A magnetic needle vibrates in a vertical plane parallel to the magnetic meridlian about a horizontal axis passing through its centre. Its frequency is `pi`. If the plane of oscillation is turned about a vertical axis by `90^(@)`, the frequency of oscillation in vertical plane will be :

Options

A. `pi`
B. zero
C. less than `pi`
D. more than `pi`

Answer

To solve the problem, we need to analyze the effect of changing the orientation of the magnetic needle on its frequency of oscillation.

### Step-by-Step Solution:

1. **Understanding the Frequency of Oscillation**:
The frequency of oscillation \( f \) of a magnetic needle can be expressed as:
\[ f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{M B \cos \phi}{I}} \]
where:
- \( M \) = magnetic moment of the needle,
- \( B \) = magnetic field strength,
- \( \phi \) = angle of dip,
- \( I \) = moment of inertia of the needle.

2. **Given Frequency**:
We are given that the initial frequency of oscillation is \( f = \pi \).

3. **Effect of Turning the Plane of Oscillation**:
When the plane of oscillation is turned by \( 90^\circ \), the angle of dip \( \phi \) changes. The new angle of dip \( \phi' \) will be different from the original angle \( \phi \).

4. **Analyzing the Change in Cosine**:
The frequency is proportional to \( \sqrt{\cos \phi} \). When the angle is turned by \( 90^\circ \):
- The new angle \( \phi' \) will be greater than \( \phi \) (since the magnetic needle is now in a different orientation).
- Consequently, \( \cos \phi' < \cos \phi \).

5. **Conclusion on Frequency**:
Since the frequency is proportional to \( \sqrt{\cos \phi} \), if \( \cos \phi' < \cos \phi \), it follows that:
\[ f' < f \]
This means the new frequency of oscillation \( f' \) will be less than the original frequency \( f \).

6. **Final Answer**:
Therefore, the frequency of oscillation in the vertical plane after turning the plane of oscillation by \( 90^\circ \) will be less than \( \pi \).