The period of oscillation of a magnet at a place is `4`seconds. When it is remagnetised, so that the pole strength becomes `1//9th` of initial value, the period of oscillation in seconds is

Options

A. `3`
B. `12`
C. `5`
D. `4`

Explanation

To solve the problem, we need to determine how the period of oscillation of a magnet changes when its pole strength is altered.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The initial period of oscillation (T) is given as 4 seconds.
- The pole strength after remagnetization becomes \( \frac{1}{9} \) of the initial value.

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

3. **Identify the Relationship Between Period and Magnetic Moment**:
The magnetic moment \( M \) is defined as:
\[ M = m \cdot L \]
where \( m \) is the pole strength and \( L \) is the magnetic length.

4. **Determine the New Magnetic Moment**:
When the pole strength becomes \( \frac{1}{9} \) of the initial value, the new magnetic moment \( M' \) can be expressed as:
\[ M' = \frac{m}{9} \cdot L \]
Thus, the new magnetic moment is:
\[ M' = \frac{M}{9} \]
where \( M \) is the initial magnetic moment.

5. **Relate the New Period of Oscillation to the Old One**:
Since the period of oscillation is inversely proportional to the square root of the magnetic moment, we can express the new period \( T' \) as:
\[ T' = T \cdot \sqrt{\frac{M}{M'}} \]
Substituting \( M' = \frac{M}{9} \):
\[ T' = T \cdot \sqrt{\frac{M}{\frac{M}{9}}} = T \cdot \sqrt{9} = T \cdot 3 \]

6. **Calculate the New Period**:
Given that the initial period \( T \) is 4 seconds:
\[ T' = 4 \cdot 3 = 12 \text{ seconds} \]

### Final Answer:
The new period of oscillation after remagnetization is **12 seconds**.

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