A battery of 6 volts is connected ot the termainals of a three meter long wire of uniform thickness and resistance of the order of `100 Omega`. The difference of potential between two points separated by `50 cm` on the wire will

Options

A. 2V
B. 3V
C. 1V
D. 1.5V

Answer

To solve the problem, we need to find the potential difference between two points separated by 50 cm on a wire that is 3 meters long and connected to a 6-volt battery. The resistance of the wire is given as 100 ohms.

### Step-by-Step Solution:

1. **Understand the Total Length and Voltage**:
The total length of the wire is 3 meters (300 cm) and the total potential difference (voltage) across this length is 6 volts.

2. **Calculate the Potential Difference per Centimeter**:
To find the potential difference per centimeter, we can use the formula:
\[ \text{Potential difference per cm} = \frac{\text{Total Voltage}}{\text{Total Length}} = \frac{6 \text{ volts}}{300 \text{ cm}} = \frac{1}{50} \text{ volts/cm} \]

3. **Calculate the Potential Difference for 50 cm**:
Now, we need to find the potential difference across a length of 50 cm. We can multiply the potential difference per centimeter by the length of interest:
\[ \text{Potential difference for 50 cm} = \text{Potential difference per cm} \times 50 \text{ cm} = \left(\frac{1}{50} \text{ volts/cm}\right) \times 50 \text{ cm} = 1 \text{ volt} \]

4. **Conclusion**:
Therefore, the potential difference between the two points separated by 50 cm on the wire is 1 volt.

### Final Answer:
The potential difference between the two points separated by 50 cm on the wire is **1 volt**.

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