One solid sphere A and another hollow spher B are of same mass and same outer radii. Their moment of inertia aobut their diameters are respectively `I_A and I_B` such that
where `d_A and d_B` are their densities,

Options

A. `I_A = I_B `
B. `I_A gtI_B `
C. `I_A lt I_B`
D. data is incomplete

Answer

To solve the problem, we need to find the relationship between the moments of inertia of a solid sphere (A) and a hollow sphere (B) given that they have the same mass and outer radii.

### Step-by-Step Solution:

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

2. **Understand the Variables**:
- Here, \( m \) is the mass of the spheres, and \( r \) is the radius of the spheres.
- Since both spheres have the same mass and radius, we can use the same \( m \) and \( r \) for both formulas.

3. **Compare the Moments of Inertia**:
- We need to compare \( I_A \) and \( I_B \):
\[ I_A = \frac{2}{5} m r^2 \]
\[ I_B = \frac{2}{3} m r^2 \]
- To compare \( I_A \) and \( I_B \), we can look at the coefficients:
- The coefficient for \( I_A \) is \( \frac{2}{5} \).
- The coefficient for \( I_B \) is \( \frac{2}{3} \).

4. **Finding a Common Denominator**:
- To compare \( \frac{2}{5} \) and \( \frac{2}{3} \), we can convert them to a common denominator:
- The least common multiple of 5 and 3 is 15.
- Convert \( \frac{2}{5} \) to \( \frac{6}{15} \).
- Convert \( \frac{2}{3} \) to \( \frac{10}{15} \).

5. **Conclusion**:
- Since \( \frac{10}{15} > \frac{6}{15} \), we conclude that:
\[ I_B > I_A \]
- Thus, the moment of inertia of the hollow sphere is greater than that of the solid sphere.

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