Two circular discs are of same thickness. The diameter of `A` is twice that of `B`. The moment of inertia of `A` as compared to that of `B` is

Options

A. A. twice as large
B. B. four times as large
C. C. eight times as large
D. D. 16 times as large

Explanation

To find the moment of inertia of disc A compared to disc B, we can follow these steps:

### Step 1: Define the Radii
Let the radius of disc B be \( r_B \). Since the diameter of disc A is twice that of disc B, the radius of disc A will be:
\[ r_A = 2r_B \]

### Step 2: Write the Moment of Inertia Formula
The moment of inertia \( I \) for a circular disc about its central axis is given by:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass and \( r \) is the radius of the disc.

### Step 3: Express the Mass of Each Disc
Since both discs have the same thickness and are circular, their masses can be expressed in terms of their areas. The area \( A \) of a disc is given by:
\[ A = \pi r^2 \]
Thus, the mass of disc A \( m_A \) and disc B \( m_B \) can be expressed as:
\[ m_A \propto \pi r_A^2 \quad \text{and} \quad m_B \propto \pi r_B^2 \]
Substituting \( r_A = 2r_B \):
\[ m_A \propto \pi (2r_B)^2 = 4\pi r_B^2 \]
So, we have:
\[ m_A = 4m_B \]

### Step 4: Calculate the Moment of Inertia for Each Disc
Now, we can calculate the moment of inertia for both discs:
\[ I_A = \frac{1}{2} m_A r_A^2 = \frac{1}{2} (4m_B) (2r_B)^2 \]
Calculating this gives:
\[ I_A = \frac{1}{2} (4m_B) (4r_B^2) = 8 m_B r_B^2 \]

For disc B:
\[ I_B = \frac{1}{2} m_B r_B^2 \]

### Step 5: Find the Ratio of Moments of Inertia
Now, we can find the ratio of the moments of inertia:
\[ \frac{I_A}{I_B} = \frac{8 m_B r_B^2}{\frac{1}{2} m_B r_B^2} = \frac{8}{\frac{1}{2}} = 16 \]

### Conclusion
Thus, the moment of inertia of disc A compared to that of disc B is:
\[ I_A = 16 I_B \]