A voltmeter has a range `O - V` with a series resistance `R`. With a series resistance `2R`, the range is `O - V'`. The correct relation between `V` and `V'`is

Options

A. V'=2V
B. V'> 2 V
C. V' >> 2V
D. V' <2V

Answer

To solve the problem, we need to establish the relationship between the voltages \( V \) and \( V' \) when the voltmeter is connected with different series resistances.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A voltmeter is connected in series with a resistance \( R \) and measures a voltage \( V \).
- When the series resistance is changed to \( 2R \), the voltmeter measures a new voltage \( V' \).

2. **Current through the Voltmeter**:
- Let \( I_g \) be the current flowing through the circuit when the voltmeter is connected with resistance \( R \).
- The relationship for the current \( I_g \) when connected with resistance \( R \) is given by:
\[ I_g = \frac{V}{R + G} \]
where \( G \) is the internal resistance of the voltmeter.

3. **Current with Series Resistance \( 2R \)**:
- When the resistance is changed to \( 2R \), the current \( I'_g \) can be expressed as:
\[ I'_g = \frac{V'}{2R + G} \]

4. **Equating the Currents**:
- Since the same current flows through both configurations, we have:
\[ I_g = I'_g \]
- Therefore, we can set the equations equal to each other:
\[ \frac{V}{R + G} = \frac{V'}{2R + G} \]

5. **Cross Multiplying**:
- Cross multiplying gives:
\[ V(2R + G) = V'(R + G) \]

6. **Expanding and Rearranging**:
- Expanding both sides:
\[ 2VR + VG = V'R + V'G \]
- Rearranging gives:
\[ 2VR - V'R = V'G - VG \]
- Factoring out common terms:
\[ R(2V - V') = G(V' - V) \]

7. **Finding the Ratio**:
- Dividing both sides by \( R \) and rearranging gives:
\[ \frac{V'}{V} = \frac{2R + G}{R + G} \]

8. **Final Relation**:
- Thus, we can derive the relationship:
\[ V' = \frac{2V(R + G)}{2R + G} \]

### Conclusion:
The correct relation between \( V \) and \( V' \) is:
\[ V' < 2V \]