A galvanometer has coil of resistance `50Omega` and shows full deflection at `100muA`. The resistance to be added for the galvanometer to work as an ammeter of range 10mA is nearly

Options

A. `0.5Omega` in series
B. `0.5Omega` in parallel
C. `5.0Omega` series
D. `5.0Omega` in parallel

Answer

To solve the problem, we need to determine the resistance that must be added in parallel to a galvanometer in order to convert it into an ammeter with a specified range. Here are the steps to find the required resistance:

### Step-by-Step Solution:

1. **Understand the Given Values:**
- Resistance of the galvanometer (R_g) = 50 Ω
- Full deflection current (I_g) = 100 µA = 100 × 10^(-6) A
- Desired range of the ammeter (I_max) = 10 mA = 10 × 10^(-3) A

2. **Calculate the Current through the Added Resistance:**
- The current flowing through the added resistance (I_r) can be calculated as:
\[ I_r = I_{max} - I_g = 10 \times 10^{-3} A - 100 \times 10^{-6} A \]
- Since 10 mA is much larger than 100 µA, we can approximate:
\[ I_r \approx 10 \times 10^{-3} A \]

3. **Apply Ohm's Law:**
- The potential difference (V) across the galvanometer and the added resistance is the same. According to Ohm's Law:
\[ V = I_g \cdot R_g = I_r \cdot R \]
- Substituting the values:
\[ 100 \times 10^{-6} A \cdot 50 Ω = (10 \times 10^{-3} A) \cdot R \]

4. **Rearranging the Equation:**
- Rearranging gives:
\[ R = \frac{100 \times 10^{-6} A \cdot 50 Ω}{10 \times 10^{-3} A} \]

5. **Calculating R:**
- Performing the calculation:
\[ R = \frac{5000 \times 10^{-6}}{10 \times 10^{-3}} = \frac{5000}{10} \times 10^{-3} = 500 \times 10^{-3} = 0.5 Ω \]

6. **Conclusion:**
- The resistance to be added in parallel to the galvanometer to convert it into an ammeter with a range of 10 mA is approximately **0.5 Ω**.

### Final Answer:
The resistance to be added is **0.5 Ω**.