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`

Explanation

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

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

### Step 2: Identify the initial conditions
Given:
- Initial resistance \( R = 4 \, \Omega \)
- Initial length \( L = L \)
- Initial cross-sectional area \( A = A_1 \)

### Step 3: Determine the new length and volume
When the wire is stretched to twice its original length:
\[ L' = 2L \]
Since the volume of the wire remains constant during stretching, we have:
\[ V = A_1 L = A_2 L' \]
where \( A_2 \) is the new cross-sectional area after stretching.

### Step 4: Relate the areas using the volume constant
Substituting the new length:
\[ A_1 L = A_2 (2L) \]
Cancelling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ A_1 = 2A_2 \]
This implies:
\[ \frac{A_1}{A_2} = 2 \]

### Step 5: Calculate the new resistance
Using the resistance formula for the stretched wire:
\[ R' = \rho \frac{L'}{A_2} = \rho \frac{2L}{A_2} \]
Substituting \( A_2 \) in terms of \( A_1 \):
\[ R' = \rho \frac{2L}{\frac{A_1}{2}} = \rho \frac{2L \cdot 2}{A_1} = \rho \frac{4L}{A_1} \]
Now, we know that:
\[ R = \rho \frac{L}{A_1} \]
Thus:
\[ R' = 4R \]

### Step 6: Substitute the initial resistance
Substituting the initial resistance value:
\[ R' = 4 \times 4 \, \Omega = 16 \, \Omega \]

### Conclusion
The resistance of the stretched wire is \( 16 \, \Omega \).

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