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. `(2 R)/(sqrt(15))`
B. `R sqrt((2)/(15))`
C. `(4 R)/(sqrt(15))`
D. `( R)/(4)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Moment of Inertia of the Sphere
The moment of inertia \( I \) of a solid sphere about an axis passing through its center of mass is given by the formula:
\[ I = \frac{2}{5} M R^2 \]
where \( M \) is the mass of the sphere and \( R \) is its radius.

### Step 2: 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. The moment of inertia \( I' \) of a disc about its center is given by:
\[ I' = \frac{1}{2} M r^2 \]
where \( r \) is the radius of the disc.

Using the parallel axis theorem, the moment of inertia about an axis through the edge of the disc is:
\[ I_{edge} = I' + M d^2 \]
where \( d \) is the distance from the center of the disc to the edge, which is equal to \( r \). Thus:
\[ I_{edge} = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2 \]

### Step 3: Set the Moments of Inertia Equal
According to the problem, the moment of inertia of the disc about its edge is equal to the moment of inertia of the sphere:
\[ \frac{3}{2} M r^2 = \frac{2}{5} M R^2 \]

### Step 4: Cancel the Mass and Solve for \( r^2 \)
We can cancel \( M \) from both sides (assuming \( M \neq 0 \)):
\[ \frac{3}{2} r^2 = \frac{2}{5} R^2 \]
Now, multiply both sides by \( 10 \) to eliminate the fractions:
\[ 15 r^2 = 4 R^2 \]
Now, divide both sides by 15:
\[ r^2 = \frac{4}{15} R^2 \]

### Step 5: Find the Radius of the Disc
Taking the square root of both sides gives us the radius \( r \) of the disc:
\[ r = \sqrt{\frac{4}{15}} R = \frac{2}{\sqrt{15}} R \]

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

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