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

Options

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

Explanation

To solve the problem of finding 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 relationship between mass, density, and volume:**
The mass \( m \) of an object can be expressed as:
\[ m = \text{density} \times \text{volume} \]
For a disc, the volume can be calculated as:
\[ \text{Volume} = \text{Area} \times \text{Thickness} \]
The area of a disc is given by:
\[ \text{Area} = \pi r^2 \]
Therefore, the volume of a disc is:
\[ V = \pi r^2 T \]
where \( T \) is the thickness.

2. **Set up the equations for the masses of the discs:**
Let the mass of the first disc be \( m_1 \) and the second disc be \( m_2 \). Then we can write:
\[ m_1 = \pi_1 \cdot (\pi r_1^2 T) \quad \text{and} \quad m_2 = \pi_2 \cdot (\pi r_2^2 T) \]
Since the masses are equal, we have:
\[ \pi_1 \cdot (\pi r_1^2 T) = \pi_2 \cdot (\pi r_2^2 T) \]

3. **Simplify the mass equation:**
We can cancel out \( \pi T \) from both sides (since thickness is the same):
\[ \pi_1 r_1^2 = \pi_2 r_2^2 \]
Rearranging gives:
\[ \frac{r_1^2}{r_2^2} = \frac{\pi_2}{\pi_1} \]

4. **Calculate the moment of inertia for each disc:**
The moment of inertia \( I \) of a disc about its central axis is given by:
\[ I = \frac{1}{2} m r^2 \]
For the first disc:
\[ I_1 = \frac{1}{2} m_1 r_1^2 \]
For the second disc:
\[ I_2 = \frac{1}{2} m_2 r_2^2 \]

5. **Set up the ratio of the moments of inertia:**
The ratio of the moments of inertia \( \frac{I_1}{I_2} \) can be expressed as:
\[ \frac{I_1}{I_2} = \frac{m_1 r_1^2}{m_2 r_2^2} \]
Since \( m_1 = m_2 \), we can simplify this to:
\[ \frac{I_1}{I_2} = \frac{r_1^2}{r_2^2} \]

6. **Substitute the ratio of the radii:**
From step 3, we found that:
\[ \frac{r_1^2}{r_2^2} = \frac{\pi_2}{\pi_1} \]
Therefore, we can substitute this into our ratio of moments of inertia:
\[ \frac{I_1}{I_2} = \frac{\pi_2}{\pi_1} \]

### Final Answer:
The ratio of the moments of inertia about the central axis of the two discs is:
\[ \frac{I_1}{I_2} = \frac{\pi_2}{\pi_1} \]