The moment of inertia of a solid sphere of mass M and radius R about its diameter is I. The moment of inertia of the same sphere about a tangent parallel to the diameter is

Options

A. 5 I
B. 7 I
C. 2.5 I
D. 3.5 I

Answer

To find the moment of inertia of a solid sphere about a tangent parallel to its diameter, we can use the parallel axis theorem. Here’s a step-by-step solution:

### Step 1: Understand the Moment of Inertia about the Diameter
The moment of inertia \( I \) of a solid sphere of mass \( M \) and radius \( R \) about its diameter is given by the formula:
\[ I = \frac{2}{5} M R^2 \]

### Step 2: Identify the New Axis
We need to find the moment of inertia about a tangent line parallel to the diameter. This new axis is parallel to the diameter and located a distance \( R \) (the radius of the sphere) away from the center of the sphere.

### Step 3: Apply the Parallel Axis Theorem
The parallel axis theorem states that the moment of inertia \( I \) about any axis parallel to an axis through the center of mass is given by:
\[ I = I_{cm} + M d^2 \]
where:
- \( I_{cm} \) is the moment of inertia about the center of mass axis (which we already calculated),
- \( M \) is the mass of the sphere,
- \( d \) is the distance between the two axes.

In our case:
- \( I_{cm} = \frac{2}{5} M R^2 \)
- \( d = R \)

### Step 4: Substitute Values into the Equation
Substituting the values into the parallel axis theorem:
\[ I = \frac{2}{5} M R^2 + M R^2 \]

### Step 5: Simplify the Expression
Combine the terms:
\[ I = \frac{2}{5} M R^2 + \frac{5}{5} M R^2 = \frac{7}{5} M R^2 \]

### Final Result
Thus, the moment of inertia of the solid sphere about a tangent parallel to the diameter is:
\[ I = \frac{7}{5} M R^2 \]