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. a 3 v
B. b 1v
C. c 1.5 v
D. d 2 v

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Identify the given values
- Voltage of the battery (V) = 6 V
- Length of the wire (L) = 3 m
- Resistance of the wire (R) = 100 Ω
- Distance between the two points on the wire (L') = 50 cm = 0.5 m

### Step 2: Calculate the total current flowing through the circuit
Using Ohm's law, the current (I) can be calculated using the formula:
\[ I = \frac{V}{R} \]
Substituting the values:
\[ I = \frac{6 \, \text{V}}{100 \, \Omega} = 0.06 \, \text{A} \]

### Step 3: Determine the resistance of the segment of the wire (R')
To find the resistance of the 50 cm segment of the wire, we can use the formula for resistance:
\[ R' = \frac{R \cdot L'}{L} \]
Where:
- \( R \) is the total resistance of the wire,
- \( L' \) is the length of the segment we are interested in (0.5 m),
- \( L \) is the total length of the wire (3 m).

Substituting the values:
\[ R' = \frac{100 \, \Omega \cdot 0.5 \, \text{m}}{3 \, \text{m}} = \frac{50}{3} \, \Omega \approx 16.67 \, \Omega \]

### Step 4: Calculate the potential difference (V') across the 50 cm segment
Using Ohm's law again, the potential difference (V') can be calculated as:
\[ V' = I \cdot R' \]
Substituting the values:
\[ V' = 0.06 \, \text{A} \cdot \left(\frac{50}{3} \, \Omega\right) \]
Calculating:
\[ V' = 0.06 \cdot 16.67 \approx 1 \, \text{V} \]

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

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