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. `L/(4sqrt(2))m`
B. `L/(2sqrt(2))m`
C. `L/(sqrt(2))m`
D. `L/(8sqrt(2))m`

Explanation

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

### Step 1: Understand the Configuration
The rod is bent at a right angle, forming two equal segments. Since the total length of the rod is 1 metre, each arm of the 'L' shape will be \( \frac{L}{2} = \frac{1}{2} \) metre.

### Step 2: Assign Masses
Let the total mass of the rod be \( M \). When bent into an 'L' shape, each arm will have a mass of \( \frac{M}{2} \).

### Step 3: Set the Coordinate System
We will set the bent point (the corner of the 'L') as the origin (0,0). The first arm (horizontal) will extend along the x-axis, and the second arm (vertical) will extend along the y-axis.

### Step 4: Calculate the Center of Mass in the X-Direction
For the horizontal arm:
- Mass \( M_1 = \frac{M}{2} \)
- The center of mass of this arm is at \( \left( \frac{L}{4}, 0 \right) \) because it is halfway along the arm.

For the vertical arm:
- Mass \( M_2 = \frac{M}{2} \)
- The center of mass of this arm is at \( \left( 0, \frac{L}{4} \right) \).

Using the formula for the center of mass in the x-direction:
\[ X_{CM} = \frac{M_1X_1 + M_2X_2}{M_1 + M_2} \]
Substituting the values:
\[ X_{CM} = \frac{\left(\frac{M}{2} \cdot \frac{L}{4}\right) + \left(\frac{M}{2} \cdot 0\right)}{\frac{M}{2} + \frac{M}{2}} = \frac{\frac{ML}{8}}{M} = \frac{L}{8} \]

### Step 5: Calculate the Center of Mass in the Y-Direction
Using the formula for the center of mass in the y-direction:
\[ Y_{CM} = \frac{M_1Y_1 + M_2Y_2}{M_1 + M_2} \]
Substituting the values:
\[ Y_{CM} = \frac{\left(\frac{M}{2} \cdot 0\right) + \left(\frac{M}{2} \cdot \frac{L}{4}\right)}{\frac{M}{2} + \frac{M}{2}} = \frac{\frac{ML}{8}}{M} = \frac{L}{8} \]

### Step 6: Calculate the Distance from the Bent Point
Now, we have the coordinates of the center of mass:
\[ (X_{CM}, Y_{CM}) = \left(\frac{L}{8}, \frac{L}{8}\right) \]
To find the distance \( D \) from the bent point (origin) to the center of mass, we use the distance formula:
\[ D = \sqrt{X_{CM}^2 + Y_{CM}^2} \]
Substituting the values:
\[ D = \sqrt{\left(\frac{L}{8}\right)^2 + \left(\frac{L}{8}\right)^2} = \sqrt{\frac{L^2}{64} + \frac{L^2}{64}} = \sqrt{\frac{2L^2}{64}} = \sqrt{\frac{L^2}{32}} = \frac{L}{4\sqrt{2}} \]

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