A small magnetic needle performs 10 oscillations/minute in the earth's horizontal magnetic field . When a bar magnet is placed near the small magnet in same position , frequency of oscillations becomes `10sqrt2` oscillations/minute . If the bar magnet be turned around end to end , the rate of oscillation of small magnet will become

Options

A. `5 sqrt2`/min
B. `20sqrt2`/min
C. `2sqrt2`/min
D. Zero

Answer

To solve the problem step by step, we will analyze the oscillations of a small magnetic needle in the presence of the Earth's magnetic field and the magnetic field due to a bar magnet.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- The small magnetic needle oscillates at a frequency of \( f_1 = 10 \) oscillations/minute in the Earth's horizontal magnetic field \( B_H \).
- Convert this frequency to Hertz (Hz):
\[ f_1 = \frac{10}{60} \text{ Hz} = \frac{1}{6} \text{ Hz} \]

2. **Using the Formula for Frequency:**
- The frequency of oscillation \( f \) of a magnetic needle is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{m B_H}{I}} \]
- Rearranging gives:
\[ B_H = \frac{(2\pi f)^2 I}{m} \]

3. **Analyzing the Case with the Bar Magnet:**
- When the bar magnet is placed near the small magnet, the frequency of oscillation becomes \( f_2 = 10\sqrt{2} \) oscillations/minute.
- Convert this frequency to Hertz:
\[ f_2 = \frac{10\sqrt{2}}{60} \text{ Hz} = \frac{\sqrt{2}}{6} \text{ Hz} \]

4. **Setting Up the New Frequency Equation:**
- The new frequency \( f_2 \) is influenced by both the Earth's magnetic field \( B_H \) and the magnetic field \( B \) of the bar magnet:
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{m (B_H + B)}{I}} \]

5. **Relating the Two Frequencies:**
- From the two frequencies, we can set up the ratio:
\[ \frac{f_1}{f_2} = \frac{B_H}{B_H + B} \]
- Substituting the values:
\[ \frac{\frac{1}{6}}{\frac{\sqrt{2}}{6}} = \frac{B_H}{B_H + B} \]
- This simplifies to:
\[ \frac{1}{\sqrt{2}} = \frac{B_H}{B_H + B} \]

6. **Solving for \( B \):**
- Cross-multiplying gives:
\[ B_H + B = \sqrt{2} B_H \]
- Rearranging gives:
\[ B = (\sqrt{2} - 1) B_H \]

7. **Considering the Bar Magnet's Orientation:**
- When the bar magnet is turned end to end, its magnetic field \( B \) reverses direction.
- The effective magnetic field acting on the small magnet becomes:
\[ B_H - B \]
- Substituting \( B \):
\[ B_H - (\sqrt{2} - 1) B_H = (2 - \sqrt{2}) B_H \]

8. **Finding the New Frequency:**
- The new frequency \( f_3 \) when the bar magnet is turned end to end is:
\[ f_3 = \frac{1}{2\pi} \sqrt{\frac{m ((2 - \sqrt{2}) B_H)}{I}} \]
- This leads to a frequency of oscillation of zero because the effective magnetic field becomes zero:
\[ f_3 = 0 \text{ Hz} \]

### Conclusion:
Thus, when the bar magnet is turned around end to end, the rate of oscillation of the small magnet will become **0 oscillations/minute**.