A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc abut an axis passing the edge and perpendicular to the plane remains l. Then R and r are related as

Options

A. `r = sqrt(2)/(15)R`
B. `r= (2)/(sqrt15) R`
C. `r = (2)/(15) R`
D. ` r=(sqrt2)/(15) R`

Explanation

To solve the problem, we need to relate the radius \( R \) of the solid sphere to the radius \( r \) of the solid disc after recasting. We will use the moment of inertia and the conservation of mass principles.

### Step-by-Step Solution:

1. **Identify the Moment of Inertia of the Sphere**:
The moment of inertia \( I \) of a solid sphere about its diameter is given by the formula:
\[ I_{\text{sphere}} = \frac{2}{5} M R^2 \]

2. **Identify the Moment of Inertia of the Disc**:
The moment of inertia \( I \) of a solid disc about an axis passing through its edge and perpendicular to its plane can be calculated using the parallel axis theorem. The moment of inertia about the center of mass is:
\[ I_{\text{disc, cm}} = \frac{1}{2} m r^2 \]
Using the parallel axis theorem, the moment of inertia about the edge is:
\[ I_{\text{disc}} = I_{\text{disc, cm}} + m \cdot d^2 \]
where \( d = \frac{r}{2} \) (the distance from the center of the disc to the edge). Thus:
\[ I_{\text{disc}} = \frac{1}{2} m r^2 + m \left(\frac{r}{2}\right)^2 = \frac{1}{2} m r^2 + \frac{1}{4} m r^2 = \frac{3}{4} m r^2 \]

3. **Conservation of Mass**:
Since the sphere is recast into a disc, the mass remains the same:
\[ M = m \]

4. **Equating the Moments of Inertia**:
Now we equate the moments of inertia of the sphere and the disc:
\[ \frac{2}{5} M R^2 = \frac{3}{4} m r^2 \]
Substituting \( M = m \):
\[ \frac{2}{5} M R^2 = \frac{3}{4} M r^2 \]

5. **Canceling Mass**:
Since \( M \) is common on both sides, we can cancel it out:
\[ \frac{2}{5} R^2 = \frac{3}{4} r^2 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ r^2 = \frac{2}{5} \cdot \frac{4}{3} R^2 = \frac{8}{15} R^2 \]

7. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ r = R \sqrt{\frac{8}{15}} = R \cdot \frac{2\sqrt{2}}{\sqrt{15}} \]

### Final Relation:
Thus, the relation between \( R \) and \( r \) is:
\[ r = \frac{2\sqrt{2}}{\sqrt{15}} R \]