Two discs of same mass and different radii are made of different materials such that their thicknesses are 1 cm and 0.5 cm respectively. The densities of materials are in the ratio 3 : 5. The moment of inertia of these discs respectively about their diameters will be in the ratio of $$\dfrac{x}{6}$$. The value of $$x$$ is ______.

Two discs of same mass $$M$$, thicknesses $$t_1 = 1$$ cm and $$t_2 = 0.5$$ cm, density ratio $$\rho_1 : \rho_2 = 3 : 5$$.

The mass of a disc is given by $$M = \pi R^2 t \rho$$.

Since $$M_1 = M_2$$:

$$\pi R_1^2 t_1 \rho_1 = \pi R_2^2 t_2 \rho_2$$ $$\frac{R_1^2}{R_2^2} = \frac{t_2 \rho_2}{t_1 \rho_1} = \frac{0.5 \times 5}{1 \times 3} = \frac{5}{6}$$

For a thin disc about its diameter, the moment of inertia is $$I = \frac{1}{4}MR^2$$.

Since both discs have the same mass:

$$\frac{I_1}{I_2} = \frac{R_1^2}{R_2^2} = \frac{5}{6}$$

Given that the ratio is $$\frac{x}{6}$$:

$$\frac{x}{6} = \frac{5}{6} \implies x = 5$$

The answer is $$\boxed{5}$$.