1. Write the standard moments of inertia for bodies of the same mass $M$ and radius $R$.

• Thin circular ring about a diameter: The M.I. of a ring about an axis through the center and perpendicular to its plane is $MR^2$. By perpendicular axis theorem, the M.I. about any diameter is $I_1 = \frac{1}{2}MR^2$

• Circular disc about an axis perpendicular to the disc through the center: $I_2 = \frac{1}{2}MR^2$

• Solid cylinder about its own axis: $I_3 = \frac{1}{2}MR^2$

• Solid sphere about its diameter: $I_4 = \frac{2}{5}MR^2$

2. Compare the values:

We have $I_1 = I_2 = I_3 = \frac{1}{2}MR^2$ and $I_4 = \frac{2}{5}MR^2$

Since $\frac{1}{2} > \frac{2}{5}$ we get $I_1 = I_2 = I_3 > I_4$

So, Option A is correct.

3. Check the other options:

• Option B: $I_1 + I_3 = \frac{1}{2}MR^2 + \frac{1}{2}MR^2 = MR^2$ $I_2 + I_4 = \frac{1}{2}MR^2 + \frac{2}{5}MR^2 = \frac{9}{10}MR^2$ So, $I_1 + I_3 < I_2 + I_4$ becomes $MR^2 < \frac{9}{10}MR^2$ which is false.

• Option C: It says $I_1 = I_2 = I_3 < I_4$, but actually $\frac{1}{2}MR^2 > \frac{2}{5}MR^2$. False.

• Option D: $I_1 + I_2 = \frac{1}{2}MR^2 + \frac{1}{2}MR^2 = MR^2$ $I_3 + \frac{5}{2}I_4 = \frac{1}{2}MR^2 + \frac{5}{2}\cdot \frac{2}{5}MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$ Not equal. False.

4. Final answer:

The correct option is $\boxed{A}$

5. Comparison with stored answer:

Stored correct answer: A

My derived answer: A

They agree.