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

Options

A. `(105)/(176)R^(2)rho`
B. `(176)/(105)R^(2)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 Moment of Inertia Definition
The moment of inertia \( I \) about an axis is defined as:
\[ I = \int r^2 \, dm \]
where \( r \) is the distance from the axis of rotation to the mass element \( dm \).

### Step 2: Identify the Moment of Inertia for a Solid Sphere
For a solid sphere, the moment of inertia about an axis through its center (which is also its diameter) is given by the formula:
\[ I_{cm} = \frac{2}{5} M R^2 \]
where \( M \) is the mass of the sphere and \( R \) is its radius.

### Step 3: Calculate the Mass of the Sphere
The mass \( M \) of the sphere can be calculated using its volume and density:
\[ M = \text{Volume} \times \text{Density} = \left(\frac{4}{3} \pi R^3\right) \rho \]

### Step 4: Substitute Mass into the Moment of Inertia Formula
Now, substitute the expression for mass \( M \) into the moment of inertia formula:
\[ I_{cm} = \frac{2}{5} \left(\frac{4}{3} \pi R^3 \rho\right) R^2 \]

### Step 5: Simplify the Expression
Now simplify the expression:
\[ I_{cm} = \frac{2}{5} \cdot \frac{4}{3} \pi R^5 \rho = \frac{8}{15} \pi R^5 \rho \]

### Step 6: Convert \( \pi \) to a Fraction
For further simplification, we can express \( \pi \) as \( \frac{22}{7} \) (for approximation):
\[ I_{cm} = \frac{8}{15} \cdot \frac{22}{7} R^5 \rho = \frac{176}{105} R^5 \rho \]

### Step 7: Conclusion
Thus, the moment of inertia of the solid sphere about its diameter is:
\[ I = \frac{176}{105} R^5 \rho \]

### Final Answer
The correct option is:
\[ \text{Option 2: } \frac{176}{105} R^5 \rho \]

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