The magnetic needle of a vibration magnetometer makes `12` oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line it makes `15` oscillations per minute. If the poles of the bar magnet are inter changed, the number of oscillations it takes per minute is

Options

A. `sqrt(61)`
B. `sqrt(63)`
C. `sqrt(65)`
D. `sqrt(67)`

Answer

To solve the problem, we will use the relationship between the frequency of oscillation of a magnetic needle and the magnetic fields acting on it.

### Step-by-step Solution:

1. **Understanding the Problem**:
- The magnetic needle makes 12 oscillations per minute in the Earth's magnetic field (denote this frequency as \( f_1 = 12 \) oscillations/min).
- When an external bar magnet is introduced, the frequency increases to 15 oscillations per minute (denote this frequency as \( f_2 = 15 \) oscillations/min).
- We need to find the frequency when the poles of the bar magnet are interchanged (denote this frequency as \( f_3 \)).

2. **Using the Relationship of Frequency**:
- The frequency of oscillation of the magnetic needle is related to the effective magnetic field acting on it. The frequency \( f \) is proportional to the square root of the magnetic field \( B \):
\[ f \propto \sqrt{B} \]

3. **Setting Up the Equations**:
- Let \( B_H \) be the horizontal component of the Earth's magnetic field and \( B \) be the magnetic field due to the bar magnet.
- When the bar magnet is present, the effective magnetic field is \( B_H + B \), thus:
\[ f_2 \propto \sqrt{B_H + B} \]
- When only the Earth's field is present:
\[ f_1 \propto \sqrt{B_H} \]

4. **Expressing Frequencies**:
- We can express the frequencies in terms of the magnetic fields:
\[ \frac{f_1}{f_2} = \frac{\sqrt{B_H}}{\sqrt{B_H + B}} \quad \text{(1)} \]

5. **Substituting Known Values**:
- Substitute \( f_1 = 12 \) and \( f_2 = 15 \) into equation (1):
\[ \frac{12}{15} = \frac{\sqrt{B_H}}{\sqrt{B_H + B}} \]
- Simplifying gives:
\[ \frac{4}{5} = \frac{\sqrt{B_H}}{\sqrt{B_H + B}} \]

6. **Squaring Both Sides**:
- Squaring both sides results in:
\[ \left(\frac{4}{5}\right)^2 = \frac{B_H}{B_H + B} \]
- This simplifies to:
\[ \frac{16}{25} = \frac{B_H}{B_H + B} \]

7. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 16(B_H + B) = 25B_H \]
- Rearranging leads to:
\[ 16B_H + 16B = 25B_H \implies 9B_H = 16B \implies B = \frac{9}{16}B_H \]

8. **Finding \( f_3 \)**:
- When the poles of the bar magnet are interchanged, the effective magnetic field becomes \( B_H - B \):
\[ f_3 \propto \sqrt{B_H - B} \]
- Substitute \( B = \frac{9}{16}B_H \):
\[ B_H - B = B_H - \frac{9}{16}B_H = \frac{7}{16}B_H \]
- Therefore, we have:
\[ f_3 \propto \sqrt{\frac{7}{16}B_H} = \frac{\sqrt{7}}{4}\sqrt{B_H} \]

9. **Relating \( f_3 \) to \( f_1 \)**:
- Since \( f_1 \propto \sqrt{B_H} \):
\[ f_3 = \frac{\sqrt{7}}{4} \cdot \frac{f_1}{\sqrt{B_H}} = \frac{\sqrt{7}}{4} \cdot 12 \]
- Thus:
\[ f_3 = 3\sqrt{7} \]

10. **Calculating the Final Value**:
- Approximating \( \sqrt{7} \approx 2.64575 \):
\[ f_3 \approx 3 \cdot 2.64575 \approx 7.93725 \text{ oscillations/min} \]

### Final Answer:
The number of oscillations per minute when the poles of the bar magnet are interchanged is approximately \( 8 \) oscillations/min.