A galvanometer of 50 ohm resistance has 25 divisions. A current of `4xx10^(-4)` ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of :

Options

A. `2500 Omega` as a shunt
B. `2450 Omega` as a shunt
C. `2550 Omega` in series
D. `2450 Omega` in series

Explanation

To solve the problem of converting a galvanometer into a voltmeter, we need to follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Resistance of the galvanometer (G) = 50 ohms
- Number of divisions = 25
- Current for 1 division = \(4 \times 10^{-4}\) A

2. **Calculate Maximum Current for Full Deflection**:
- The maximum current (I_max) for full deflection (25 divisions) can be calculated as:
\[ I_{\text{max}} = 25 \times (4 \times 10^{-4}) = 100 \times 10^{-4} = 10^{-2} \text{ A} = 0.01 \text{ A} \]

3. **Determine the Voltage Range**:
- We want to convert this galvanometer into a voltmeter with a maximum voltage range of 25 volts.

4. **Use Ohm's Law**:
- According to Ohm's Law, the voltage (V) across a resistor is given by:
\[ V = I \times R \]
- Here, the total resistance in the circuit when the galvanometer is connected in series with an additional resistance (R) is \(R + 50 \, \text{ohms}\).

5. **Set Up the Equation**:
- For the voltmeter to read 25 volts at maximum current:
\[ 25 = I_{\text{max}} \times (R + 50) \]
- Substituting \(I_{\text{max}} = 0.01 \, \text{A}\):
\[ 25 = 0.01 \times (R + 50) \]

6. **Solve for R**:
- Rearranging the equation gives:
\[ R + 50 = \frac{25}{0.01} = 2500 \]
\[ R = 2500 - 50 = 2450 \, \text{ohms} \]

### Final Answer:
The resistance that should be connected in series with the galvanometer to convert it into a voltmeter with a range of 25 volts is **2450 ohms**.

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