Particles of masses 1g, 2g, 3g, ……. 100g re kept at the marks 1 cm, 2cm, 3cm,………. 100 cm, respectively on a metre scale. Find the moment of inertia of the system of particle about a perpendicular bisector of the metre scale.

Explanation

To find the moment of inertia of a system of particles with masses from 1g to 100g placed at distances from 1 cm to 100 cm on a meter scale about the perpendicular bisector (which is at 50 cm), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Positions**:
- The masses are \( m_1 = 1 \, \text{g}, m_2 = 2 \, \text{g}, \ldots, m_{100} = 100 \, \text{g} \).
- The positions of these masses are \( x_1 = 1 \, \text{cm}, x_2 = 2 \, \text{cm}, \ldots, x_{100} = 100 \, \text{cm} \).

2. **Determine the Axis of Rotation**:
- The perpendicular bisector of the meter scale is at \( x = 50 \, \text{cm} \).

3. **Calculate the Moment of Inertia**:
- The moment of inertia \( I \) about an axis is given by:
\[ I = \sum m_i r_i^2 \]
where \( r_i \) is the distance from the axis of rotation to the mass \( m_i \).

4. **Calculate for Left-Side Masses (1g to 49g)**:
- For masses \( m_1 \) to \( m_{49} \):
- The distances from the axis (50 cm) are:
- \( r_1 = 50 - 1 = 49 \, \text{cm} \)
- \( r_2 = 50 - 2 = 48 \, \text{cm} \)
- ...
- \( r_{49} = 50 - 49 = 1 \, \text{cm} \)
- Thus, the moment of inertia for the left-side masses is:
\[ I_L = \sum_{i=1}^{49} m_i (50 - i)^2 \]

5. **Calculate for Right-Side Masses (51g to 100g)**:
- For masses \( m_{51} \) to \( m_{100} \):
- The distances from the axis (50 cm) are:
- \( r_{51} = 51 - 50 = 1 \, \text{cm} \)
- \( r_{52} = 52 - 50 = 2 \, \text{cm} \)
- ...
- \( r_{100} = 100 - 50 = 50 \, \text{cm} \)
- Thus, the moment of inertia for the right-side masses is:
\[ I_R = \sum_{i=51}^{100} m_i (i - 50)^2 \]

6. **Combine Both Moments of Inertia**:
- The total moment of inertia is:
\[ I = I_L + I_R \]

7. **Calculate Each Sum**:
- For \( I_L \):
\[ I_L = 10^{-4} \sum_{i=1}^{49} i (49 - i + 1)^2 \]
- For \( I_R \):
\[ I_R = 10^{-4} \sum_{i=51}^{100} i (i - 50)^2 \]

8. **Final Calculation**:
- After calculating both sums, combine them to find \( I \).

### Final Result:
After performing the calculations, the moment of inertia about the perpendicular bisector is found to be approximately \( 0.43 \, \text{kg m}^2 \).

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