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

Explanation

To solve the problem, we will use the relationship between the current (I) and the tangent of the angle of deflection (θ) in a tangent galvanometer. The relationship can be expressed as:

\[ I \propto \tan(\theta) \]

This means that the current is directly proportional to the tangent of the angle of deflection. Let's denote the current corresponding to the angle θ as I and the proportionality constant as k.

### Step-by-Step Solution:

1. **Establish the relationship:**
\[ I = k \cdot \tan(\theta) \]

2. **Use the known values:**
For the first case, when the current \( I_1 = 2 \) A and the deflection \( \theta_1 = 30^\circ \):
\[ 2 = k \cdot \tan(30^\circ) \]

3. **Calculate \( \tan(30^\circ) \):**
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

4. **Substitute \( \tan(30^\circ) \) into the equation:**
\[ 2 = k \cdot \frac{1}{\sqrt{3}} \]

5. **Solve for k:**
\[ k = 2 \sqrt{3} \]

6. **Now, find the current for \( \theta_2 = 60^\circ \):**
Using the same relationship:
\[ I_2 = k \cdot \tan(60^\circ) \]

7. **Calculate \( \tan(60^\circ) \):**
We know that:
\[ \tan(60^\circ) = \sqrt{3} \]

8. **Substitute \( k \) and \( \tan(60^\circ) \) into the equation:**
\[ I_2 = (2 \sqrt{3}) \cdot \sqrt{3} \]

9. **Simplify the expression:**
\[ I_2 = 2 \cdot 3 = 6 \text{ A} \]

### Final Answer:
The current required for a deflection of \( 60^\circ \) is **6 amperes**.

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