A thin magnetic needle vibrates in the horizontal plane with a period of 4s. The needle is cut into two halves by a plane normal to the magnetic axis of the needle. Then, the period of vibration of each half needle is approximately

Options

A. 4 s
B. 2 s
C. 8 s
D. 1 s

Answer

To solve the problem, we need to determine the new period of vibration of each half of the magnetic needle after it has been cut into two equal halves.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The original period of the magnetic needle (T) is given as 4 seconds.
- The needle is cut into two equal halves by a plane normal to its magnetic axis.

2. **Formula for the Period of Vibration:**
The period of vibration (T) of a magnetic needle is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mB}} \]
where:
- \( I \) is the moment of inertia,
- \( m \) is the magnetic moment,
- \( B \) is the horizontal magnetic field.

3. **Calculate the Moment of Inertia for the Original Needle:**
For a thin rod (the needle) of length \( L \) and mass \( m \), the moment of inertia about an axis through one end is:
\[ I = \frac{1}{3} m L^2 \]
However, since the magnetic needle is vibrating about its center, we use:
\[ I = \frac{1}{12} m L^2 \]

4. **Cutting the Needle:**
When the needle is cut into two equal halves, each half will have:
- Length = \( \frac{L}{2} \)
- Mass = \( \frac{m}{2} \)

5. **Calculate the Moment of Inertia for Each Half:**
The moment of inertia for each half about its center is:
\[ I' = \frac{1}{12} \left(\frac{m}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{12} \cdot \frac{m}{2} \cdot \frac{L^2}{4} = \frac{m L^2}{96} \]

6. **New Period of Vibration for Each Half:**
Substitute \( I' \) into the period formula:
\[ T' = 2\pi \sqrt{\frac{I'}{mB}} = 2\pi \sqrt{\frac{\frac{mL^2}{96}}{mB}} = 2\pi \sqrt{\frac{L^2}{96B}} \]

7. **Relate the New Period to the Original Period:**
Since the original period \( T = 2\pi \sqrt{\frac{L^2}{12B}} \), we can relate \( T' \) to \( T \):
\[ T' = T \cdot \sqrt{\frac{12}{96}} = T \cdot \sqrt{\frac{1}{8}} = T \cdot \frac{1}{\sqrt{8}} = T \cdot \frac{1}{2\sqrt{2}} \]
Given \( T = 4 \) seconds:
\[ T' = 4 \cdot \frac{1}{2\sqrt{2}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 2.828 \text{ seconds} \]

8. **Final Approximation:**
The approximate period of vibration for each half of the needle is:
\[ T' \approx 2.828 \text{ seconds} \approx 2 \text{ seconds (to the nearest whole number)} \]

### Conclusion:
The period of vibration of each half needle is approximately **2 seconds**.