A galvanometer having a coil resistance `200 Omega` gives a full scale deflection when a current of 1mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 50 V?

Options

A. `19.9 k Omega`
B. `48.9 k Omega`
C. `49.8 k Omega`
D. `10 k Omega`

Explanation

To solve the problem, we need to determine the resistance that should be connected in series with the galvanometer to convert it into a voltmeter that gives a full-scale deflection for a potential difference of 50 V.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer, \( R_g = 200 \, \Omega \)
- Full-scale deflection current, \( I_g = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \)
- Desired voltage for full-scale deflection, \( V = 50 \, \text{V} \)

2. **Understand the Circuit Configuration:**
- The galvanometer (with resistance \( R_g \)) and the unknown resistance \( R \) are connected in series.
- The total voltage across both resistances when the full-scale deflection occurs is \( V = 50 \, \text{V} \).

3. **Apply Ohm's Law:**
- According to Ohm's law, the total voltage \( V \) across the series combination of resistances is given by:
\[ V = I \times R_{\text{total}} \]
- Where \( R_{\text{total}} = R_g + R \).

4. **Substitute Known Values:**
- We can substitute the known values into the equation:
\[ 50 = (1 \times 10^{-3}) \times (200 + R) \]

5. **Rearrange the Equation:**
- Rearranging gives:
\[ 50 = 0.001 \times (200 + R) \]
- Dividing both sides by \( 0.001 \):
\[ 50000 = 200 + R \]

6. **Solve for \( R \):**
- Now, isolate \( R \):
\[ R = 50000 - 200 = 49800 \, \Omega \]

7. **Convert to Kilohms:**
- To express \( R \) in kilohms:
\[ R = 49.8 \, \text{k}\Omega \]

### Final Answer:
The value of the resistance that can convert the galvanometer into a voltmeter giving full-scale deflection for a potential difference of 50 V is \( 49.8 \, \text{k}\Omega \).