A circular disc of radius R and thickness `r/6` has moment of inertia I about its axis. If it is melted and casted into a solid sphere then what will be its moment of inertia, about its diametric axis?

Options

A. `I/5`
B. `(5I)/12`
C. `I/10`
D. `I`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the Moment of Inertia of the Disc
The moment of inertia \( I \) of a circular disc about its axis is given by the formula:
\[ I = \frac{1}{2} m R^2 \]
where \( m \) is the mass of the disc and \( R \) is its radius.

### Step 2: Calculate the Volume of the Disc
The volume \( V \) of the disc can be calculated using the formula:
\[ V = \text{Area} \times \text{Thickness} \]
The area \( A \) of the disc is:
\[ A = \pi R^2 \]
The thickness of the disc is given as \( \frac{R}{6} \). Therefore, the volume of the disc is:
\[ V = \pi R^2 \times \frac{R}{6} = \frac{\pi R^3}{6} \]

### Step 3: Set the Volume of the Sphere Equal to the Volume of the Disc
When the disc is melted and cast into a solid sphere, the volume of the sphere \( V_s \) must equal the volume of the disc:
\[ V_s = \frac{4}{3} \pi r'^3 \]
Setting the volumes equal gives:
\[ \frac{4}{3} \pi r'^3 = \frac{\pi R^3}{6} \]
We can cancel \( \pi \) from both sides:
\[ \frac{4}{3} r'^3 = \frac{R^3}{6} \]

### Step 4: Solve for the Radius of the Sphere
Multiplying both sides by 6:
\[ 8 r'^3 = R^3 \]
Dividing both sides by 8:
\[ r'^3 = \frac{R^3}{8} \]
Taking the cube root:
\[ r' = \frac{R}{2} \]

### Step 5: Calculate the Moment of Inertia of the Sphere
The moment of inertia \( I' \) of a solid sphere about its diametric axis is given by:
\[ I' = \frac{2}{5} m r'^2 \]
Substituting \( r' = \frac{R}{2} \):
\[ I' = \frac{2}{5} m \left(\frac{R}{2}\right)^2 = \frac{2}{5} m \frac{R^2}{4} = \frac{1}{10} m R^2 \]

### Step 6: Relate the Mass of the Sphere to the Mass of the Disc
From the moment of inertia of the disc, we have:
\[ I = \frac{1}{2} m R^2 \]
Thus, we can express \( m \) in terms of \( I \):
\[ m = \frac{2I}{R^2} \]

### Step 7: Substitute for Mass in the Moment of Inertia of the Sphere
Substituting \( m \) into the equation for \( I' \):
\[ I' = \frac{1}{10} \left(\frac{2I}{R^2}\right) R^2 = \frac{1}{10} \times 2I = \frac{I}{5} \]

### Final Answer
The moment of inertia of the solid sphere about its diametric axis is:
\[ I' = \frac{I}{5} \]