A short bar magnet having magnetic moment `4 Am^2` , placed in a vibrating magnetometer, vibrates with a time period of 8 s. Another short bar magnet having a magnetic moment `8Am^2` vibrates with a time period of 6 s. If the moment of inertia of the second magnet is` 9 xx 10^(-2) kgm^2` , the moment of inertia of the first magnet is (assuming that both magnets are kept in the same uniform magnetic induction field):

Options

A. `9 xx 10^(-2) kg m^2`
B. `8 xx 10^(-2) kg m^2`
C. `5.33 xx 10^(-2) kg m^2`
D. `12.2 xx 10^(-2) kg m^2`

Explanation

To solve the problem, we will use the relationship between the time period of a vibrating magnet and its moment of inertia and magnetic moment. The time period \( T \) of a magnet in a magnetic field is given by the formula:

\[ T = 2\pi \sqrt{\frac{I}{M}} \]

where:
- \( T \) is the time period,
- \( I \) is the moment of inertia,
- \( M \) is the magnetic moment.

Given two magnets, we can set up the ratio of their time periods:

\[ \frac{T_1}{T_2} = \sqrt{\frac{I_1}{I_2}} \cdot \sqrt{\frac{M_2}{M_1}} \]

### Step 1: Write down the known values
- For the first magnet:
- Magnetic moment \( M_1 = 4 \, \text{Am}^2 \)
- Time period \( T_1 = 8 \, \text{s} \)
- Moment of inertia \( I_1 = ? \)

- For the second magnet:
- Magnetic moment \( M_2 = 8 \, \text{Am}^2 \)
- Time period \( T_2 = 6 \, \text{s} \)
- Moment of inertia \( I_2 = 9 \times 10^{-2} \, \text{kg m}^2 \)

### Step 2: Set up the ratio of the time periods
Using the formula, we can express the ratio of the time periods:

\[ \frac{T_1}{T_2} = \sqrt{\frac{I_1}{I_2}} \cdot \sqrt{\frac{M_2}{M_1}} \]

### Step 3: Square both sides
Squaring both sides gives:

\[ \left(\frac{T_1}{T_2}\right)^2 = \frac{I_1}{I_2} \cdot \frac{M_2}{M_1} \]

### Step 4: Rearrange to find \( I_1 \)
Rearranging the equation to solve for \( I_1 \):

\[ I_1 = I_2 \cdot \left(\frac{T_1}{T_2}\right)^2 \cdot \frac{M_1}{M_2} \]

### Step 5: Substitute known values
Substituting the known values into the equation:

- \( I_2 = 9 \times 10^{-2} \, \text{kg m}^2 \)
- \( T_1 = 8 \, \text{s} \)
- \( T_2 = 6 \, \text{s} \)
- \( M_1 = 4 \, \text{Am}^2 \)
- \( M_2 = 8 \, \text{Am}^2 \)

Now substituting these values:

\[ I_1 = 9 \times 10^{-2} \cdot \left(\frac{8}{6}\right)^2 \cdot \frac{4}{8} \]

### Step 6: Calculate the values
Calculating \( \left(\frac{8}{6}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \) and \( \frac{4}{8} = \frac{1}{2} \):

\[ I_1 = 9 \times 10^{-2} \cdot \frac{16}{9} \cdot \frac{1}{2} \]

### Step 7: Simplify the expression
Now simplify:

\[ I_1 = 9 \times 10^{-2} \cdot \frac{16}{18} = 9 \times 10^{-2} \cdot \frac{8}{9} \]

The \( 9 \) cancels out:

\[ I_1 = 8 \times 10^{-2} \, \text{kg m}^2 \]

### Final Answer
Thus, the moment of inertia of the first magnet is:

\[ I_1 = 8 \times 10^{-2} \, \text{kg m}^2 \]