A circular disc of radius R and thickness `(R)/(6)` has moment of inertia about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is

Options

A. `I/5`
B. `I/6`
C. `I/32`
D. `I/64`

Answer

To solve the problem, we need to find the moment of inertia of a solid sphere that is formed by melting a circular disc of given dimensions. We will follow these steps:

### Step 1: Calculate the moment of inertia of the circular disc.
The moment of inertia \( I \) of a circular disc about an axis passing through its center and perpendicular to its plane 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 circular disc.
The volume \( V \) of the circular disc can be calculated using the formula:
\[ V = \text{Area} \times \text{Thickness} \]
The area of the disc is \( A = \pi R^2 \) and the thickness is \( \frac{R}{6} \). Therefore, the volume \( V \) is:
\[ V = \pi R^2 \times \frac{R}{6} = \frac{\pi R^3}{6} \]

### Step 3: Calculate the mass of the circular disc.
The mass \( m \) of the disc can be expressed in terms of its volume and density \( \rho \):
\[ m = \rho V = \rho \left(\frac{\pi R^3}{6}\right) \]

### Step 4: Calculate the radius of the solid sphere.
When the disc is melted and recast into a solid sphere, the volume remains the same. The volume \( V \) of the sphere is given by:
\[ V = \frac{4}{3} \pi r'^3 \]
Setting the volumes equal gives:
\[ \frac{\pi R^3}{6} = \frac{4}{3} \pi r'^3 \]
Canceling \( \pi \) from both sides:
\[ \frac{R^3}{6} = \frac{4}{3} r'^3 \]
Multiplying both sides by \( 6 \):
\[ R^3 = 8 r'^3 \]
Dividing 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 solid sphere.
The moment of inertia \( I' \) of a solid sphere about its diameter 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 \( m \) to the moment of inertia \( I \) of the disc.
From Step 1, we know:
\[ I = \frac{1}{2} m R^2 \]
Thus, we can express \( m \) in terms of \( I \):
\[ m = \frac{2I}{R^2} \]

### Step 7: Substitute \( m \) back into the moment of inertia of the sphere.
Substituting \( m \) into the equation for \( I' \):
\[ I' = \frac{1}{10} \left(\frac{2I}{R^2}\right) R^2 = \frac{1}{10} \cdot 2I = \frac{1}{5} I \]

### Final Answer:
The moment of inertia of the sphere about its diameter is:
\[ I' = \frac{1}{5} I \]