A system consists of two identical sphers each of mass 1.5 kg and radius 50 cm at the ends of a light rod .The distane between the centres of the spheres is 5 m .What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint .

Options

A. `18.75kgm^(2)`
B. `1.905xx10^(5)kgm^(2)`
C. `1.875xx10^(5)kgm^(2)`
D. `19.05kgm^(2)`

Answer

To find the moment of inertia of the system consisting of two identical spheres about an axis perpendicular to the rod passing through its midpoint, we can follow these steps:

### Step 1: Identify the parameters
- Mass of each sphere, \( m = 1.5 \, \text{kg} \)
- Radius of each sphere, \( r = 0.5 \, \text{m} \) (converted from 50 cm)
- Distance between the centers of the spheres, \( d = 5 \, \text{m} \)
- Distance from the midpoint of the rod to the center of each sphere, \( \frac{d}{2} = \frac{5}{2} = 2.5 \, \text{m} \)

### Step 2: Calculate the moment of inertia of one sphere about its own center
The moment of inertia \( I \) of a solid sphere about its own center is given by the formula:
\[ I_{\text{sphere}} = \frac{2}{5} m r^2 \]
Substituting the values:
\[ I_{\text{sphere}} = \frac{2}{5} \times 1.5 \, \text{kg} \times (0.5 \, \text{m})^2 = \frac{2}{5} \times 1.5 \times 0.25 = \frac{2 \times 1.5 \times 0.25}{5} = \frac{0.75}{5} = 0.15 \, \text{kg} \cdot \text{m}^2 \]

### Step 3: Use the parallel axis theorem to find the moment of inertia about the midpoint
According to the parallel axis theorem, the moment of inertia about an axis parallel to one through the center of mass is given by:
\[ I' = I + m d^2 \]
Where \( d \) is the distance from the center of mass to the new axis. Here, \( d = 2.5 \, \text{m} \).

For each sphere, we have:
\[ I'_{\text{sphere}} = I_{\text{sphere}} + m \left(\frac{d}{2}\right)^2 = 0.15 + 1.5 \times (2.5)^2 \]
Calculating \( (2.5)^2 \):
\[ (2.5)^2 = 6.25 \]
Now substituting:
\[ I'_{\text{sphere}} = 0.15 + 1.5 \times 6.25 = 0.15 + 9.375 = 9.525 \, \text{kg} \cdot \text{m}^2 \]

### Step 4: Calculate the total moment of inertia for both spheres
Since there are two identical spheres, the total moment of inertia \( I_{\text{total}} \) is:
\[ I_{\text{total}} = 2 \times I'_{\text{sphere}} = 2 \times 9.525 = 19.05 \, \text{kg} \cdot \text{m}^2 \]

### Final Answer
The moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint is:
\[ \boxed{19.05 \, \text{kg} \cdot \text{m}^2} \]