A solid sphere of mass `M`, radius `R` and having moment of inertia about as axis passing through the centre of mass as `I`, is recast into a disc of thickness `t`, whose moment of inertia about an axis passing through its edge and perpendicular to its plance remains `I`. Then, radius of the disc will be.

Options

A. `r=sqrt(2/15)R`
B. `r=2sqrt(1/15)R`
C. `r=2/15R`
D. `r=2/(sqrt(5))R`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the moment of inertia of the solid sphere.
The moment of inertia \( I \) of a solid sphere about an axis passing through its center of mass is given by the formula:
\[ I_{\text{sphere}} = \frac{2}{5} M R^2 \]

### Step 2: Write down the moment of inertia of the disc.
When the solid sphere is recast into a disc, we need to find the moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane. According to the parallel axis theorem, the moment of inertia \( I \) about the edge of the disc can be expressed as:
\[ I = I_0 + M r^2 \]
where \( I_0 \) is the moment of inertia about the center of mass of the disc, and \( r \) is the distance from the center of mass to the new axis.

For a disc, the moment of inertia about its center of mass is:
\[ I_0 = \frac{1}{2} M r^2 \]
Thus, substituting this into the equation gives:
\[ I = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2 \]

### Step 3: Set the moments of inertia equal to each other.
Since the moment of inertia of the disc is equal to the moment of inertia of the sphere, we have:
\[ \frac{2}{5} M R^2 = \frac{3}{2} M r^2 \]

### Step 4: Cancel out the mass \( M \) from both sides.
Assuming \( M \neq 0 \), we can simplify the equation:
\[ \frac{2}{5} R^2 = \frac{3}{2} r^2 \]

### Step 5: Rearrange the equation to solve for \( r^2 \).
Multiplying both sides by \( 10 \) to eliminate the fractions gives:
\[ 4 R^2 = 15 r^2 \]
Now, rearranging for \( r^2 \):
\[ r^2 = \frac{4}{15} R^2 \]

### Step 6: Take the square root to find \( r \).
Taking the square root of both sides gives:
\[ r = R \sqrt{\frac{4}{15}} = \frac{2R}{\sqrt{15}} \]

### Final Result:
Thus, the radius of the disc is:
\[ r = \frac{2R}{\sqrt{15}} \]

---