The perio of oscillation of a magnet in vibration magnetometer is 2 sec. The period of oscillation of a magnet whosr magnetic moment is four times that of the first magnet is

Options

A. 1 sec
B. 4 sec
C. 8 sec
D. 0.5 sec

Answer

To solve the problem, we need to determine the period of oscillation of a magnet whose magnetic moment is four times that of the first magnet. We start with the known period of oscillation of the first magnet and use the relationship between the period and the magnetic moment.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The period of oscillation of the first magnet (T) is 2 seconds.
- The magnetic moment of the second magnet (M') is four times that of the first magnet (M).
- Therefore, M' = 4M.

2. **Recall the Formula for the Period of Oscillation:**
The period of oscillation (T) of a magnet in a vibration magnetometer is given by:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot B}} \]
where:
- \( I \) is the moment of inertia,
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field strength.

3. **Understand the Relationship:**
The period of oscillation is inversely proportional to the square root of the magnetic moment:
\[ T \propto \frac{1}{\sqrt{M}} \]
This means that if the magnetic moment increases, the period of oscillation decreases.

4. **Set Up the Ratio of Periods:**
Let T' be the period of oscillation for the second magnet. Using the relationship:
\[ \frac{T}{T'} = \sqrt{\frac{M'}{M}} \]

5. **Substitute the Values:**
Since \( M' = 4M \):
\[ \frac{T}{T'} = \sqrt{\frac{4M}{M}} = \sqrt{4} = 2 \]
Therefore:
\[ \frac{T}{T'} = 2 \implies T' = \frac{T}{2} \]

6. **Calculate T':**
Substitute \( T = 2 \) seconds:
\[ T' = \frac{2}{2} = 1 \text{ second} \]

### Final Answer:
The period of oscillation of the magnet whose magnetic moment is four times that of the first magnet is **1 second**.

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