The magnet of vibration magnetometer is heated so as to reduce its magnetic moment by 36%. By doing this the periodic time of the magnetometer will

Options

A. Increases by 67%
B. Increases by 25%
C. Increases by 12%
D. Is not changed

Answer

To solve the problem, we need to analyze how the reduction in the magnetic moment affects the periodic time of the vibration magnetometer.

### Step-by-Step Solution:

1. **Understand the relationship between time period and magnetic moment**:
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,
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field.

From this equation, we can see that the time period \( T \) is inversely proportional to the square root of the magnetic moment \( m \).

2. **Define initial and final magnetic moments**:
Let the initial magnetic moment be \( m_1 \) and the final magnetic moment after heating be \( m_2 \). According to the problem, the magnetic moment is reduced by 36%. Therefore:
\[ m_1 = 100 \quad \text{(assuming initial magnetic moment is 100 units)} \]
\[ m_2 = m_1 - 0.36 \cdot m_1 = 100 - 36 = 64 \]

3. **Calculate the ratio of the time periods**:
The ratio of the time periods \( T_1 \) (initial) and \( T_2 \) (final) can be expressed as:
\[ \frac{T_1}{T_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{64}{100}} = \sqrt{0.64} = \frac{8}{10} = 0.8 \]

4. **Express \( T_2 \) in terms of \( T_1 \)**:
Rearranging the ratio gives:
\[ T_2 = \frac{10}{8} T_1 = 1.25 T_1 \]

5. **Determine the percentage increase in the time period**:
The increase in the time period can be calculated as:
\[ \text{Percentage Increase} = \left(\frac{T_2 - T_1}{T_1}\right) \times 100\% = \left(\frac{1.25 T_1 - T_1}{T_1}\right) \times 100\% = (0.25) \times 100\% = 25\% \]

### Conclusion:
The periodic time of the magnetometer will increase by 25%.

---