When length of wire is increased by 10% , then increase in resistance is

Options

A. 0.1
B. 0.21
C. 0.25
D. 0.35

Explanation

To solve the problem of how the resistance of a wire changes when its length is increased by 10%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between resistance and length**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

2. **Calculate the new length**:
If the length of the wire is increased by 10%, the new length \( L' \) can be calculated as:
\[ L' = L + 0.1L = 1.1L \]

3. **Assume the volume of the wire remains constant**:
Since the volume of the wire remains constant during the stretching, we have:
\[ \text{Initial Volume} = \text{Final Volume} \]
This gives us:
\[ A \cdot L = A' \cdot L' \]
where \( A' \) is the new cross-sectional area.

4. **Express the new area in terms of the old area**:
Substituting \( L' \) into the volume equation:
\[ A \cdot L = A' \cdot (1.1L) \]
Rearranging gives:
\[ A' = \frac{A}{1.1} \]

5. **Calculate the new resistance**:
The new resistance \( R' \) can now be calculated using the new length and new area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (1.1L)}{\frac{A}{1.1}} = \frac{\rho (1.1L) \cdot 1.1}{A} = \frac{1.21 \rho L}{A} \]
Thus, we can express the new resistance in terms of the original resistance \( R \):
\[ R' = 1.21 R \]

6. **Determine the percentage increase in resistance**:
The change in resistance \( \Delta R \) is:
\[ \Delta R = R' - R = 1.21R - R = 0.21R \]
The percentage increase in resistance is given by:
\[ \text{Percentage Increase} = \left(\frac{\Delta R}{R}\right) \times 100 = \left(\frac{0.21R}{R}\right) \times 100 = 21\% \]

### Final Answer:
The percentage increase in resistance when the length of the wire is increased by 10% is **21%**.

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