Moment of inertia of a solid sphere about its diameter is `I_(0)` . Then moment of inertia about an axis parallel to its diameter at a distance equal to half of its radius is

Options

A. `8I_(0)//13`
B. `13I_(0)//8`
C. `7I_(0)//2`
D. `2I_(0)//7`

Explanation

To find the moment of inertia of a solid sphere about an axis parallel to its diameter and at a distance equal to half of its radius, we can use the Parallel Axis Theorem. Here’s the step-by-step solution:

### Step 1: Understand the Given Information
We know that the moment of inertia of a solid sphere about its diameter is given as \( I_0 \). The formula for the moment of inertia of a solid sphere about its diameter is:

\[ I_0 = \frac{2}{5} m r^2 \]

where \( m \) is the mass of the sphere and \( r \) is its radius.

### Step 2: Identify the New Axis
We need to find the moment of inertia about an axis that is parallel to the diameter and at a distance of \( \frac{r}{2} \) from the diameter.

### Step 3: Apply the Parallel Axis Theorem
The Parallel Axis Theorem states that:

\[ I = I_{cm} + m d^2 \]

where:
- \( I \) is the moment of inertia about the new axis,
- \( I_{cm} \) is the moment of inertia about the center of mass axis (which is \( I_0 \) in this case),
- \( m \) is the mass of the sphere,
- \( d \) is the distance between the two axes.

Here, \( d = \frac{r}{2} \).

### Step 4: Substitute the Values
Substituting the known values into the equation:

\[ I = I_0 + m \left(\frac{r}{2}\right)^2 \]

### Step 5: Calculate \( m \left(\frac{r}{2}\right)^2 \)
Calculating \( m \left(\frac{r}{2}\right)^2 \):

\[ m \left(\frac{r}{2}\right)^2 = m \cdot \frac{r^2}{4} = \frac{m r^2}{4} \]

### Step 6: Substitute \( I_0 \)
Now substituting \( I_0 = \frac{2}{5} m r^2 \):

\[ I = \frac{2}{5} m r^2 + \frac{m r^2}{4} \]

### Step 7: Find a Common Denominator
To combine these fractions, we need a common denominator. The common denominator of 5 and 4 is 20:

\[ I = \frac{8}{20} m r^2 + \frac{5}{20} m r^2 = \frac{13}{20} m r^2 \]

### Step 8: Express in Terms of \( I_0 \)
Since \( I_0 = \frac{2}{5} m r^2 \), we can express \( m r^2 \) in terms of \( I_0 \):

\[ m r^2 = \frac{5}{2} I_0 \]

Now substituting this back into the equation for \( I \):

\[ I = \frac{13}{20} \cdot \frac{5}{2} I_0 = \frac{65}{40} I_0 = \frac{13}{8} I_0 \]

### Final Answer
Thus, the moment of inertia about the desired axis is:

\[ I = \frac{13}{8} I_0 \]