The period of oscillation of compass needle is 8 s at a place where dip angle is `30^(@)` and magnetic field is `B_(1)` . At another place where dip angle is `60^(@)` and magnetic field is `B_(2)` , the period of oscillation is 4 is then `(B_(2))/(B_(1))` is

Options

A. `4sqrt3`
B. `3sqrt3`
C. `(5)/(7)`
D. `(8)/(7)`

Answer

To solve the problem, we need to find the ratio \( \frac{B_2}{B_1} \) using the given information about the periods of oscillation and the angles of dip.

### Step-by-step Solution:

1. **Understand the formula for the period of oscillation**:
The period \( T \) of a compass needle is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mB_h}} \]
where \( I \) is the moment of inertia, \( m \) is the magnetic moment, and \( B_h \) is the horizontal component of the magnetic field.

2. **Relate the horizontal component of the magnetic field to the dip angle**:
The horizontal component \( B_h \) can be expressed in terms of the total magnetic field \( B \) and the dip angle \( \theta \):
\[ B_h = B \cos(\theta) \]
Therefore, for the two cases:
- For the first case (dip angle \( 30^\circ \)):
\[ B_{h1} = B_1 \cos(30^\circ) = B_1 \cdot \frac{\sqrt{3}}{2} \]
- For the second case (dip angle \( 60^\circ \)):
\[ B_{h2} = B_2 \cos(60^\circ) = B_2 \cdot \frac{1}{2} \]

3. **Set up the equations for the periods**:
Using the periods given in the problem:
- For the first case:
\[ T_1 = 2\pi \sqrt{\frac{I}{m B_{h1}}} = 8 \text{ s} \]
- For the second case:
\[ T_2 = 2\pi \sqrt{\frac{I}{m B_{h2}}} = 4 \text{ s} \]

4. **Divide the two equations**:
Dividing the first equation by the second:
\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{I}{m B_{h1}}}}{2\pi \sqrt{\frac{I}{m B_{h2}}}} = \frac{\sqrt{B_{h2}}}{\sqrt{B_{h1}}} \]
This simplifies to:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_{h2}}{B_{h1}}} \]

5. **Substitute the known values**:
Substituting \( T_1 = 8 \) s and \( T_2 = 4 \) s:
\[ \frac{8}{4} = \sqrt{\frac{B_{h2}}{B_{h1}}} \implies 2 = \sqrt{\frac{B_{h2}}{B_{h1}}} \]
Squaring both sides gives:
\[ 4 = \frac{B_{h2}}{B_{h1}} \]

6. **Substitute the expressions for \( B_{h1} \) and \( B_{h2} \)**:
\[ 4 = \frac{B_2 \cdot \frac{1}{2}}{B_1 \cdot \frac{\sqrt{3}}{2}} \implies 4 = \frac{B_2}{B_1 \sqrt{3}} \]

7. **Rearranging gives the ratio**:
\[ B_2 = 4B_1 \sqrt{3} \]
Therefore, the ratio \( \frac{B_2}{B_1} \) is:
\[ \frac{B_2}{B_1} = 4\sqrt{3} \]

### Final Answer:
\[ \frac{B_2}{B_1} = 4\sqrt{3} \]