A piece of wire of resistance `4Omega` is bent through `180^(@)` at its midpoint and the two halves are twisted together.Then the resistance is

Options

A. `8Omega`
B. `1Omega`
C. `2Omega`
D. `5Omega`

Explanation

To solve the problem, we need to determine the resistance of a wire that has been bent and twisted. Here’s a step-by-step solution:

### Step 1: Understand the initial resistance
The initial resistance of the wire is given as \( R = 4 \, \Omega \).

### Step 2: Determine the length of the wire
Let the total length of the wire be \( L \). Since the wire is bent at its midpoint, each half of the wire will have a length of \( \frac{L}{2} \).

### Step 3: Calculate the resistance of each half
Using the formula for resistance, \( R = \rho \frac{L}{A} \), where:
- \( \rho \) is the resistivity (constant for the material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

Since the area of cross-section remains constant, the resistance of each half of the wire can be calculated as follows:
- The resistance of each half is:
\[ R' = \rho \frac{L/2}{A} = \frac{1}{2} \rho \frac{L}{A} = \frac{R}{2} = \frac{4 \, \Omega}{2} = 2 \, \Omega \]

### Step 4: Combine the two halves
Now, the two halves of the wire, each with a resistance of \( 2 \, \Omega \), are twisted together. When two resistances are connected in parallel, the equivalent resistance \( R_{eq} \) can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
where \( R_1 = 2 \, \Omega \) and \( R_2 = 2 \, \Omega \).

### Step 5: Calculate the equivalent resistance
Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1 \]
Thus,
\[ R_{eq} = 1 \, \Omega \]

### Conclusion
The final resistance of the twisted wire is \( 1 \, \Omega \).