Resistance of a wire is `8 Omega `. It is drawn in such away that it experiences a longitudinal strain of 400 %. The new resistance is

Options

A. `100 Omega `
B. `200 Omega `
C. `300 Omega `
D. `400 Omega `

Answer

To find the new resistance of the wire after it has been drawn and experiences a longitudinal strain of 400%, we can follow these steps:

### Step 1: Understand the given data
- Initial resistance \( R = 8 \, \Omega \)
- Longitudinal strain \( \epsilon = 400\% = 4 \)

### Step 2: Calculate the new length of the wire
The longitudinal strain is defined as the change in length per unit original length. Thus, if the original length is \( L \), the new length \( L' \) can be calculated as:
\[ L' = L + \epsilon \cdot L = L + 4L = 5L \]

### Step 3: Use the volume conservation principle
Since the volume of the wire remains constant during the drawing process, we can express this as:
\[ A \cdot L = A' \cdot L' \]
where \( A \) is the original cross-sectional area and \( A' \) is the new cross-sectional area. Rearranging gives:
\[ A' = \frac{A \cdot L}{L'} \]

### Step 4: Substitute the new length into the area equation
Substituting \( L' = 5L \) into the equation for \( A' \):
\[ A' = \frac{A \cdot L}{5L} = \frac{A}{5} \]

### Step 5: Calculate the new resistance
The resistance of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
For the new resistance \( R' \):
\[ R' = \rho \frac{L'}{A'} = \rho \frac{5L}{\frac{A}{5}} = \rho \frac{5L \cdot 5}{A} = 25 \cdot \rho \frac{L}{A} \]
Thus, we can express the new resistance in terms of the original resistance:
\[ R' = 25R \]

### Step 6: Substitute the original resistance
Given \( R = 8 \, \Omega \):
\[ R' = 25 \cdot 8 = 200 \, \Omega \]

### Final Answer
The new resistance \( R' \) is \( 200 \, \Omega \).

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