A vibration magnetometer consist of two identical bar magnets placed one over the other such that they are mutually perpendicular and bisect each other. The time period of oscillations of combination in a horizontal magnetic field is `4s`. If one of the magnets is removed, then the period of oscillations of the other in the same field is

Options

A. 5.78s
B. 3.36 s
C. 4.36s
D. 5.36s

Answer

To solve the problem, we need to analyze the behavior of the vibration magnetometer with two identical bar magnets and then determine the time period of oscillation when one magnet is removed.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two identical bar magnets placed perpendicular to each other. They bisect each other, which means they are aligned at their midpoints.
- The time period of oscillation for this combination in a horizontal magnetic field is given as \( T_1 = 4 \, \text{s} \).

2. **Resultant Magnetic Moment**:
- The magnetic moment of each magnet is denoted as \( M \).
- Since the magnets are perpendicular, the resultant magnetic moment \( M_R \) of the combination can be calculated using the Pythagorean theorem:
\[ M_R = \sqrt{M^2 + M^2} = \sqrt{2M^2} = M\sqrt{2} \]

3. **Moment of Inertia**:
- The moment of inertia \( I_R \) of the combination of the two magnets is:
\[ I_R = 2I \]
- Here, \( I \) is the moment of inertia of one magnet.

4. **Time Period Formula**:
- The time period \( T \) of oscillation for a magnetic moment in a magnetic field is given by:
\[ T = 2\pi \sqrt{\frac{I}{MB}} \]
- For the combination of two magnets, substituting \( I_R \) and \( M_R \):
\[ T_1 = 2\pi \sqrt{\frac{2I}{M\sqrt{2}B}} = 2\pi \sqrt{\frac{I}{MB/\sqrt{2}}} \]

5. **Removing One Magnet**:
- When one magnet is removed, the resultant magnetic moment becomes \( M \) and the moment of inertia becomes \( I \):
\[ T_2 = 2\pi \sqrt{\frac{I}{MB}} \]

6. **Finding the Ratio of Time Periods**:
- We can find the ratio of the time periods \( T_2 \) and \( T_1 \):
\[ \frac{T_2}{T_1} = \frac{\sqrt{I/(MB)}}{\sqrt{2I/(M\sqrt{2}B)}} = \frac{1}{\sqrt{2}} \cdot \sqrt{2} = \frac{1}{\sqrt{2}} \cdot \sqrt{2} = \frac{1}{2^{1/4}} \]
- Therefore, we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = T_1 \cdot 2^{-1/4} \]

7. **Substituting the Known Value**:
- Now substituting \( T_1 = 4 \, \text{s} \):
\[ T_2 = 4 \cdot 2^{-1/4} \]
- Evaluating \( 2^{-1/4} \):
\[ 2^{-1/4} \approx 0.8409 \quad \text{(using a calculator)} \]
- Thus:
\[ T_2 \approx 4 \cdot 0.8409 \approx 3.36 \, \text{s} \]

### Final Answer:
The time period of oscillations of the remaining magnet is approximately \( 3.36 \, \text{s} \).