`A 8 Omega` resistance wire is bent through `180^(@)` at its mid point and the two halves are twisted together. Then the resistance is

Options

A. `16 Omega`
B. `4 Omega`
C. `2 Omega`
D. `1 Omega`

Answer

To solve the problem of finding the new resistance after bending an 8 ohm resistance wire through 180 degrees at its midpoint and twisting the two halves together, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have a resistance wire of resistance \( R = 8 \, \Omega \).
- The wire is bent at its midpoint, creating two equal halves.

2. **Determine the Length of Each Half**:
- Let the total length of the wire be \( L \).
- After bending, each half of the wire will have a length of \( \frac{L}{2} \).

3. **Cross-Sectional Area**:
- When the two halves are twisted together, the effective cross-sectional area of the wire doubles.
- If the original cross-sectional area is \( A \), the new cross-sectional area \( A' \) when twisted together is \( 2A \).

4. **Using the Resistance Formula**:
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
- For each half of the wire, the resistance \( R_{half} \) can be calculated as:
\[ R_{half} = \frac{\rho \left(\frac{L}{2}\right)}{A} = \frac{\rho L}{2A} \]

5. **Calculating the New Resistance**:
- When the two halves are twisted together, the new resistance \( R' \) can be calculated using the new length and area:
\[ R' = \frac{\rho \left(\frac{L}{2}\right)}{2A} = \frac{\rho L}{4A} \]
- Notice that \( \frac{\rho L}{A} \) is the original resistance \( R \):
\[ R' = \frac{R}{4} \]

6. **Substituting the Original Resistance**:
- Now substituting \( R = 8 \, \Omega \):
\[ R' = \frac{8 \, \Omega}{4} = 2 \, \Omega \]

### Final Answer:
The new resistance after bending and twisting the wire is \( R' = 2 \, \Omega \).

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