The time period of a vibration magnetometer is `T_(0)`. Its magnet is replaced by another magnet whose moment of inertia is 3 times and magnetic moment is `1//3` of the initial magnet. The time period now will

Options

A. `3T_(0)`
B. `(T_(0))/(sqrt3)`
C. `T_(0)`
D. `(T_(0))/(3)`

Answer

To solve the problem, we need to analyze how the time period of the vibration magnetometer changes when the moment of inertia and magnetic moment of the magnet are altered.

### Step-by-Step Solution:

1. **Understanding the Time Period Formula**:
The time period \( T \) of a vibration magnetometer is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{m \cdot B}} \]
where:
- \( I \) is the moment of inertia of the magnet,
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field.

2. **Initial Conditions**:
Let the initial moment of inertia be \( I_0 \) and the initial magnetic moment be \( m_0 \). The initial time period is given as \( T_0 \):
\[ T_0 = 2\pi \sqrt{\frac{I_0}{m_0 \cdot B}} \]

3. **New Conditions**:
According to the problem, the moment of inertia is increased to three times the initial value:
\[ I = 3I_0 \]
The magnetic moment is decreased to one-third of the initial value:
\[ m = \frac{1}{3}m_0 \]

4. **Substituting New Values into the Time Period Formula**:
Now, we substitute the new values of \( I \) and \( m \) into the time period formula:
\[ T' = 2\pi \sqrt{\frac{I}{m \cdot B}} = 2\pi \sqrt{\frac{3I_0}{\frac{1}{3}m_0 \cdot B}} \]

5. **Simplifying the Expression**:
Simplifying the expression inside the square root:
\[ T' = 2\pi \sqrt{\frac{3I_0}{\frac{1}{3}m_0 \cdot B}} = 2\pi \sqrt{\frac{3I_0 \cdot 3}{m_0 \cdot B}} = 2\pi \sqrt{\frac{9I_0}{m_0 \cdot B}} \]

6. **Relating New Time Period to Initial Time Period**:
We can relate this new time period \( T' \) to the initial time period \( T_0 \):
\[ T' = 3 \cdot 2\pi \sqrt{\frac{I_0}{m_0 \cdot B}} = 3T_0 \]

7. **Final Result**:
Thus, the new time period \( T' \) is:
\[ T' = 3T_0 \]

### Conclusion:
The time period now will be \( 3T_0 \).