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, we need to understand how the resistance of a wire changes when its cross-sectional area is altered while keeping the volume constant. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Volume Conservation**:
When a resistance wire is drawn through a die, its volume remains constant. The volume \( V \) of the wire can be expressed as:
\[ V = A \times L \]
where \( A \) is the cross-sectional area and \( L \) is the length of the wire.

2. **Expressing Changes in Area and Length**:
Since the volume is constant, if the cross-sectional area decreases, the length must increase to keep the volume the same. This relationship can be expressed as:
\[ A_1 \times L_1 = A_2 \times L_2 \]
where \( A_1 \) and \( L_1 \) are the initial area and length, and \( A_2 \) and \( L_2 \) are the new area and length.

3. **Percentage Change in Area and Length**:
Given that the area decreases by 1%, we can express this as:
\[ \frac{\Delta A}{A} = -0.01 \]
From the conservation of volume, we can relate the change in area to the change in length:
\[ \frac{\Delta A}{A} = -\frac{\Delta L}{L} \]
Thus, if the area decreases by 1%, the length must increase by 1%:
\[ \frac{\Delta L}{L} = 0.01 \]

4. **Resistance Formula**:
The resistance \( R \) of the wire is given by:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material.

5. **Calculating Change in Resistance**:
To find the change in resistance, we can express the relative change in resistance as:
\[ \frac{\Delta R}{R} = \frac{\Delta \rho}{\rho} + \frac{\Delta L}{L} - \frac{\Delta A}{A} \]
Since the resistivity \( \rho \) remains constant, \( \frac{\Delta \rho}{\rho} = 0 \). Therefore, we have:
\[ \frac{\Delta R}{R} = \frac{\Delta L}{L} - \frac{\Delta A}{A} \]

6. **Substituting Values**:
Substituting the values we found:
\[ \frac{\Delta R}{R} = 0.01 - (-0.01) = 0.01 + 0.01 = 0.02 \]
This means the change in resistance is:
\[ \Delta R = 0.02 R \]
Thus, the percentage change in resistance is:
\[ \Delta R \text{ (in percentage)} = 0.02 \times 100 = 2\% \]

7. **Conclusion**:
Since the change is positive, the resistance increases by 2%. Therefore, the final answer is:
\[ \text{The change in resistance is } +2\%. \]