A dip needle vibrates in a vertical plane with time period of 3 second . The same needle is suspended horizontally and made to vibrate is a horizontal plane . The time period is again 3 sec. The angle of dip at that place is .

Options

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

Explanation

To solve the problem step by step, we will analyze the given information about the dip needle and apply the relevant physics concepts.

### Step 1: Understand the Problem
We are given that a dip needle vibrates in both vertical and horizontal planes with the same time period of 3 seconds. We need to find the angle of dip (θ) at that location.

### Step 2: Use the Formula for Time Period
The time period (T) of a magnetic needle suspended in a magnetic field is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{MB}} \]
where:
- \( I \) is the moment of inertia of the needle,
- \( M \) is the magnetic moment of the needle,
- \( B \) is the magnetic field strength.

### Step 3: Set Up Equations for Vertical and Horizontal Vibrations
1. For vertical vibrations (when the needle is vibrating in a vertical plane), we denote the vertical component of the Earth's magnetic field as \( B_v \):
\[ T_1 = 2\pi \sqrt{\frac{I}{M B_v}} \]
Given \( T_1 = 3 \) seconds, we can write:
\[ 3 = 2\pi \sqrt{\frac{I}{M B_v}} \] (Equation 1)

2. For horizontal vibrations (when the needle is vibrating in a horizontal plane), we denote the horizontal component of the Earth's magnetic field as \( B_h \):
\[ T_2 = 2\pi \sqrt{\frac{I}{M B_h}} \]
Given \( T_2 = 3 \) seconds, we can write:
\[ 3 = 2\pi \sqrt{\frac{I}{M B_h}} \] (Equation 2)

### Step 4: Divide the Two Equations
Now, we divide Equation 1 by Equation 2:
\[ \frac{T_1}{T_2} = \frac{\sqrt{B_h}}{\sqrt{B_v}} \]
Substituting \( T_1 = T_2 = 3 \):
\[ \frac{3}{3} = \frac{\sqrt{B_h}}{\sqrt{B_v}} \implies 1 = \frac{\sqrt{B_h}}{\sqrt{B_v}} \]
Squaring both sides gives:
\[ 1 = \frac{B_h}{B_v} \]
This implies:
\[ B_h = B_v \]

### Step 5: Relate the Components to the Angle of Dip
The angle of dip \( \theta \) is related to the vertical and horizontal components of the magnetic field by the formula:
\[ \tan \theta = \frac{B_v}{B_h} \]
Since we found \( B_h = B_v \), we can substitute:
\[ \tan \theta = \frac{B_v}{B_v} = 1 \]

### Step 6: Calculate the Angle of Dip
Now, we can find \( \theta \):
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Conclusion
The angle of dip at that place is \( 45^\circ \).