A uniform metre rod is bent into `L` shape with the bent arms at `90^@` to each other. The distance of the centre of mass from the bent point is

Options

A. on the bisector of the angle, `(1/(sqrt2))`m from vertex
B. on the bisector of angle, `(1/(2sqrt2))`m from vertex
C. on the bisector of the angle, 1/2 m from vertex
D. on the bisector of the angle, `(1/(4sqrt(2)))`m from vertex

Explanation

To find the distance of the center of mass from the bent point of a uniform meter rod bent into an 'L' shape, we can follow these steps:

### Step 1: Understand the Configuration
The uniform meter rod is bent into an 'L' shape, where each arm of the 'L' is 0.5 meters long (since the total length of the rod is 1 meter).

### Step 2: Identify the Center of Mass of Each Arm
1. **Horizontal Arm**: The center of mass of the horizontal arm (0.5 m) is located at its midpoint, which is at (0.25 m, 0).
2. **Vertical Arm**: The center of mass of the vertical arm (0.5 m) is also at its midpoint, which is at (0, 0.25 m).

### Step 3: Assign Masses
Assuming the mass of each half of the rod is \( m \) (since the rod is uniform), we have:
- Mass of the horizontal part, \( m_1 = m \)
- Mass of the vertical part, \( m_2 = m \)

### Step 4: Calculate the Center of Mass of the Composite System
The coordinates of the center of mass (CM) can be calculated using the formula:
\[ x_{CM} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \]
\[ y_{CM} = \frac{m_1 \cdot y_1 + m_2 \cdot y_2}{m_1 + m_2} \]
Substituting the values:
- For \( x_{CM} \):
\[ x_{CM} = \frac{m \cdot 0.25 + m \cdot 0}{m + m} = \frac{0.25m}{2m} = \frac{0.25}{2} = \frac{1}{8} \text{ m} \]
- For \( y_{CM} \):
\[ y_{CM} = \frac{m \cdot 0 + m \cdot 0.25}{m + m} = \frac{0.25m}{2m} = \frac{0.25}{2} = \frac{1}{8} \text{ m} \]

### Step 5: Determine the Coordinates of the Center of Mass
The coordinates of the center of mass are:
\[ \left( \frac{1}{8}, \frac{1}{8} \right) \]

### Step 6: Calculate the Distance from the Bent Point
To find the distance \( d \) from the bent point (origin) to the center of mass, we use the Pythagorean theorem:
\[ d = \sqrt{x_{CM}^2 + y_{CM}^2} = \sqrt{\left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2} = \sqrt{\frac{1}{64} + \frac{1}{64}} = \sqrt{\frac{2}{64}} = \sqrt{\frac{1}{32}} = \frac{1}{4\sqrt{2}} \text{ m} \]

### Final Answer
The distance of the center of mass from the bent point is:
\[ \frac{1}{4\sqrt{2}} \text{ meters} \]

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