The real angle of dip, if a magnet is suspended at an angle of `30^(@)` to the magnetic meridian and the dip needle makes an angle of `45^(@)` with horizontal, is:

Options

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

Answer

To find the real angle of dip when a magnet is suspended at an angle of \(30^\circ\) to the magnetic meridian and the dip needle makes an angle of \(45^\circ\) with the horizontal, we can follow these steps:

### Step 1: Understand the Components of the Magnetic Field
Let:
- \(B_v\) = vertical component of the Earth's magnetic field
- \(B_h\) = horizontal component of the Earth's magnetic field

The angle of dip \(\theta\) is given by:
\[ \tan \theta = \frac{B_v}{B_h} \]

### Step 2: Analyze the Situation
When the magnet is suspended at an angle of \(30^\circ\) to the magnetic meridian, the dip needle makes an angle of \(45^\circ\) with the horizontal.

### Step 3: Determine the Components of the Magnetic Field
In this case, the vertical component \(B_v\) remains the same, but the horizontal component \(B_h\) in the direction of the magnetic meridian is affected. The horizontal component in this case can be expressed as:
\[ B_h' = B_h \cos(30^\circ) \]

### Step 4: Set Up the Equation
From the geometry of the situation, we can write:
\[ \tan(45^\circ) = \frac{B_v}{B_h \cos(30^\circ)} \]
Since \(\tan(45^\circ) = 1\), we have:
\[ 1 = \frac{B_v}{B_h \cos(30^\circ)} \]

### Step 5: Rearranging the Equation
This can be rearranged to give:
\[ B_v = B_h \cos(30^\circ) \]

### Step 6: Substitute into the Original Equation
Now, substituting \(B_v\) into the original equation for \(\tan \theta\):
\[ \tan \theta = \frac{B_h \cos(30^\circ)}{B_h} \]
This simplifies to:
\[ \tan \theta = \cos(30^\circ) \]

### Step 7: Calculate \(\theta\)
Now, we know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Thus:
\[ \tan \theta = \frac{\sqrt{3}}{2} \]
To find \(\theta\), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \]

### Step 8: Find the Value of \(\theta\)
Calculating \(\tan^{-1}\left(\frac{\sqrt{3}}{2}\right)\) gives approximately:
\[ \theta \approx 40.9^\circ \]

### Conclusion
The real angle of dip is approximately \(40.9^\circ\).

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