What is the moment of inertia of a solid sphere of density `rho` and radius R about its diameter ?

Options

A. `(105)/(176) R^(5) rho`
B. `(176)/(105) R^(5) rho`
C. `(105)/(176) R^(2) rho`
D. `(176)/(105) R^(2) rho`

Explanation

To find the moment of inertia of a solid sphere of density \( \rho \) and radius \( R \) about its diameter, we can follow these steps:

### Step 1: Understand the formula for moment of inertia
The moment of inertia \( I \) of a solid sphere about its diameter 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: Calculate the mass of the sphere
The mass \( m \) of the sphere can be calculated using the formula:
\[ m = \text{Volume} \times \text{Density} \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass \( m \) becomes:
\[ m = \frac{4}{3} \pi R^3 \times \rho \]

### Step 3: Substitute mass into the moment of inertia formula
Now, substitute the expression for mass \( m \) into the moment of inertia formula:
\[ I = \frac{2}{5} \left(\frac{4}{3} \pi R^3 \rho\right) R^2 \]

### Step 4: Simplify the expression
Now simplify the expression:
\[ I = \frac{2}{5} \times \frac{4}{3} \pi R^3 \rho \times R^2 \]
\[ I = \frac{2 \times 4}{5 \times 3} \pi \rho R^{3+2} \]
\[ I = \frac{8}{15} \pi \rho R^5 \]

### Final Result
Thus, the moment of inertia of a solid sphere of density \( \rho \) and radius \( R \) about its diameter is:
\[ I = \frac{8}{15} \pi \rho R^5 \]