Answer :
To find the potential difference between two points on the wire separated by 50 cm, let's follow these steps:
Firstly, understand that the circuit consists of a 6 V battery, a 3 m long wire with 100 ohms resistance, and another 100 ohm resistor in series.
Total resistance in the circuit:
The total resistance of the circuit is the sum of the resistance of the wire and the additional resistor:
[tex]R_{total} = 100 \,\Omega + 100 \,\Omega = 200 \,\Omega[/tex]
Current in the circuit:
Using Ohm's law, we find the current [tex]I[/tex] in the circuit:
[tex]V = IR \quad \Rightarrow \quad I = \frac{V}{R_{total}}[/tex]
[tex]I = \frac{6 \, V}{200 \,\Omega} = 0.03 \, A[/tex]
Potential drop along the wire:
The potential drop [tex]V_{wire}[/tex] across the entire wire (3 m) is calculated using:
[tex]V_{wire} = I \times R_{wire} = 0.03 \, A \times 100 \, \Omega = 3 \, V[/tex]
Potential drop for 50 cm of the wire:
Since the potential drop is uniform along the wire, we calculate the potential difference across 50 cm (or 0.5 m) using proportion:
[tex]\frac{V_{0.5m}}{V_{wire}} = \frac{0.5}{3} \quad \Rightarrow \quad V_{0.5m} = \frac{0.5}{3} \times 3 \, V = 0.5 \, V[/tex]
Therefore, the potential difference between points on the wire separated by a distance of 50 cm is 0.5 V.
The correct multiple-choice option is (4) 0.5 V.
