A bar magnet , held horizontally , is set into angular oscillations in the earth's magnetic field its time periods are `T_(1)` and `T_2` at two places where the angles of dip are `theta_(1)` and `theta_2` , respectively . The ratio of the resultant magnetic fields at these two places will be

Options

A. `T_1 sin theta_1 : T_2 sin theta_2`
B. `T_(1) cos theta_1 : T_2 cos theta_2`
C. `T_(2)^(2) sin theta_2 : T_(1)^(2) sin theta_1`
D. `T_(2)^(2) cos theta_2 : T_(1)^(2) cos theta_1`

Explanation

To solve the problem, we will analyze the relationship between the time periods of the oscillating bar magnet and the resultant magnetic fields at two different locations with different angles of dip.

### Step-by-Step Solution:

1. **Understanding the Time Period of a Bar Magnet:**
The time period \( T \) of a bar magnet oscillating in a magnetic field is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mB \cos \theta}} \]
where:
- \( I \) is the moment of inertia of the magnet,
- \( m \) is the magnetic moment of the magnet,
- \( B \) is the magnitude of the magnetic field,
- \( \theta \) is the angle of dip.

2. **Setting Up the Ratios for Two Locations:**
For two locations with angles of dip \( \theta_1 \) and \( \theta_2 \), we can write:
\[ T_1 = 2\pi \sqrt{\frac{I}{mB_1 \cos \theta_1}} \]
\[ T_2 = 2\pi \sqrt{\frac{I}{mB_2 \cos \theta_2}} \]

3. **Taking the Ratio of the Time Periods:**
Dividing the two equations gives:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2 \cos \theta_2}{B_1 \cos \theta_1}} \]

4. **Squaring Both Sides:**
Squaring both sides results in:
\[ \left(\frac{T_1}{T_2}\right)^2 = \frac{B_2 \cos \theta_2}{B_1 \cos \theta_1} \]

5. **Rearranging to Find the Ratio of Magnetic Fields:**
Rearranging gives:
\[ \frac{B_1}{B_2} = \frac{T_2^2 \cos \theta_2}{T_1^2 \cos \theta_1} \]

6. **Final Result:**
Therefore, the ratio of the resultant magnetic fields at the two places is:
\[ \frac{B_1}{B_2} = \frac{T_2^2 \cos \theta_2}{T_1^2 \cos \theta_1} \]

### Conclusion:
From the options provided, the correct answer corresponds to option D:
\[ \frac{B_1}{B_2} = \frac{T_2^2 \cos \theta_2}{T_1^2 \cos \theta_1} \]