If a `0.1 %` increase in length due to stretching, the percentage increase in its resistance will be

Options

A. `0.2%`
B. `2%`
C. `1%`
D. `0.1%`

Answer

To find the percentage increase in resistance due to a 0.1% increase in length of a wire, 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:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity (constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

### Step 2: Determine the new length and area after stretching
If the length increases by 0.1%, the new length \( L' \) can be expressed as:
\[ L' = L + 0.001L = 1.001L \]

Since the volume of the wire remains constant during stretching, we have:
\[ V = A \cdot L = A' \cdot L' \]
where \( A' \) is the new cross-sectional area. Therefore:
\[ A \cdot L = A' \cdot (1.001L) \]
From this, we can derive the new area:
\[ A' = \frac{A \cdot L}{1.001L} = \frac{A}{1.001} \]

### Step 3: Calculate the new resistance
Now we can find the new resistance \( R' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (1.001L)}{\frac{A}{1.001}} = \frac{\rho (1.001^2 L)}{A} = R \cdot 1.001^2 \]

### Step 4: Find the percentage increase in resistance
The percentage increase in resistance can be calculated as:
\[ \text{Percentage Increase} = \left( \frac{R' - R}{R} \right) \times 100 = \left( \frac{R \cdot 1.001^2 - R}{R} \right) \times 100 \]
This simplifies to:
\[ \text{Percentage Increase} = (1.001^2 - 1) \times 100 \]

### Step 5: Calculate \( 1.001^2 \)
Calculating \( 1.001^2 \):
\[ 1.001^2 = 1.002001 \]
Thus,
\[ 1.001^2 - 1 = 0.002001 \]

### Step 6: Final percentage increase
Now, substituting this back into the percentage increase formula:
\[ \text{Percentage Increase} = 0.002001 \times 100 = 0.2001\% \]
Rounding this gives approximately:
\[ \text{Percentage Increase} \approx 0.2\% \]

### Conclusion
Therefore, the percentage increase in resistance due to a 0.1% increase in length is approximately **0.2%**.

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