When `2` amperes current is passed through a tangent galvanometer, it gives a deflection of `30^(@)`. For `60^(@)` deflection, the current must be

Options

A. `1 amp`
B. `2sqrt(3) amp`
C. `4 amp`
D. `6 amp`

Explanation

To solve the problem, we will use the relationship between the current passing through a tangent galvanometer and the angle of deflection it produces. The tangent galvanometer operates on the principle that the deflection angle (θ) is directly proportional to the current (I) flowing through the coil.

### Step-by-step Solution:

1. **Understand the relationship**: The deflection angle (θ) is related to the current (I) by the formula:
\[ \tan(\theta) \propto I \]
This means that if the angle of deflection doubles, the current must also change in a specific way.

2. **Set up the known values**:
- For the first case, we have:
- \( I_1 = 2 \, \text{A} \)
- \( \theta_1 = 30^\circ \)
- For the second case, we want to find \( I_2 \) when:
- \( \theta_2 = 60^\circ \)

3. **Use the tangent function**:
- We can express the relationship using the tangent function:
\[ \frac{I_1}{I_2} = \frac{\tan(\theta_1)}{\tan(\theta_2)} \]

4. **Calculate the tangent values**:
- Calculate \( \tan(30^\circ) \) and \( \tan(60^\circ) \):
- \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \)
- \( \tan(60^\circ) = \sqrt{3} \)

5. **Set up the equation**:
\[ \frac{2}{I_2} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} \]

6. **Simplify the equation**:
\[ \frac{2}{I_2} = \frac{1}{3} \]

7. **Cross-multiply to solve for \( I_2 \)**:
\[ 2 \cdot 3 = I_2 \implies I_2 = 6 \, \text{A} \]

8. **Conclusion**: The current required for a deflection of \( 60^\circ \) is \( 6 \, \text{A} \).

### Final Answer:
The current must be \( 6 \, \text{A} \) for a deflection of \( 60^\circ \).

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