1. Given data

Let the two discs have:

• equal masses: $M_1 = M_2$
• thicknesses: $t_1 = 1\text{ cm}$, $t_2 = 0.5\text{ cm}$
• densities in ratio: $\rho_1 : \rho_2 = 3:5$

We need the ratio of their moments of inertia about their diameters.

2. Use mass of a disc

For a uniform disc, $M = \rho \times \text{volume} = \rho \pi R^2 t$

Since the masses are equal, $\rho_1 \pi R_1^2 t_1 = \rho_2 \pi R_2^2 t_2$

So, $\rho_1 R_1^2 t_1 = \rho_2 R_2^2 t_2$

Substitute the given values: $3 \cdot R_1^2 \cdot 1 = 5 \cdot R_2^2 \cdot 0.5$ $3R_1^2 = \frac{5}{2}R_2^2$

Hence, $\frac{R_1^2}{R_2^2} = \frac{5}{6}$

3. Moment of inertia of a disc about a diameter

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

Since both discs have the same mass, $\frac{I_1}{I_2} = \frac{\frac14 M R_1^2}{\frac14 M R_2^2} = \frac{R_1^2}{R_2^2}$

Therefore, $\frac{I_1}{I_2} = \frac{5}{6}$

This is given as $\frac{x}{6}$, so $\frac{x}{6} = \frac{5}{6}$

Thus, $x = 5$

4. Comparison with stored answer

Derived answer: $5$

Stored correct answer: $5$

They match.