Two small spheres of mass `5kg` and `15kg ` are joined by a rod of length `0.5m` and of negligible mass. The M.I. of the system about an axis passing through centre of rod and normal to it is

Options

A. `10kgm^(2)`
B. `1.25kgm^(2)`
C. `20kgm^(2)`
D. `8kgm^(2)`

Answer

To find the moment of inertia (M.I.) of the system consisting of two small spheres of mass 5 kg and 15 kg joined by a rod of length 0.5 m, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**: We have two spheres with masses \( m_1 = 5 \, \text{kg} \) and \( m_2 = 15 \, \text{kg} \) connected by a rod of length \( L = 0.5 \, \text{m} \).

2. **Determine the Center of Mass**: The center of the rod is the midpoint, which is at \( 0.25 \, \text{m} \) from either mass since the rod is \( 0.5 \, \text{m} \) long.

3. **Calculate the Distances from the Axis**:
- The distance from the center of the rod to the first mass \( m_1 \) (5 kg) is:
\[ r_1 = 0.25 \, \text{m} \]
- The distance from the center of the rod to the second mass \( m_2 \) (15 kg) is also:
\[ r_2 = 0.25 \, \text{m} \]

4. **Use the Moment of Inertia Formula**: The moment of inertia \( I \) about an axis perpendicular to the rod and passing through the center is given by:
\[ I = m_1 r_1^2 + m_2 r_2^2 \]

5. **Substitute the Values**:
- For \( m_1 = 5 \, \text{kg} \):
\[ I_1 = 5 \times (0.25)^2 = 5 \times 0.0625 = 0.3125 \, \text{kg m}^2 \]
- For \( m_2 = 15 \, \text{kg} \):
\[ I_2 = 15 \times (0.25)^2 = 15 \times 0.0625 = 0.9375 \, \text{kg m}^2 \]

6. **Calculate the Total Moment of Inertia**:
\[ I = I_1 + I_2 = 0.3125 + 0.9375 = 1.25 \, \text{kg m}^2 \]

### Final Answer:
The moment of inertia of the system about the specified axis is:
\[ \boxed{1.25 \, \text{kg m}^2} \]