A deflection magnetometer is adjusted in the usual way . When a magnet is introduced , the deflection observed is `theta = 60^(@)` and the period of oscillation of the needle in magnetometer is T . When the magnet is removed the time period of oscillation is `T_(0)` . The relation between T and `T_(0)` is

Options

A. `2 T^(2) = T_(0)^(2)`
B. `(2T^(2))/(sqrt3) = T_(0)^(2)`
C. `T = T_(0)`
D. `T^(2) = (T_(0)^(2))/(sqrt2)`

Answer

To solve the problem, we need to derive the relationship between the time periods \( T \) and \( T_0 \) of the deflection magnetometer when a magnet is introduced and when it is removed, respectively.

### Step-by-Step Solution:

1. **Understanding the Deflection Magnetometer**:
The deflection magnetometer measures the magnetic field produced by a magnet and the horizontal component of the Earth's magnetic field. When a magnet is introduced, it causes a deflection \( \theta \) in the needle.

2. **Magnetic Field Components**:
When the magnet is introduced, the net magnetic field \( B \) acting on the needle is given by:
\[ B = \sqrt{F^2 + H^2} \]
where \( F \) is the magnetic field due to the magnet and \( H \) is the horizontal component of the Earth's magnetic field.

3. **Time Period of Oscillation**:
The time period \( T \) of the oscillation of the needle when the magnet is present is given by:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot B}} \]
Substituting for \( B \):
\[ T = 2\pi \sqrt{\frac{I}{M \cdot \sqrt{F^2 + H^2}}} \]

4. **Removing the Magnet**:
When the magnet is removed, the magnetic field \( F \) becomes zero, and the time period \( T_0 \) is given by:
\[ T_0 = 2\pi \sqrt{\frac{I}{M \cdot H}} \]

5. **Relating \( F \) and \( H \) using \( \theta \)**:
For a deflection magnetometer, the relationship between \( F \) and \( H \) is given by:
\[ \frac{F}{H} = \tan(\theta) \]
Therefore, we can express \( F \) as:
\[ F = H \tan(\theta) \]

6. **Substituting \( F \) in the Expression for \( T \)**:
Substituting \( F \) in the equation for \( T \):
\[ T = 2\pi \sqrt{\frac{I}{M \cdot \sqrt{(H \tan(\theta))^2 + H^2}}} \]
Simplifying this, we get:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot H \sqrt{\tan^2(\theta) + 1}}} \]

7. **Using Trigonometric Identity**:
We know from trigonometric identities that:
\[ \tan^2(\theta) + 1 = \sec^2(\theta) \]
Thus, we can rewrite \( T \) as:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot H \cdot \sec(\theta)}} \]

8. **Relating \( T \) and \( T_0 \)**:
Since \( T_0 = 2\pi \sqrt{\frac{I}{M \cdot H}} \), we can express \( T \) in terms of \( T_0 \):
\[ T = T_0 \cdot \frac{1}{\sqrt{\sec(\theta)}} \]

9. **Substituting \( \theta = 60^\circ \)**:
For \( \theta = 60^\circ \), we have \( \sec(60^\circ) = \frac{2}{1} \):
\[ T = T_0 \cdot \frac{1}{\sqrt{2}} \]

10. **Squaring Both Sides**:
Squaring both sides gives:
\[ T^2 = \frac{T_0^2}{2} \]
Rearranging gives:
\[ 2T^2 = T_0^2 \]

### Final Relation:
The relation between \( T \) and \( T_0 \) is:
\[ 2T^2 = T_0^2 \]