Using the formula for the moment of inertia of a uniform sphere, find the moment of inertia of a thin spherical layer of mass `m` and radius `R` relative to the axis passing through its centre.

Options

A. `MR^(2)`
B. `MR^(2)//2`
C. `2//3 MR^(2)`
D. `2//5 MR^(2)`

Answer

To find the moment of inertia of a thin spherical layer of mass \( m \) and radius \( R \) relative to the axis passing through its center, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Moment of Inertia of a Sphere**:
The moment of inertia \( I \) of a solid uniform sphere about an axis through its center 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.

2. **Consider a Thin Spherical Layer**:
We consider a thin spherical layer of thickness \( dr \) at radius \( r \) from the center of the sphere. The total mass of this thin layer is \( m \).

3. **Calculate the Volume of the Thin Layer**:
The volume \( dV \) of the thin spherical layer can be calculated using the formula for the volume of a sphere:
\[ dV = \text{Surface Area} \times \text{Thickness} = 4\pi r^2 \cdot dr \]

4. **Relate Mass to Density**:
The mass \( m \) of the thin layer can be expressed in terms of its density \( \rho \):
\[ m = \rho \cdot dV = \rho \cdot (4\pi r^2 dr) \]

5. **Express Density in Terms of Mass**:
Rearranging the equation gives:
\[ \rho = \frac{m}{4\pi r^2 dr} \]

6. **Moment of Inertia of the Thin Layer**:
The moment of inertia \( I' \) of the thin spherical layer can be derived from the moment of inertia of a sphere of radius \( r + dr \) minus the moment of inertia of a sphere of radius \( r \):
\[ I' = I(r + dr) - I(r) \]
Using the formula for the moment of inertia of a sphere, we get:
\[ I(r + dr) = \frac{2}{5} M (r + dr)^2 \]
\[ I(r) = \frac{2}{5} M r^2 \]

7. **Expanding the Moment of Inertia**:
Expanding \( I(r + dr) \) using the binomial expansion:
\[ I(r + dr) = \frac{2}{5} M \left( r^2 + 2r \cdot dr + (dr)^2 \right) \]
Thus,
\[ I' = \frac{2}{5} M \left( r^2 + 2r \cdot dr + (dr)^2 \right) - \frac{2}{5} M r^2 \]
Simplifying gives:
\[ I' = \frac{2}{5} M (2r \cdot dr + (dr)^2) \]

8. **Neglecting Higher Order Terms**:
Since \( dr \) is very small, we can neglect \( (dr)^2 \):
\[ I' \approx \frac{2}{5} M (2r \cdot dr) = \frac{4}{5} M r \cdot dr \]

9. **Final Expression**:
Therefore, the moment of inertia of the thin spherical layer is:
\[ I' = \frac{4}{5} m r^2 \]