When a resistance wire is passed through a die the cross-section area decreases by `1%`, the change in resistance of the wire is

Options

A. 1% decreases
B. 1% increases
C. 2% decreases
D. 2% increases

Explanation

To solve the problem of how the resistance of a wire changes when its cross-sectional area decreases by 1%, we can follow these steps:

### 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,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

### Step 2: Determine the new cross-sectional area
If the cross-sectional area decreases by 1%, the new area \( A' \) can be calculated as:
\[ A' = A - 0.01A = 0.99A \]

### Step 3: Use the volume conservation principle
Since the volume of the wire remains constant, we can relate the initial and new dimensions:
\[ A \cdot L = A' \cdot L' \]
Substituting \( A' \):
\[ A \cdot L = 0.99A \cdot L' \]
From this, we can solve for the new length \( L' \):
\[ L' = \frac{L}{0.99} \]

### Step 4: Calculate the new resistance
Now, substituting \( A' \) and \( L' \) into the resistance formula:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{0.99}\right)}{0.99A} \]
This simplifies to:
\[ R' = \frac{\rho L}{A} \cdot \frac{1}{0.99^2} = R \cdot \frac{1}{0.99^2} \]

### Step 5: Calculate \( \frac{1}{0.99^2} \)
Calculating \( \frac{1}{0.99^2} \):
\[ 0.99^2 = 0.9801 \quad \Rightarrow \quad \frac{1}{0.9801} \approx 1.0204 \]
Thus, we can say:
\[ R' \approx 1.0204 R \]

### Step 6: Determine the change in resistance
The change in resistance can be calculated as:
\[ \Delta R = R' - R = (1.0204 R - R) = 0.0204 R \]

### Step 7: Calculate the percentage change in resistance
To find the percentage change in resistance:
\[ \text{Percentage Change} = \left(\frac{\Delta R}{R}\right) \times 100\% = \left(\frac{0.0204 R}{R}\right) \times 100\% \approx 2.04\% \]

### Conclusion
The resistance of the wire increases by approximately 2.04%, which we can round to 2%.

### Final Answer
The change in resistance of the wire is approximately **2% increase**.

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