A wire of resistance `4 Omega` is stretched to twice its original length. The resistance of stretched wire would be

Options

A. `2 Omega`
B. `4 Omega`
C. `8 Omega`
D. `16 Omega`

Answer

To solve the problem of finding the resistance of a wire that has been stretched to twice its original length, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the original length of the wire be \( L \).
- The original resistance of the wire is given as \( R = 4 \, \Omega \).

2. **Determine the New Length**:
- When the wire is stretched to twice its original length, the new length \( L' \) becomes:
\[ L' = 2L \]

3. **Volume Conservation**:
- The volume of the wire before stretching is equal to the volume after stretching. The volume \( V \) of a wire is given by:
\[ V = A \cdot L \]
- Let \( A \) be the original cross-sectional area. After stretching, the volume can be expressed as:
\[ V' = A' \cdot L' \]
- Since the volume remains constant:
\[ A \cdot L = A' \cdot L' \]

4. **Substituting for New Length**:
- Substitute \( L' = 2L \) into the volume equation:
\[ A \cdot L = A' \cdot (2L) \]
- Simplifying gives:
\[ A = 2A' \]
- This means the new cross-sectional area \( A' \) is:
\[ A' = \frac{A}{2} \]

5. **Using the Resistance Formula**:
- The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho L}{A} \]
- For the new wire, the resistance \( R' \) can be expressed as:
\[ R' = \frac{\rho L'}{A'} \]
- Substitute \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[ R' = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L}{\frac{A}{2}} = \frac{2 \cdot 2 \rho L}{A} = \frac{4 \rho L}{A} \]

6. **Relating New Resistance to Original Resistance**:
- Since \( R = \frac{\rho L}{A} \), we can express \( R' \) in terms of \( R \):
\[ R' = 4R \]
- Given that \( R = 4 \, \Omega \):
\[ R' = 4 \cdot 4 = 16 \, \Omega \]

### Final Answer:
The resistance of the stretched wire is \( 16 \, \Omega \).