\text{total (self)}} = N_1 \Phi_{\text{single (self)}} = n_1 l \left(\mu_0 n_1 I \,\pi r_1^2\right) = \mu_0 n_1^2 l \,\pi r_1^2 \, I.$$

By definition, the self-inductance $$L_1$$ of the inner coil is

$$L_1 = \frac{\text{total flux linkage}}{I} = \frac{\Phi_{\text{total (self)}}}{I} = \mu_0 n_1^2 l \,\pi r_1^2.$$

Now we compute the mutual inductance $$M$$. The same current $$I$$ in the inner coil produces the same field $$B_1=\mu_0 n_1 I$$ everywhere inside radius $$r_1$$. This field also threads the outer coil, but only through the region of radius $$r_1$$; outside that region the field is (ideally) zero. Thus, the flux through each single turn of the outer solenoid is exactly the same $$\Phi_{\text{single (mutual)}} = \mu_0 n_1 I \,\pi r_1^2.$$

The total number of turns in the outer solenoid is

$$N_2 = n_2 l.$$

Hence, the total flux linked with the outer solenoid because of the current in the inner one is

$$\Phi_{\text{total (mutual)}} = N_2 \Phi_{\text{single (mutual)}} = n_2 l \left(\mu_0 n_1 I \,\pi r_1^2\right) = \mu_0 n_1 n_2 l \,\pi r_1^2 \, I.$$

By definition, the mutual inductance $$M$$ is

$$M = \frac{\text{total mutual flux linkage}}{I} = \frac{\Phi_{\text{total (mutual)}}}{I} = \mu_0 n_1 n_2 l \,\pi r_1^2.$$

We can now form the desired ratio:

$$\frac{M}{L_1} = \frac{\mu_0 n_1 n_2 l \,\pi r_1^2}{\mu_0 n_1^2 l \,\pi r_1^2} = \frac{n_2}{n_1}.$$

All constant factors $$\mu_0$$, $$l$$ and $$\pi r_1^2$$ cancel out, leaving a very simple result that depends only on the turn densities.

Hence, the correct answer is Option D.