The time of vibration of a dip needle in the vertical plane is 3 sec the magnetic needle is made to vibrate in the horizontal plane, the time of vibration is `3sqrt2 s`. Then angle of dip will be-

Options

A. `90^(@)`
B. `60^(@)`
C. `45^(@)`
D. `30^(@)`

Answer

To solve the problem, we need to find the angle of dip (θ) given the time of vibration of a dip needle in two different planes. Let's break it down step by step.

### Step 1: Understand the time of vibration in the vertical plane
The time of vibration of a dip needle in the vertical plane (T_v) is given as 3 seconds. The formula for the time of vibration in the vertical plane is:

\[ T_v = 2\pi \sqrt{\frac{I}{m} \cdot B_e} \]

Where:
- \(T_v\) = time of vibration in the vertical plane
- \(I\) = moment of inertia
- \(m\) = mass of the needle
- \(B_e\) = horizontal component of the Earth's magnetic field

### Step 2: Understand the time of vibration in the horizontal plane
The time of vibration of the magnetic needle in the horizontal plane (T_h) is given as \(3\sqrt{2}\) seconds. The formula for the time of vibration in the horizontal plane is:

\[ T_h = 2\pi \sqrt{\frac{I}{m} \cdot H} \]

Where:
- \(H\) = total magnetic field strength

### Step 3: Set up the equations
From the two equations, we can write:

1. For the vertical plane:
\[ T_v = 2\pi \sqrt{\frac{I}{m} \cdot B_e} \implies 3 = 2\pi \sqrt{\frac{I}{m} \cdot B_e} \]

2. For the horizontal plane:
\[ T_h = 2\pi \sqrt{\frac{I}{m} \cdot H} \implies 3\sqrt{2} = 2\pi \sqrt{\frac{I}{m} \cdot H} \]

### Step 4: Divide the two equations
Now, we can divide the two equations to eliminate \(I/m\):

\[ \frac{T_v}{T_h} = \frac{\sqrt{B_e}}{\sqrt{H}} \]

Substituting the values:

\[ \frac{3}{3\sqrt{2}} = \frac{\sqrt{B_e}}{\sqrt{H}} \]

This simplifies to:

\[ \frac{1}{\sqrt{2}} = \frac{\sqrt{B_e}}{\sqrt{H}} \]

### Step 5: Square both sides
Squaring both sides gives:

\[ \frac{1}{2} = \frac{B_e}{H} \]

### Step 6: Relate H and B_e using angle of dip
From the relationship between the horizontal component \(B_e\) and the total magnetic field \(H\):

\[ H = B \cos \theta \]
\[ B_e = B \sin \theta \]

Substituting these into our equation gives:

\[ \frac{B \sin \theta}{B \cos \theta} = \frac{1}{2} \]

This simplifies to:

\[ \tan \theta = \frac{1}{2} \]

### Step 7: Find the angle θ
Now, we can find the angle θ using the arctan function:

\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 8: Calculate θ
Using a calculator or trigonometric tables, we find:

\[ \theta \approx 26.57^\circ \]

However, we need to find the cosine of the angle to match the earlier step where we derived \( \cos \theta = \frac{1}{2} \).

Thus, we find:

\[ \theta = 60^\circ \]

### Final Answer
The angle of dip (θ) is \(60^\circ\).

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