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, we will follow these steps:

### Step 1: Understand the Time Period Formula
The time period \( T \) of a vibrating dip needle can be expressed as:
\[ T = 2\pi \sqrt{\frac{I}{mB}} \]
where:
- \( I \) is the moment of inertia of the needle,
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field strength.

### Step 2: Set Up the Equations for Vertical and Horizontal Vibrations
1. For vertical vibrations, the time period is given as \( T_V = 3 \) seconds. The magnetic field acting on the needle is the vertical component of the Earth's magnetic field, denoted as \( B_V \).
\[ T_V = 2\pi \sqrt{\frac{I}{mB_V}} \quad \text{(1)} \]

2. For horizontal vibrations, the time period is also \( T_H = 3 \) seconds. The magnetic field acting on the needle is the horizontal component of the Earth's magnetic field, denoted as \( B_H \).
\[ T_H = 2\pi \sqrt{\frac{I}{mB_H}} \quad \text{(2)} \]

### Step 3: Equate the Time Periods
Since both time periods are equal:
\[ T_V = T_H \]
Substituting the expressions from equations (1) and (2):
\[ 2\pi \sqrt{\frac{I}{mB_V}} = 2\pi \sqrt{\frac{I}{mB_H}} \]

### Step 4: Simplify the Equation
By squaring both sides and canceling common terms:
\[ \frac{I}{mB_V} = \frac{I}{mB_H} \]
This implies:
\[ B_H = B_V \]

### Step 5: Relate the Components of the Earth's Magnetic Field
The angle of dip \( \theta \) is defined such that:
\[ \tan \theta = \frac{B_V}{B_H} \]
Since we found that \( B_H = B_V \), we can substitute:
\[ \tan \theta = \frac{B_V}{B_V} = 1 \]

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

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

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