When length of a metal wire is doubled and area of cross-section is reduced to half then the resistance will be

Options

A. Double
B. Four times
C. No change
D. Half

Answer

To solve the problem, we need to analyze how the resistance of a metal wire changes when its length is doubled and its area of cross-section is reduced to half. We will use the formula for resistance and apply the changes step by step.

### Step-by-Step Solution:

1. **Understand the Formula for Resistance**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance
- \( \rho \) = resistivity of the material (constant for a given material)
- \( L \) = length of the wire
- \( A \) = area of cross-section of the wire

2. **Initial Resistance**:
Let the initial length of the wire be \( L \) and the initial area of cross-section be \( A \). The initial resistance \( R_1 \) can be expressed as:
\[ R_1 = \frac{\rho L}{A} \]

3. **Changes in Length and Area**:
According to the problem:
- The length of the wire is doubled: \( L' = 2L \)
- The area of cross-section is reduced to half: \( A' = \frac{A}{2} \)

4. **Calculate the New Resistance**:
Now, we can calculate the new resistance \( R_2 \) using the new values of length and area:
\[ R_2 = \frac{\rho L'}{A'} = \frac{\rho (2L)}{\frac{A}{2}} \]

5. **Simplifying the Expression for \( R_2 \)**:
Substituting the values:
\[ R_2 = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L}{\frac{A}{2}} = \frac{2\rho L \cdot 2}{A} = \frac{4\rho L}{A} \]

6. **Relate New Resistance to Initial Resistance**:
Now, we can relate \( R_2 \) to \( R_1 \):
\[ R_2 = 4 \cdot \frac{\rho L}{A} = 4R_1 \]

7. **Conclusion**:
Thus, the new resistance \( R_2 \) is four times the initial resistance \( R_1 \):
\[ R_2 = 4R_1 \]

### Final Answer:
The resistance will be **4 times the initial resistance**.

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