A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2 sec. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2 seconds. Then the angle of dip is

Options

A. `0^(@)`
B. `30^(@)`
C. `45^(@)`
D. `90^(@)`

Explanation

To solve the problem, we need to analyze the behavior of a dip needle in both vertical and horizontal planes and relate it to the angle of dip.

### Step-by-Step Solution:

1. **Understanding the Dip Needle**: A dip needle is a magnetized needle that can pivot freely. It aligns itself with the Earth's magnetic field. The angle of dip (δ) is the angle that the needle makes with the horizontal plane.

2. **Time Period of Vibration**: The time period (T) of a dip needle vibrating in a magnetic field can be expressed as:
\[ T = 2\pi \sqrt{\frac{I}{mB}} \]
where:
- \(I\) = moment of inertia of the needle,
- \(m\) = magnetic moment,
- \(B\) = magnetic field strength.

3. **Case 1 - Vertical Plane**: When the dip needle vibrates in the vertical plane, the effective magnetic field acting on it is \(B \cos(δ)\), where δ is the angle of dip. Thus, the time period in the vertical plane is given by:
\[ T_v = 2\pi \sqrt{\frac{I}{mB \cos(δ)}} \]

4. **Case 2 - Horizontal Plane**: When the dip needle vibrates in the horizontal plane, the effective magnetic field is \(B \sin(δ)\). The time period in the horizontal plane is:
\[ T_h = 2\pi \sqrt{\frac{I}{mB \sin(δ)}} \]

5. **Equating Time Periods**: According to the problem, the time periods for both cases are the same:
\[ T_v = T_h \]
This leads to the equation:
\[ 2\pi \sqrt{\frac{I}{mB \cos(δ)}} = 2\pi \sqrt{\frac{I}{mB \sin(δ)}} \]

6. **Simplifying the Equation**: We can cancel out the common terms \(2\pi\) and \(\sqrt{\frac{I}{mB}}\) from both sides:
\[ \frac{1}{\cos(δ)} = \frac{1}{\sin(δ)} \]

7. **Cross-Multiplying**: This gives us:
\[ \sin(δ) = \cos(δ) \]

8. **Finding the Angle of Dip**: The equation \(\sin(δ) = \cos(δ)\) implies:
\[ \tan(δ) = 1 \]
Therefore, the angle of dip \(δ\) is:
\[ δ = 45^\circ \]

### Final Answer:
The angle of dip is \(45^\circ\).

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