At a place, a freely suspended magnet shows a dip angle of `30^(@)`. If the plane of the magnet is at an angle of `45^(@)` from the magnetic meridian, then the real dip of the magnet will be

Options

A. `cot^ (-1) ((sqrt(3))/(2))`
B. `cot^(-1) sqrt(3)`
C. `cot^(-1) ((1)/(sqrt(6)))`
D. `cot^(-1) sqrt(6)`

Answer

To solve the problem, we need to find the real dip angle of a freely suspended magnet that shows a dip angle of \(30^\circ\) while being at an angle of \(45^\circ\) from the magnetic meridian.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The dip angle of the magnet (\(\delta'\)) is given as \(30^\circ\).
- The angle of the magnet with respect to the magnetic meridian (\(\theta\)) is \(45^\circ\).
- We need to find the real dip angle (\(\delta\)).

2. **Using the Relationship Between Dip Angles**:
- The relationship between the dip angles can be expressed using the tangent function:
\[ \tan(\delta') = \frac{V}{H} \]
- Here, \(V\) is the vertical component of the magnetic field, and \(H\) is the horizontal component.

3. **Adjusting for the Angle with the Magnetic Meridian**:
- When the magnet is tilted at an angle \(\theta\) from the magnetic meridian, the horizontal component of the magnetic field changes while the vertical component remains the same.
- The modified relationship becomes:
\[ \tan(\delta') = \frac{V}{H \cos(\theta)} \]

4. **Substituting the Known Values**:
- We substitute \(\delta' = 30^\circ\) and \(\theta = 45^\circ\):
\[ \tan(30^\circ) = \frac{V}{H \cos(45^\circ)} \]
- We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\) and \(\cos(45^\circ) = \frac{1}{\sqrt{2}}\):
\[ \frac{1}{\sqrt{3}} = \frac{V}{H \cdot \frac{1}{\sqrt{2}}} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ V = \frac{H}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}} = \frac{H}{\sqrt{6}} \]

6. **Finding the Real Dip Angle**:
- Now we can express \(\tan(\delta)\):
\[ \tan(\delta) = \frac{V}{H} = \frac{H/\sqrt{6}}{H} = \frac{1}{\sqrt{6}} \]
- Therefore, we have:
\[ \delta = \tan^{-1}\left(\frac{1}{\sqrt{6}}\right) \]

7. **Final Result**:
- The real dip angle is:
\[ \delta = \cot^{-1}(\sqrt{6}) \]

### Conclusion:
The real dip of the magnet is \(\cot^{-1}(\sqrt{6})\).

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