The resistance of wire of length `l` and area of cross-section a is x ohm. If the wire is stretched to double its length, its resistance would become:

Options

A. 2 x ohm
B. 0.5 x ohm
C. 4 x ohm
D. 6 x ohm

Explanation

To find the new resistance of a wire when it is stretched to double its length, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for resistance**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( R \) is the resistance, \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the area of cross-section.

2. **Identify the initial conditions**:
According to the problem, the initial resistance of the wire is \( x \) ohms, with length \( l \) and area \( a \):
\[ x = \frac{\rho l}{a} \]

3. **Determine the new length and area after stretching**:
When the wire is stretched to double its length, the new length \( L' \) becomes:
\[ L' = 2l \]
When a wire is stretched, its volume remains constant. Therefore, the new area \( A' \) can be found using the volume conservation:
\[ \text{Volume} = \text{Length} \times \text{Area} \Rightarrow l \cdot a = L' \cdot A' \]
Substituting \( L' = 2l \):
\[ l \cdot a = 2l \cdot A' \Rightarrow A' = \frac{a}{2} \]

4. **Calculate the new resistance**:
Now, we can find the new resistance \( R' \) using the new length and area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2l)}{(a/2)} = \frac{2\rho l}{(a/2)} = \frac{2\rho l \cdot 2}{a} = \frac{4\rho l}{a} \]
Since we know \( \frac{\rho l}{a} = x \):
\[ R' = 4x \]

5. **Conclusion**:
Therefore, the resistance of the wire when stretched to double its length becomes:
\[ R' = 4x \text{ ohms} \]

### Final Answer:
The resistance would become \( 4x \) ohms.

---