The resistance of a wire R `Omega` . The wire is stretched to double its length keeping volume constant. Now, the resistance of the wire will become

Options

A. `4ROmega`
B. `2ROmega`
C. `1ROmega`
D. `1/2ROmega`

Answer

To solve the problem step-by-step, we will analyze the situation where a wire is stretched to double its length while keeping its volume constant. We will derive the new resistance of the wire based on the given conditions.

### Step 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 (constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Relate volume to length and area
The volume \( V \) of the wire is given by:
\[ V = A \cdot L \]
Since the volume is kept constant, we can express this relationship as:
\[ A_1 L_1 = A_2 L_2 \]
where \( A_1 \) and \( L_1 \) are the initial area and length, and \( A_2 \) and \( L_2 \) are the final area and length.

### Step 3: Set up the initial and final conditions
Initially, let the length of the wire be \( L \) and the area be \( A \). After stretching, the new length \( L' \) becomes:
\[ L' = 2L \]
Since the volume is constant, we can write:
\[ A \cdot L = A' \cdot (2L) \]
From this, we can solve for the new area \( A' \):
\[ A' = \frac{A}{2} \]

### Step 4: Calculate the new resistance
Now we can find the new resistance \( R' \) using the new length \( L' \) and new area \( A' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L \cdot 2}{A} = \frac{4\rho L}{A} \]
This shows that:
\[ R' = 4R \]
where \( R = \frac{\rho L}{A} \) is the original resistance.

### Conclusion
Thus, the new resistance of the wire when it is stretched to double its length while keeping the volume constant is:
\[ R' = 4R \]

### Final Answer
The resistance of the wire will become \( 4R \, \Omega \).

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