Two galvanometers A and B require 3 mA and 5 mA respectively to produce the same deflection of `i_(0)` division. Then

Options

A. A is more sensitive than B
B. B is more sensitive than A
C. A and B are equally sensitive
D. sensitiveness of B is 5/3 times of that of A

Answer

To solve the problem, we need to determine the sensitivity of the two galvanometers A and B based on the currents required to produce the same deflection.

### Step-by-Step Solution:

1. **Understanding Sensitivity**:
The sensitivity \( S \) of a galvanometer is defined as the current required to produce a unit deflection (1 division) in the galvanometer. Mathematically, it can be expressed as:
\[ S = \frac{1}{I} \]
where \( I \) is the current required for a unit deflection.

2. **Given Values**:
- For galvanometer A, the current \( I_A = 3 \, \text{mA} \).
- For galvanometer B, the current \( I_B = 5 \, \text{mA} \).

3. **Calculating Sensitivities**:
- The sensitivity of galvanometer A (\( S_A \)) can be calculated as:
\[ S_A = \frac{1}{I_A} = \frac{1}{3 \, \text{mA}} = \frac{1}{3} \, \text{mA}^{-1} \]
- The sensitivity of galvanometer B (\( S_B \)) can be calculated as:
\[ S_B = \frac{1}{I_B} = \frac{1}{5 \, \text{mA}} = \frac{1}{5} \, \text{mA}^{-1} \]

4. **Comparing Sensitivities**:
To compare the sensitivities \( S_A \) and \( S_B \):
\[ \frac{S_A}{S_B} = \frac{\frac{1}{3}}{\frac{1}{5}} = \frac{5}{3} \]
This means that the sensitivity of A is greater than the sensitivity of B.

5. **Conclusion**:
Since \( S_A > S_B \), we conclude that galvanometer A is more sensitive than galvanometer B.

### Final Answer:
Galvanometer A is more sensitive than galvanometer B.