Two circular discs `A` and `B` of equal masses and thicknesses. But are made of metals with densities `d_A and d_B (d_A gt d_B)`. If their moments of inertia about an axis passing through the centre and normal to the circular faces be `I_A and I_B`, then.

Options

A. `I_(A)=I_(B)`
B. `I_(A)gtI_(B)`
C. `I_(A)ltI_(B)`
D. `I_(A)geI_(B)`

Explanation

To solve the problem, we need to analyze the relationship between the moments of inertia of two circular discs made of different materials but having equal masses and thicknesses.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- We have two circular discs, A and B.
- Both discs have equal masses (m) and equal thicknesses (t).
- The densities of the discs are different, with \( d_A > d_B \).

2. **Relating Mass and Density:**
- The mass of each disc can be expressed in terms of its density and volume. The volume \( V \) of a circular disc is given by:
\[ V = \pi r^2 t \]
- Therefore, the mass \( m \) of each disc can be written as:
\[ m = d \cdot V = d \cdot (\pi r^2 t) \]
- For disc A:
\[ m_A = d_A \cdot (\pi r_A^2 t) \]
- For disc B:
\[ m_B = d_B \cdot (\pi r_B^2 t) \]

3. **Setting the Masses Equal:**
- Since the masses are equal:
\[ d_A \cdot (\pi r_A^2 t) = d_B \cdot (\pi r_B^2 t) \]
- We can simplify this to:
\[ d_A r_A^2 = d_B r_B^2 \]

4. **Calculating the Moment of Inertia:**
- The moment of inertia \( I \) of a disc about an axis through its center and normal to its face is given by:
\[ I = \frac{1}{2} m r^2 \]
- For disc A:
\[ I_A = \frac{1}{2} m_A r_A^2 = \frac{1}{2} (d_A \cdot \pi r_A^2 t) r_A^2 = \frac{1}{2} d_A \cdot \pi t r_A^4 \]
- For disc B:
\[ I_B = \frac{1}{2} m_B r_B^2 = \frac{1}{2} (d_B \cdot \pi r_B^2 t) r_B^2 = \frac{1}{2} d_B \cdot \pi t r_B^4 \]

5. **Expressing \( r_A^2 \) and \( r_B^2 \):**
- From the mass equality derived earlier:
\[ r_A^2 = \frac{d_B}{d_A} r_B^2 \]
- Substitute this into the expressions for \( I_A \) and \( I_B \):
\[ I_A = \frac{1}{2} d_A \cdot \pi t \left(\frac{d_B}{d_A} r_B^2\right)^2 = \frac{1}{2} d_A \cdot \pi t \cdot \frac{d_B^2}{d_A^2} r_B^4 \]
- Thus:
\[ I_A = \frac{1}{2} \cdot \frac{d_B^2}{d_A} \cdot \pi t r_B^4 \]

6. **Comparing \( I_A \) and \( I_B \):**
- Now we can compare \( I_A \) and \( I_B \):
\[ I_B = \frac{1}{2} d_B \cdot \pi t r_B^4 \]
- Since \( d_A > d_B \), we can conclude:
\[ I_A < I_B \]

### Conclusion:
Thus, the moment of inertia of disc A is less than that of disc B:
\[ I_A < I_B \]