A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness `t//4`. The relation between the moments of inertia `I_(A)` and `I_(B)` is (about an axis passing through centre and perpendicular to the disc)

Options

A. `l_(A)gtl_(B)`
B. `l_(A)=l_(B)`
C. `l_(A)ltl_(B)`
D. depends on the actul values of t and r

Answer

To solve the problem, we need to find the relationship between the moments of inertia \( I_A \) and \( I_B \) for the two circular discs A and B.

### Step 1: Calculate the Moment of Inertia for Disc A
The moment of inertia \( I \) of a disc about an axis passing through its center and perpendicular to the disc is given by the formula:

\[ I = \frac{1}{2} m R^2 \]

Where:
- \( m \) is the mass of the disc
- \( R \) is the radius of the disc

For disc A:
- Radius \( R_A = r \)
- Thickness \( t \)

First, we need to calculate the mass \( m_A \) of disc A. The mass can be found using the formula:

\[ m_A = \text{Volume} \times \text{Density} = \text{Area} \times \text{Thickness} \times \text{Density} \]

The area of disc A is:

\[ \text{Area} = \pi R_A^2 = \pi r^2 \]

Thus, the volume of disc A is:

\[ \text{Volume} = \text{Area} \times \text{Thickness} = \pi r^2 t \]

So, the mass of disc A is:

\[ m_A = \pi r^2 t \cdot \rho \]

Now substituting \( m_A \) into the moment of inertia formula:

\[ I_A = \frac{1}{2} m_A R_A^2 = \frac{1}{2} (\pi r^2 t \cdot \rho) r^2 = \frac{1}{2} \pi r^4 t \rho \]

### Step 2: Calculate the Moment of Inertia for Disc B
For disc B:
- Radius \( R_B = 4r \)
- Thickness \( t_B = \frac{t}{4} \)

Using the same process to find the mass \( m_B \):

The area of disc B is:

\[ \text{Area} = \pi R_B^2 = \pi (4r)^2 = 16\pi r^2 \]

Thus, the volume of disc B is:

\[ \text{Volume} = \text{Area} \times \text{Thickness} = 16\pi r^2 \cdot \frac{t}{4} = 4\pi r^2 t \]

So, the mass of disc B is:

\[ m_B = 4\pi r^2 t \cdot \rho \]

Now substituting \( m_B \) into the moment of inertia formula:

\[ I_B = \frac{1}{2} m_B R_B^2 = \frac{1}{2} (4\pi r^2 t \cdot \rho) (4r)^2 = \frac{1}{2} (4\pi r^2 t \cdot \rho) (16r^2) = \frac{32}{2} \pi r^4 t \rho = 16 \pi r^4 t \rho \]

### Step 3: Find the Relation between \( I_A \) and \( I_B \)
Now we have:

\[ I_A = \frac{1}{2} \pi r^4 t \rho \]
\[ I_B = 16 \pi r^4 t \rho \]

To find the ratio of \( I_A \) to \( I_B \):

\[ \frac{I_A}{I_B} = \frac{\frac{1}{2} \pi r^4 t \rho}{16 \pi r^4 t \rho} \]

Cancelling out common terms:

\[ \frac{I_A}{I_B} = \frac{1}{2} \cdot \frac{1}{16} = \frac{1}{32} \]

Thus, we find:

\[ I_A = \frac{1}{32} I_B \]

### Conclusion
From the above calculations, we conclude that:

\[ I_B = 32 I_A \]