A wire of resistance `1 Omega` is stretched so as to change its diameter by `0.25 %.` The percentage change in its resistance is

Options

A. `1.0%`
B. `2.0%`
C. `4.0%`
D. `8.0%`

Explanation

To solve the problem, we need to find the percentage change in resistance when the diameter of a wire is changed by 0.25%.

### Step-by-Step Solution:

1. **Understand the relationship between resistance and diameter**:
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 of the wire.

2. **Express the area in terms of diameter**:
The area \( A \) of the wire can be expressed in terms of its diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]

3. **Determine how stretching affects length and area**:
When the wire is stretched, its length increases and its diameter decreases. The volume of the wire remains constant, so:
\[ V = L \times A = \text{constant} \]

4. **Relate changes in diameter to changes in resistance**:
Since resistance is inversely proportional to the area, we can express the change in resistance in terms of the change in diameter:
\[ R \propto \frac{1}{d^4} \]
Therefore, the percentage change in resistance can be expressed as:
\[ \frac{\Delta R}{R} = -4 \frac{\Delta d}{d} \]

5. **Calculate the percentage change in diameter**:
Given that the diameter changes by \( 0.25\% \):
\[ \frac{\Delta d}{d} = \frac{0.25}{100} = 0.0025 \]

6. **Substitute the percentage change in diameter into the resistance formula**:
Now substituting this value into the equation for percentage change in resistance:
\[ \frac{\Delta R}{R} = -4 \times 0.0025 = -0.01 \]

7. **Convert to percentage**:
To express this as a percentage:
\[ \Delta R = -0.01 \times 100 = -1\% \]

### Conclusion:
The percentage change in resistance is \( -1\% \), indicating that the resistance decreases by 1%.

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