If a copper wire is stretched so that its length increases by `20%` then what is the percentage increase in its resistance (assuming its volume remaing constant) ?

Options

A. `10%`
B. `21%`
C. `44%`
D. `120%`

Explanation

To solve the problem of finding the percentage increase in the resistance of a copper wire when its length is increased by 20%, we can follow these steps:

### Step-by-Step Solution:

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 \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire.

2. **Express the volume of the wire**:
The volume \( V \) of the wire remains constant and is given by:
\[ V = A \times L \]

3. **Determine the new length after a 20% increase**:
If the original length is \( L_0 \), then after a 20% increase, the new length \( L_1 \) is:
\[ L_1 = L_0 + 0.2L_0 = 1.2L_0 \]

4. **Find the new area using the constant volume**:
Since the volume remains constant, we can express the new area \( A_1 \) in terms of the new length:
\[ V = A_0 L_0 = A_1 L_1 \]
Rearranging gives:
\[ A_1 = \frac{A_0 L_0}{L_1} = \frac{A_0 L_0}{1.2L_0} = \frac{A_0}{1.2} \]

5. **Calculate the initial and final resistances**:
The initial resistance \( R_0 \) is:
\[ R_0 = \frac{\rho L_0}{A_0} \]
The new resistance \( R_1 \) is:
\[ R_1 = \frac{\rho L_1}{A_1} = \frac{\rho (1.2L_0)}{\frac{A_0}{1.2}} = \frac{\rho (1.2L_0) \cdot 1.2}{A_0} = \frac{1.44 \rho L_0}{A_0} = 1.44 R_0 \]

6. **Determine the percentage increase in resistance**:
The percentage increase in resistance can be calculated as:
\[ \text{Percentage Increase} = \frac{R_1 - R_0}{R_0} \times 100 = \frac{1.44 R_0 - R_0}{R_0} \times 100 = (1.44 - 1) \times 100 = 0.44 \times 100 = 44\% \]

### Final Answer:
The percentage increase in resistance is **44%**.