If a circular concentric hole is made on a disc then about an axis passing through the centre of the disc and perpendicular to its plane

Options

A. moment of inertia decreases
B. moment of inertia increases
C. radius of gyration increases
D. radius of gyration decreases

Answer

To solve the problem, we need to analyze the effect of cutting a concentric circular hole in a disc on its moment of inertia and radius of gyration. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
- We have a solid disc with mass \( M \) and radius \( R \).
- The moment of inertia \( I_1 \) of the solid disc about an axis passing through its center and perpendicular to its plane is given by:
\[ I_1 = \frac{1}{2} M R^2 \]

### Step 2: Define the Hole
- A circular hole of radius \( r \) is cut out from the disc. Let the mass of the hole be \( m \).
- The mass of the remaining part of the disc after cutting the hole is:
\[ M' = M - m \]

### Step 3: Calculate the Moment of Inertia After Cutting the Hole
- The moment of inertia \( I_2 \) of the remaining disc can be calculated by subtracting the moment of inertia of the hole from the moment of inertia of the original disc.
- The moment of inertia of the hole about the same axis is:
\[ I_{\text{hole}} = \frac{1}{2} m r^2 \]
- Therefore, the moment of inertia \( I_2 \) of the remaining disc is:
\[ I_2 = I_1 - I_{\text{hole}} = \frac{1}{2} M R^2 - \frac{1}{2} m r^2 \]

### Step 4: Analyze the Change in Moment of Inertia
- Since \( m \) is a positive quantity (the mass of the hole), it follows that:
\[ I_2 < I_1 \]
- Thus, the moment of inertia decreases when the hole is cut out.

### Step 5: Calculate the Radius of Gyration
- The radius of gyration \( k \) is defined as:
\[ I = m' k^2 \]
- For the remaining disc, we have:
\[ I_2 = (M - m) k_2^2 \]
- Rearranging gives:
\[ k_2^2 = \frac{I_2}{M - m} \]

### Step 6: Compare the Radius of Gyration Before and After
- The original radius of gyration \( k_1 \) is:
\[ k_1^2 = \frac{I_1}{M} = \frac{\frac{1}{2} M R^2}{M} = \frac{1}{2} R^2 \]
- After substituting \( I_2 \) into the equation for \( k_2^2 \):
\[ k_2^2 = \frac{\frac{1}{2} M R^2 - \frac{1}{2} m r^2}{M - m} \]
- Since \( m \) is positive, the term \( \frac{1}{2} m r^2 \) reduces the moment of inertia, leading to a larger value for \( k_2^2 \) compared to \( k_1^2 \). Thus:
\[ k_2 > k_1 \]

### Conclusion
- The moment of inertia decreases when a hole is cut out of the disc.
- The radius of gyration increases as a result of cutting the hole.

### Final Answers
- Moment of Inertia: **Decreases**
- Radius of Gyration: **Increases**