The moment of inertia of a squaer lamin about the perpendicluar axis though its centre of mass is ` 20 kg- m^(2)` . Then its moment of inertia about an axis toching its side and in the plane of the lamina will be : -

Options

A. `10 kg - m^(2)`
B. `30 kg- m^(2)`
C. `40 k g- m^(2)`
D. `25 kg m^(2)`

Explanation

To find the moment of inertia of a square lamina about an axis touching its side and in the plane of the lamina, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The moment of inertia of the square lamina about the perpendicular axis through its center of mass (Iz) is given as \( Iz = 20 \, \text{kg m}^2 \).

2. **Use the Perpendicular Axis Theorem**:
- The perpendicular axis theorem states that for a planar lamina, the moment of inertia about an axis perpendicular to the plane (Iz) is the sum of the moments of inertia about two perpendicular axes in the plane of the lamina (Ix and Iy):
\[ Iz = Ix + Iy \]
- Since the lamina is square and uniform, we can assume:
\[ Ix = Iy \]
- Therefore, we can write:
\[ Iz = 2Iy \]
- From this, we can find \( Iy \):
\[ Iy = \frac{Iz}{2} = \frac{20}{2} = 10 \, \text{kg m}^2 \]

3. **Identify the Axis of Interest**:
- We need to find the moment of inertia about an axis (let's call it Iab) that touches one side of the square and lies in the plane of the lamina.

4. **Apply the Parallel Axis Theorem**:
- The parallel axis theorem states:
\[ I_{ab} = I_y + m \cdot d^2 \]
- Where:
- \( I_y \) is the moment of inertia about the axis through the center of mass (which we found to be \( 10 \, \text{kg m}^2 \)).
- \( m \) is the mass of the lamina.
- \( d \) is the distance from the center of mass axis to the new axis (which is half the side length of the square, \( \frac{L}{2} \)).
- The distance \( d \) can be expressed in terms of the side length \( L \).

5. **Relate Mass and Side Length**:
- The moment of inertia about the center of mass axis perpendicular to the plane can be expressed as:
\[ I_z = \frac{mL^2}{6} = 20 \, \text{kg m}^2 \]
- From this, we can find \( mL^2 \):
\[ mL^2 = 120 \, \text{kg m}^2 \]

6. **Calculate the Moment of Inertia about the New Axis**:
- Substitute \( m \) and \( d \) into the parallel axis theorem:
\[ I_{ab} = Iy + m \cdot \left(\frac{L}{2}\right)^2 \]
- We know \( Iy = 10 \, \text{kg m}^2 \) and \( mL^2 = 120 \, \text{kg m}^2 \):
\[ I_{ab} = 10 + m \cdot \frac{L^2}{4} \]
- Since \( m = \frac{120}{L^2} \):
\[ I_{ab} = 10 + \frac{120}{L^2} \cdot \frac{L^2}{4} = 10 + 30 = 40 \, \text{kg m}^2 \]

7. **Final Answer**:
- The moment of inertia about the axis touching the side and in the plane of the lamina is:
\[ I_{ab} = 40 \, \text{kg m}^2 \]