A wire of resistance R is stretched so that its length increases by 10%. The resistance of the wire increases by

Options

A. 0.11
B. 0.15
C. 0.21
D. 0.28

Explanation

To solve the problem of how much the resistance of a wire increases when it is stretched by 10%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the initial length of the wire be \( L \).
- Let the initial area of cross-section be \( A \).
- Let the initial resistance be \( R \).

2. **Calculate the New Length**:
- The wire is stretched so that its length increases by 10%.
- Therefore, the new length \( L' \) is given by:
\[ L' = L + 0.1L = 1.1L \]

3. **Volume Conservation**:
- Since the volume of the wire remains constant during stretching, we can express this as:
\[ \text{Volume} = \text{Length} \times \text{Area} \]
- Thus, the initial volume \( V \) is:
\[ V = L \times A \]
- The final volume \( V' \) is:
\[ V' = L' \times A' = 1.1L \times A' \]
- Setting these equal gives:
\[ L \times A = 1.1L \times A' \]
- Dividing both sides by \( L \) (assuming \( L \neq 0 \)):
\[ A = 1.1A' \]
- Rearranging gives:
\[ A' = \frac{A}{1.1} \]

4. **Calculate the New Resistance**:
- The resistance of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
- The new resistance \( R' \) can be calculated using the new length and new area:
\[ R' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{\frac{A}{1.1}} = \rho \frac{1.1^2 L}{A} = 1.21 \cdot \rho \frac{L}{A} = 1.21R \]

5. **Calculate the Increase in Resistance**:
- The increase in resistance is given by:
\[ \Delta R = R' - R = 1.21R - R = 0.21R \]
- The percentage increase in resistance is:
\[ \text{Percentage Increase} = \frac{\Delta R}{R} \times 100 = \frac{0.21R}{R} \times 100 = 21\% \]

### Final Answer:
The resistance of the wire increases by \( 21\% \).

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