The moment of interia of a spherical shell about a tangent is 20 kg `m^(2)` . What is the minimum moment of intertia about any axis ?

Options

A. `12 kg- m^(2)`
B. `10 kg - m^(2)`
C. `20 kg-m^(2)`
D. `8 kgm^(2)`

Answer

To find the minimum moment of inertia of a spherical shell about any axis, we can use the given information and apply the parallel axis theorem.

### Step-by-Step Solution:

1. **Understanding the Moment of Inertia of a Spherical Shell**:
The moment of inertia \( I \) of a hollow spherical shell about its center is given by the formula:
\[ I_{\text{center}} = \frac{2}{3} m r^2 \]
where \( m \) is the mass of the shell and \( r \) is its radius.

2. **Using the Parallel Axis Theorem**:
The parallel axis theorem states that if you know the moment of inertia about an axis through the center of mass, you can find the moment of inertia about any parallel axis by:
\[ I' = I_{\text{center}} + md^2 \]
where \( d \) is the distance between the two axes.

3. **Given Information**:
We are given that the moment of inertia about a tangent to the shell is \( I' = 20 \, \text{kg m}^2 \). The distance \( d \) from the center to the tangent is equal to the radius \( r \) of the shell.

4. **Setting Up the Equation**:
Substituting the known values into the parallel axis theorem:
\[ 20 = \frac{2}{3} m r^2 + m r^2 \]
Here, \( d = r \), so \( md^2 = m r^2 \).

5. **Combining Terms**:
Combine the terms on the right side:
\[ 20 = \frac{2}{3} m r^2 + \frac{3}{3} m r^2 = \frac{5}{3} m r^2 \]

6. **Solving for \( m r^2 \)**:
Rearranging the equation gives:
\[ m r^2 = \frac{20 \times 3}{5} = 12 \, \text{kg m}^2 \]

7. **Finding the Minimum Moment of Inertia**:
The minimum moment of inertia about any axis (which is through the center) is:
\[ I_{\text{min}} = \frac{2}{3} m r^2 = \frac{2}{3} \times 12 = 8 \, \text{kg m}^2 \]

### Final Answer:
The minimum moment of inertia about any axis is \( 8 \, \text{kg m}^2 \).