A wire of resistance `2.20 Omega` has a length `2m`. Calculate the length of the similar wire which connected in parallel with `2m` length wire will give a resistance of `2.0 Omega`

Explanation

To solve the problem, we need to find the length of a similar wire that, when connected in parallel with a 2-meter wire of resistance 2.20 Ω, results in a total resistance of 2.0 Ω.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Resistance of the first wire (R1) = 2.20 Ω
- Length of the first wire (L1) = 2 m
- Desired total resistance (R_total) = 2.0 Ω

2. **Calculate the Resistance per Unit Length**:
The resistance per unit length (R_per_unit_length) can be calculated using the formula:
\[ R_{\text{per unit length}} = \frac{R_1}{L_1} \]
Substituting the known values:
\[ R_{\text{per unit length}} = \frac{2.20 \, \Omega}{2 \, \text{m}} = 1.1 \, \Omega/\text{m} \]

3. **Use the Formula for Parallel Resistance**:
When two resistors (R1 and R2) are in parallel, the total resistance (R_total) can be calculated using the formula:
\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Here, R2 is the resistance of the second wire, which we need to find.

4. **Express R2 in Terms of Length**:
The resistance of the second wire (R2) can be expressed in terms of its length (L2):
\[ R_2 = R_{\text{per unit length}} \times L_2 = 1.1 \, \Omega/\text{m} \times L_2 \]

5. **Substitute R1 and R2 into the Parallel Resistance Formula**:
Substitute R1 and R2 into the parallel resistance formula:
\[ \frac{1}{2.0} = \frac{1}{2.20} + \frac{1}{1.1 \times L_2} \]

6. **Solve for L2**:
Rearranging the equation gives:
\[ \frac{1}{1.1 \times L_2} = \frac{1}{2.0} - \frac{1}{2.20} \]
Finding a common denominator for the right-hand side:
\[ \frac{1}{2.0} = \frac{11}{22}, \quad \frac{1}{2.20} = \frac{10}{22} \]
So,
\[ \frac{1}{1.1 \times L_2} = \frac{11 - 10}{22} = \frac{1}{22} \]
Therefore,
\[ 1.1 \times L_2 = 22 \quad \Rightarrow \quad L_2 = \frac{22}{1.1} = 20 \, \text{m} \]

### Final Answer:
The length of the similar wire (L2) that needs to be connected in parallel is **20 meters**.

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