One solid sphere `A` and another hollow sphere `B` are of same mass and same outer radii. Their moment of inertia about their diameters are respectively `I_(A)` and `I_(B)` such that.

Options

A. `I_(A) lt I_(B)`
B. `I_(A) gt I_(B)`
C. `I_(A) = I_(B)`
D. `I_(A)/I_(B) = d_(A)/d_(B)`

Explanation

To solve the problem, we need to find the relationship between the moments of inertia of a solid sphere and a hollow sphere, both having the same mass and outer radius.

### Step-by-Step Solution:

1. **Identify the Moment of Inertia Formulas**:
- For a solid sphere, the moment of inertia about its diameter is given by:
\[ I_A = \frac{2}{5} m r^2 \]
- For a hollow sphere (thin spherical shell), the moment of inertia about its diameter is given by:
\[ I_B = \frac{2}{3} m r^2 \]

2. **Set Up the Given Information**:
- Both spheres have the same mass \( m \) and the same outer radius \( r \).

3. **Substitute the Values**:
- Substitute the mass and radius into the formulas:
\[ I_A = \frac{2}{5} m r^2 \]
\[ I_B = \frac{2}{3} m r^2 \]

4. **Compare the Two Moments of Inertia**:
- To compare \( I_A \) and \( I_B \), we can look at the coefficients of \( m r^2 \):
- For \( I_A \): Coefficient is \( \frac{2}{5} \)
- For \( I_B \): Coefficient is \( \frac{2}{3} \)

5. **Find a Common Denominator**:
- The common denominator of 5 and 3 is 15. Convert both fractions:
\[ I_A = \frac{2}{5} m r^2 = \frac{6}{15} m r^2 \]
\[ I_B = \frac{2}{3} m r^2 = \frac{10}{15} m r^2 \]

6. **Conclusion**:
- From the comparison, we see that:
\[ I_B > I_A \]
- Therefore, the relationship between the moments of inertia is:
\[ I_B > I_A \]

### Final Answer:
The moment of inertia of the hollow sphere \( I_B \) is greater than that of the solid sphere \( I_A \).

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