The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x form it is, 'I(x)'. Which one of the graphs represents the variation of 'I(x) with x correctly ?

Options

A.
B.
C.
D.

Explanation

To solve the problem of finding the moment of inertia \( I(x) \) of a solid sphere about an axis parallel to its diameter and at a distance \( x \) from it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Moment of Inertia**:
The moment of inertia \( I \) about an axis is a measure of how difficult it is to change the rotational motion of an object about that axis. For a solid sphere, the moment of inertia about an axis through its center (CM) is given by:
\[ I_{CM} = \frac{2}{5} m R^2 \]
where \( m \) is the mass of the sphere and \( R \) is its radius.

2. **Use the Parallel Axis Theorem**:
To find the moment of inertia about an axis that is parallel to the diameter and at a distance \( x \) from it, we can use the Parallel Axis Theorem:
\[ I(x) = I_{CM} + m d^2 \]
where \( d \) is the distance from the center of mass axis to the new axis. In this case, \( d = x \).

3. **Substitute the Values**:
Substitute \( I_{CM} \) and \( d \) into the equation:
\[ I(x) = \frac{2}{5} m R^2 + m x^2 \]

4. **Rearranging the Equation**:
The equation can be rearranged as:
\[ I(x) = m x^2 + \frac{2}{5} m R^2 \]
This shows that \( I(x) \) is a quadratic function of \( x \).

5. **Identify the Characteristics of the Graph**:
The equation \( I(x) = m x^2 + \frac{2}{5} m R^2 \) is a parabola that opens upwards. The y-intercept (when \( x = 0 \)) is \( \frac{2}{5} m R^2 \), which is a positive value. This means the graph will not pass through the origin.

6. **Conclusion**:
The correct graph representing the variation of \( I(x) \) with \( x \) will be a parabola that starts from the point \( \left(0, \frac{2}{5} m R^2\right) \) and opens upwards.