Two discs have same mass and thickness. Their materials are of densities `rho_(1)` and `rho_(2)`. The ratio of their moment of inertia about central axis will be

Options

A. `1:rho_(1)rho_(2)`
B. `rho_(1)rho_(2):1`
C. `rho_(1):rho_(2)`
D. `rho_(2):rho_(1)`

Explanation

To find the ratio of the moments of inertia of two discs with the same mass and thickness but different densities, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Moment of Inertia Formula**:
The moment of inertia \( I \) of a disc about its central axis is given by the formula:
\[ I = \frac{1}{2} m R^2 \]
where \( m \) is the mass and \( R \) is the radius of the disc.

2. **Relate Mass, Density, and Volume**:
The mass \( m \) of a disc can be expressed in terms of its volume and density:
\[ m = \text{Volume} \times \text{Density} = \pi R^2 T \rho \]
where \( T \) is the thickness of the disc and \( \rho \) is the density.

3. **Find the Radius in Terms of Mass and Density**:
Rearranging the mass equation gives:
\[ R^2 = \frac{m}{\pi T \rho} \]
Thus, we can express the radius \( R \) as:
\[ R = \sqrt{\frac{m}{\pi T \rho}} \]

4. **Calculate the Moment of Inertia for Each Disc**:
For Disc 1 (density \( \rho_1 \)):
\[ I_1 = \frac{1}{2} m R_1^2 = \frac{1}{2} m \left(\frac{m}{\pi T \rho_1}\right) = \frac{m^2}{2 \pi T \rho_1} \]
For Disc 2 (density \( \rho_2 \)):
\[ I_2 = \frac{1}{2} m R_2^2 = \frac{1}{2} m \left(\frac{m}{\pi T \rho_2}\right) = \frac{m^2}{2 \pi T \rho_2} \]

5. **Find the Ratio of Moments of Inertia**:
Now, we can find the ratio \( \frac{I_1}{I_2} \):
\[ \frac{I_1}{I_2} = \frac{\frac{m^2}{2 \pi T \rho_1}}{\frac{m^2}{2 \pi T \rho_2}} = \frac{\rho_2}{\rho_1} \]

6. **Final Result**:
Therefore, the ratio of the moments of inertia of the two discs is:
\[ \frac{I_1}{I_2} = \frac{\rho_2}{\rho_1} \]

### Conclusion:
The answer is that the ratio of their moments of inertia about the central axis is \( \rho_2 : \rho_1 \).