The resistance of a galvanometer is `10 Omega`. It gives full-scale deflections when `1 mA` current is passed. The resistance connected in series for converting it into a voltmeter of `2.5 V` will be

Options

A. `24.9 Omega`
B. `249 Omega`
C. `2490Omega`
D. `24900 Omega`

Explanation

To solve the problem of finding the resistance that needs to be connected in series with a galvanometer to convert it into a voltmeter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer, \( G = 10 \, \Omega \)
- Full-scale deflection current, \( I_G = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \)
- Maximum voltage to be measured, \( V = 2.5 \, \text{V} \)

2. **Understand the Configuration:**
- When converting a galvanometer into a voltmeter, a high resistance \( R \) is connected in series with the galvanometer. The total voltage across the combination should equal the maximum voltage \( V \).

3. **Use the Formula:**
- The formula for the resistance \( R \) to be connected in series is given by:
\[ R = \frac{V}{I_G} - G \]
- Here, \( \frac{V}{I_G} \) gives the total resistance needed to allow the full-scale deflection at the given voltage.

4. **Calculate \( \frac{V}{I_G} \):**
- Substitute the values into the formula:
\[ \frac{V}{I_G} = \frac{2.5 \, \text{V}}{1 \times 10^{-3} \, \text{A}} = 2500 \, \Omega \]

5. **Substitute into the Resistance Formula:**
- Now substitute \( \frac{V}{I_G} \) and \( G \) into the formula for \( R \):
\[ R = 2500 \, \Omega - 10 \, \Omega = 2490 \, \Omega \]

6. **Conclusion:**
- The resistance that needs to be connected in series with the galvanometer to convert it into a voltmeter of 2.5 V is:
\[ R = 2490 \, \Omega \]

### Final Answer:
The resistance connected in series for converting the galvanometer into a voltmeter of 2.5 V will be **2490 Ω**.