Moment of inertia of a solid sphere about its diameter is I . If that sphere is recast into 8 identical small spheres, then moment of inertia of such small sphere about its diameter is

Options

A. `I//8`
B. `I//16`
C. `I//24`
D. `I//32`

Explanation

To solve the problem, we need to find the moment of inertia of each of the 8 identical small spheres that are formed from the original solid sphere.

### Step-by-Step Solution:

1. **Understand the Moment of Inertia of the Original Sphere**:
The moment of inertia \( I \) of a solid sphere about its diameter is given by the formula:
\[ I = \frac{2}{5} m R^2 \]
where \( m \) is the mass of the sphere and \( R \) is its radius.

2. **Determine the Mass of Each Small Sphere**:
When the original sphere is recast into 8 identical small spheres, the mass of each small sphere \( m_s \) is:
\[ m_s = \frac{m}{8} \]

3. **Determine the Radius of Each Small Sphere**:
The volume of the original sphere is:
\[ V = \frac{4}{3} \pi R^3 \]
The volume of one small sphere is:
\[ V_s = \frac{1}{8} V = \frac{1}{8} \left(\frac{4}{3} \pi R^3\right) = \frac{1}{6} \pi R^3 \]
Let \( r \) be the radius of each small sphere. The volume of a small sphere can also be expressed as:
\[ V_s = \frac{4}{3} \pi r^3 \]
Setting the two expressions for the volume equal gives:
\[ \frac{4}{3} \pi r^3 = \frac{1}{6} \pi R^3 \]
Canceling \( \pi \) and multiplying through by \( \frac{3}{4} \):
\[ r^3 = \frac{1}{8} R^3 \]
Taking the cube root:
\[ r = \frac{R}{2} \]

4. **Calculate the Moment of Inertia of Each Small Sphere**:
Using the formula for the moment of inertia for a solid sphere about its diameter, we can find the moment of inertia \( I_s \) of each small sphere:
\[ I_s = \frac{2}{5} m_s r^2 \]
Substituting \( m_s = \frac{m}{8} \) and \( r = \frac{R}{2} \):
\[ I_s = \frac{2}{5} \left(\frac{m}{8}\right) \left(\frac{R}{2}\right)^2 \]
\[ I_s = \frac{2}{5} \left(\frac{m}{8}\right) \left(\frac{R^2}{4}\right) \]
\[ I_s = \frac{2}{5} \cdot \frac{m R^2}{32} \]
\[ I_s = \frac{m R^2}{80} \]

5. **Relate \( I_s \) to \( I \)**:
Recall that \( I = \frac{2}{5} m R^2 \). Therefore, we can express \( I_s \) in terms of \( I \):
\[ I_s = \frac{1}{16} I \]

### Final Answer:
The moment of inertia of each small sphere about its diameter is:
\[ I_s = \frac{1}{16} I \]