If `I_(1),I_(2)` and `I_(3)` are the moments of inertia about the natural axies of solid sphere, hollow sphere and a spherical shell of same mass and radii, the correct result of the following is

Options

A. `I_(1)gt I_(2)gt I_(3)`
B. `I_(3)gt I_(2) gt I_(1)`
C. `I_(2)gt I_(1)gt I_(3)`
D. `I_(2)=I_(3)gt I_(1)`

Explanation

To solve the problem, we need to determine the moments of inertia for a solid sphere, a hollow sphere, and a spherical shell, and then compare their values.

### Step-by-Step Solution:

1. **Identify the Moments of Inertia**:
- For a solid sphere, the moment of inertia \( I_1 \) is given by:
\[ I_1 = \frac{2}{5} m R^2 \]
- For a hollow sphere, the moment of inertia \( I_2 \) is given by:
\[ I_2 = \frac{2}{3} m R^2 \]
- For a spherical shell, the moment of inertia \( I_3 \) is also given by:
\[ I_3 = \frac{2}{3} m R^2 \]

2. **Compare the Moments of Inertia**:
- We need to compare \( I_1 \), \( I_2 \), and \( I_3 \).
- From the formulas:
- \( I_1 = \frac{2}{5} m R^2 \)
- \( I_2 = \frac{2}{3} m R^2 \)
- \( I_3 = \frac{2}{3} m R^2 \)

3. **Convert to 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 \( \frac{2}{5} \):
\[ \frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15} \]
- Convert \( \frac{2}{3} \):
\[ \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \]

4. **Determine the Order**:
- Now we can see:
- \( I_1 = \frac{6}{15} m R^2 \)
- \( I_2 = \frac{10}{15} m R^2 \)
- \( I_3 = \frac{10}{15} m R^2 \)
- Thus, we have:
\[ I_1 < I_2 = I_3 \]

5. **Conclusion**:
- The correct relationship is:
\[ I_1 < I_2 = I_3 \]
- Therefore, the answer is that \( I_2 \) and \( I_3 \) are greater than \( I_1 \).

### Final Answer:
The correct result is \( I_2 = I_3 > I_1 \).