A deflection magnetometer is placed with its arm along the east-west direction (tan A position) and a short bar magnet is placed symmetrically along its axis at some distance with its north pole pointing towards east. In this position the needle of the magnetometer shows a deflection of `60^(@)`. If we double the distance of the bar magnet, then the deflection will be

Options

A. `sin^(-1)[(sqrt3)/(8)]`
B. `cos^(-1)[(sqrt3)/(8)]`
C. `tan^(-1)[(sqrt3)/(8)]`
D. `cot^(-1)[(sqrt3)/(8)]`

Answer

To solve the problem, we will use the relationship between the deflection angle of the magnetometer and the distance from the bar magnet. The key principle here is that the tangent of the deflection angle is inversely proportional to the cube of the distance from the magnet.

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- The deflection magnetometer is placed in the east-west direction.
- A short bar magnet is placed such that its north pole points east, causing the needle to deflect by \(60^\circ\).

2. **Apply the Tangent Law**:
- According to the tangent law, we have:
\[ \tan A \propto \frac{1}{d^3} \]
- Here, \(A\) is the angle of deflection, and \(d\) is the distance from the magnet.

3. **Set Up the Initial Condition**:
- Let \(d_1\) be the initial distance from the magnet where the deflection is \(60^\circ\).
- Thus, we can write:
\[ \tan(60^\circ) = \sqrt{3} \]
- Therefore, we have:
\[ \tan(60^\circ) \propto \frac{1}{d_1^3} \implies \sqrt{3} \propto \frac{1}{d_1^3} \]

4. **Change the Distance**:
- Now, we double the distance, so \(d_2 = 2d_1\).
- We need to find the new angle of deflection \(A_2\) when the distance is doubled.

5. **Apply the New Condition**:
- Using the relationship again:
\[ \tan A_2 \propto \frac{1}{d_2^3} = \frac{1}{(2d_1)^3} = \frac{1}{8d_1^3} \]

6. **Relate the Two Conditions**:
- From the initial condition, we know:
\[ \tan A_1 = \sqrt{3} \propto \frac{1}{d_1^3} \]
- Therefore, we can relate the two angles:
\[ \frac{\tan A_2}{\tan A_1} = \frac{1/8d_1^3}{1/d_1^3} = \frac{1}{8} \]
- This implies:
\[ \tan A_2 = \frac{\tan A_1}{8} = \frac{\sqrt{3}}{8} \]

7. **Calculate the New Angle**:
- To find \(A_2\), we take the arctangent:
\[ A_2 = \tan^{-1}\left(\frac{\sqrt{3}}{8}\right) \]

### Final Answer:
The new deflection angle when the distance is doubled is:
\[ A_2 = \tan^{-1}\left(\frac{\sqrt{3}}{8}\right) \]