The moment of inertia of two spheres of equal masses about their diameters are equal. If one of them is solid and the other is hollow, then the ratio of their radii is
[M.I. of hollow sphere = `(2)/(3)MR_(h)^(2)` ]

Options

A. `5:3`
B. `3:5`
C. `sqrt5:sqrt3`
D. `sqrt3:sqrt5`

Explanation

To solve the problem, we need to find the ratio of the radii of a solid sphere and a hollow sphere given that their moments of inertia about their diameters are equal.

### Step-by-Step Solution:

1. **Identify the Moment of Inertia Formulas:**
- For a solid sphere, the moment of inertia \( I_s \) about its diameter is given by:
\[ I_s = \frac{2}{5} M R_s^2 \]
- For a hollow sphere, the moment of inertia \( I_h \) about its diameter is given by:
\[ I_h = \frac{2}{3} M R_h^2 \]

2. **Set the Moments of Inertia Equal:**
Since the problem states that the moments of inertia are equal, we can set them equal to each other:
\[ \frac{2}{5} M R_s^2 = \frac{2}{3} M R_h^2 \]

3. **Cancel the Mass \( M \):**
Since the masses are equal and non-zero, we can divide both sides by \( M \):
\[ \frac{2}{5} R_s^2 = \frac{2}{3} R_h^2 \]

4. **Eliminate the Coefficient 2:**
We can simplify the equation by multiplying both sides by \( \frac{1}{2} \):
\[ \frac{1}{5} R_s^2 = \frac{1}{3} R_h^2 \]

5. **Cross-Multiply to Solve for the Radii:**
Cross-multiplying gives us:
\[ 3 R_s^2 = 5 R_h^2 \]

6. **Express the Ratio of the Radii:**
Rearranging the equation to find the ratio \( \frac{R_s}{R_h} \):
\[ \frac{R_s^2}{R_h^2} = \frac{5}{3} \]
Taking the square root of both sides:
\[ \frac{R_s}{R_h} = \sqrt{\frac{5}{3}} = \frac{\sqrt{5}}{\sqrt{3}} \]

7. **Final Ratio:**
Therefore, the ratio of the radii \( R_s : R_h \) is:
\[ R_s : R_h = \sqrt{5} : \sqrt{3} \]

### Final Answer:
The ratio of the radii of the solid sphere to the hollow sphere is \( \sqrt{5} : \sqrt{3} \).