The maximum potential that can be measured with a voltmeter of resistance 1000 `Omega ` is 6V. Resistance that must be connected to measure a potential of 30V with it is

Options

A. `4000 Omega` in series
B. `6000 Omega ` in series
C. `12000 Omega` in series
D. `2000 Omega` in series

Explanation

To solve the problem, we need to determine the resistance that must be connected to a voltmeter in order to measure a potential of 30V, given that the voltmeter can measure a maximum potential of 6V with a resistance of 1000 ohms.

### Step-by-Step Solution:

1. **Understand the given values**:
- Maximum potential (V1) = 6V
- Resistance of the voltmeter (R1) = 1000 ohms
- Desired potential (V2) = 30V

2. **Use the relationship between the voltages and resistances**:
The relationship can be expressed as:
\[ \frac{V1}{V2} = \frac{R1}{R1 + R2} \]
Here, R2 is the resistance that we need to find.

3. **Substitute the known values into the equation**:
\[ \frac{6}{30} = \frac{1000}{1000 + R2} \]

4. **Cross-multiply to eliminate the fraction**:
\[ 6(1000 + R2) = 30 \times 1000 \]

5. **Expand and simplify the equation**:
\[ 6000 + 6R2 = 30000 \]
\[ 6R2 = 30000 - 6000 \]
\[ 6R2 = 24000 \]

6. **Solve for R2**:
\[ R2 = \frac{24000}{6} = 4000 \text{ ohms} \]

7. **Conclusion**:
The resistance that must be connected to measure a potential of 30V with the voltmeter is **4000 ohms**.