When a current of 1A passes through a tangent galvanometer, angle of defection is deg 60. When current is reduced to 0.5A, then the new angle of deflection will be

Options

A. `deg30`
B. `deg45`
C. `deg60`
D. `deg15`

Answer

To solve the problem, we need to use the relationship between the angle of deflection (θ) and the current (I) in a tangent galvanometer, which is given by the equation:

\[ \theta = k \cdot I \]

where \( k \) is a constant specific to the galvanometer.

### Step-by-Step Solution:

1. **Identify the given values**:
- Current \( I_1 = 1 \, \text{A} \)
- Angle of deflection \( \theta_1 = 60^\circ \)

2. **Convert the angle from degrees to radians**:
- To use the formula, we need to convert \( 60^\circ \) to radians:
\[ \theta_1 = 60^\circ = \frac{\pi}{3} \, \text{radians} \]

3. **Substitute the values into the equation to find \( k \)**:
- Using the first case:
\[ \frac{\pi}{3} = k \cdot 1 \implies k = \frac{\pi}{3} \]

4. **Now consider the second case with reduced current**:
- Current \( I_2 = 0.5 \, \text{A} \)

5. **Use the same formula to find the new angle of deflection \( \theta_2 \)**:
\[ \theta_2 = k \cdot I_2 = \frac{\pi}{3} \cdot 0.5 = \frac{\pi}{6} \]

6. **Convert \( \theta_2 \) back to degrees**:
- To convert \( \frac{\pi}{6} \) radians to degrees:
\[ \theta_2 = \frac{\pi}{6} \times \frac{180}{\pi} = 30^\circ \]

### Final Answer:
The new angle of deflection when the current is reduced to 0.5 A is \( 30^\circ \).

---