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 will calculate the moments of inertia \( I_A \) and \( I_B \) for the two discs A and B, and then find the relation between them.

### Step 1: Calculate the mass of disc A
The mass \( m_A \) of disc A can be calculated using the formula:
\[ m_A = \text{density} \times \text{area} \times \text{thickness} \]
For disc A:
- Radius \( r \)
- Thickness \( t \)
- Area \( A_A = \pi r^2 \)

Thus, the mass of disc A is:
\[ m_A = \rho \times \pi r^2 \times t \]

### Step 2: Calculate the mass of disc B
Similarly, for disc B:
- Radius \( 4r \)
- Thickness \( \frac{t}{4} \)
- Area \( A_B = \pi (4r)^2 = 16\pi r^2 \)

The mass of disc B is:
\[ m_B = \rho \times 16\pi r^2 \times \frac{t}{4} = 4\rho \pi r^2 t \]

### Step 3: Express mass of disc B in terms of mass of disc A
From the above calculations, we can express the mass of disc B in terms of the mass of disc A:
\[ m_B = 4 m_A \]

### Step 4: Calculate the moment of inertia of disc A
The moment of inertia \( I_A \) of disc A about an axis passing through the center and perpendicular to the disc is given by:
\[ I_A = \frac{1}{2} m_A r^2 \]
Substituting \( m_A \):
\[ I_A = \frac{1}{2} \left( \rho \pi r^2 t \right) r^2 = \frac{1}{2} \rho \pi t r^4 \]

### Step 5: Calculate the moment of inertia of disc B
Similarly, the moment of inertia \( I_B \) of disc B is:
\[ I_B = \frac{1}{2} m_B (4r)^2 \]
Substituting \( m_B \):
\[ I_B = \frac{1}{2} \left( 4\rho \pi r^2 t \right) (16r^2) = \frac{1}{2} \times 4 \rho \pi t \times 16 r^4 = 32 \rho \pi t r^4 \]

### Step 6: Relate \( I_A \) and \( I_B \)
Now we can relate \( I_A \) and \( I_B \):
\[ I_B = 32 \rho \pi t r^4 \]
\[ I_A = \frac{1}{2} \rho \pi t r^4 \]

To find the ratio \( \frac{I_B}{I_A} \):
\[ \frac{I_B}{I_A} = \frac{32 \rho \pi t r^4}{\frac{1}{2} \rho \pi t r^4} = \frac{32}{\frac{1}{2}} = 64 \]

### Conclusion
Thus, the relation between the moments of inertia \( I_A \) and \( I_B \) is:
\[ I_B = 64 I_A \]