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

Options

A. `I`
B. `2I//3`
C. `I//5`
D. `I//10`

Explanation

To solve the problem, we need to find the moment of inertia of a solid sphere formed by melting a circular disc of radius \( R \) and thickness \( \frac{R}{6} \).

### Step-by-Step Solution:

1. **Calculate the Volume of the Disc:**
The volume \( V \) of the disc can be calculated using the formula for the volume of a cylinder:
\[ V = \text{Area of base} \times \text{Height} = \pi R^2 \times \frac{R}{6} = \frac{\pi R^3}{6} \]

2. **Determine the Mass of the Disc:**
Assuming the density of the disc is \( \rho \), the mass \( m \) of the disc can be expressed as:
\[ m = \rho V = \rho \cdot \frac{\pi R^3}{6} \]

3. **Calculate the Volume of the Sphere:**
When the disc is melted and recast into a sphere, the volume of the sphere \( V_s \) will be equal to the volume of the disc:
\[ V_s = \frac{4}{3} \pi r^3 \]
Setting the volumes equal gives:
\[ \frac{4}{3} \pi r^3 = \frac{\pi R^3}{6} \]

4. **Solve for the Radius of the Sphere:**
Canceling \( \pi \) from both sides and solving for \( r^3 \):
\[ \frac{4}{3} r^3 = \frac{R^3}{6} \]
\[ r^3 = \frac{R^3}{6} \cdot \frac{3}{4} = \frac{R^3}{8} \]
Taking the cube root gives:
\[ r = \frac{R}{2} \]

5. **Calculate the Moment of Inertia of the Sphere:**
The moment of inertia \( I_s \) of a solid sphere about its diameter is given by:
\[ I_s = \frac{2}{5} m r^2 \]
Substituting \( m \) and \( r \):
\[ I_s = \frac{2}{5} \left(\rho \cdot \frac{\pi R^3}{6}\right) \left(\frac{R}{2}\right)^2 \]
Simplifying this:
\[ I_s = \frac{2}{5} \cdot \frac{\rho \pi R^3}{6} \cdot \frac{R^2}{4} = \frac{2 \rho \pi R^5}{120} = \frac{\rho \pi R^5}{60} \]

6. **Relate the Moment of Inertia of the Sphere to the Disc:**
The moment of inertia of the disc \( I \) is given by:
\[ I = \frac{1}{2} m R^2 = \frac{1}{2} \left(\rho \cdot \frac{\pi R^3}{6}\right) R^2 = \frac{\rho \pi R^5}{12} \]
Now, we can relate \( I_s \) to \( I \):
\[ I_s = \frac{2}{5} \cdot \frac{I}{2} = \frac{I}{5} \]

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