A current of 2amp is flowing in the primary circuit of a potentiometer wire having resistance of `0.2 ohm//m`. In a coil when current of 1A flows, then potential difference across its terminals are balanced at 2.5m of the potentiometer wire. The resistance of the coil is

Options

A. 1 ohm
B. 2.5 ohm
C. 0.4 ohm
D. 5.0 ohm

Explanation

To solve the problem, we need to find the resistance of the coil using the information provided about the potentiometer wire and the current flowing through the circuit.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current in the primary circuit (I₁) = 2 A
- Resistance of the potentiometer wire (R₁) = 0.2 ohm/m
- Length of the potentiometer wire used for balancing (L) = 2.5 m
- Current in the coil (I₂) = 1 A

2. **Calculate the Potential Difference (V₁) across the potentiometer wire:**
- The potential difference across the potentiometer wire can be calculated using Ohm's Law:
\[ V₁ = I₁ \times R_{\text{total}} \]
- The total resistance of the potentiometer wire for 2.5 m is:
\[ R_{\text{total}} = R₁ \times L = 0.2 \, \text{ohm/m} \times 2.5 \, \text{m} = 0.5 \, \text{ohm} \]
- Therefore, the potential difference (V₁) is:
\[ V₁ = 2 \, \text{A} \times 0.5 \, \text{ohm} = 1 \, \text{V} \]

3. **Relate the Potential Difference to the Coil's Resistance:**
- According to Ohm's Law, the potential difference across the coil when 1 A flows through it is:
\[ V₂ = I₂ \times R_{\text{coil}} \]
- Since we know that V₁ = V₂, we can set the equations equal:
\[ 1 \, \text{V} = 1 \, \text{A} \times R_{\text{coil}} \]

4. **Solve for the Resistance of the Coil (Rₐ):**
- Rearranging the equation gives:
\[ R_{\text{coil}} = \frac{1 \, \text{V}}{1 \, \text{A}} = 1 \, \text{ohm} \]

5. **Conclusion:**
- The resistance of the coil is 1 ohm.

### Final Answer:
The resistance of the coil is **1 ohm**.