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

Explanation

To solve the problem, we need to find the potential difference between two points separated by 50 cm on a 3-meter long wire that has a total resistance of 100 ohms when connected to a 6-volt battery.

### Step-by-step Solution:

1. **Identify the Total Length and Resistance**:
- The total length of the wire (L) = 3 meters = 300 cm.
- The total resistance (R) = 100 ohms.

2. **Determine the Voltage Across the Entire Wire**:
- The battery provides a voltage (V) = 6 volts across the entire length of the wire.

3. **Calculate the Resistance per Unit Length**:
- The resistance per meter (R_per_meter) can be calculated as:
\[ R_{\text{per meter}} = \frac{R}{L} = \frac{100 \, \Omega}{3 \, \text{m}} \approx 33.33 \, \Omega/\text{m} \]

4. **Calculate the Resistance for 50 cm**:
- The length of interest (l) = 50 cm = 0.5 meters.
- The resistance for this length (R_50cm) is:
\[ R_{50 \, \text{cm}} = R_{\text{per meter}} \times l = 33.33 \, \Omega/\text{m} \times 0.5 \, \text{m} = 16.67 \, \Omega \]

5. **Use Ohm's Law to Find the Voltage Across 50 cm**:
- The voltage across the 50 cm segment can be calculated using Ohm's Law:
\[ V_{50 \, \text{cm}} = \frac{R_{50 \, \text{cm}}}{R} \times V = \frac{16.67 \, \Omega}{100 \, \Omega} \times 6 \, \text{V} \]
- Simplifying this gives:
\[ V_{50 \, \text{cm}} = 0.1667 \times 6 \approx 1 \, \text{V} \]

6. **Conclusion**:
- The potential difference between the two points separated by 50 cm on the wire is approximately **1 volt**.