A 6 V battery is connected to the terminals of a 3 m long wire of uniform thickness and resistance of `100 Omega`. The difference of potential between two points on the wire separated by a distance of 50 cm will be

Options

A. 2 V
B. 3 V
C. 1 V
D. 15 V

Answer

To solve the problem, we need to find the potential difference between two points on a wire that is connected to a 6 V battery. The wire has a length of 3 m and a total resistance of 100 Ω. We are interested in the potential difference across a segment of the wire that is 50 cm long.

### Step-by-Step Solution:

1. **Determine the total resistance of the wire:**
The total resistance \( R \) of the wire is given as \( 100 \, \Omega \).

2. **Calculate the current flowing through the circuit:**
Using Ohm's law, we can find the current \( I \) flowing through the circuit when a voltage \( V \) is applied across the resistance \( R \):
\[ I = \frac{V}{R} = \frac{6 \, \text{V}}{100 \, \Omega} = 0.06 \, \text{A} \]

3. **Calculate the resistance of the 50 cm segment of the wire:**
The wire is 3 m long, and we need to find the resistance of a 50 cm (0.5 m) segment. The resistance of a segment of wire is proportional to its length:
\[ R_{segment} = R \times \frac{\text{length of segment}}{\text{total length}} = 100 \, \Omega \times \frac{0.5 \, \text{m}}{3 \, \text{m}} = \frac{100 \times 0.5}{3} = \frac{50}{3} \, \Omega \]

4. **Calculate the potential drop across the 50 cm segment:**
The potential drop \( V_{drop} \) across the segment can be calculated using Ohm's law:
\[ V_{drop} = I \times R_{segment} = 0.06 \, \text{A} \times \frac{50}{3} \, \Omega \]
Simplifying this:
\[ V_{drop} = 0.06 \times \frac{50}{3} = \frac{3}{3} = 1 \, \text{V} \]

5. **Conclusion:**
The potential difference between the two points on the wire separated by a distance of 50 cm is **1 V**.