Ratio of densities of materials of two circular discs of same mass and thickness is `5:6`. The ratio of their M.I about natural axes is

Options

A. `5 : 6`
B. `6 : 5`
C. `25 : 36`
D. `1: 1`

Answer

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

### Step-by-Step Solution:

1. **Define Densities**:
Let the density of the first disc be \( \rho_1 = 5x \) and the density of the second disc be \( \rho_2 = 6x \), where \( x \) is a constant.

2. **Express Mass in Terms of Density and Volume**:
Since both discs have the same mass \( m \) and thickness \( t \), we can express the mass of each disc in terms of its density and volume. The volume \( V \) of a disc is given by:
\[ V = \pi r^2 t \]
Therefore, the mass of the first disc is:
\[ m = \rho_1 \cdot V_1 = 5x \cdot (\pi r_1^2 t) \]
And for the second disc:
\[ m = \rho_2 \cdot V_2 = 6x \cdot (\pi r_2^2 t) \]

3. **Set the Masses Equal**:
Since the masses are equal, we can set the equations equal to each other:
\[ 5x \cdot (\pi r_1^2 t) = 6x \cdot (\pi r_2^2 t) \]
Canceling out \( x \), \( \pi \), and \( t \) from both sides, we get:
\[ 5 r_1^2 = 6 r_2^2 \]

4. **Find the Ratio of Radii**:
Rearranging the equation gives:
\[ \frac{r_1^2}{r_2^2} = \frac{6}{5} \]

5. **Moment of Inertia Formula**:
The moment of inertia \( I \) of a disc about its natural axis is given by:
\[ I = \frac{1}{2} m r^2 \]
Therefore, for the first disc:
\[ I_1 = \frac{1}{2} m r_1^2 \]
And for the second disc:
\[ I_2 = \frac{1}{2} m r_2^2 \]

6. **Ratio of Moments of Inertia**:
The ratio of the moments of inertia \( \frac{I_1}{I_2} \) can be expressed as:
\[ \frac{I_1}{I_2} = \frac{\frac{1}{2} m r_1^2}{\frac{1}{2} m r_2^2} = \frac{r_1^2}{r_2^2} \]
Substituting the ratio we found earlier:
\[ \frac{I_1}{I_2} = \frac{6}{5} \]

### Final Answer:
The ratio of the moments of inertia of the two discs about their natural axes is:
\[ \frac{I_1}{I_2} = \frac{6}{5} \]